JEE Mains 2015 10 th April (online) Physics Single Correct Answer Type: 1. In an ideal at temperature T, the average force that a molecule applies on the walls of a closed container depends on . A good estimate for q is: (A) 2 (B) 1 2 (C) 1 (D) 1 4 Answer: (C) Solution: Average linear for collision to occur = 2 Change in momentum in 1 collision Δ = 2 ∴ average force in collision = Δ = root mean square speed = 2 2 × ⇒ ∝ 2 ∴ 2 ∝ ⇒ × ⇒ =1 2. In an unbiased n – p junction electrons diffuse from n-region to p-region because: (A) Electrons travel across the junction due to potential difference (B) Only electrons move from n to p region and not the vice – versa (C) Electron concentration in n – region is more as compared to that in p – region (D) Holes in p – region attract them Answer: (C) Solution: In a − junction diffusion occurs due to spontaneous movement of majority charge carrier from the region of high concentration to low concentration so option 3 in correct. 3. A 10V battery with internal resistance 1Ω 15 battery with internal resistance 0.6Ω are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to:
Transcript
JEE Mains 2015 10th April (online)
Physics
Single Correct Answer Type
1 In an ideal at temperature T the average force that a molecule applies on the walls of a closed
container depends on 119879 119886119904 119879119902 A good estimate for q is
(A) 2 (B) 1
2 (C) 1 (D)
1
4
Answer (C)
Solution
Average linear for collision to occur
119905 =2119889
119906
Change in momentum in 1 collision
Δ119901 = 2 119898119906
there4 average force in collision
= Δ119901
119905
119906 = root mean square speed
=2 119898119906
2119889 times 119906
rArr 119891 prop 1199062
there4 1199062 prop 119879
rArr 119891 times 119879
rArr 119902 = 1
2 In an unbiased n ndash p junction electrons diffuse from n-region to p-region because
(A) Electrons travel across the junction due to potential difference
(B) Only electrons move from n to p region and not the vice ndash versa
(C) Electron concentration in n ndash region is more as compared to that in p ndash region
(D) Holes in p ndash region attract them
Answer (C)
Solution
In a 119901 minus 119899 junction diffusion occurs due to spontaneous movement of majority charge carrier from
the region of high concentration to low concentration so option 3 in correct
3 A 10V battery with internal resistance 1Ω 119886119899119889 119886 15119881 battery with internal resistance 06Ω are
connected in parallel to a voltmeter (see figure) The reading in the voltmeter will be close to
10 119909 119886119899119889 119910 displacements of a particle are given as 119909(119905) = 119886 sin120596119905 119886119899119889 119910(119905) = 119886 sin 2120596119905 Its
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
10 119909 119886119899119889 119910 displacements of a particle are given as 119909(119905) = 119886 sin120596119905 119886119899119889 119910(119905) = 119886 sin 2120596119905 Its
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
10 119909 119886119899119889 119910 displacements of a particle are given as 119909(119905) = 119886 sin120596119905 119886119899119889 119910(119905) = 119886 sin 2120596119905 Its
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
10 119909 119886119899119889 119910 displacements of a particle are given as 119909(119905) = 119886 sin120596119905 119886119899119889 119910(119905) = 119886 sin 2120596119905 Its
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
10 119909 119886119899119889 119910 displacements of a particle are given as 119909(119905) = 119886 sin120596119905 119886119899119889 119910(119905) = 119886 sin 2120596119905 Its
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
10 119909 119886119899119889 119910 displacements of a particle are given as 119909(119905) = 119886 sin120596119905 119886119899119889 119910(119905) = 119886 sin 2120596119905 Its
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
⟹ 119886 prop1
119877
27 If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2
radic120587 119888119898 then the
Reynolds number for the flow is (density of water =103 119896119892 1198983 frasl 119886119899119889 119907119894119904119888119900119904119894119905119910 119900119891 119908119886119905119890119903 =
10minus3 119875119886 119904) close to
(A) 5500 (B) 550 (C) 1100 (D) 11000
Answer (A)
Solution
Reynolds number
119877 =119878119881119863
120578
119863 = Diameter of litre
Also rate of flow = 119881119900119897119906119898119890
119905119894119898119890= 119860 119881
119881
119905= 120587 1198632
4times 119881 rArr 119881 =
4119881
1205871198632119905
there4 119877 = 119878 119863
120578times4 119881
120587 1198632 119905
=4 119878 119881
120587 120578 119863 119905
=4 times 103 times 15 times 10minus3
120587 times 10minus3 times 2 times 5 times 60 radic120587 times 102
=10000
radic120587 asymp 5500
28 If one were to apply Bohr model to a particle of mass lsquomrsquo and charge lsquoqrsquo moving in a plane
under the influence of a magnetic field lsquoBrsquo the energy of the charged particle in the 119899119905ℎ level
will be
(A) 119899 (ℎ119902119861
120587119898) (B) 119899 (
ℎ119902119861
4120587119898) (C) 119899 (
ℎ119902119861
2120587119898) (D) 119899 (
ℎ119902119861
8120587119898)
Answer (B)
Solution
For a charge q moving in a +r uniform magnetic field B
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
15 If the principal quantum number n = 6 the correct sequence of filling of electrons will be
(A) ns rarr (n minus 1) d rarr (n minus 2) f rarr np
(B) ns rarr np rarr (n minus 1)d rarr (n minus 2)f
(C) ns rarr (n minus 2)f rarr np rarr (n minus 1)d
(D) ns rarr (n minus 2)f rarr (n minus 1)d rarr np
Solution (D) As per (n + ℓ) rule when n = 6
ns subshell rArr 6+ 0 = 6
(n ndash 1) d subshell rArr 5+ 2 = 7
(n ndash 2) f subshell rArr 4 + 3 = 7
np subshell rArr 6+ 1 = 7
When n + ℓ values are same the one have lowest n value filled first
ns (n minus 2)f (n minus 1)d np
(n + ℓ) values rArr 7 7 7
n value rArr 4 5 6
16 The cation that will not be precipitated by H2S in the presence of dil HCl is
(A) Co2+
(B) As3+
(C) Pb2+
(D) Cu2+
Solution (A) Co2+ precipitated by H2S in presence of NH4OH in group IV as CoS (Black ppt)
Other are precipitated as sulphide in presence of dil HCl in group II
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
17 The geometry of XeOF4 by VSEPR theory is
(A) Trigonal bipyramidal
(B) Square pyramidal
(C) Pentagonal planar
(D) Octahedral
Solution (B) H =1
2(V + Mminus C + A)
=1
2(8 + 4) = 6
sp3d2 Hybridization
4 BP + 1 BP (Double bonded) + 1 LP
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on apical position
18 The correct order of thermal stability of hydroxides is
(A) Mg(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Ba(OH)2
(B) Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Ba(OH)2
(C) Ba(OH)2 lt Sr(OH)2 lt Ca(OH)2 lt Mg(OH)2
(D) Ba(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt Mg(OH)2
Solution (B) Thermal stabilities of hydroxides of group II A elements increase from
Be(OH)2 to Ba(OH)2 because going down the group the cation size increases amp covalent
character decreases amp ionic character increases ie Mg(OH)2 lt Ca(OH)2 lt Sr(OH)2 lt
Ba(OH)2
19 Photochemical smog consists of excessive amount of X in addition to aldehydes ketones
peroxy acetyl nitrile (PAN) and so forth X is
(A) CH4
(B) CO2
(C) O3
(D) CO
Solution (C) Photochemical smog is the chemical reaction of sunlight nitrogen oxides and VOCs in
the atmosphere
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
NO2hvrarr NO + O
O + O2 rarr O3
So it consists of excessive amount of ozone molecules as atomic oxygen reacts with one of the
abundant oxygen molecules producing ozone
20 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of
hydration is removed The dried sample weighed 52 g The formula of the hydrated salt is
(atomic mass Ba = 137 amu Cl = 355 amu)
(A) BaCl2 ∙ H2O
(B) BaCl2 ∙ 3H2O
(C) BaCl2 ∙ 4H2O
(D) BaCl2 ∙ 2H2O
Solution (D) BaCl2 ∙ xH2O rarr BaCl2 + x H2O
(137 + 2 times 355 + 18x)
= (208 + 18x) gmole
208 + 18 x
208=61
52
10816 + 936 x = 12688
936 x = 1872
x = 2
Formula is BaCl2 ∙ 2H2O
21 The following statements relate to the adsorption of gases on a solid surface Identify the
incorrect statement among them
(A) Entropy of adsorption is negative
(B) Enthalpy of adsorption is negative
(C) On adsorption decrease in surface energy appears as heat
(D) On adsorption the residual forces on the surface are increased
Solution (D) Adsorption is spontaneous process ∆G is ndashve
During adsorption randomness of adsorbate molecules reduced ∆S is ndashve
∆G = ∆H minus T∆S
∆H = ∆G + T∆S
∆H is highly ndashve and residual forces on surface are satisfied
22 In the isolation of metals calcination process usually results in
(A) Metal oxide
(B) Metal carbonate
(C) Metal sulphide
(D) Metal hydroxide
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
Solution (A) Calcination used for decomposition of metal carbonates
M CO3 ∆rarrMO+ CO2 uarr
23 A variable opposite external potential (Eext) is applied to the cell Zn | Zn2+ (1M) ∥
Cu2+ (1 M)| Cu of potential 11 V When Eext lt 11 V and Eext gt 11 V respectively electrons flow from
(A) Anode to cathode in both cases
(B) Anode to cathode and cathode to anode
(C) Cathode to anode and anode to cathode
(D) Cathode to anode in both cases
Solution (B) For the Daniel cell
Ecell = 034 minus (minus076) = 110 V
When Eext lt 110 V electron flow from anode to cathode in external circuit
When Eext gt 110 V electrons flow from cathode to anode in external circuit (Reverse
Reaction)
24 Complete hydrolysis of starch gives
(A) Galactose and fructose in equimolar amounts
(B) Glucose and galactose in equimolar amouunts
(C) Glucose and fructose in equimolar amounts (D) Glucose only
Solution (D) On complete hydrolysis of starch glucose is formed Amylase is an enzyme that
catalyses the hydrolysis of starch into sugars
25 Match the polymers in column-A with their main uses in column-B and choose the correct
answer
Column - A Column - B A Polystyrene i Paints and lacquers B Glyptal ii Rain coats C Polyvinyl chloride
chloride iii Manufacture of toys
D Bakelite iv Computer discs
(A) A ndash iii B ndash i C ndash ii D ndash iv (B) A ndash ii B ndash i C ndash iii D ndash iv
(C) A ndash ii B ndash iv C ndash iii D ndash i
(D) A ndash iii B ndash iv C ndash ii D ndash i
Solution (A) A ndash iii B ndash i C ndash ii D ndash iv
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
26 Permanent hardness in water cannot be cured by
(A) Treatment with washing soda
(B) Ion exchange method
(C) Calgonrsquos methos
(D) Boiling
Solution (D) Permanent hardness due to SO42minus Clminus of Ca2+ and Mg2+ cannot be removed by boiling
27 In the long form of periodic table the valence shell electronic configuration of 5s25p4
corresponds to the element present in
(A) Group 16 and period 5
(B) Group 17 and period 5
(C) Group 16 and period 6
(D) Group 17 and period 6
Solution (A) 5s2 5p4 configuration is actually 36[Kr]5s2 4d10 5p4 ie 5th period and group 16 and
element Tellurium
28 The heat of atomization of methane and ethane are 360 kJmol and 620 kJmol respectively The longest wavelength of light capable of breaking the C minus C bond is (Avogadro number =
6023 times 1023 h = 662 times 10minus34 J s)
(A) 248 times 104 nm
(B) 149 times 104 nm
(C) 248 times 103 nm
(D) 149 times 103 nm
Solution (D) 4 BE (C minus H) bond = 360 kJ
BE (C minus H) bond = 90 kJmole
In C2H6 rArr B E(CminusC) + 6B E(CminusH) = 620 kJ
B E(CminusC) bond = 620 minus 6 times 90 = 80 kJ molefrasl
B E(CminusC) bond =80
9648= 083 eV bondfrasl
λ(Photon in Å) for rupture of
C minus C bond =12408
083= 14950Å
= 1495 nm
asymp 149 times 103 nm
29 Which of the following is not an assumption of the kinetic theory of gases
(A) Collisions of gas particles are perfectly elastic
(B) A gas consists of many identical particles which are in continual motion
(C) At high pressure gas particles are difficult to compress
(D) Gas particles have negligible volume
Solution (C) At high pressures gas particles difficult to compress rather they are not compressible at
all
30 After understanding the assertion and reason choose the correct option
Assertion In the bonding molecular orbital (MO) of H2 electron density is increases between
the nuclei
Reason The bonding MO is ψA +ψB which shows destructive interference of the combining
electron waves
(A) Assertion and Reason are correct but Reason is not the correct explanation for the Assertion
(B) Assertion and Reason are correct and Reason is the correct explanation for the Assertion
(C) Assertion is incorrect Reason is correct
(D) Assertion is correct Reason is incorrect
Solution (D) Electron density between nuclei increased during formation of BMO in H2
BMO is ψA +ψB (Linear combination of Atomic orbitals) provides constructive interference
JEE Mains 2015 10th April (online)
Mathematics
1 If the coefficient of the three successive terms in the binomial expansion of (1 + 119909)119899 are in the
ratio 1 7 42 then the first of these terms in the expansion is
1 9119905ℎ
2 6119905ℎ
3 8119905ℎ
4 7119905ℎ
Answer (4)
Solution Let 119899119862119903 be the first term then 119899119862119903119899119862119903+1
=1
7
rArr 119903 + 1
119899 minus 119903=1
7
rArr 7119903 + 7 = 119899 minus 119903
119899 minus 8119903 = 7 hellip(i)
Also 119899119862119903+1119899119862119903+2
=7
42=1
6
rArr 119903 + 2
119899 minus 119903 minus 1=1
6
rArr 6119903 + 12 = 119899 minus 119903 minus 1
119899 minus 7119903 = 13 helliphellip(ii)
Solving
119899 minus 8119903 = 7 hellip(i)
119899 minus 7119903 = 13 hellip(ii)
____________
minus119903 = minus6
119903 = 6
Hence 7119905ℎ term is the answer
2 The least value of the product 119909119910119911 for which the determinant |11990911 11199101 11119911| is non ndash negative is
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
11 The area (in square units) of the region bounded by the curves 119910 + 21199092 = 0 119886119899119889 119910 + 31199092 = 1 is equal to
1 3
4
2 1
3
3 3
5
4 4
3
Answer (4)
Solution
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
Point of intersection
Put 119910 = minus21199092 119894119899 119910 + 31199092 = 1
1199092 = 1
119909 = plusmn 1
The desired area would be
int (1199101 minus 1199102) 119889119909 = int ((1 minus 31199092) minus (minus21199092)) 1198891199091
minus1
1
minus1
int (1 minus 1199092)1198891199091
minus1
(119909 minus 1199093
3)minus1
1
= ((1 minus1
3) minus (minus1 +
1
3))
2
3minus (
minus2
3)
=4
3
12 If 119910 + 3119909 = 0 is the equation of a chord of the circle 1199092 + 1199102 minus 30119909 = 0 then the equation of
the circle with this chord as diameter is
1 1199092 + 1199102 + 3119909 minus 9119910 = 0
2 1199092 + 1199102 minus 3119909 + 9119910 = 0
3 1199092 + 1199102 + 3119909 + 9119910 = 0
4 1199092 + 1199102 minus 3119909 minus 9119910 = 0
Answer (2)
Solution
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
119910 = minus3119909
41199092 + 1199102 minus 30119909 = 0
Point of intersection
1199092 + 91199092 minus 30119909 = 0
101199092 minus 30119909 = 0
10119909 (119909 minus 3) = 0
119909 = 0 or 119909 = 3
Therefore y = 0 if x = 0 and y =-9 if x = 3
Point of intersection (0 0) (3 -9)
Diametric form of circle
119909 (119909 minus 3) + 119910(119910 + 9) = 0
1199092 + 1199102 minus 3119909 + 9119910 = 0
13 The value of sum (119903 + 2) (119903 minus 3)30119903=16 is equal to
1 7775
2 7785
3 7780
4 7770
Answer (3)
Solution sum (119903 + 2) (119903 minus 3)30119903=16
= sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301
Put r = 30
in (119903(119903+1) (2119903+1)
6minus
119903(119903+1)
2minus 6119903)
30 ∙ (31)(61)
6minus 15(31) minus 6(30)
9455 minus 465 minus 180
8810
And on putting 119903 = 15
We get 15∙(16) (31)
6minus
15∙16
2minus 6 ∙ (15)
= (7) ∙ (8) ∙ (31) minus 15 ∙16
2minus 6 ∙ (15)
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
= 1240 minus 120 minus 90
= 1030
Therefore sum (1199032 minus 119903 minus 6) minus sum (1199032 minus 119903 minus 6)151
301 = 8810 minus 1030
= 7780
14 Let L be the line passing through the point P(1 2) such that its intercepted segment between
the co-ordinate axes is bisected at P If 1198711 is the line perpendicular to L and passing through the
point (-2 1) then the point of intersection of L and 1198711 is
1 (3
523
10)
2 (4
512
5)
3 (11
2029
10)
4 (3
1017
5)
Answer (2)
Solution
If P is the midpoint of the segment between the axes them point A would be (2 0) and B would be (0
4) The equation of the line would be 119909
2+119910
4= 1
That is 2119909 + 119910 = 4 hellip(i)
The line perpendicular to it would be 119909 minus 2119910 = 119896
Since it passes through (-2 1) minus2minus 2 = 119896
minus4 = 119896
there4 Line will become 119909 minus 2119910 = minus4 hellip(ii)
Solving (i) and (ii) we get (4
512
5)
15 The largest value of r for which the region represented by the set 120596 isin119862
|120596minus4minus119894| le 119903 is contained in
the region represented by the set 119911 isin119862
|119911minus1| le |119911+119894| is equal to
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
1 2radic2
2 3
2 radic2
3 radic17
4 5
2 radic2
Answer (4)
Solution
|119911 minus 1| le |119911 + 119894|
The region in show shaded right side of the line 119909 + 119910 = 0
The largest value of r would be the length of perpendicular from A (4 1) on the line 119909 + 119910 = 0
|4 + 1
radic2| =
5
radic2
= 5
2 radic2
16 Let the sum of the first three terms of an AP be 39 and the sum of its last four terms be 178 If
the first term of this AP is 10 then the median of the AP is
1 265
2 295
3 28
4 31
Answer (2)
Solution Let the AP be a a + d a + 2d helliphelliphelliphelliphelliphelliphellipℓ minus 3119889 ℓ minus 2119889 ℓ minus 119889 ℓ
Where a is the first term and ℓ is the last term
Sum of 1119904119905 3 terms is 39
3119886 + 3119889 = 39
30 + 3119889 = 30 as 119886 = 10 (Given)
119889 =9
3= 3
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
Sum of last 4 terms is 178
4ℓ minus 6119889 = 178
4ℓ minus 18 = 178
4ℓ = 196
ℓ = 49
10 13 16 19helliphellip46 49
Total number of the 10 + (n ndash 1) 3 - 49
n ndash 1 = 13
n = 14
So the median of the series would be mean of 7119905ℎ 119886119899119889 8119905ℎ term 10+6∙(3)+10+7∙3
2
28 + 31
2 =59
2 = 295
Alternate way
The median would be mean of 10 and 49 That is 295
17 For 119909 gt 0 let 119891(119909) = intlog 119905
1+119905 119889119905
119909
1 Then 119891(119909) + 119891 (
1
119909) is equal to
1 1
2 (log 119909)2
2 log 119909
3 1
4log 1199092
4 1
4 (log 119909)2
Answer (1)
Solution
119891(119909) = intlog 119905
1 + 119905
119909
1
∙ 119889119905
And 119891 (1
119909) = int
log 119905
1+119905 ∙ 119889119905
1
1199091
Put 119905 =1
119911
119889119905 = minus1
1199112 119889119905
minus1
1199092 119889119909 = 119889119905
119891(119909) = intlog 119911
1199112 (1 + 1119911)
119911
1
∙ 119889119911
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
119891(119909) = intlog 119911
119911(1 + 119911) 119889119911
119911
1
119891(119909) + 119891 (1
119909) = int log 119911 [
1
1 + 119911+
1
2(1 + 119911)] 119889119911
119909
1
= int1
119911log 119911 119889119911
119909
1
Put log 119911 = 119875 1
119911 119889119911 = 119889119901
int119875 ∙ 119889119901
119909
1
(1198752
2)1
119909
=1
2 (log 119911)1
119909 = (log 119909)2
2
18 In a certain town 25 of the families own a phone and 15 own a car 65 families own
neither a phone nor a car and 2000 families own both a car and a phone Consider the
following three statements
(a) 5 families own both a car and a phone
(b) 35 families own either a car or a phone
(c) 40 000 families live in the town
Then
1 Only (b) and (c) are correct
2 Only (a) and (b) are correct
3 All (a) (b) and (c) are correct
4 Only (a) and (c) are correct
Answer (3)
Solution Let set A contains families which own a phone and set B contain families which own a car
If 65 families own neither a phone nor a car then 35 will own either a phone or a car
there4 (119860⋃119861) = 35
Also we know that
119899(119860 cup 119861) = 119899(119860) + 119899(119861) minus 119899(119860 cap 119861)
35 = 25 + 15 - 119899(119860 cap 119861)
119899(119860 cap 119861) = 5
5 families own both phone and car and it is given to be 2000
there4 5 119900119891 119909 = 2000 5
100 119909 = 2000
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
X = 40000
Hence correct option is (a) (b) and (c) are correct
19 IF 119860 = [01 minus10] then which one of the following statements is not correct
1 1198603 + 119868 = 119860(1198603 minus 119868)
2 1198604 minus 119868 = 1198602 + 119868
3 1198602 + 119868 = 119860(1198602 minus 119868)
4 1198603 minus 119868 = 119860(119860 minus 119868)
Answer (3)
Solution A = [0 minus11 0
]
1198602 = [0 minus11 0
] [0 minus11 0
] = [minus1 00 minus1
]
1198603 = [minus1 00 minus1
] [0 minus11 0
] = [0 1minus1 0
]
1198604 = [0 1minus1 0
] [0 minus11 0
] [1 00 1
]
Option (1) 1198603 + 119868 = 119860 (1198603 minus 119868)
[01 minus10] [minus1minus1 1minus1] = [
1minus1 11]
[1minus1 11] = [
1minus1 11] hellipCorrect
Option (2) 1198604 minus 119868 = 1198602 + 119868
[0 00 0
] = [0 00 0
] hellipCorrect
Option (3) [0 00 0
] = [0 minus11 0
] [minus2 00 minus2
] = [0 2minus2 0
] hellipIncorrect
Option 4
1198603 minus 119868 = 119860(119860 minus 119868)
[minus1 minus1minus1 minus1
] = [0 minus11 0
] [minus1 minus11 minus1
] [minus1 1minus1 1
]
1198603 minus 119868 = 1198604 minus 119860
[1 1minus1 1
] = [1 00 1
] minus [0 minus11 0
]
= [1 1minus1 1
] helliphellipCorrect
20 Let X be a set containing 10 elements and P(X) be its power set If A and B are picked up at
random from P(X) with replacement then the probability that A and B have equal number of
elements is
1 (210minus1)
220
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
2 2011986210
220
3 2011986210
210
4 (210minus1)
210
Answer (2)
Solution The power set of x will contain 210 sets of which 101198620 will contain 0 element 101198621 will contain 1 element 101198622 will contain 2 element
⋮
⋮ 1011986210 will contain 10 element
So total numbers of ways in which we can select two sets with replacement is 210 times 210 = 220
And favorable cases would be 101198620 ∙101198620 +
101198621 101198621 + helliphellip
1011986210 1011986210 =
2011986210
Hence Probability would be = 2011986210
220
Hence 2011986210
220 in the correct option
21 If 2 + 3119894 is one of the roots of the equation 21199093 minus 91199092 + 119896119909 minus 13 = 0 119896 isin 119877 then the real
root of this equation
1 Exists and is equal to 1
2
2 Does not exist
3 Exists and is equal to 1
4 Exists and is equal to minus1
2
Answer (1)
Solution If 2 + 3119894 in one of the roots then 2 minus 3119894 would be other
Since coefficients of the equation are real
Let 120574 be the third root then product of roots rarr 120572 120573 120574 =13
2
(2 + 3119894) (2 minus 3119894) ∙ 120574 =13
2
(4 + 9) ∙ 120574 =13
2
120574 =1
2
The value of k will come if we
Put 119909 =1
2 in the equation
2 ∙1
8minus9
4+ 119896 ∙
1
2minus 13 = 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
119896
2= 15
119896 = 30
there4 Equation will become
21199093 minus 91199092 + 30119909 minus 13 = 0
120572120573 + 120573120574 + 120574120572 =30
2= 15
(2 + 3119894)1
2+ (2 minus 3119894)
1
2+ (2 + 3119894) (2 minus 3119894) = 15
1 +119894
2+ 1 minus
119894
2+ 13 = 15
15 = 15
Hence option (1) is correct lsquoExists and is equal to 1
2 lsquo
22 If the tangent to the conic 119910 minus 6 = 1199092 at (2 10) touches the circle 1199092 + 1199102 + 8119909 minus 2119910 = 119896 (for some fixed k) at a point (120572 120573) then (120572 120573) is
1 (minus7
176
17)
2 (minus8
172
17)
3 (minus6
1710
17)
4 (minus4
171
17)
Answer (2)
Solution The equation of tangent (T = 0) would be 1
2 (119910 + 10) minus 6 = 2119909
4119909 minus 119910 + 2 = 0
The centre of the circle is (minus4 1) and the point of touch would be the foot of perpendicular from
(minus4 1) on 4119909 minus 119910 + 2 = 0 119909 + 4
4=119910 minus 1
minus1= minus(
minus16 minus 1 + 2
42 + 12)
119909+4
4=15
17 and
119910minus1
minus1=15
17
119909 = minus8
17 119910 =
minus15
17+ 1 =
2
17
Hence option (minus8
172
17) is correct
23 The number of ways of selecting 15 teams from 15 men and 15 women such that each team
consists of a man and a woman is
1 1960
2 1240
3 1880
4 1120
Answer (2)
Solution No of ways of selecting 1119904119905 team from 15 men and 15 women 151198621
151198621 = 152
2119899119889 team- 141198621 141198621 14
2 and so on
So total number of way
12 + 22helliphelliphellip152
= 15 (16) (31)
6
= (5) ∙ (8) ∙ (31)
1240
Hence option 1240 is correct
24 If the shortest distance between the line 119909minus1
120572=
119910+1
minus1=119911
1 (120572 ne minus1) and 119909 + 119910 + 119911 + 1 = 0 =
2119909 minus 119910 + 119911 + 3 119894119904 1
radic3 then a value of 120572 is
1 minus19
16
2 32
19
3 minus16
19
4 19
32
Answer (2)
Solution Let us change the line into symmetric form
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
30 If 119910 (119909) is the solution of the differential equation (119909 + 2)119889119910
119889119909= 1199092 + 4119909 minus 9 119909 ne minus2 and
119910(0) = 0 then 119910(minus4) is equal to
1 -1
2 1
3 0
4 2
Answer (3)
Solution
(119909 + 2) ∙119889119910
119889119909= 1199092 + 4119909 + 4 minus 13
119889119910
119889119909= (119909 + 2)2
(119909 + 2)minus
13
(119909 + 2)
119889119910 = ((119909 + 2) minus13
119909119898)
119889119909
119910 =1199092
2+ 2119909 minus 13 log119890|(119909 + 2)| + 119862
If 119909 = 0 then 119910 = 0
0 = 0 + 0 minus 13 119897119900119892|2| + 119862
119888 ∶ 13 log(2)
If 119909 = minus4 then 119910
119910 =16
2minus 8 minus 13 log|minus2| + 13 log |2|
119910 = 0
Hence as is option 0
25 The distance from the origin of the normal to the curve 119909 = 2 cos 119905 + 2119905 sin 119905 119910 =
2 sin 119905 minus 2119905 cos 119905 119886119905 119905 =120587
4 is
1 radic2
2 2radic2 3 4
4 2
Answer (4)
Solution at 119905 =120587
4
119909 = 21
radic2+ 2
120587
4 = (radic2 +
120587
2radic2) = (
8 + 120587
2radic2)
119910 = 21
radic2minus 2
120587
4 ∙ 1
radic2 = (radic2 minus
120587
2radic2) minus (
8 minus 120587
2radic2)
119889119910
119889119909= 2 cos 119905 minus 2 [cos 119905 + 119905 (minus sin 119905)] = 2119905 sin 119905
119889119909
119889119905= minus2 sin 119905 + 2 [sin 119905 + 119905 ∙ cos 119905] = 2119905 cos 119905
119889119910
119889119909= tan 119905 119886119899119889 119905 =
120587
4 119886119899119889 tan
120587
4= 1
119889119910
119889119909= 1 Slope of tangent is 1 amp therefore slope of normal would be -1
Equation of normal 119910 minus (8minus120587
2radic 2) = minus1 (119909 minus (
8+120587
2radic2))
119909 + 119910 = 119905(8 + 120587)
2radic2+ (
8 minus 120587
2radic2)
119909 + 119910 =16
2radic2 and distance from origin
16
2radic2 radic2 = 4
26 An ellipse passes through the foci of the hyperbola 91199092 minus 41199102 = 36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively If the product of
eccentricities of the two conics is 1
2 then which of the following points does not lie on the
ellipse
1 (radic39
2 radic3)
2 (1
2 radic13
radic3
2)
3 (radic13
2 radic6)
4 (radic13 0)
Answer (2)
Solution Equation of the hyperbola
1199092
4minus1199102
9= 1
Focus of hyperbola (ae 0) and (-ae 0)
a = 2 119890 = radic1 +9
4=
radic13
2
there4 Focus would be (+radic13
2 0) 119886119899119889 (minus
radic13
2 0)
Product of eccentricity would be
radic13
2 ∙ 1198901 =
1
2
there4 1198901 = 1
radic13
As the major amp minor axis of the ellipse coin side with focus of the hyperbola then the value of a for
ellipse would be radic13
119890 = radic1 minus1198872
1198862
1198872
13=12
13
1
radic3= radic1 minus
1198872
13
1198872 = 12
1
13= 1 minus
1198872
13
there4 Equation of the ellipse would be
1199092
13+1199102
12= 1
Option (i) 39
4 ∙(13)+
3
12= 1
Satisfies the equation hence it lies on the ellipse
Option (ii) 13
4 (13)+
3
412= 1
does not lie on the ellipse
Option (iii) 13
2(13)+
6
12= 1 satisfy
Option (iv) 13
13+ 0 = 1 satisfy
So option (1
2 radic13
radic3
2) is the answer
27 The points (08
3) (1 3) 119886119899119889 (82 30)
1 Form an obtuse angled triangle
2 Form an acute angled triangle
3 Lie on a straight line
4 Form a right angled triangle
Answer (3)
Solution The options
A B C
(08
2) (1 3) (82 30)
Are collinear as slope f AB is equal to slope of BC
3 minus83
1 minus 0= 30 minus 3
82 minus 1
1
3=27
81=1
3
Hence option (Lie on a straight line) is correct
28 If 119891(119909) minus 2 tanminus1 119909 + sinminus1 (2119909
