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SAMPLE TEST - 1

Time Allotted: 3 Hours

Maximum Marks: 360

Do not open this Test Booklet until you are asked to do so. Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

Important Instructions:

1. Immediately fill in the particulars on this page of the Test Booklet with Blue / Black Ball Point Pen. Use of pencil is strictly prohibited.

2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out

the Answer Sheet and fill in the particulars carefully. 3. The test is of 3 hours duration. 4. The Test Booklet consists of 90 questions. The maximum marks are 360. 5. There are three sections in the question paper I, II, III consisting of Physics, Chemistry and Mathematics

having 30 questions in each section of equal weightage. Each question is allotted 4 (four) marks for correct response.

6. Candidates will be awarded marks as stated above in instruction No.5 for correct response of each question.

¼ (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.

7. There is only one correct response for each question. Filling up more than one response in any question will

be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.

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FIITJEE - JEE (Main)

JEE-MAIN -SAMPLE TEST -1-PCM-2


Useful Data Chemistry:

Gas Constant R = 8.314 J K1 mol1 = 0.0821 Lit atm K1 mol1 = 1.987 2 Cal K1 mol1 Avogadro's Number Na = 6.023 1023 Planck’s Constant h = 6.626 10–34 Js = 6.25 x 10-27 erg.s 1 Faraday = 96500 Coulomb 1 calorie = 4.2 Joule 1 amu = 1.66 x 10-27 kg 1 eV = 1.6 x 10-19 J

Atomic No : H=1, D=1, Li=3, Na=11, K=19, Rb=37, Cs=55, F=9, Ca=20, He=2, O=8, Au=79. Atomic Masses: He=4, Mg=24, C=12, O=16, N=14, P=31, Br=80, Cu=63.5, Fe=56, Mn=55, Si = 28 Pb=207, u=197, Ag=108, F=19, H=2, Cl=35.5

Useful Data Physics:

Acceleration due to gravity g = 10 2m / s



Section – I (Physics) 1. A particle is projected vertically upwards with a velocity of 20 m/sec. Find the time at which the

distance travelled is twice the displacement (A) 2 + 3/4 sec. (B) 1 sec. (C) 2 + 4/3 (D) 3 sec 1. A Sol. For AB

2 2 222sv u g

30 20 20 2 10 .2s

20 20 4010 3 3

S

Now, 212

s ut gt

240 120 103 2

t t

On solving 423

t

s/2

s

A

B

C

2. Two men who can swim with a speed v1 in still water start from the middle of a river of width d

and move in opposite directions always swimming at an angle with the banks. What is the distance between them along the river when they reach the opposite banks, if the velocity of the river is v2

(A) 2

1

dvdv

cot (B) 21

1

vvcosdv

(C)

1

2

vdv

tan (D) 2

1 sinv d

v

2. D

Sol. 11

/ 2sin

dtv

Distance = 2 1 2 11

cos cos2 sin

dv v v vv

2

1 sinv d

v

3. A uniform chain of mass m hangs from a light pulley, with unequal lengths of the chain hanging

from the two sides of pulley the force exerted by moving chain on the pulley is (A) = mg (B) > mg (C) < mg (D) either b or c depending upon acceleration of

chain 3. C Sol. Force is less than mg as it is accelerated



4. In the figure ball A is released from rest, when the spring is at its natural length. For the block B of mass M to leave contact with the ground at some stage, the minimum mass of A must be

(A) 2M (B) M (C) M/2 (D) a function of M and force constant of spring.

A

M B 4. C Sol. For B

Just to loose contact MgMg Kx xK

Applying WET

g sWD WD KE

21 02

Kx mgx

2Mm

Kx

MgN

5. A system is shown in figure pulleys and strings are ideal system

is released from rest a1 acceleration of 2 kg, a2 = acceleration of 3 kg

(i) a1 = 2a2 (ii) a2 = 2a1 (iii) a1 = a2 = 0 (iv) Tension T in the string = 15 N (v) frictional force between 2 kg & incline = 5N (vi) frictional force between 2 kg & incline = 15N

2 kg

300

T

3 kg

s = 3 / 2

(A) (ii), (iv) & (vi) are correct (B) (iii), (v) are correct (C) (iii), (iv) & (v) are correct (D) (i), (iv) & (vi) are correct 5. D Sol. Using constraint

12

02

a a

1 22a a ….(1) Applying Newton’s law and solving

15T f N

T

3g

a23

N2

f2 sing

2 cosg

1a

6. A bullet of mass 10 gm is fired from a rifle with a velocity of 800 m/s. After passing through a mud

wall 180 cm thick, the velocity drops to 100 m/s. The average resistance of the wall is (A) 750 N (B) 1250 N (C) 1750 N (D) 2250 N 6. C Sol. Using Kinematics find time taken by bullet

Then, 1750P F Nt

7. A body is gently dropped on a conveyor belt moving 3 m/s. If = 0.5 how far will the body move

relative to the belt before coming to rest? (g = 10m/s2) (A) 0.3 m (B) 0.6 m (C) 0.9 m (D) 0.8 m 7. C



Sol. With respect to belt 2 2 2v u as 20 3 2 g s

9s m 8. Consider the disc kept on a rough horizontal surface as

shown in the diagram. If a horizontal force ‘F’ has to be applied such that the disc starts pure rolling, what should be the value of ‘h’ ?

F

R

h

(A) R (B) R/3 (C) R / 2 (D) Body can’t start pure rolling for any value of

‘h’ 8. C Sol. For Pure rolling CMa Rh

2

.

32

R F h RFmRm

On solving, 2Rh

9. Moment of inertia of a half shell of mass ‘M’ about an

axis tangential to it, as shown, would be

R

O

(A) 2MR32

(B) 2MR31

(C) MR2 (D) none of these

9. B Sol. Applying parallel axis theorem

213

I MR

10. Consider ‘3’ bodies namely a disc ‘A’, a sphere ‘B’ and a hollow cylinder all

with same mass and radius being released from top of fixed inclined plane. If tA, tB & tc be the time they take to reach the bottom, find the correct alternative for each of given situations.

In absence of any friction (A) tC > tA > tB (B) tC < tA < tB (C) tC = tA = tB (D) none of these

10. C Sol. In absence of friction only mg is acting through COM. So, this case of sliding in all cases and time taken will be same as CMa is same.



11. A small ball starts rolling on an inclined track which becomes loop if radius R in vertical plane.

h=2RR

(A) speed of the ball at highest point is zero but and highest point is 2R above the ground. (B) speed of the ball at highest point is non zero but highest point is 2R above the ground (C) speed of the ball is along horizontal at highest point and highest point is less than the 2R

above the ground. (D) speed of the ball is along horizontal at highest point but height of highest point above the

ground can not be calculated. 11. C Sol. In Pure rolling, energy is conserved . .PE T KE R KE 12. A double star consists of two stars having mass M and 2M. The distance between their centre is

equal to r. They revolve under their mutual gravitational interaction. Then which of the following statements is/are correct.

(A) heavier star revolves in orbit of radius r/3 (B) kinetic energy of heavier star is twice of that of the other star.

(C) lighter star revolves in orbit of radius 2r5

(D) all above are correct. 12. A Sol. Taking M as origin applying COM

0 2 2

3 3CM

M m rx r

M

So, from heavier black 3r

13. Three small identical bodies each of mass m are moving in circular orbit around a fixed point with

same angular velocity under their gravitational interaction. If the separation between any two bodies is R, the total energy possessed by the system is given by

(A) R2

GM3 2 (B)

R4GM3 2

(C) 30cosR2GM3 2

(D) R

GM3 2

13. A

Sol. Total potential energy = 23GM

R

Total K.E. = 23

2GM

R

So, total energy = 23.

2GMKE P E

R

14. A man of mass M stands at one end of a plank of length L which lies at rest on a frictionless

surface. The man walks to the other end of the plank. If the mass of the plank is M/3, the distance that the man moves relative to the ground is

(A) 4L3

(B) 5L4

(C) 4L

(D) 3L

14. C



Sol. Let the distance moved by plank be x. Now, there is no external force so COM Will not move

1 1 2 2

1 2CM

m n m nxm m

30 4

3

MM L x x

M

3MxML Mx

44.3

ML Mx

34Lx

So, distance moved by person wrt. 4L

15. The centre of mass of a half disc shown is at C while O is the centre. Thus OC

is (A) R/2 (B) 2R/

(C) 3R4

(D) none of the above

R

O

C

15. C

Sol. COM of sector = 4 sin

23

R

Here

43

Rr

16. If the tension in a stretched string fixed at both ends in changed by 20%, the fundamental

frequency is found to change by 15 Hz. Then the (A) original frequency is 150 Hz (B) velocity of propagation of the transverse wave along the string changes by 5 % (C) velocity of propagation of the transverse wave along the string change by 10 % (D) fundamental wavelength on the string does not change. 16. B Sol. '

2 1.2T T

Tv

17. Beats are produced by two progressive waves. Maximum loudness at the waxing is x times the

loudness of each wave. The value of x is (A) 1 (B) 2 (C) 2 (D) 4 17. D Sol. At maximum net amplitude = 2A Loudness 2A



18. A thermodynamic system undergoes a cyclic

process as shown in the figure. The cycle consists of two closed loops. Over one complete cycle, the system performs

(A) positive work (B)negative work (C) zero work (D) nothing can be predicted V

P

18. B Sol. Area under PV graph is greater is anticlockwise direction. So, net W.D. is negative 19. A half ring of radius R is charged with a linear charged density . The field at the centre is (A) 0 (B) k/R (C) 2k/R (D) k/R 19. C Sol. dl Rd

.dq R

2

.K dqdE

R

/2

2/2

2. cosdq KE KR R

d

20. The maximum electric field strength E due to a uniformly charged ring of radius r, happens at a

distance x, where value of x is (x is measured from the centre of the ring)

(A) x = R (B) x = R/2 (C) x = 2

R

(D) x = 2 R

20. C

Sol. 3/22 2

KQxER x

For maximum 0dEdx

So, 2

Rx

21. Two charges +q and q are placed fixed on the corner of a

massless rigid rod of length 2L.Calculate the potential energy of the dipole thus formed.

+q q

2L

(A) 2

2

o L4q

41

(B) L2

q4

1 2

o (C)

L2q

21 2

o (D) None of these.

21. B

Sol. PE = 2

0

14 2

KqV qr L



22. If the above discussed dipole is placed in a uniform electric field E as shown, calculate the proper potential energy of the dipole.

(A) L2

q4

1 2

o (B)

L2q

41qLE4

2

o

(C) L2

q4

1qLE22

o (D)

2

o

1 q4qLE4 2L

+q

2L

E

22. D Sol. Net PE = PE due to EF + PE of system

2

0

144 2

qqLEL

23. In the shown network current through 20 resistor equals

(A) 32

A (B) 9 A2

(C) 1A (D) 2 A3

10

20 V10 V

20 23. A Sol. Applying KVL

32

I A

24. Equivalent resistance between the points A and B is (A) 1 (B) 2 (C) 3 (D) 4

BA 3 3

3 3

3

3

3

3

3

3

24. D Sol. Using node removal method eqR h 25. The two rails of a railway track; insulated from each other and the ground, are connected to

millivoltmeter. What is the reading of the millivoltmeter when a train passes at a speed of 180 km/hr along the track, given that the horizontal component of earth’s magnetic field is 0.2 10-4 wb/m2 and rails are separated by 1 metre

(A) 10-2 V (B) 10 mV (C) 1 V (D) 1 mV 25. D

Sol. . 1dl B v mV

26. In an R-L-C circuit v = 20 sin (314 t + 5/6) and i = 10 sin (314 t + 2/3) The power factor of the circuit is (A) 0.5 (B) 0.966 (C) 0.866 (D) 1 26. C Sol. Conceptual based 27. A circular current carrying coil has a radius a. The distance from the centre of coil, on its axis,

where the magnetic induction will be 1/8 th of its value at centre of coil is (A) 3a (B) + a3 (C) a3 (D) 3/a2 27. C



Sol.

20 0

3/22 2 22

IR IBRR x

B

3x a 28. In photoelectric effect, the photo current (A) increases with increase of frequency of incident photon

(B) decreases with increase of frequency of incident photon (C) does not depend on the frequency of the photon but depends only on intensity of incident light. (D) depends both on intensity and frequency of photon.

28. C Sol. Conceptual based 29. If refractive index of water is 4/3 and glass is 5/3 then critical angle so that light travelling form

glass to water is completely reflected is (A) sin-1 (4/5) (B) sin-1 (5/4) (C) sin-1 (3/5) (D) sin-1 (5/3) 29. A Sol. Applying 1 2sin sini r

4sin5

30. When a ray of light enters a glass slab from air (A) its wavelength decreases (B) its wavelength increases (C) its frequency increases (D) neither wavelength nor frequency changes 30. A Sol. v f

reqf remains constant

Here volume increases, so decreases

space for rough work



Section – II (Chemistry) 1. For the equations 2 4 12 ;C diamond H g CH g H

2 4 22 ;C graphite H g CH g H Predict whether (A) 1 2H H (B) 1 2H H (C) 1 2H H (D) 1 2 23vap dissH H H H H 1. B Sol. This is because diamond is more stable than graphite and has more energy content. 2. Which of the following can behave as both electrophile and nucleophile? (A) 3CH C N (B) 3CH OH (C) 2 3CH CH CH (D) 3 2CH NH 2. A Sol. 3CH CN can behave as both electrophile and nucleophile. 3. Steam reacts with iron at high temperature as follows: 2 3 4 23 4 4Fe s H O g Fe O s H g The correct expression for the equilibrium constant (KP) is:

(A) 2

22

2

H

H O

p

p (B)

2

2

4

4

H

H O

p

p (C)

2

2

4

3 4

4 3

H

H O

p Fe O

p Fe (D)

3 4Fe OFe

3. B Sol. Molar concentration or partial pressure of solid = 1. 4. In which of the following solvents, AgBr has the maximum solubility? (A) 10-3 M NaBr (B) 10-3 M NH4OH (C) Pure water (D) 310 M HBr 4. B Sol. AgBr reacts with NH4OH to form a soluble complex. 4 3 22

2 2 .AgBr NH OH Ag NH Br H O 5. The electrode potential of a copper wire dipped in 0.1 M CuSO4 solution at 250C (the standard

reduction potential of copper is 0.34V): (A) 0.34V (B) 0.31V (C) 0.349 V (D) 0.28 V 5. B Sol. The electrode reaction written as reduction reaction is 2 2 2Cu e Cu n

Applying Nernst equation, E = 02

0.0591 1log 12

E Cu sCu

0.0591 10.34 log2 0.1

0.34 0.0295 1 0.3105V 6. The strongest reducing agent among the following is (A) F- (B) Cl- (C) Br- (D) I- 6. D Sol. I- can losses electrons easily (undergoes oxidation easily). So strong reducing agent



7. Diazonium salt decomposes as: 6 5 2 6 5 2C H N Cl C H Cl N At 00 ,C the evolution of N2 becomes two times faster when the initial concentration of the salt is

doubled., Therefore, it is (A) a first order reaction (B) a second order reaction (C) independent of the initial concentration of the salt (D) a zero order reaction. 7. A Sol. Because rate 1conc 8. If 32.68 10 mole of a solution containing an ion nA requires 31.61 10 mole of 4MnO for

the oxidation of nA to 3AO in an acidic medium, then what is the value of n? (A) 3 (B) 2 (C) 5 (D) 4 8. B Sol. 2

4 3nA MnO AO Mn

mole 3 32.68 10 1.61 10

3 32.68 10 5 1.61 10 5n

5 3n 2n 9. The equivalent weight of MnSO4 is half its molecular weight when it is converted to (A) 2 3Mn O (B) 2MnO (C) 4MnO (D) 2

4MnO 9. B Sol. 2 4 2Mn Mn e 2nf

..2

M wtE wt

10. The largest number of molecules is in (A) 36 g of H2O (B) 28 g of CO (C) 46 g of CH3CH2OH (D) 54 g of N2O5 (Use atomic weight: O = 16, C = 12, N = 14, H = 1) 10. A

Sol. 2

36 218H O AV AVN N N

2828CO AV AVN N N

2 5

4646C H OH AV AVN N N

2 5

54 0.528 80N O AV AVN N N

11. Consider a titration of 2 2 7K Cr O with acidified Mohr’s salt solution 4 4 4 22. .6FeSO NH SO H O

using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is

2 2 3 32 7

HFe Cr O Fe Cr

(A) 3 (B) 4 (C) 5 (D) 6



11. D Sol. neq Mohr salt = neq 2

2 7Cr O

1 1 6n 6n 12. The normality of 0.3 M phosphorus acid 3 3H PO is (A) 0.1 (B) 0.9 (C) 0.3 (D) 0.6 12. D Sol.

OH P OH

H

O

2nf 0.3 2 0.6N 13. The energy of hydrogen atom in the ground state is 13.6 eV. Its energy corresponding to the

quantum number n = 5 is (A) 0.54eV (B) 5.40 eV (C) 0.85eV (D) 2.72eV ] 13. A

Sol. 2

13.6nE

n

14. If r1 is the radius of first orbit of hydrogen atom, then the radii of second, third and fourth orbit in

terms of r1 are (A) 1 1 18 , 27 ,64r r r (B) 1 1 12 ,6 ,8r r r (C) 1 1 14 ,9 ,16r r r (D) 1 1 1, 2 ,3r r r 14. C Sol. 2

1nr r n 15. If kinetic energy of an electron is increased nine times, the de-Broglie wavelength associated with

it would become

(A) 3 times (B) 9 times (C) 13

times (D) 19

times

15. C

Sol. 1KE

16. Which electronic level allows the hydrogen atom to absorb a photon but not emit a photon? (A) 2s (B) 2p (C) 1s (D) 3d 16. C Sol. 1s is the lowest energy level

17.

CH3

CH3

Br 3

2 reaction

NaOMeCH OHE

Product. Product of the reaction is:



(A)

(B)

(C) OMe

(D) No reaction

17. D Sol. Since there is no H present anti to Br, hence 2E doesn’t occur.

18.

Cl

CH3

NaOEt

Major product of the reaction is:

(A)

CH3

(B)

CH3

(C)

CH3

(D)

CH3

18. B

Sol. The product obtained are:

CH3 CH3

and

(major) (minor) 19. Which of the given options best describes the product of the following reaction?

Ph

CH3

CD3Br

2

K t BuOE

Product

(A) Absolute configuration has been inverted (B) Absolute configuration has been retained (C) Racemisation (loss of absolute configuration) (D) Loss of chirality has occurred (the product is achiral) 19. D Sol.

Ph

CH3

CD3Br

2

K t BuOE

Ph

CH2

CD3



20. Which one of the following undergoes nucleophlic aromatic substitution at the fastest rate?

(A)

Cl

NO2

(B)

Cl

Cl

(C)

Cl

CH3

(D)

Cl

OMe 20. A Sol. The electron – withdrawing groups make nucleophilic aromatic substitution proceed at the fastest

rate.

21.

Br

NO2

Br

OH A

Product of the following reaction is:

(A)

Br

OH

NO2

(B)

OH

Br

NO2

(C)

Br

Br

NO2OH

(D)

Br

NO2

OH

21. B Sol. The electron-withdrawing groups at ortho or para position to the leaving group help nucleophilic

aromatic substitution proceed at a faster rate.

22. OH OH

Ph

Ph 2 4

99% yield

H SO P

The product (P) is:

(A) O

Ph

Ph

(B) C

Ph

Ph

(C) O

Ph

Ph

(D)

CPh

Ph

OH

22. A

Sol. OH OH

Ph

Ph H 2H O

OH

Ph

Ph H

OPh

Ph

23. An organic compound ‘B’ is formed by the reaction of ethyl magnesium iodide 3 2CH CH MgI

with a substance ‘A’, followed by treatment with dil. aqueous acid. Compound ‘B’ doesnot react with PCC. Identify A?



(A)

CH3 C

O

H

(B)

CH3CH2 C

O

CH3

(C) 2CH O (D)

CH2 CH2

O

23. B

Sol.

CH3CH2 C

O

CH3 EtMgI CH3CH2 C

O(MgI)

CH3

Et

dil aqacid CH3CH2 C

OH

CH3

Et

No reaction

A B

PCC

24. Ph C

O

OH CH3 O H 2 .H X H O

18

Identify X

(A) X Ph C O

O

CH3 (Esterification)

(B) X Ph C O

O

CH3 (Esterification)18

(C) X Ph C O

O

CH3 (Saponification)18

(D) X Ph C O

O

CH3 (Hydrolysis)

24. B

Sol.

Ph C

OH

OH CH3 O H 18

Ph C

OH

O

OH H

CH318 Ph C

OH2

O

OH

CH318

Ph C O

O

CH3

18

X

25.

32

2

/2 2 2 2 pyridine

4 7

O ZnSOClH OCH CH CH CH CH OH A B

C H ClO

Compound (B) is



(A)

CH3 C

O

CH2 CH2

CH3 (B)

HC CH2 CH2 CH2

O

CH2 Cl (C)

HC CH2 CH2 CH2

O

Cl (D)

HC CH2 CH

Cl

CH3

O

25. C Sol.

3

2

/2 2 2 2 2 2 2

A B

O ZnH OCH CH CH CH CH Cl HCHO CHO CH CH CH Cl

26. Equimolar solutions of two non – electrolytes in the same solvent have (A) same b.pt but different f.pt (B) same f.pt but different b.pt (C) same b.pt and same f.pt (D) different b.pt and different f.pt 26. C Sol. fT and bT are constant for a solvent. 27. The degree of dissociation of weak electrolyte x yA B is related to van’t Hoff’s factor (i) by

the expression:

(A)

11

ix y

(B)

1

1i

x y

(C)

11

x yi

(D) 1

1x y

i

27. A Sol. y x

x yA B xA yB

1 x y

1i x y

1 1i x y

11

ix y

28. When 20 gm of naphthoic acid 11 8 2C H O is dissolved in 50 gm of benzene 1.72fk , a

freezing point depression of 2k is observed. The van’t Hoff factor (i) is (A) 0.5 (B) 1 (C) 2 (D) 3 28. A Sol. f fT K molality i

20 10002 1.72

172 50i

0.5i



29. A 0.004 M solution of 2 4Na SO is isotonic with a 0.01 M solution of glucose at same temperature.

The apparent degree of dissociation of 2 4Na SO is (A) 25% (B) 50% (C) 75% (D) 85% 29. C Sol.

2 4 cosNa SO Glu e

1 2CRT CRT

0.004 1 2 0.01

0.75 75% 30. If equal volumes of 2BaCl and NaF solutions are mixed, which of these combination will not

give a precipitate. 72 1.7 10 of BaFspK

(A) 3210 M BaCl and 22 10 M NaF (B) 3

210 M BaCl and 21.5 10 M NaF

(C) 221.5 10 M BaCl and 210 M NaF (D) 2

22 10 M BaCl and 22 10 M NaF 30. C Sol. For precipitation sp spQ K

pace for rough work



Section – III (Mathematics) 1.

n n 1n

r r

r 1

C Cr 1 n 1

equals to

(A) 1n 1

(B) nn 1

(C) (n 2)n 1

(D) 1n 1

1. B

Sol.

n n 1n

r r

r 1

C Cr 1 n 1

= n 1 n 1n

r 1 r

r 1

C Cn 1 n 1

= n

n 1 n 1r 1 r

r 1

1 ( C C )n 1

=

1 n(1 n 1)n 1 n 1

.

2. Statement -1: If a, b, c are distinct and , ,x y z are not all zero, then 0ax by cz 0bx cy az 0cx ay bz Gives a + b + c 0 Statement – 2: 2 2 2a b c ab bc ca if a, b, c are distinct. (A) Statement-1 is true, Statement-2 is true, but Statement-2 is a correct explanation for

Statement-1. (B) Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation for

Statement-1. (C) Statement-1 is true, Statement-2 is false. (D) Statement-1 is false, Statement-2 is true. 2. D

Sol. Let D = a b cb c ac a b

= 3 3 3 3a b c abc

2 2 2a b c a b c ab bc ca

Now 2 2 2 0a b c ab bc ca the given system of equations can have non-zero solution (x, y, z are not all zero) If D = 0, . .i e If 0a b c Statement -1 is false. Again 2 2 2a b c ab bc ca

2 2 21 2 2 2 2 2 22

a b c ab bc ca

2 2 21 02

a b b c c a

a, b, c are all different

2 2 2a b c ab bc ca Statement-2 is true. D holds.

3. If f x is an odd periodic function with period 2, then 4f equals (A) 0 (B) 2 (C) 4 (D) - 4



3. A Sol. 4 2 0 0f f f

4. The solution of the differential equation 2 3dyx ydx

represents

(A) straight lines (B) circle (C) parabola (D) ellipse 4. C

Sol. 3

2 2dy ydx x x

1. .I Fx

3y c x

5. If [x] denotes the greatest integer less than or equal to x, then the value of 5

1

3x dx is

(A) 1 (B) 2 (C) 4 (D) 8 5. B

Sol. 5 2

1 2

3

x dx x dx

2

0

2 x dx

6. A tower AB leans towards west making an angle with the vertical. The angular elevation of B,

the top most point of the tower, is , as observed from a point C due east of A at a distance d from A. If the angular elevation of B from a point due east of C is at a distance 2d from C is , then

(A) 2tan 2cot cot (B) 2tan 3cot cot (C) tan cot cot (D) None of these 6. B Sol.

Let h be height of tower then

tan 90 ... 1hx

tan ... 2hx d

tan ... 33

hd x

Solve (1), (2) and (3) 2 tan 3cot cot

7. Let 3 2 25sinf x x ax bx x be an increasing function in the set of real number R. Then (A) 2 3 15 0a b (B) 2 3 15 0a b (C) 2 3 15 0a b (D) 0a and 0b 7. C

West A

B

C

x d 2d

East

h

90



Sol. 23 2 10sin cosdy x ax b x xdx

and discriminant < 0

8. If 1 2 3, , ,...a a a are in H.P. and 1

,n

r kr

f k a a

then

31 2, , ,...,1 2 3

na aa af f f f n

are in

(A) A.P. (B) G.P. (C) H.P. (D) None of these 8. C

Sol. Let 1

n

r nr

a S

1 2 3

1 1 1 1, , ,....na a a a

in A.P.

1 2 3

, , ,.....,n n n n

n

S S S Sa a a a

in A.P.

1 2 3

1, 1, 1...n n nS S Sa a a

in A.P.

1 2 3

1 1 1, , ......1 1 1n n nS S S

a a a

H.P.

31 2, , ,.....1 2 3

aa af f f

in H.P.

9. A problem in mathematics is given to 3 students whose chances of solving individually are 1 1,2 3

and 1 .4

The probability that the problem will be solved atleast by one is

(A) 14

(B) 124

(C) 2334

(D) 34

9. D Sol. 1 – Nobody solves the problem

1 1 11 1 1 12 3 4

10. If a I and the equation 10 1 0x a x has integral roots, then the values of a are (A) 6, 8 (B) 8, 10 (C) 10, 12 (D) 8, 12 10. D Sol. 10 1 1x a x

so either 1x a and 10 1x or 10 1x and 1x a 11. If 2 6 2x y touches the hyperbola 2 22 4x y , then the point of contact is

(A) 2, 6 (B) 5,2 6 (C) 1 1,2 6

(D) 4, 6

11. D



Sol. Put 2 6

2yx

in equation of hyperbola

6y 4x 12. Consider a circle with its centre lying on the focus of the parabola 2 2y px such that it touches

the directrix of the parabola. Then a point of intersection of the circle and the parabola is

(A) , or ,2 2p pp p

(B) ,

2 2p p

(C) ,2p p

(D) ,

2 2p p

12. A Sol. If focal chord is diameter of the circle then it touches the directrix of the parabola

13. If 2,a a falls inside the angle made by the linear equations , 02xy x and 3 , 0y x x , then

a belongs to

(A) 13,2

(B) 10,2

(C) 3, (D) 1 ,32

13. D Sol.

Locus of point 2a,a is 2y x

2y x and 2xy

10,2

x

2 , 3y x y x 0,3x

1 32 a

14. The straight lines whose direction cosines satisfy 0, 0al bm cn fmn gnl hlm are

perpendicular if

(A) 0af bg ch (B) 2 2 2

0a b cf g h

(C) 0f g ha b c (D) 2 2 2 0a f b g c h

14. C

Sol. Put bm cnl

a

in the another equation,

0

bm cnfmn gn hma

2 2 0 bhm mn bg ch af cgn

1 2 1 2 1 2

/ / /

m m n n l lg b h c f a

15. Domain of derivative of the function f (x) = |sin–1 (2x2 – 1)| is

3y x

2y x

2xy

1 1,2 4

(3, 9)



(A) [–1, 1] (B) [–1, 1] ~ 10,2

(C) [–1, 1] ~ {0} (D) [–1, 1] ~ 12

15. B Sol. y = |sin–1 (2x2 – 1)|

y = 1 2

1 2 2

sin (2x 1) 4xsin (2x 1) | 2x | 1 x

x 0 and sin–1 (2x2 – 1) 0 and |2x2 – 1| 1 |x| 1

x 0, x 12

.

16. 12

r 1

1tanr 5r 7

equals to

(A) tan–1 3 (B) 4

(C) 1 1sin10

(D) cot–1 2

16. C

Sol. an = 1 11 (n 3) (n 2)tan tan1 (n 3)(n 2) 1 (n 3)(n 2)

= tan–1 (n + 3) – tan–1 (n + 2) Sn = tan–1 (n + 3) – tan–1 3

S = 1 1 1 1tan 3 cot 3 sin2 10

.

17. If P1, P2, P3 be the lengths of perpendiculars from the vertices of the triangle ABC to the opposite sides, then

(A) P1P2P3 = abc (B) P1P2P3 = 8R3

(C) P1P2P3 =2 2 2

3a b c

R (D) P1P2P3 =

2 2 2

3a b c

8R

(where R is circumradius of triangle ABC). 17. D Sol. P1 = c sin B, P2 = a sin C, P3 = b sin A

P1P2P3 = abc sin A sin B sin C = 2 2 2

3a b c

8R.

18. A unit vector is orthogonal to ˆ ˆ ˆ5i 2j 6k and is coplanar to ˆ ˆ ˆ2i j k and ˆ ˆ ˆi j k , then the

vector is

(A) ˆ ˆ3 j k10 (B)

ˆ ˆ2i 5 j29

(C) ˆ ˆ6 i 5k

61 (D)

ˆ ˆ ˆ2i 2 j k3

18. A Sol. Let a

be the unit vector



ˆ ˆ ˆ ˆ ˆ ˆa 2i j k i j k

| a |

= 1 (2 + )2 + ( – )2 + ( + )2 = 1

62 + 4 + 32 = 1 a

is orthogonal to ˆ ˆ ˆ5i 2j 6k so, 5(2 + ) + 2( – ) + 6( + ) = 0

18 + 9 = 0 2 = – 62 – 82 + 122 = 1 102 = 1

= 110

= 210

ˆ ˆ3 j ka10

.

19. If the curve y = x2 + bx + c touches the straight line y = x at the point (1, 1), then b and c are given by

(A) –1, 1 (B) –1, 2 (C) 2, 1 (D) 1, 1 19. A Sol. y = x touches y = x2 + bx + c 1 = 1 + b + c b + c = 0 ……(1)

(1, 1)

dy 2x b 2 b 1 b 1dx

, c = 1.

(b, c) = (–1, 1). 20. In the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 ....... , where n consecutive terms have the value n, the

150th term is (A) 17 (B) 16 (C) 18 (D) none of these 20. A Sol. 1, 22,333, 4444,........ 1 2 3 4 .......

when last time 16 will appear that will be 16 17

2

= 136th term

so after that 17 will repeat from 137th term to 153th term Ans is 17

21. The value of

245 n2

, n N where {.} denotes the fractional part of x, is

(A) 5/24 (B) 9/24 (C) 1/24 (D) None of these 21. C

Sol. 24 1

24

n

22. If z1 and z2 are two complex numbers satisfying the equation 1zzzz

21

21

, then 2

1

zz

is a number

which is (A) Positive real (B) Negative real (C) Zero or purely imaginary (D) None of these 22. C



Sol. Let 1

2

1, 1 1 11

z tt t tz t

locus of t is line y axis 23. The number of ways of switching the network such that the bulb glows is

(A) 61 (B) 60 (C) 63 (D) None of these

23. A Sol. 3 3 3 3

0 1 2 3 3 1 3 1 1 3 61C C C C

24. If 2 sin .sec 3 = tan 3 -tan , then 2[sin .sec 3+sin 3.sec 32 +..+sin 3n –1 .sec 3n] =

(A) tan 3n -tan (B) tan 3n -ntan (C) tan 3n -tan 3n –1 (D) 12

(tan 3n -tan )

24. A Sol. Using method of difference, 2 22sin 3 sec3 tan 3 tan 3 and so on.

25. The determinate cosC tan A 0sin B 0 tan A

0 sin B cos C has the value where A, B, C are angles of a triangle

(A) 0 (B) 1 (C) sinA. sinB (D) cosA cosB cosC 25. A Sol. Expand along the 1st row cos tan sin tan sin cosC A B A B C 26. The equation of the image of the circle 2 2 16 24 183 0x y x y by the line mirror 4 7 13 0x y is (A) 2 2 32 4 235 0x y x y (B) 2 2 32 4 235 0x y x y (C) 2 2 32 4 235 0x y x y (D) None of these 26. A Sol. Radius will remain same, image of center (-8, 12) is (-16, -2) w.r.t. line 4 7 13 0x y

27. For hyperbola 2 2

2 2 1,cos sin

x y

which of the following remains constant with change in ' ' ?

(A) abscissae of vertices (B) abscissae of foci (C) eccentricity (D) directrix 27. B Sol. eccentricity = sec so foci are constant



28. If

sin, 0

0, 0

xx

xf xx

Where, [x] denotes the greatest integer less than or equal to x, then 0

limx

f x

equals:

(A) 1 (B) 0 (C) -1 (D) None of these 28. D

Sol.

sin, 0

0 , 0

xx

xf xx

0

lim 0

x

f x

0

sin 1lim sin1

1

xf x

29. For a real number y, let [y] denote the greatest integer less than or equal to y. Then the function

2

tan

1

xf x

x

is:

(A) discontinuous at some x (B) continuous at all x, but the derivative 'f x does not exist for some x

(C) 'f x exists for all x, but the derivative ''f x does not exist for some x

(D) ''f x exists for all x 29. D Sol. It is a zero function, 0f x

30. 2 / 3

1/ 34 / 3

1 (sin x) d(sinx)1 (sin x) is equal to

(A)

2 / 3

1 / 3

sin x 11 c2 2 sin x

(B)

2 / 31

1 / 3

sin x 11 tan c2 2 sinx

(C)

1 / 31

2 / 3

sin x 11 tan c2 2 sin x

(D) none of these

30. B

Sol. Let 11/33sin sinx t d x dt

Using algebraic twins, 2

1

1 2

d tt

tt

space for rough work

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