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SOLUTIONS Joint Entrance Exam | IITJEE-2019
08th APRIL 2019 | Morning Session
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Joint Entrance Exam | JEE Mains 2019
PART-A PHYSICS
1.(3) Refer NCERT
2.(1) Let us assume x grams of water vaporizer. amount of water frozen Heat gained by vaporized water = Heat lost by frozen water
3.(3) Let us assume and
4.(1) When catapult is released, elastic potential energy stored in cord gets converted to KE of stone. EPE lost by cord = KE gained by stone
SI units
5.(3)
Given
6.(4)
Þ (150 )= - x
Þ (150 )´ = - ´V fx L x L5 521 10 (150 ) 3.36 10´ ´ = - ´ ´x x
Þ 21 504 3.36= -x x Þ504 20.7 2024.36
= = »x g g
0bV V= =aV xV
1 2 3 0+ + =i i i
Þ 31 2
1 2 10
2 2x Ex E x E
R R R-- =
+ + =
2 4 4 02 2 2- - -
+ + =x x x
Þ 3 10 0- =x Þ 3.3=x V0 3.3- = - =a bV V x V
2 21 12 2æ öD =ç ÷è ø
!!
YA mv
Þ( )23
2 23 10
(0.2) 0.02 (20)0.42
-p ´´ ´ = ´Y Þ 6
0.02 400 0.429 10 0.04
Y -´ ´
=p´ ´ ´
63 10» ´
2max 2 1
min 2 1
I a aI a a
æ ö+= ç ÷-è ø
1
2
13
=aa
Þ 2 1
2 1
3 1 23 1
+ += =
- -a aa a
Þ max
min4I
I=
VdRn
r=
3100 / min 0.1 / min= =Q l m30.1 /
60= m s
20.1 160 (0.05)
= = ´p´
QVA
Þ 32
3
0.110 0.160 (0.05)
10-
= ´ ´´p´
R4
410
60 25 10-=
´p´ ´42 10» ´
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7.(1)
8.(1)
9.(3)
Conserver momentum,
along x :
along y :
Squaring and adding,
10.(2) Rate of heat dissipation
Rate of energy storage in inductor
Assuming initially current was zero, after time t,
,
11.(3) KE of particle = work done by force = area under F–s curve = 6.5J
2loadStressarea
mgr
= =p
3 24 3.1(2 10 )-´ p
=p´ ´
6 23.1 10 /= ´ N m
m Bt = ´ˆ=B Bi
2 ˆ[ ( ) ]= pm N r i j
Þ 2 ˆ[ . ]N r i B Rt = - p
Þ 2| | r NiBt = p
10 cos+ = q
l lh h
20 sin+ = q
l lh h
2
22
2 21 2
1æ ö=ç ÷ç ÷ ll + lè ø
hh Þ 2 22 1 2
1 1 1= +
l l l
2= i RdiiLdt
=
2 dii R iLdt
= Þ =di R idt L
1-æ ö
ç ÷= -ç ÷è ø
RtLEi e
R-
=RtLdi E e
dt L
Þ 1-
-æ öç ÷= ´ -ç ÷è ø
Rt RtL LE R Ee e
L L R
Þ12
-=
RtLe Þ ln 2=
RtL
202 210
Lt ln lnR
= =
2 2ln=
12 2 (2 3) 12
= ´ + ´ + ´
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12.(4)
(1) For first refraction by lens
(2) For reflection by mirror
(3) For second refraction by lens
image is at same position as object image is of same size and inverted.
13.(3) Using Snell’s law,
Length of fibre transverse in one refraction Length of wire = 2m
Number of refractions
14.(3)
Direction of magnetic field will be perpendicular to both electric field and direction of propagation of wave
4 40 2= - = -cm fÞ 2 40= + = +v f cm
1vmu
= = -
(60 20) 40 2= - - = - = -u cm fÞ 2 40v f cm= - = -
1vmu
= - = -
(60 20) 40 2u cm f= - - = - = -
2 40v f cmÞ = =
1vmu
Þ = = -
Þ ( 1) ( 1) ( 1) 1netm = - × - × - = -
Þ
21 sin 40 1.31sin× ° = q
Þ 2sin 0.5q » Þ 2 30q = °6 5
2cot 20 10 3 3.46 10d m- -= × q = ´ ´ » ´
Þ 52 57000
3.4 10-= »
´
=E cB
Þ 86 2 103 108
-= = ´´
B T
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15.(1)
16.(2)
slope of A will greater in magnitude at t = 0
17.(4) We know
18. BONUS Force per molecule = 2mv
Total force
19.(3) Energy from radiation
For helium
2 to 4
20.(2)
,
21.(4) Assuming no diode potential drop across resistor = 7.2 V
Zener breakdown has occurred Current through resistor
Current through resistor
Current through Zener diode
1 1 2 2 3 3 4 4
1 2 3 4
+ + +=
+ + +cMm a m a m a m aa
m m m m
! !( ) 2 ( ) 3 ( ) 4 ( )2 3 4
- + + + -=
+ + +
! !m ai m a j m ai m a jm m m m
!!2 2 ( )
10 5-
= = -!
!ai a j a i j
28 10= ´ ´AMass V 310= ´Bmass V
\ >B Amass mass
( )0 0msd d kkdt dt msq q= q-q Þ = q-q
28 10 2000= ´ ´ ´A Am s V 310 4000= ´ ´B Bm s V
\ >B B A Am s m s \
00
00 0
1c c Î= Þ Î =
Î µµ
21 2
0
14
q qFr
=Îp
2 21 3 4 21 2
0 2 2 2[ ] [ ][ ] [ ][ ][ ] [ [ ]q q A T M L T Af r MLT L
- --Î = = =
\ 1 1 3 4 2 1 2 3 20[ ] [ ][ ] [ ]- - - - -Î = =c LT M L T A M L T A
26 4 222 10 10 2 10- -= ´ ´ = ´22 22 210 2 10 2 /-= ´ ´ = N m
2 21 113.6 (1) 10.21 2æ ù= ´ - =ç úè û
eV
(2 to )n
2 21 110.2 13.6 (4) 42
æ ö= ´ - Þ =ç ÷è ø
nn
\
cosT mgq =6
3200 5 10 1tan
22 10 10Eqmg
-
-´ ´
q = = =´ ´
sinT Eqq = Þ 1tan 0.5-q =
800W
Þ Þ 800W 5.6800
A=
Þ 200W 9 5.6 3.4200 200-
= =
Þ3.4 5.6 8 10200 800 800
mA= - = Þ
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22.(1) means means Red, Black, Brown. If Red replaced by green
23.(4) Frequency will be same in both
Tension will be same in both but mass per unit length will be 4 times in 2 (as radius is twice)
24.(3) radius of the circle
FBD of any one particle
25.(4) If velocity of strip is v emf in it is
Current in it is
Resistive force exerted by
Amplitude reduces as If amplitude reduces by factor of e
The new frequency
New time period Number of oscillations
26.(3)
Using parallel axis theorem
200W 120 10´ 150 10 500´ = W
1 21 22 2
pc qc pc qcl l= Þ =
µ
\ 2 14µ = µ \1 2
T Tp q=µ µ
\ 1
2
12
pq
µ= =
µ
2a r= =
2 2 2
2 22 1
2 2 / 2GM GM Mva a a
+ = Þ1 1 1.162 2
GM GMa a
+ =
BvlBvlR
\2 2B vlB BilR
= =
20( )
b tmA t A e
-=
34
2 2 2 22 2 2 50 10 101 10
2 (0.1) (0.1)bt m mRt sm b B l
-´ ´ ´= = = = =
2 2 2
3 4 40.5 1 1' 10 10
2 50 10 10 10k bm m -
æ ö æ ö æ öw = - = - = -ç ÷ ç ÷ ç ÷´è ø è ø è ø!
2 2' 2' 10
T p p= = » »
w\
410 50002
= =
52 2 0
00
2(2 )( )5
R
CMRI dm r r dr r r pr
= = p r =ò ò3
00
0
223
R RM dm rdr r pr= = p r =ò ò
2CMI I MR= +
5 5 50 0 0
2 2 165 3 15
R R R= pr + pr = pr
5 20
3 16 2 82 15 3 5
R MR= ´ ´ pr =
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27.(4) Initial
Final
28.(1) Time period
Time to go from
29.(4)
Position of B as function of time
Position of A as function of time
Relative position of B as fraction of time
Taking magnitude square Time derivative
30.(3)
PART-B CHEMISTRY 1.(1) Ellingham diagram tells us about values (feasibility) of thermal reduction of an ore using suitable
reducing agents.
2.(4) From given data
Rate
Now, … (i)
… (ii)
… (iii) From equation (i) and (ii) we get,
1 2
KQ KQpd VR R
= - =
1 2 2 2
4 4KQ KQ KQ KQpdR R R R
æ ö æ ö= - - -ç ÷ ç ÷è ø è ø
1 2
KQ KQ VR R
= - =
2 2 1100 50
Tp p= = = =w p
1 3.32 4 12 6 300A T T Tto A s ms= - = = =
ˆ ˆ ˆ80 150 10i j ti= + -
ˆ ˆ30 50t i t j+
\ ˆ ˆ ˆ ˆ80 150 40 50i j t i t j= + - -2 2(80 40 ) (150 50 )t t= - + -
2(80 40 )( 40) 2(150 50 )( 50) 0t t= - - + - - =
Þ (8 4 )( 4) (15 5 )( 5) 0t t- - + - - = Þ 16 25 32 75t t+ = +
Þ 41 107t = Þ107 2.641
t hrs= =
max maxE d V=
4500 5 10106
d m-= = ´
12 40
12 40
15 10 5 10 8.58.86 10 10
k A CdC kd A
- -
- -
Î ´ ´ ´= Þ = = »
Î ´ ´
GD
a bk[A] [B]=a b0.045 k[0.05] +=
a b a0.090 k[0.05] 2+=a b 2a b0.72 k[0.05] 2+ += ×
a1 1 a 12 2
Þ = Þ =
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And from equation (i) and (iii) we get
3.(3)
No. of 3 0 1 2 Unpaired electrons (all complex are inner orbital complex because ligands are strong) as, Magnetic moment
n = no. of unpaired electrons More the number of unpaired electron more will the value of spin only magnetic moment
4.(1) For iso-electronic species size is governed by proton to electron ratio i.e. ratio
More the value of ratio, smaller will be its size as more number of proton will have an ability to
hold electrons more strongly resulting in decrement in ionic size. Hence, size is affected by nuclear charge i.e. no. of proton.
5.(4) In acidic medium reduces to
oxidizes to
oxides to For
gm equivalents of = gm equivalents of gm equivalents of gm
equivalent of gm of
6.(1) ; moles
0.01 moles in 100 ml water. equivalent in 100 ml water 0.02 equivalent of in 100 ml water
0.01 moles of in 100 ml water of
Hardness
2 b0.045 10.72 2 +Þ = 2 b
45 1720 2 +Þ =
(2 b)720 245
+Þ = 4 2 b2 2 +Þ = B 2Þ =
4 4 3 26 6 3 6 3 6[V(CN) ] , [Fe(CN) ] , [Ru(NH ) ] , [Cr(NH ) ]- - + +
Þ 2V + 2Fe + 3Ru + 2Cr +
Þ 3s d° 6s d° 5s d° 4s d°
n(n 2),µ = +
Pe-æ öç ÷è ø
Pe-æ öç ÷è ø
4KMnO 2Mn +
22 4C O -
2CO2Fe + 3Fe +
4KMnO 2 4FeC O + 2 2 4 3Fe (C O ) +
4FeSO + 2 4 3Fe (SO ) .
4 2 4 2 2 4 3 4 2 4 3f KMnO f FeC O f Fe (C O ) f FeSO f Fe (SO )(moles n ) (moles n ) (moles n ) (moles n ) (moles n )´ = ´ + ´ + ´ + ´
4KMnO(moles) 5 (1 3) (1 6) (1 1) (1 0)´ = ´ + ´ + ´ + ´
4KMnO(moles) 2=
3 2 3 2Ca (HCO ) Ca(HCO )0.81 1g 0.81g, n162 200
= = = 3 2 3 2Mg(HCO ) Mg(HCO )1g 0.73g, n200
= =
T1 1n 0.01200 200
= + =
(0.01 2)´ \ 3CaCO\ 3CaCO
0.01 100g´ 3CaCO
Þ 31 10 mg 1000 10,000ppm100L
´ ´ =
2 3 3 42 4
3 42 2 4 3
2 24
4 3
Fe Fe , C CFeC O 1 (1 2)Fe (C O ) 2 3 6 C CFeSO 1 Fe FeFe(SO ) 0 No oxidation
+ + + +
+ +
+ +
® ®+ ´´ = ®
®
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7.(3) Given :
As, [hence solution formed is ideal solution]
And Using Raoult’s law:
, , And in vapor phase:
[from Raoult’s law] …(i)
[from Dalton’s law of partial pressure] Equating (i) and (ii), we get
8.(1)
9.(1)
10.(3) value E : Energy of electron in a particular subshell n : principal quantum number l : azimuthal quantum number (I) (II) (III) (IV) According to Aufbau’s rule, lower the value of (n + l), lower will be its energy. In case if (n + l) value
are same for two different subshell then subshell having lower value of n will have lower energy. So, correct increasing order of energy is: (IV) < (II) < (III) < (I)
oAP 400 mmHg=oBP 600mmHg=
A B solutionV V V+ =
B Ax 0.5, x 0.5= =
o oT A A B BP x P x P= + TP 200 300= + TP 500mmHg=
oA A AP x P=
A A v TP (x ) P=
A v0.5 400 4(x ) 0.5500 5´
= = ´2 0.410
= =
B v(x ) 1 0.4 0.6= - =
4 33 4 4 4
3s 4sa sZr (PO ) 3Zr 4PO+ -
-+
4 3 3 4sp 4K [Zr ] [PO ]+ -=
3 4[3s] [4s]=3 427s 256s= ´
76912 s=1/7
spKS6912æ ö
= ç ÷è ø
E (n l)µ +
n l 6+ = n l 5 (n 3)+ = =
n 1 5(n 4)+ = = n l 4+ =
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11.(3) As alkylation of nitrogen increases, Basicity of amines increase due to (+I) effect of Alkyl groups which results in more electron cloud density over nitrogen atom (available toward donation). Hence correct order is [Gaseous phase]
12.(3) Gabriel pthalimide reaction is used to prepare only 1° aliphatic amines.
In above reaction, R – Br undergoes reaction, Hence electrophilicity of α-carbon should be high as
well as its hindrance should be low for greater ease of reaction.
Hence, ease of formation of n-butyl amine is higher. Therefore, it is most probable answer
13.(1) x should be a weak acid as it is soluble in 10% NaOH only.
Oleic acid > Benzamide > o-toluidine > m-cresol Order of decreasing acidic strength x is m-cresol (weakest of all)
2 5 2 2 5 2 3(C H ) NH C H NH NH> >
2SN
\
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14.(4) Electronic configuration of following lanthanide ions are given below. (colorless, no unpaired electron)
(colorless, no unpaired electron) colourless (yellow, 5 unpaired electrons) Ions having f electrons show colour. of these ions may be attributed to presence of f electrons.
15.(3)
Carboxylic group has high priority than hydroxyl group so numbering starts from carbon of carboxylic group. 3-hydroxy-4-methylpentanoic acid.
16.(1)
17.(1) Benzene diazonium chloride react with that ring of 1-naphthol which contain group as it is activating and also it will undergo coupling at p-position w.r.t. group of 1-napthol.
18.(2) … (i)
Slope of v/s graph gives value of 1/n
From graph
Putting in (i) ,
19.(3) Four donor atoms are present in it hence it act as a tetradentate.
3 141u : 4f3 01a : 4f+
3 7Gd : 4f+
3 5Sm : 4f+
3
3 2
CH OH5 4 | 3| 2 1
H C CH CH CH COOH- - - -
OH-OH-
1nx k (P)
m=
x 1log logk logPm n= +
xlogm
logP
\1 2n 3=
1n
2/3x (P)mµ
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20.(2) According to first law of thermodynamics:
For adiabatic process, q = 0 Hence,
21.(2) Consider the following reactions:
22.(4)
23.(4) Maltose is a disaccharide made up two D-glucose units. On treatment with dil. HCl is undergoes hydrolysis to give two D-glucose units. (Monosaccharide)
24.(2)
Here 1 and 2 are conc. HBr sensitive regions Alkene, will undergo electrophilic addition reaction [Markonikov’s addtion] Ether, will undergo forced [Acid catalysed] nucleophilic substitution reaction mechanism]
U q wD = +
U wD =
Oxidation2 6 2 2 3 2
3B H O B O 3H2
+ ¾¾¾¾® +
Hydrolysis2 6 2 3 3 2B H 6H O 2H BO 6H+ ¾¾¾¾¾® +
PH nC dTD = ò1000
300
3 (23 0.01T)dT= +ò10002
300
0.01T3 23T2
é ù= +ê ú
ë û
220.01 0.01(300)3 23000 (1000) 23(300)
2 2é ù
= + - -ê úë û
[ ]3 23000 5000 6900 450 61950J= + - - =
62kJ!
1º2 º 2[SN
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25.(2)
26.(3) Let number of B atoms = N
Number of atoms of
Number of oxygen atoms = 2N A : B : O
is the formula
27.(1) Greater is the reduction potential, stronger is the oxidizing agent.
28.(2) Fact (refer of NCERT) Chemistry ; Class XI, Page No – 405 & 406.
29.(3) Hydration enthalpy charge on an ion
Hence, correct order of hydration enthalpy is:
30.(3) Using plastic bags is wrong as plastic bags cause environmental pollution.
NA2
=
N :N : 2N21 :1: 221: 2 : 4
2 4AB O
µ1
size of an ionµ
Li Na K Rb Cs+ + + + +> > > >
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PART-C MATHEMATICS
1.(2)
Liner differential equation
Integrating factor
Let
,
2.(1)
For non-trivial solutions
D = 0
, Greatest value of ‘c’ is 1/2.
3.(4)
4.(2) Case I
2 2 2( 1) 2 ( 1) 1dyx x x ydx
+ + + =
2 2 22 11 ( 1)
dy x ydx x x
æ öÞ + =ç ÷+ +è ø
221x dxxe +=
ò2 1 2x t xdx dt+ = Þ =
ln 2 1dt
tte e t x= = = = +ò
2 12( 1) tan1
dxy x x Cx
-Þ + = = ++ò 2 1( 1) tany x x C-Þ + = +
0 0x y= Þ =!
10 tan 0 0C C-= + Þ =1
2 12
tan( 1) tan1xy x x y
x
--+ = Þ =
+
(1)8
y p= (1)
32a y p
=
1 18 32 4 16
a a ap pÞ ´ = Þ = Þ =
0x cy cz- - =
0cx y cz- + =
0cx cy z+ - =
0x y zD D D= = =! \
2 2 21
1 1(1 ) ( ) ( )1
c cD c c c c c c c c c
c c
- -= - = - + - - - +
-2 2 3 3 21 c c c c c= - - - - -3 22 3 1 0c c= - - + =
3 22 3 1 0c c+ - =2(2 1)( 1) 0c cÞ - + = 1, 1, 1/ 2c = - -
\
2
2
2
212 1ln 21 11
xx xf xx
x
æ ö-ç ÷æ ö += ç ÷ç ÷+è ø ç ÷+ç ÷+è ø2
2( 1) 1ln 2ln
1( 1)x x
xx
æ ö- -= =ç ÷ç ÷ ++è ø
12ln 2 ( )1x f xx
-æ ö= =ç ÷+è ø
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Case II
or (Not possible since )
5.(1)
Sum of coefficients of all even degree term is
6.(4)
… (i)
… (ii)
Adding (i) and (ii) we get,
7.(3) are roots
,
Least value of n = 4
2x ³4xÞ ³
2 4 2 0x x x- + - + =
3xÞ =9x =
2x <4xÞ <
2 4 2 0x x x- + - + =
4 5 0x x+ - =
( )( )4 1 0x x- - =
1x = 4x = 2x <1xÞ = Sum 9 1 10Þ = + =
( ) ( )6 63 31 1x x x x+ - + - -
6 4 3 2 3 2 3 30 2 4 62 ( 1) ( 1) ( 1)C x C x x C x x C xé ù= + - + - + -ë û
[ ]0 2 4 4 6 62 3C C C C C C- + + - -
[ ]2 1 15 15 15 4 24- + + - =
4tan( )3
a +b =5tan( )12
a -b =
1 4( ) tan3
- æ öÞ a +b = ç ÷è ø
1 5( ) tan12
- æ öÞ a -b = ç ÷è ø
1 14 52 tan tan3 12
- -æ ö æ öÞ a = + ç ÷ç ÷è øè ø
4 5633 12tan 2 20 161
36
+a = =
-
2 2 2 0x x- + = &a b
2 4 12
i+ -a = = +
2 4 12
i- -b = = -
11i ii
a += =
b -
( ) 1n
niaæ ö = =ç ÷bè ø
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8.(4)
9.(4)
Parallel vector to line is
10.(1) Vector perpendicular to plane containing vectors and is parallel to vector
Required magnitude of projection is
11.(4) Then contrapositive of this is If you are not a citizen of India, then you are not born in India.
12.(3) ;
;
Vertex = (2,0) ;
Tangent of slope 1 on parabola
;
Distance of this line from y = x
2020
0(3 2) r
rr C
=
+å20 20
20
0 03 2r rr rr C C
= =
= +å å20
19 201
03 20 2 2rr
C -=
= ´ + ´å19 203 20(2) 2 (2)= ´ + ´
21 252 (15 1) 2= + =
ˆˆ ˆ5 3 4AP i j k= - +!!!"
ˆˆ ˆ10 7i j k- +
ˆ ˆˆ ˆ ˆ ˆ(5 3 4 ) (10 7 )cos150
i j k i j k- + × - +q =
75 3 1cos sin2 250 3
Þ q = = Þ q =
5 2 sinPN = q
5 2 1 52 2´
= =
ˆˆ ˆi j k+ + ˆˆ ˆ2 3i j k+ +
ˆˆ ˆˆˆ ˆ1 1 1 2
1 2 3
i j ki j k= - +
\ˆ ˆˆ ˆ ˆ ˆ(2 3 ) ( 2 ) 3
26i j k i j k+ + × - +
=
p qÞ
q pÞ! !
2 2y x= - y x=
2 ( 2)y x= - 2 14 ( 2)4
y x= ´ -
14
a =
11 ( 2)4
y x= ´ - +74
y x= -
7 74 1 1 4 2
d = =´ +
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13.(2)
14.(1) For point to be equidistant from axes it must lie on the or line
2nd quadrant
15.(4) Odd digits are 1, 1, 3 Even places 2nd , 4th , 6th, 8th
No. of ways
16.(2)
Sign scheme of is Local minimum at –2 and 1 Local maximum at 0
17.(1) Let the point P be (h, k)
5sin2
sin2
x
dxxò
52 sin cos2 2
2sin cos2 2
x x
x x
×Þ ò
52 sin cos2 2sin
x x
x
×Þ ò
sin3 sin 2sinx xx
+Þ ò
33sin 4sin 2sin cossin
x x x xx
- +Þ ò
2(3 4sin 2cos )x x dxÞ - +ò(3 2(1 cos2 ) 2cos )x x dx= - - +ò(2cos2 2cos 1)x x dx= + +òsin 2 2sinx x x C= + + +
y x= y x= -
3 5 15x y+ =
8 15y x x= Þ =
1515 158 ,
15 8 88
xPt
y
ü= ïï æ öý ç ÷
è øï=ïþ
y x= - 3 5 15x y+ =
3 5( ) 15x x+ - =
2 15x- =15 15,2 2
x y-= =
15 15,2 2-æ ö
ç ÷è ø
33! 6!4 1802! 4!2!C= ´ ´ =
3 2'( ) 36 36 72f x x x x= + -3 236( 2 )x x x= + -236 ( 2)x x x= + -
36 ( 2)( 1)x x x= + -
'( )f x
1 2{ 2,1}, {0}S S= - =
Vidyamandir Classes
VMC | JEE Mains-2019 18 Solutions | Morning Session
(required locus)
18.(1) Mean
Variance = 16
19.(3) Slope of the tangent at for ellipse is
Slope of tangent at (1,2) is –2 Perpendicular tangents slope will be 1/2 Slope of tangent at
For is
;
20.(2)
2 2 2 21 ( 1) 4h k h k\ + + + + - =
2 2 2 2( 1) 3h k h kÞ + - = - +
2 2 2 2 2 22 1 9 6h k k h k h kÞ + - + = + + - +
2 23 4h k kÞ + = +2 28 9 8 16 0k h kÞ + - - =2 29 8 8 16x y yÞ + - =
2 4 10 12 14 428 87 7
a b a b+ + + + + + + += = Þ =
42 56a b+ + =2
2xVNå
= -µ 14a b+ =
2 2 2 2 2 2 222 4 10 12 14 8
7a bV + + + + + +
= -
2 2 46016 647
+ += -a b
2 2 460807
a b+ +=
2 2560 460a b= + +2 2 100a b+ =2 2 100a b+ =
2( ) 2 100a b ab+ - =2(14) 2 100ab- =
196 2 100ab- =196 100 2ab- =96 482
ab ab= Þ =
1 1( , )x y2 2
12 8x y
+ = 1 1
1 1
8 42x xy y
- ´ = -
\
( cos , sin )a bq q2 2
2 2 1x ya b
+ = cot 2cotba-
q = - q
12cot2
Þ - q =1 1cot cos4 17-
q = Þ q = ±
( )22 2 22 cos 2cos17
aÞ = q = q =
3 4cos sin5 5
a = a =
Vidyamandir Classes
VMC | JEE Mains-2019 19 Solutions | Morning Session
21.(2) n is between 100 & 200 HCF (91, n) > 1 (Sum of natural no divisible by 7) + (Sum of natural no divisible by 13) – (Sum of no divisible by 91)
22.(3)
; ;
23.(4) will be intersecting
if
Length (Distance of from origin)
(radius)
24.(4)
3 1cos sin10 10
b = b =
sin( ) sin cos cos sina -b = a b- a b
12 35 10 5 10
-9
5 10=
1 9sin5 10
- æ öa -b = ç ÷è ø
14 8
1 1(98 7 ) (91 13 ) (182)
r rr r
= =
+ + + -å å1372 735 728 468 (182) 3121+ + + - =
2
1cos
62 cotcos
3
xy
x
-
æ öæ öpæ ö-ç ÷ç ÷ç ÷è øç ÷ç ÷=pæ öç ÷ç ÷+ç ÷ç ÷ç ÷è øè øè ø
2
1cos
62 cotsin
6
xy
x
-
æ öæ öpæ ö-ç ÷ç ÷ç ÷è øç ÷ç ÷=pæ öç ÷ç ÷-ç ÷ç ÷ç ÷è øè øè ø
212 cot cot
6y x-æ öæ öpæ ö= -ç ÷ç ÷ç ÷è øè øè ø
22
6y xpæ ö= -ç ÷
è ø2 ' 2
6y xpæ ö= - -ç ÷
è ø'
6y x p= -
x y n+ =2 2 16x y+ = 1,2,3,4,5n = ( )n NÎ
2nOL = x y n+ =
4OM =2
2 162nMN = -
22( ) 4 16
2nMN
æ ö= -ç ÷
è ø5 2
1
5 6 114 16 4 16 5 320 110 2102 2 6
=
æ ö ´ ´é ù- = ´ - = - =ç ÷ ê úè ø ´ë ûån
n
cos sinsin cos
Aa - aé ù
= ê úa aë û
2 cos sin cos sinsin cos sin cos
Aa - a a - aé ù é ù
= ê ú ê úa a a aë û ë û
Vidyamandir Classes
VMC | JEE Mains-2019 20 Solutions | Morning Session
25.(2) is function If is decreasing If is increasing
26.(3)
27.(1)
Required area
28.(4)
29.(2) Plane passing through given two planes can be written as
2 cos2 sin 2sin 2 cos2
Aa - aé ù
= ê úa aë û
32 cos32 sin32 0 1sin32 cos32 1 0
Aa - a -é ù é ù
= =ê ú ê úa aë û ë û
322p
a =64p
Þ a =
''( ) 0 (0,2)f x x> " Î
'( )f xÞ '( ) '( ) '(2 )x f x f xf = - -
(0,1)xÎ'( ) 0 ( )x xf < Þ f
(1,2)xÎ'( ) 0 ( )x xf > Þ f
2
0
sinlim2 1 cos
2x
xx® æ ö-ç ÷
è ø
2 2
0 2
4sin cos2 2lim
2 2 sin4
x
x x
x®=
2
220 0
4 sin4lim cos lim 12
2 216
x x
xx x
xx® ®
é ù= =ê úë û4 2=
12
0
11 59( 3 ) 4 2 86 6
x x dx= + + ´ = + =ò
( ) ( )( / )( ) ( )
P A B P AP A BP B P BÇ
= =
( ) ( )( )P A P AP B
³
(2 4) ( 2 4) 0x y y z- - + l + - =
3 ( 3) 0- + l - =
1l = -
Vidyamandir Classes
VMC | JEE Mains-2019 21 Solutions | Morning Session
30.(3)
2 4 1( 2 4) 0x y y z- - - + - =
0x y zÞ - - =
2 cos( ( )) ln2 cosx xg f xx x
-æ ö= ç ÷+è ø
4
4
2 cosln2 cosx xI dxx x
p
p-
-æ ö= ç ÷+è øò
4
4
2 cosln2 cosx xI dxx x
p
p-
+æ ö= ç ÷-è øò
( )4
4
2 ln 1 0I dx
p
p-
Þ = =ò 0 log 1eIÞ = Þ