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Mock Test - 2 (Code-E) (Answers) All India Aakash Test Series for JEE (Main)-2019
1/14
1. (3)
2. (1)
3. (3)
4. (4)
5. (3)
6. (2)
7. (3)
8. (2)
9. (2)
10. (1)
11. (2)
12. (1)
13. (2)
14. (3)
15. (3)
16. (3)
17. (3)
18. (3)
19. (3)
20. (4)
21. (3)
22. (1)
23. (2)
24. (3)
25. (3)
26. (4)
27. (4)
28. (2)
29. (2)
30. (2)
PHYSICS CHEMISTRY MATHEMATICS
31. (4)
32. (2)
33. (4)
34. (3)
35. (2)
36. (1)
37. (2)
38. (1)
39. (1)
40. (3)
41. (3)
42. (2)
43. (2)
44. (2)
45. (1)
46. (4)
47. (4)
48. (1)
49. (4)
50. (3)
51. (4)
52. (4)
53. (3)
54. (1)
55. (3)
56. (2)
57. (1)
58. (3)
59. (4)
60. (3)
61. (4)
62. (1)
63. (3)
64. (2)
65. (3)
66. (1)
67. (2)
68. (1)
69. (1)
70. (3)
71. (2)
72. (2)
73. (2)
74. (2)
75. (4)
76. (3)
77. (3)
78. (3)
79. (1)
80. (2)
81. (3)
82. (3)
83. (1)
84. (2)
85. (2)
86. (3)
87. (2)
88. (3)
89. (4)
90. (4)
Test Date : 03/03/2019
ANSWERS
MOCK TEST - 2 - Code-E
All India Aakash Test Series for JEE (Main)-2019
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-E) (Hints & Solutions)
2/14
1. Answer (3)
Hint :dx
vdt
Sol. : cosv A t
cosv
tA
sinx
tA
2 22 2
2 2cos sin
v xt t
A A
2 2 2v x A
2. Answer (1)
Hint : There will be static friction between 2 kg and
3 kg blocks.
Sol. : (2 3)G
F f a 0.1 50 5 NGf
10 – 5 = 5a a = 1 m/s2
2 kgf
2 1 2 Nf
3. Answer (3)
Hint : Thermal stress = Y T
Sol. : Stressmg
Y TA
5 11 5
100 10
10 2 10 10
mgT
AY
50 CT T = 27 – 50 = –23°C
4. Answer (4)
Hint : sin( )y A t kx
Sol. : 2 2 200 400f
100100 m/s
0.01
Tv
100 1m
200 2
v
f
12
2 24k
2 mm sin(400 4 )y t x
at t = 0, at x = 0 y = –2 mm
2 mm 2 mmsin 3
2
PART - A (PHYSICS)
5. Answer (3)
Hint :1 1
ML T ⎡ ⎤ ⎣ ⎦
Sol. :1 1
ML T ⎡ ⎤ ⎣ ⎦
1
1 12 3
4
⎡ ⎤ ⎢ ⎥⎣ ⎦
8
3
6. Answer (2)
Hint : inputBE
B
VR
I
Sol. :4
6
(1.4 1.1) 0.310
70 A 40 A 30 10
VR
7. Answer (3)
Hint : Particle should reach a point where forces on
it are balanced.
Sol. :2 2
44
(12 )
GMm G Mmx R
x R x ⇒
For v0 minimum
2
0
4 1 4
11 2 4 8
GMm G Mm GMm G Mmmv
R R R R
0
27
22
GMv
R
8. Answer (2)
Hint : Mechanical Energy Conservation
Sol. : Major axismin max
2a r r
2 2 3a r r r
2
max
1
2 2
GMm GMmmv
a r
2
3 2 2
GM v GM
r r
3
GMv
r
Mock Test - 2 (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
3/14
9. Answer (2)
Hint :primary
AB
VV r
R R
⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠
Sol. : dxdR
A
1
0
0
9xdx
dRA
∫ ∫
09
2A
2 2010 22 9
(6 9) 2 3V x x
A
⎛ ⎞ ⇒ ⎜ ⎟ ⎝ ⎠
1
3x
10. Answer (1)
Hint :1 1 1
f v u
0
ih v
h u
Sol. : For M1, u = –30 cm, f = –20 cm
1 1 1
20 30v
60 cmv
( 60)
1mm 30
ih
2mmi
h
For M2
, u = –40 cm so v = –40 cm
03 mmh so 3 mm
ih
11. Answer (2)
Hint : E dlt
∫
� �
�
Sol. : 2
0B r
2
0 02
2
B r B rE r E
t t
⇒
0( 2 )2
B rr F rqE r r
t
��
�
Angular impulse = 3
0t B r
12. Answer (1)
Hint :2 m
TqB
Sol. :
R
90
2
R
3
2
R
12sin
2
R
R 30
Total angle 2
90 303
0
2
3
mt
eB
13. Answer (2)
Hint and Sol. :
Hinge HingeI
22
4 12 4
L mL Lmg m
⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
12
7
g
L
CM CM
12 3
4 7 7
L g ga R a
L ⇒
14. Answer (3)
Hint :max c m
min c m
Sol. :max
10100 c = 10000
min9900 f
c = 5000 Hz
m100
50 Hz2
m
mf
15. Answer (3)
Hint :rms rms
cosP V I
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-E) (Hints & Solutions)
4/14
Sol. :rms
13V
2V rms
rms
VI
Z
rms2 2
13 1
22 12 (15 10)I A
2 2
12 12
1312 (15 10)cos
13 1 126 W
132 2P
16. Answer (3)
Hint : A charge moving with constant velocity
produces magnetic field that does not change
with time. So it does not produce time varying
electric field.
Sol. : A charge moving with constant velocity
produces magnetic field that does not change
with time. So it does not produce time varying
electric field.
17. Answer (3)
Hint : string CM2
Rv v
Sol. :
CM
2
Rv
CMv R 2
R
string CM
3
2 2 2
R Rv v R R
CMv R
CM
string
2
3 3
2
v R
Rv
CM string
2
3v v
18. Answer (3)
Hint : 2xxT
mgd
�
Sol. : For physical pendulum
2xxT
mgd
�22
12 2xx
mm⎛ ⎞ ⎜ ⎟⎝ ⎠
� ��
27
12
m �
27
2
122
mT
mg
�
�
7 22
12T
g �
19. Answer (3)
Hint : 2
2 2
1 2
1 113.6E Z
n n
⎛ ⎞ ⎜ ⎟⎜ ⎟
⎝ ⎠
Sol. : 2 2
1 113.6
1 3H
E⎛ ⎞ ⎜ ⎟⎝ ⎠
2
2 2
1 113.6( )
3E Z
n
⎛ ⎞ ⎜ ⎟⎝ ⎠
2
2 2 2
1 1 1 113.6 13.6
13 3Z
n
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2
2 2
13.6 1 1 1 113.6
9 1 1 3
3
Z
n
⎛ ⎞⎜ ⎟
⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
Comparing Z = 3, n = 9
20. Answer (4)
Hint : Intensitynh
At
Sol. :nh
IAt
double
So becomes halfn
So saturation current becomes half.
21. Answer (3)
Hint : Take a semicircular ring as an element and
integrate it
Mock Test - 2 (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
5/14
Sol. :
dr
r
0 0
CM
0 0
2( ) ( )
R R
R R
rdm y dA
y
dm dA
∫ ∫
∫ ∫
0
0
0
0
2( )
R
R
rr rdr
r rdr
⎛ ⎞ ⎜ ⎟⎝ ⎠
∫
∫
CM
3
2
Ry
22. Answer (1)
Hint : Minimum for pure rolling
min 2
CM
tan
1MR
I
Sol. : min 2
2
tan 2tan
51
2
3
MR
MR
min
So there is pure rolling.
So work done by friction is zero.
23. Answer (2)
Hint : 0B C
Q
Sol. : ( ) 3 ( )1
A B p B A B A
n RQ nC T T T nR T T
0 03
A BQ PV
1 10,
B C B CB CQ T V T V
1.5 1 1.5 1
0(2 )(2 )
A A CT V T V
08
CV V
0
0 0
0
ln ln8
A
C A A
C
VVQ nRT P V
V V
⎛ ⎞ ⎜ ⎟
⎝ ⎠
0 03 ln2P V
0 02.08PV
0 0 0 0 0 03 0 2.08 0.92
ABCAQ P V P V P V
24. Answer (3)
Hint : 1 2
0 02 2
E
Sol. :
+
+
+
vd
+
+
+
+
+
+
+
+
+
1
2
e
F = eE
1 2
0
( )
2
eF ma
1 2
0
( )
2
ea
m
21
2y y y
S u t a t
2
1 2
0
( )1
2 2 2
e td
m
0
1 2
2
( )
mdt
e
x xS u t
0
1 2
2
( )
mdl v
e
25. Answer (3)
Hint : Net 1 2
0
q
Sol. : Through both the cones
q
x
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-E) (Hints & Solutions)
6/14
Net 1 2
0
q
1
0
3
7
q
So 2
0
4
7
q
26. Answer (4)
Hint : N = No
e–t
Sol. : 60% decay 40% remaining
90% decay 10% remaining
So remaining portion becomes 1th
4
So 2 1
2t t T
27. Answer (4)
Hint : ( 1)x t n
Sol. : (1.525 1)10 m 5000Ån 6 7
0.525 10 10 5 10n
52.510.5
5n
11th dark at path difference of 10.5 .
28. Answer (2)
Hint :/R M R M
V V V � � �
Sol. : Take ˆi along east, ˆj along north, ˆk along
upwards
Case 1:/R M R M
V V V � � �
ˆ ˆ10 5R
k V i �
ˆ ˆ5 10R
V i k �
Case 2: ˆ5M
V j�
/ˆ ˆ ˆ5 10 5
R MV i k j �
Angle with ˆk 2 2 2
10 10cos
1505 5 10
2cos
3
29. Answer (2)
Hint :
22 ( )
9T
r gV
Sol. : VT
4
3
2 10 (1000 700)10
9 10
200m/s 66.7 m/s
3
30. Answer (2)
Hint :sep
app
1v
e
v
APP2 2v gR
Sol. :2 1
2 2M gR m gR mV MV ....(1)
2 12 2M gR MV V M ....(2)
MV1
mV2
2
2 35
gR M mV gR
M m
( 5 2)
(3 2 5)
mM
PART - B (CHEMISTRY)
31. Answer (4)
Hint : Magnetic nature of complex compound.
Sol. : Only IV has unpaired electron.
32. Answer (2)
Hint : Hg22+ ions disproportionate in presence of
cyanide.
Sol. : 2
2 2Hg 2CN Hg Hg(CN)
33. Answer (4)
Hint : Alkali metal hydrides are denser than parent
metal.
Sol. : Metallic hydrides have a lattice which is
different from that of the parent metal.
Mock Test - 2 (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
7/14
34. Answer (3)
Hint : Oxymercuration of alkene.
Sol. :
OHOH
HgOAc+
+
Hg
CH2 CH2
HgOAc
NaBH4
+OAc O
+
O
H
35. Answer (2)
Hint : Formation of hydrocarbons.
Sol. :
CHC OH
O
Reaction I: +CH
2
CH2
OHC
CH
O
Reaction II: +C
H
C
C O CH3
O
CH
C O CH3
C
O
36. Answer (1)
Hint : Combustion of hydrocarbon.
Sol. : In Reaction-I, highest heat per gram of the
hydrocarbon is liberated.
37. Answer (2)
Hint : Reaction of Alkyne
Sol. :
C H2 5
C C H3 7
CO
3
Zn, H O2
C H2 5
C C
O O
C H3 7
38. Answer (1)
Hint : Elimination reaction; addition of H2O.
Sol. :
CH OH3
heat, KOH
Major
Br
CH3
CH3
BH –THF3
H O ; OH2 2
–
CH3
CH3
O Et
CH3
(i) Na
(ii) CH – 3 CH – Br2
OH
39. Answer (1)
Hint : Preparation of phenol from cumene.
Sol. :
(i) O2
(ii) H3O
+
C
C
C OH
O C
CH3
CH3
A B
+
CH3
H3C H
HCl
CH3
CH3
O C
CH3
CH3
2 OH
HO C OH
+
40. Answer (3)
Hint : I and III are mirror images
Sol. : (I, III) enantiomers
(I, IV), (III, IV) diasteromers
II is a positional isomer of the others.
41. Answer (3)
Hint : Alkylation reaction of Amine.
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-E) (Hints & Solutions)
8/14
Sol. :
CH I3
N N+
N+
N
CH3
Ag2O
CH3
CH3
I–
OH–
42. Answer (2)
Hint : Oxidation of 1, 4-Dihydroxy benzene.
Sol. :
O O
OO
H
H
::
:
:
Na Cr O2 2 7
H SO /H O2 4 2
43. Answer (2)
Hint : Reaction of carbonyl compound.
Sol. :
C CH CH3
C C CH3
Ph
CH3
CH2
O
Ph
CH3
O
(i) LDA
(ii) Ph CH2 Br
44. Answer (2)
Hint : Reaction due to –H
Sol. :
CH2
O OEt
C C
O O
C O CH2
CH2
CH3
CH3
OCH2
CH2
CH3
CH3
CH
C
O
O
CO
O
OEt– –
O
45. Answer (1)
Hint : Limiting Reagent
Sol. :4 8 2 2 2
C H 6O 4CO 4H O
moles of O2 available = 6
⇒ Volume = 6 × 22.4 = 134.4 litre
46. Answer (4)
Hint : HCP is ABAB type of packing
Sol. :
T
T
O
T
T
O
T
T
O
T
T
O
T
T
T T T T T
T T T T T T
O
T
O
O O O O O O
47. Answer (4)
Hint : Elevation in boiling point.
Sol. : Elevation in boiling point is proportional to the
no. of solute particle.
NaCl Na Cl
2
2CaCl Ca 2Cl
48. Answer (1)
Hint : Order of reactivity on the basis of (ECS).
Sol. : E°red
K < Mg < Zn < Au
49. Answer (4)
Hint : Ln+3 ions have unpaired e– in f orbital.
Sol. : Most of the Ln+3 ions are coloured both in the
solid state and in the aqueous solutions.
50. Answer (3)
Hint : Normality of 1 litre solution = 0.99
Sol. :0
10xd 10 70 1.54M
M 98
= 11
Now, meq of H3PO
4 in V ml = meq of H
3PO
4 in 1 litre
11 × V × 3 = 0.99 × 1000
V = 30 ml.
Mock Test - 2 (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
9/14
51. Answer (4)
Hint : Calculation of pressure in equilibrium mixture.
Sol. : Since pressure of the gases are same in both
the containers. Therefore the final pressure will
not be changed.
52. Answer (4)
Hint : G = H – TS
Sol. : As S increases, the value of G becomes
more negative.
53. Answer (3)
Hint : Equilibrium constant and Gibb's free energy
Sol. : G°= –RT ln Keq
.
54. Answer (1)
Hint : Conc. of reactant and eq. const. (Le-Chatelier's
principle).
Sol. :
2
2 2
HI 0.4 0.4Q 8 ; Q K
H I 0.1 0.2
55. Answer (3)
Hint : Slaked lime and formation of bleaching powder.
Sol. : MnO2 + 4HCl MnCl
2 + Cl
2 + 2H
2O
Cl2 + Ca(OH)
2 CaOCl
2 + H
2O
(Y)
CaOCl2
+ 2HCl CaCl2
+ Cl2
+ H2O
56. Answer (2)
Hint : Chemical properties (epoxidation) of alkene.
Sol. :
CH2
CH2
LiAlH4
O
C
OH
CC
O
O
Cl
O H
57. Answer (1)
Hint : Factual
Sol. : Insulin is an hormone.
58. Answer (3)
Hint : Fact Based
Sol. : Soluble fluoride is added to drinking water to
bring the concentration to 1 ppm.
59. Answer (4)
Hint : Lassaigne solution (test of nitrogen)
Sol. : Cyanide ions are formed on treatment with Na
metal which from a complex with Fe+2 ions.
Some Fe2+ ion also get oxidised to Fe+3 and
thus prussian blue is formed.
60. Answer (3)
Hint : Cellulose – Hydrolysis
Sol. : Cellulose is a polymer of -D-Glucose.
PART - C (MATHEMATICS)
61. Answer (4)
Hint :3
2
0
2( )
3 ∫
bb
f x dx b
Sol. :3
2
0
2( )
3
b bf x dx b ∫
d. w.r. t. b
f(b) = 2b – 2b2 f(x) = 2x – 2x2
f(x) = 2 1 2
24 4
x x⎡ ⎤ ⎢ ⎥⎣ ⎦
=
21 1
22 2
x⎛ ⎞ ⎜ ⎟⎝ ⎠
f(x)max.
= 1
2 when x =
1
2
62. Answer (1)
Hint : Form : 1
Sol. : l =
1
1 cot
0
sinlim
x x
x
x
x
⎛ ⎞⎜ ⎟⎝ ⎠
form : 1
∵0
sinlimx
x
x
⎛ ⎞⎜ ⎟⎝ ⎠
= 1
0
lim ( cot )x
x x
= 0
limtanx
x
x = 1
l =
sin 1lim 1
1 cot0
x
x x xxe
⎛ ⎞ ⎜ ⎟ ⎝ ⎠
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-E) (Hints & Solutions)
10/14
=
sin tanlim
tan0
x x x
x x xxe
⎛ ⎞⎜ ⎟⎝ ⎠
=
sin tanlim
tan0
x x x
x x xxe
⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
=
11
2e
=
1
e
63. Answer (3)
Hint : Take log on both sides then convert in definite
integral
Sol. :
logl =1 1 2
lim log log log ... logn
n n n n n
n n n n n
⎛ ⎞⎡ ⎤ ⎛ ⎞ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎝ ⎠
=
21lim log
n
nr n
r
n n
⎛ ⎞⎜ ⎟⎝ ⎠
∑ = 2
1logxdx∫
= [xlogx – x]12 = 2log2 – 2 + 1 = 2log2 – 1
= 4
loge
⎛ ⎞⎜ ⎟⎝ ⎠
l = 4
e
64. Answer (2)
Hint : Draw graph
Sol. :
x
y
AB
C
x = 1
y a =
(1, 1)
O
y = 2x – x2 (x – 1)2 = –(y – 1)
If y = a, then x = 1 1 a
A1 =
2 2
0(2 )x x dx∫ =
23
2
03
xx
⎡ ⎤⎢ ⎥
⎣ ⎦
= 8 4
43 3
A(OABC) = 1
3
1 1 2
0
1(2 ) 1
3
a
x x dx a a
∫
2 31 11 1 1 1 1
3 3a a a a
Put 1 – a = t2
2t3 = 1 t6 = 1
4 = 2–2
t2 =
2
32
a = 1 – t2 = 2
31 2
65. Answer (3)
Hint : 3sinx + 2cosx
= 3cos 2sin (3cos 2sin ) dx x x x
dx
Sol. :
I = 3sin 2cos
3cos 2sin
∫
x xdx
x x
=
(3cos 2sin )3cos 2sin
3cos 2sin 3cos 2sin
∫ ∫
dx x
x x dxA dx B dx
x x x x
3sinx + 2cosx = A(3cosx + 2sinx)
+ B(–3sinx + 2cosx)
Comparing coefficients of sinx and cosx on both
sides
A = 12
13, B =
5
13
121
1 13
5
13
⎛ ⎞
⎜ ⎟⎝ ⎠
A
B = –5
66. Answer (1)
Hint : Parabolic curve
Sol. : 0dy
dx
3ax2 + 6 0
y ax = 3 + 62
3a > 0, D < 0
Mock Test - 2 (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
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a > 0
0 – 4(3a)6 < 0
a > 0
at a = 0, y = 6x, an increasing function
67. Answer (2)
Hint : Plane contains points A, B, C
Sol. : Plane containing all medians is the plane
passing through the vertices A, B and C
Equation of the plane is
PA BA CA
⎡ ⎤⎢ ⎥⎣ ⎦
= 0 where P(x, y, z)
1 1 1
1 0 1
x y z
= 0
x – 2y + z = 0
68. Answer (1)
Hint : Multiply and divide by
1 1
2 2(sec ) (tan )n n
x x
Sol. :
R =
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
11
1 1 11
2 2 22(sec ) (tan ) (sec ) tan
...(sec tan )
n n n
nx x x x
x x
R =
2 2
22
1 1
2 2
11
22
1 1
2 2
(sec ) (tan )
(sec ) tan ...(sec tan )
(sec ) (tan )
n n
n
n
n n
x x
x x x x
x x
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟ ⎝ ⎠
= 1 1
2 2
1
(sec ) (tan )n n
x x
=
1
2
1
2
(cos )
1 (sin )
n
n
x
x
69. Answer (1)
Hint :�
a is perpendicular to c�
Sol. : a c� �
= a c� �
a c� �
ˆ ˆ ˆ( )c i j k �
a c� �
= b�
ˆ ˆ ˆ
ˆ ˆ1 1 0
i j k
i j
= 0, = 1
c�
= ˆk
70. Answer (3)
Hint : AA–1 = I
Sol. : A–1A = I, then compare corresponding
elements.
71. Answer (2)
Hint : (d) = r
Sol. : Equation of tangent of x2 = – 4y is
tx = –y + t2 ...(i)
(i) is also tangent to x2 + y2 = 4
2
2
| |
1
t
t = 2
t4 = 4 + 4t2 t2 = 2 2 2
m = – t m2 = 2 2 2 [m2] = 4
72. Answer (2)
Hint : S1 – S
2 = 0
Sol. : c1(–g, 0), (r
1 = |g|, c
2(0, – f), r
2 = |f |, equation
of common chord is
A
B
MC
1
(– , 0)g
r1
gx fy – = 0
gx fy – = 0
C1M =
2
2 2
g
g f
AM2 = r12 – (C
1M)2 =
42
2 2
gg
g f
=
2 2
2 2
g f
g f
AM = 2 2
| |gf
g f
AB = 2 2
2 | |gf
g f
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-E) (Hints & Solutions)
12/14
73. Answer (2)
Hint : y = x is axis of both hyperbolas
Sol. :
x
y y x =
xy = 4
xy = 1 A(1, 1)
(2, 2)B
AB = 2
74. Answer (2)
Hint : 2 2 2 y mx a m b
Sol. : Equations of common tangents to both
hyperbolas are
y = 2 2
x a b
For 4 common tangents, a > b
75. Answer (4)
Hint :2
a b c
⎡ ⎤⎢ ⎥⎣ ⎦
=
a a a b a c
b a b b b c
c a c b c c
⎡ ⎤ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥
⎣ ⎦
Sol. :2
a b c
⎡ ⎤⎢ ⎥⎣ ⎦
=
a a a b a c
b a b b b c
c a c b c c
⎡ ⎤ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥
⎣ ⎦
=
1 3
1 2
3 2
1
1
1
⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦
= 1 + 21
2
3 –
12 –
22 –
32
a b c
⎡ ⎤⎢ ⎥⎣ ⎦ =
2 2 21 2 3 1 2 31 2
76. Answer (3)
Hint : = 0
Sol. : = 0
5 31
2 2
50
2
30 4
2
k
= 0
5 3 3
4 (10)2 2 2
k k⎛ ⎞ ⎜ ⎟⎝ ⎠
= 0
94
4k k = 25
25
4k = 25 k = 4
tan = 2
2 h ab
a b
= 3
5
= 1 3
tan5
⎛ ⎞⎜ ⎟⎝ ⎠
77. Answer (3)
Hint : = 0
Sol. : = 0
1 1
1 2
1 1 0
= 0
2 – 2 + (–– 1) = 0
2 + = 0
= 0 or –1
78. Answer (3)
Hint : If n(A) = n, then number of reflexive relations
on the set A = 2n(n–1)
Sol. : If n(A) = n, then number of reflexive relations
on the set A = 2n(n–1)
79. Answer (1)
Hint : Make all cases
Sol. : I, I, I, I, M, P, P, S, S, S, S
1(I)
1(M)
+
1(I)
1(P)
+
1(I)
1(S)
Mock Test - 2 (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
13/14
Number of words = 9! 9! 9!
3!2!4! 3! 4! 3! 2!3!
= 9!
74!3!2!
= 1260 × 7
80. Answer (2)
Hint : A.M. G.M.
Sol. :
1
3(2 ) (4 ) (8 )
((2 )·(4 ) (8 ))3
xy yz zxxy yz zx
1
2 2 2 327
(64 )3
x y z
64x2y2z2 93
2
2 2 2 27
8x y z
⎛ ⎞ ⎜ ⎟⎝ ⎠
27
8xyz Greatest possible value of xyz
is 3.375
81. Answer (3)
Hint : Sum of G.P.
Sol. :
S = 2 3
12 12 121 ...i i i
e e e
upto 25 terms
=
2525
12 12
1212
1 1
11
i i
ii
e e
ee
=
212
12
1
1
i
i
e
e
⎛ ⎞ ⎜ ⎟⎝ ⎠
= 12
12
1
1
i
i
e
e
= 1
82. Answer (3)
Hint : 2 3 2
2 3 2 2 3 3(32) | | | | z z z z z z
Sol. : |(32)2z1 + (16)2z
2 + (8)2z
3|
= |(|z2| |z
3|)2z
1 + (|z
1| |z
3|)2z
2 + (|z
1||z
2|)2z
3|
= 2 2 3 3 1 1 1 3 3 2 1 1 2 2 3z z z z z z z z z z z z z z z
= 1 2 3 2 3 3 1 1 2z z z z z z z z z
= 1 2 3 2 3 3 1 1 2| | | | | |z z z z z z z z z
= 2 × 4 × 8 |z1z
2 + z
2z
3 + z
3z
1|
= 64
83. Answer (1)
Hint : (A + B)(A – B) = A2 – B2
Sol. : (1 + x + x2)35 (1 – x + x2)35
= ((x2 + 1)2 – x2)35
= (x4 + x2 + 1)35
Coefficient of x15 is zero
84. Answer (2)
Hint : Factorization
Sol. : 22 3 3 2 1z z i z z = 0
22 2 3 3z zi z i i z = 0
2z(z – i) + 1(z – i) – 3 ( )i z i = 0
( ) 2 1 3z i z i = 0
z = i, z = 1 3
2 2i =
18 + 18 + 1 = i18 + 18 + 1
= –1 + 1 + 1 = 1
85. Answer (2)
Hint : Find n, L.D. eq
dx
dy
Sol. :1
y
y
y
dx xee x
dy e
dx
dy + x = e–y
I.F = 1dy
e∫ = ey
Solution of differential equation is
x.ey = .
y ye e dy c ∫xey = y + c passes then (0, 0)
c = 0
Equation of curve is y = xey
86. Answer (3)
Hint :
2
2 2( )ix
xN
∑
Sol. :1 2 3 ... 10 10(11) 11
10 2 10 2x
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-E) (Hints & Solutions)
14/14
Variance =
22
2 ix
xN
∑
=
21 10(10 1) (2 10 1) 11
10 6 2
⎛ ⎞ ⎜ ⎟⎝ ⎠
2 11 21 11 11
6 4
= 11 7 11 11 1114 11
2 4 4
2 33
4 =
33
2
87. Answer (2)
Hint : Required probability
= P(odd, odd) + P(even, even)
Sol. : Let p be the probability of showing an odd
number
Then 3p = 1 p = 1
3
Required probability
= P(odd, odd) + P(even, even)
= p2 + 4p2
= 5p2
= 5
9
88. Answer (3)
Hint : Line is parallel to the vector 1 2n n��� ���
Sol. : Line is parallel to the vector 1 2n n��� ���
Where � �
1 22 2 , 3 4 6n i j k n i j k
��� ���� � � �
1 2 3. 0n n n
��� ��� ���
(where �
34 5 8n i j k
���� � )
Line is perpendicular to normal to the plane
Line is parallel to the plane
89. Answer (4)
Hint : Draw graph
Sol. :
y = 2
x
y =
x320
y =
x
y = x
y =
x +
2 y
x
=
2
y x = cos (cos )1
y
Number of real solution(s) of the equation
cos–1(cosx) + x = 2 is intersection point of
y = cos–1(cosx) and y = 2 – x
y = cos–1(cosx) and y = 2 – x intersect at
infinitely many points.
90. Answer (4)
Hint : Truth table
Sol. : Truth values of r and s are T and T respectively.
� � �
Mock Test - 2 (Code-F) (Answers) All India Aakash Test Series for JEE (Main)-2019
1/14
1. (2)
2. (2)
3. (2)
4. (4)
5. (4)
6. (3)
7. (3)
8. (2)
9. (1)
10. (3)
11. (4)
12. (3)
13. (3)
14. (3)
15. (3)
16. (3)
17. (3)
18. (2)
19. (1)
20. (2)
21. (1)
22. (2)
23. (2)
24. (3)
25. (2)
26. (3)
27. (4)
28. (3)
29. (1)
30. (3)
PHYSICS CHEMISTRY MATHEMATICS
31. (3)
32. (4)
33. (3)
34. (1)
35. (2)
36. (3)
37. (1)
38. (3)
39. (4)
40. (4)
41. (3)
42. (4)
43. (1)
44. (4)
45. (4)
46. (1)
47. (2)
48. (2)
49. (2)
50. (3)
51. (3)
52. (1)
53. (1)
54. (2)
55. (1)
56. (2)
57. (3)
58. (4)
59. (2)
60. (4)
61. (4)
62. (4)
63. (3)
64. (2)
65. (3)
66. (2)
67. (2)
68. (1)
69. (3)
70. (3)
71. (2)
72. (1)
73. (3)
74. (3)
75. (3)
76. (4)
77. (2)
78. (2)
79. (2)
80. (2)
81. (3)
82. (1)
83. (1)
84. (2)
85. (1)
86. (3)
87. (2)
88. (3)
89. (1)
90. (4)
Test Date : 03/03/2019
ANSWERS
MOCK TEST - 2 - Code-F
All India Aakash Test Series for JEE (Main)-2019
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-F) (Hints & Solutions)
2/14
1. Answer (2)
Hint :sep
app
1v
e
v
APP2 2v gR
Sol. :2 1
2 2M gR m gR mV MV ....(1)
2 12 2M gR MV V M ....(2)
MV1
mV2
2
2 35
gR M mV gR
M m
( 5 2)
(3 2 5)
mM
2. Answer (2)
Hint :
22 ( )
9T
r gV
Sol. : VT
4
3
2 10 (1000 700)10
9 10
200m/s 66.7 m/s
3
3. Answer (2)
Hint :/R M R M
V V V � � �
Sol. : Take ˆi along east, ˆj along north, ˆk along
upwards
Case 1:/R M R M
V V V � � �
ˆ ˆ10 5R
k V i �
ˆ ˆ5 10R
V i k �
Case 2: ˆ5M
V j�
/ˆ ˆ ˆ5 10 5
R MV i k j �
Angle with ˆk 2 2 2
10 10cos
1505 5 10
2cos
3
PART - A (PHYSICS)
4. Answer (4)
Hint : ( 1)x t n
Sol. : (1.525 1)10 m 5000Ån
6 70.525 10 10 5 10n
52.510.5
5n
11th dark at path difference of 10.5 .
5. Answer (4)
Hint : N = No
e–t
Sol. : 60% decay 40% remaining
90% decay 10% remaining
So remaining portion becomes 1th
4
So 2 1
2t t T
6. Answer (3)
Hint : Net 1 2
0
q
Sol. : Through both the cones
q
x
Net 1 2
0
q
1
0
3
7
q
So 2
0
4
7
q
7. Answer (3)
Hint : 1 2
0 02 2
E
Mock Test - 2 (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
3/14
Sol. :
+
+
+
vd
+
+
+
+
+
+
+
+
+
1
2
e
F = eE
1 2
0
( )
2
eF ma
1 2
0
( )
2
ea
m
21
2y y y
S u t a t
2
1 2
0
( )1
2 2 2
e td
m
0
1 2
2
( )
mdt
e
x xS u t
0
1 2
2
( )
mdl v
e
8. Answer (2)
Hint : 0B C
Q
Sol. : ( ) 3 ( )1
A B p B A B A
n RQ nC T T T nR T T
0 03
A BQ PV
1 10,
B C B CB CQ T V T V
1.5 1 1.5 1
0(2 )(2 )
A A CT V T V
08
CV V
0
0 0
0
ln ln8
A
C A A
C
VVQ nRT P V
V V
⎛ ⎞ ⎜ ⎟
⎝ ⎠
0 03 ln2P V
0 02.08PV
0 0 0 0 0 03 0 2.08 0.92
ABCAQ P V P V P V
9. Answer (1)
Hint : Minimum for pure rolling
min 2
CM
tan
1MR
I
Sol. : min 2
2
tan 2tan
51
2
3
MR
MR
min
So there is pure rolling.
So work done by friction is zero.
10. Answer (3)
Hint : Take a semicircular ring as an element and
integrate it
Sol. :
dr
r
0 0
CM
0 0
2( ) ( )
R R
R R
rdm y dA
y
dm dA
∫ ∫
∫ ∫
0
0
0
0
2( )
R
R
rr rdr
r rdr
⎛ ⎞ ⎜ ⎟⎝ ⎠
∫
∫
CM
3
2
Ry
11. Answer (4)
Hint : Intensitynh
At
Sol. :nh
IAt
double
So becomes halfn
So saturation current becomes half.
12. Answer (3)
Hint : 2
2 2
1 2
1 113.6E Z
n n
⎛ ⎞ ⎜ ⎟⎜ ⎟
⎝ ⎠
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-F) (Hints & Solutions)
4/14
Sol. : 2 2
1 113.6
1 3H
E⎛ ⎞ ⎜ ⎟⎝ ⎠
2
2 2
1 113.6( )
3E Z
n
⎛ ⎞ ⎜ ⎟⎝ ⎠
2
2 2 2
1 1 1 113.6 13.6
13 3Z
n
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2
2 2
13.6 1 1 1 113.6
9 1 1 3
3
Z
n
⎛ ⎞⎜ ⎟
⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
Comparing Z = 3, n = 9
13. Answer (3)
Hint : 2xxT
mgd
�
Sol. : For physical pendulum
2xxT
mgd
�22
12 2xx
mm⎛ ⎞ ⎜ ⎟⎝ ⎠
� ��
27
12
m �
27
2
122
mT
mg
�
�
7 22
12T
g �
14. Answer (3)
Hint : string CM2
Rv v
Sol. :
CM
2
Rv
CMv R 2
R
string CM
3
2 2 2
R Rv v R R
CMv R
CM
string
2
3 3
2
v R
Rv
CM string
2
3v v
15. Answer (3)
Hint : A charge moving with constant velocity
produces magnetic field that does not change
with time. So it does not produce time varying
electric field.
Sol. : A charge moving with constant velocity
produces magnetic field that does not change
with time. So it does not produce time varying
electric field.
16. Answer (3)
Hint :rms rms
cosP V I
Sol. :rms
13V
2V rms
rms
VI
Z
rms2 2
13 1
22 12 (15 10)I A
2 2
12 12
1312 (15 10)cos
13 1 126 W
132 2P
17. Answer (3)
Hint :max c m
min c m
Sol. :max
10100 c = 10000
min9900 f
c = 5000 Hz
m100
50 Hz2
m
mf
18. Answer (2)
Hint and Sol. :
Hinge HingeI
22
4 12 4
L mL Lmg m
⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
Mock Test - 2 (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
5/14
12
7
g
L
CM CM
12 3
4 7 7
L g ga R a
L ⇒
19. Answer (1)
Hint :2 m
TqB
Sol. :
R
90
2
R
3
2
R
12sin
2
R
R 30
Total angle 2
90 303
0
2
3
mt
eB
20. Answer (2)
Hint : E dlt
∫
� �
�
Sol. : 2
0B r
2
0 02
2
B r B rE r E
t t
⇒
0( 2 )2
B rr F rqE r r
t
��
�
Angular impulse = 3
0t B r
21. Answer (1)
Hint :1 1 1
f v u
0
ih v
h u
Sol. : For M1, u = –30 cm, f = –20 cm
1 1 1
20 30v
60 cmv
( 60)
1mm 30
ih
2mmi
h
For M2
, u = –40 cm so v = –40 cm
03 mmh so 3 mm
ih
22. Answer (2)
Hint :primary
AB
VV r
R R
⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠
Sol. : dxdR
A
1
0
0
9xdx
dRA
∫ ∫
09
2A
2 2010 22 9
(6 9) 2 3V x x
A
⎛ ⎞ ⇒ ⎜ ⎟ ⎝ ⎠
1
3x
23. Answer (2)
Hint : Mechanical Energy Conservation
Sol. : Major axismin max
2a r r
2 2 3a r r r
2
max
1
2 2
GMm GMmmv
a r
2
3 2 2
GM v GM
r r
3
GMv
r
24. Answer (3)
Hint : Particle should reach a point where forces on
it are balanced.
Sol. :2 2
44
(12 )
GMm G Mmx R
x R x ⇒
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-F) (Hints & Solutions)
6/14
PART - B (CHEMISTRY)
31. Answer (3)
Hint : Cellulose – Hydrolysis
Sol. : Cellulose is a polymer of -D-Glucose.
32. Answer (4)
Hint : Lassaigne solution (test of nitrogen)
Sol. : Cyanide ions are formed on treatment with Na
metal which from a complex with Fe+2 ions.
Some Fe2+ ion also get oxidised to Fe+3 and
thus prussian blue is formed.
For v0 minimum
2
0
4 1 4
11 2 4 8
GMm G Mm GMm G Mmmv
R R R R
0
27
22
GMv
R
25. Answer (2)
Hint : inputBE
B
VR
I
Sol. :4
6
(1.4 1.1) 0.310
70 A 40 A 30 10
VR
26. Answer (3)
Hint :1 1
ML T ⎡ ⎤ ⎣ ⎦
Sol. :1 1
ML T ⎡ ⎤ ⎣ ⎦
1
1 12 3
4
⎡ ⎤ ⎢ ⎥⎣ ⎦
8
3
27. Answer (4)
Hint : sin( )y A t kx
Sol. : 2 2 200 400f
100100 m/s
0.01
Tv
100 1m
200 2
v
f
12
2 24k
2 mm sin(400 4 )y t x
at t = 0, at x = 0 y = –2 mm
2 mm 2 mmsin 3
2
28. Answer (3)
Hint : Thermal stress = Y T
Sol. : Stressmg
Y TA
5 11 5
100 10
10 2 10 10
mgT
AY
50 CT T = 27 – 50 = –23°C
29. Answer (1)
Hint : There will be static friction between 2 kg and
3 kg blocks.
Sol. : (2 3)G
F f a 0.1 50 5 NGf
10 – 5 = 5a a = 1 m/s2
2 kgf
2 1 2 Nf
30. Answer (3)
Hint :dx
vdt
Sol. : cosv A t
cosv
tA
sinx
tA
2 22 2
2 2cos sin
v xt t
A A
2 2 2v x A
33. Answer (3)
Hint : Fact Based
Sol. : Soluble fluoride is added to drinking water to
bring the concentration to 1 ppm.
34. Answer (1)
Hint : Factual
Sol. : Insulin is an hormone.
35. Answer (2)
Hint : Chemical properties (epoxidation) of alkene.
Mock Test - 2 (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
7/14
Sol. :
CH2
CH2
LiAlH4
O
C
OH
CC
O
O
Cl
O H
36. Answer (3)
Hint : Slaked lime and formation of bleaching powder.
Sol. : MnO2 + 4HCl MnCl
2 + Cl
2 + 2H
2O
Cl2 + Ca(OH)
2 CaOCl
2 + H
2O
(Y)
CaOCl2
+ 2HCl CaCl2
+ Cl2
+ H2O
37. Answer (1)
Hint : Conc. of reactant and eq. const. (Le-Chatelier's
principle).
Sol. :
2
2 2
HI 0.4 0.4Q 8 ; Q K
H I 0.1 0.2
38. Answer (3)
Hint : Equilibrium constant and Gibb's free energy
Sol. : G°= –RT ln Keq
.
39. Answer (4)
Hint : G = H – TS
Sol. : As S increases, the value of G becomes
more negative.
40. Answer (4)
Hint : Calculation of pressure in equilibrium mixture.
Sol. : Since pressure of the gases are same in both
the containers. Therefore the final pressure will
not be changed.
41. Answer (3)
Hint : Normality of 1 litre solution = 0.99
Sol. :0
10xd 10 70 1.54M
M 98
= 11
Now, meq of H3PO
4 in V ml = meq of H
3PO
4 in 1 litre
11 × V × 3 = 0.99 × 1000
V = 30 ml.
42. Answer (4)
Hint : Ln+3 ions have unpaired e– in f orbital.
Sol. : Most of the Ln+3 ions are coloured both in the
solid state and in the aqueous solutions.
43. Answer (1)
Hint : Order of reactivity on the basis of (ECS).
Sol. : E°red
K < Mg < Zn < Au
44. Answer (4)
Hint : Elevation in boiling point.
Sol. : Elevation in boiling point is proportional to the
no. of solute particle.
NaCl Na Cl 2
2CaCl Ca 2Cl
45. Answer (4)
Hint : HCP is ABAB type of packing
Sol. :
T
T
O
T
T
O
T
T
O
T
T
O
T
T
T T T T T
T T T T T T
O
T
O
O O O O O O
46. Answer (1)
Hint : Limiting Reagent
Sol. :4 8 2 2 2
C H 6O 4CO 4H O
moles of O2 available = 6
⇒ Volume = 6 × 22.4 = 134.4 litre
47. Answer (2)
Hint : Reaction due to –H
Sol. :
CH2
O OEt
C C
O O
C O CH2
CH2
CH3
CH3
OCH2
CH2
CH3
CH3
CH
C
O
O
CO
O
OEt– –
O
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-F) (Hints & Solutions)
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48. Answer (2)
Hint : Reaction of carbonyl compound.
Sol. :
C CH CH3
C C CH3
Ph
CH3
CH2
O
Ph
CH3
O
(i) LDA
(ii) Ph CH2 Br
49. Answer (2)
Hint : Oxidation of 1, 4-Dihydroxy benzene.
Sol. :
O O
OO
H
H
::
:
:
Na Cr O2 2 7
H SO /H O2 4 2
50. Answer (3)
Hint : Alkylation reaction of Amine.
Sol. :
CH I3
N N+
N+
N
CH3
Ag2O
CH3
CH3
I–
OH–
51. Answer (3)
Hint : I and III are mirror images
Sol. : (I, III) enantiomers
(I, IV), (III, IV) diasteromers
II is a positional isomer of the others.
52. Answer (1)
Hint : Preparation of phenol from cumene.
Sol. :
(i) O2
(ii) H3O
+
C
C
C OH
O C
CH3
CH3
A B
+
CH3
H3C H
HCl
CH3
CH3
O C
CH3
CH3
2 OH
HO C OH
+
53. Answer (1)
Hint : Elimination reaction; addition of H2O.
Sol. :
CH OH3
heat, KOH
Major
Br
CH3
CH3
BH –THF3
H O ; OH2 2
–
CH3
CH3
O Et
CH3
(i) Na
(ii) CH – 3 CH – Br2
OH
54. Answer (2)
Hint : Reaction of Alkyne
Sol. :
C H2 5
C C H3 7
CO
3
Zn, H O2
C H2 5
C C
O O
C H3 7
55. Answer (1)
Hint : Combustion of hydrocarbon.
Sol. : In Reaction-I, highest heat per gram of the
hydrocarbon is liberated.
Mock Test - 2 (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
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PART - C (MATHEMATICS)
61. Answer (4)
Hint : Truth table
Sol. : Truth values of r and s are T and T
respectively.
62. Answer (4)
Hint : Draw graph
Sol. :
y = 2
x
y =
x320
y =
x
y = x
y =
x +
2 y
x
=
2
y x= cos (cos )1
y
56. Answer (2)
Hint : Formation of hydrocarbons.
Sol. :
CHC OH
O
Reaction I: +CH
2
CH2
OHC
CH
O
Reaction II: +C
H
C
C O CH3
O
CH
C O CH3
C
O
57. Answer (3)
Hint : Oxymercuration of alkene.
Sol. :
OHOH
HgOAc+
+
Hg
CH2 CH2
HgOAc
NaBH4
+OAc O
+
O
H
58. Answer (4)
Hint : Alkali metal hydrides are denser than parent
metal.
Sol. : Metallic hydrides have a lattice which is
different from that of the parent metal.
59. Answer (2)
Hint : Hg22+ ions disproportionate in presence of
cyanide.
Sol. : 2
2 2Hg 2CN Hg Hg(CN)
60. Answer (4)
Hint : Magnetic nature of complex compound.
Sol. : Only IV has unpaired electron.
Number of real solution(s) of the equation
cos–1(cosx) + x = 2 is intersection point of
y = cos–1(cosx) and y = 2 – x
y = cos–1(cosx) and y = 2 – x intersect at
infinitely many points.
63. Answer (3)
Hint : Line is parallel to the vector 1 2n n��� ���
Sol. : Line is parallel to the vector 1 2n n��� ���
Where � �
1 22 2 , 3 4 6n i j k n i j k
��� ���� � � �
1 2 3. 0n n n
��� ��� ���
(where �
34 5 8n i j k
���� � )
Line is perpendicular to normal to the plane
Line is parallel to the plane
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-F) (Hints & Solutions)
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64. Answer (2)
Hint : Required probability
= P(odd, odd) + P(even, even)
Sol. : Let p be the probability of showing an odd
number
Then 3p = 1 p = 1
3
Required probability
= P(odd, odd) + P(even, even)
= p2 + 4p2
= 5p2
= 5
9
65. Answer (3)
Hint :
2
2 2( )ix
xN
∑
Sol. :1 2 3 ... 10 10(11) 11
10 2 10 2x
Variance =
22
2 ix
xN
∑
=
21 10(10 1) (2 10 1) 11
10 6 2
⎛ ⎞ ⎜ ⎟⎝ ⎠
2 11 21 11 11
6 4
= 11 7 11 11 1114 11
2 4 4
2 33
4 =
33
2
66. Answer (2)
Hint : Find n, L.D. eq
dx
dy
Sol. :1
y
y
y
dx xee x
dy e
dx
dy + x = e–y
I.F = 1dy
e∫ = ey
Solution of differential equation is
x.ey = .
y ye e dy c ∫xey = y + c passes then (0, 0)
c = 0
Equation of curve is y = xey
67. Answer (2)
Hint : Factorization
Sol. : 22 3 3 2 1z z i z z = 0
22 2 3 3z zi z i i z = 0
2z(z – i) + 1(z – i) – 3 ( )i z i = 0
( ) 2 1 3z i z i = 0
z = i, z = 1 3
2 2i =
18 + 18 + 1 = i18 + 18 + 1
= –1 + 1 + 1 = 1
68. Answer (1)
Hint : (A + B)(A – B) = A2 – B2
Sol. : (1 + x + x2)35 (1 – x + x2)35
= ((x2 + 1)2 – x2)35
= (x4 + x2 + 1)35
Coefficient of x15 is zero
69. Answer (3)
Hint : 2 3 2
2 3 2 2 3 3(32) | | | | z z z z z z
Sol. : |(32)2z1 + (16)2z
2 + (8)2z
3|
= |(|z2| |z
3|)2z
1 + (|z
1| |z
3|)2z
2 + (|z
1||z
2|)2z
3|
= 2 2 3 3 1 1 1 3 3 2 1 1 2 2 3z z z z z z z z z z z z z z z
= 1 2 3 2 3 3 1 1 2z z z z z z z z z
= 1 2 3 2 3 3 1 1 2| | | | | |z z z z z z z z z
= 2 × 4 × 8 |z1z
2 + z
2z
3 + z
3z
1|
= 64
70. Answer (3)
Hint : Sum of G.P.
Mock Test - 2 (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
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Sol. :
S = 2 3
12 12 121 ...i i i
e e e
upto 25 terms
=
2525
12 12
1212
1 1
11
i i
ii
e e
ee
=
212
12
1
1
i
i
e
e
⎛ ⎞ ⎜ ⎟⎝ ⎠
= 12
12
1
1
i
i
e
e
= 1
71. Answer (2)
Hint : A.M. G.M.
Sol. :
1
3(2 ) (4 ) (8 )
((2 )·(4 ) (8 ))3
xy yz zxxy yz zx
1
2 2 2 327
(64 )3
x y z
64x2y2z2 93
2
2 2 2 27
8x y z
⎛ ⎞ ⎜ ⎟⎝ ⎠
27
8xyz Greatest possible value of xyz
is 3.375
72. Answer (1)
Hint : Make all cases
Sol. : I, I, I, I, M, P, P, S, S, S, S
1(I)
1(M)
+
1(I)
1(P)
+
1(I)
1(S)
Number of words = 9! 9! 9!
3!2!4! 3! 4! 3! 2!3!
= 9!
74!3!2!
= 1260 × 7
73. Answer (3)
Hint : If n(A) = n, then number of reflexive relations
on the set A = 2n(n–1)
Sol. : If n(A) = n, then number of reflexive relations
on the set A = 2n(n–1)
74. Answer (3)
Hint : = 0
Sol. : = 0
1 1
1 2
1 1 0
= 0
2 – 2 + (–– 1) = 0
2 + = 0
= 0 or –1
75. Answer (3)
Hint : = 0
Sol. : = 0
5 31
2 2
50
2
30 4
2
k
= 0
5 3 3
4 (10)2 2 2
k k⎛ ⎞ ⎜ ⎟⎝ ⎠
= 0
94
4k k = 25
25
4k = 25 k = 4
tan = 2
2 h ab
a b
= 3
5
= 1 3
tan5
⎛ ⎞⎜ ⎟⎝ ⎠
76. Answer (4)
Hint :2
a b c
⎡ ⎤⎢ ⎥⎣ ⎦
=
a a a b a c
b a b b b c
c a c b c c
⎡ ⎤ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥
⎣ ⎦
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Sol. :2
a b c
⎡ ⎤⎢ ⎥⎣ ⎦
=
a a a b a c
b a b b b c
c a c b c c
⎡ ⎤ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥
⎣ ⎦
=
1 3
1 2
3 2
1
1
1
⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦
= 1 + 21
2
3 –
12 –
22 –
32
a b c
⎡ ⎤⎢ ⎥⎣ ⎦ =
2 2 21 2 3 1 2 31 2
77. Answer (2)
Hint : 2 2 2 y mx a m b
Sol. : Equations of common tangents to both
hyperbolas are
y = 2 2
x a b
For 4 common tangents, a > b
78. Answer (2)
Hint : y = x is axis of both hyperbolas
Sol. :
x
y y x =
xy = 4
xy = 1 A(1, 1)
(2, 2)B
AB = 2
79. Answer (2)
Hint : S1 – S
2 = 0
Sol. : c1(–g, 0), (r
1 = |g|, c
2(0, – f), r
2 = |f |, equation
of common chord is
A
B
MC
1
(– , 0)g
r1
gx fy – = 0
gx fy – = 0
C1M =
2
2 2
g
g f
AM2 = r12 – (C
1M)2 =
42
2 2
gg
g f
=
2 2
2 2
g f
g f
AM = 2 2
| |gf
g f
AB = 2 2
2 | |gf
g f
80. Answer (2)
Hint : (d) = r
Sol. : Equation of tangent of x2 = – 4y is
tx = –y + t2 ...(i)
(i) is also tangent to x2 + y2 = 4
2
2
| |
1
t
t = 2
t4 = 4 + 4t2 t2 = 2 2 2
m = – t m2 = 2 2 2 [m2] = 4
81. Answer (3)
Hint : AA–1 = I
Sol. : A–1A = I, then compare corresponding
elements.
82. Answer (1)
Hint :�
a is perpendicular to c�
Sol. : a c� �
= a c� �
a c� �
ˆ ˆ ˆ( )c i j k �
a c� �
= b�
ˆ ˆ ˆ
ˆ ˆ1 1 0
i j k
i j
= 0, = 1
c�
= ˆk
83. Answer (1)
Hint : Multiply and divide by
1 1
2 2(sec ) (tan )n n
x x
Mock Test - 2 (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
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Sol. :
R =
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
11
1 1 11
2 2 22(sec ) (tan ) (sec ) tan
...(sec tan )
n n n
nx x x x
x x
R =
2 2
22
1 1
2 2
11
22
1 1
2 2
(sec ) (tan )
(sec ) tan ...(sec tan )
(sec ) (tan )
n n
n
n
n n
x x
x x x x
x x
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟ ⎝ ⎠
= 1 1
2 2
1
(sec ) (tan )n n
x x
=
1
2
1
2
(cos )
1 (sin )
n
n
x
x
84. Answer (2)
Hint : Plane contains points A, B, C
Sol. : Plane containing all medians is the plane
passing through the vertices A, B and C
Equation of the plane is
PA BA CA
⎡ ⎤⎢ ⎥⎣ ⎦
= 0 where P(x, y, z)
1 1 1
1 0 1
x y z
= 0
x – 2y + z = 0
85. Answer (1)
Hint : Parabolic curve
Sol. : 0dy
dx
3ax2 + 6 0
y ax = 3 + 62
3a > 0, D < 0
a > 0
0 – 4(3a)6 < 0
a > 0
at a = 0, y = 6x, an increasing function
86. Answer (3)
Hint : 3sinx + 2cosx
= 3cos 2sin (3cos 2sin ) dx x x x
dx
Sol. :
I = 3sin 2cos
3cos 2sin
∫
x xdx
x x
=
(3cos 2sin )3cos 2sin
3cos 2sin 3cos 2sin
∫ ∫
dx x
x x dxA dx B dx
x x x x
3sinx + 2cosx = A(3cosx + 2sinx)
+ B(–3sinx + 2cosx)
Comparing coefficients of sinx and cosx on both
sides
A = 12
13, B =
5
13
121
1 13
5
13
⎛ ⎞
⎜ ⎟⎝ ⎠
A
B = –5
87. Answer (2)
Hint : Draw graph
Sol. :
x
y
AB
C
x = 1
y a =
(1, 1)
O
y = 2x – x2 (x – 1)2 = –(y – 1)
If y = a, then x = 1 1 a
A1 =
2 2
0(2 )x x dx∫ =
23
2
03
xx
⎡ ⎤⎢ ⎥
⎣ ⎦
= 8 4
43 3
A(OABC) = 1
3
1 1 2
0
1(2 ) 1
3
a
x x dx a a
∫
2 31 11 1 1 1 1
3 3a a a a
All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-F) (Hints & Solutions)
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� � �
Put 1 – a = t2
2t3 = 1 t6 = 1
4 = 2–2
t2 =
2
32
a = 1 – t2 = 2
31 2
88. Answer (3)
Hint : Take log on both sides then convert in definite
integral
Sol. :
logl =1 1 2
lim log log log ... logn
n n n n n
n n n n n
⎛ ⎞⎡ ⎤ ⎛ ⎞ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎝ ⎠
=
21lim log
n
nr n
r
n n
⎛ ⎞⎜ ⎟⎝ ⎠
∑ = 2
1logxdx∫
= [xlogx – x]12 = 2log2 – 2 + 1 = 2log2 – 1
= 4
loge
⎛ ⎞⎜ ⎟⎝ ⎠
l = 4
e
89. Answer (1)
Hint : Form : 1
Sol. : l =
1
1 cot
0
sinlim
x x
x
x
x
⎛ ⎞⎜ ⎟⎝ ⎠
form : 1
∵0
sinlimx
x
x
⎛ ⎞⎜ ⎟⎝ ⎠
= 1
0
lim ( cot )x
x x
= 0
limtanx
x
x = 1
l =
sin 1lim 1
1 cot0
x
x x xxe
⎛ ⎞ ⎜ ⎟ ⎝ ⎠
=
sin tanlim
tan0
x x x
x x xxe
⎛ ⎞⎜ ⎟⎝ ⎠
=
sin tanlim
tan0
x x x
x x xxe
⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
=
11
2e
=
1
e
90. Answer (4)
Hint :3
2
0
2( )
3 ∫
bb
f x dx b
Sol. :3
2
0
2( )
3
b bf x dx b ∫
d. w.r. t. b
f(b) = 2b – 2b2 f(x) = 2x – 2x2
f(x) = 2 1 2
24 4
x x⎡ ⎤ ⎢ ⎥⎣ ⎦
=
21 1
22 2
x⎛ ⎞ ⎜ ⎟⎝ ⎠
f(x)max.
= 1
2 when x =
1
2