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Test - 2 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2020
All India Aakash Test Series for JEE (Main)-2020
Test Date : 11/08/2019
ANSWERS
1/12
TEST - 2 - Code-A
PHYSICS CHEMISTRY MATHEMATICS
1. (4)
2. (3)
3. (3)
4. (3)
5. (3)
6. (3)
7. (3)
8. (2)
9. (2)
10. (2)
11. (4)
12. (1)
13. (3)
14. (2)
15. (4)
16. (1)
17. (2)
18. (4)
19. (4)
20. (3)
21. (3)
22. (3)
23. (3)
24. (2)
25. (4)
26. (3)
27. (2)
28. (1)
29. (1)
30. (2)
31. (2)
32. (1)
33. (3)
34. (4)
35. (3)
36. (2)
37. (4)
38. (4)
39. (3)
40. (2)
41. (1)
42. (4)
43. (2)
44. (1)
45. (4)
46. (3)
47. (4)
48. (1)
49. (1)
50. (4)
51. (2)
52. (1)
53. (3)
54. (3)
55. (2)
56. (1)
57. (3)
58. (1)
59. (4)
60. (2)
61. (2)
62. (4)
63. (2)
64. (2)
65. (4)
66. (1)
67. (4)
68. (2)
69. (3)
70. (4)
71. (2)
72. (3)
73. (2)
74. (3)
75. (4)
76. (4)
77. (2)
78. (2)
79. (2)
80. (1)
81. (2)
82. (1)
83. (1)
84. (3)
85. (1)
86. (4)
87. (1)
88. (3)
89. (1)
90. (1)
All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-A) (Hints & Solutions)
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1. Answer (4)
Hint : 01 2(cos cos )
4
= +
IB
d
Sol. : 2 3
d =
= 30°
03 (cos30 cos30 )
42 3
IB
= +
( )0 2 33 3
4
I =
09
2
I=
2. Answer (3)
Hint : 2= I
TMB
Sol. : 0
603 s
20T = =
Now, 01
0 2 02
BBT
T B B= =
0 3s
2 2
TT = =
3. Answer (3)
Hint : Use Ampere’s circulation law
Sol. : 0 inB dl i = 1forB r r R
30 forB r R=
4. Answer (3)
Hint: Identify the equipotential points and
redraw the circuit.
Sol. : It can be reduced as
RAB = 7
12R by simple calculation
5. Answer (3)
Hint : , 2 ( )mv
r mv m KEqB
= =
Sol.: 2 ( )m KEmv
rqB qB
= =
2 ( . )
,ee
m K Er
exB =
2 ( . )p
p
m K Er
eB=
2 ( )2 ( )
(2 )
pm KEm KEr
e B eB
= =
Clearly, rp > re and rp = r
6. Answer (3)
Hint :
0 (4 2)
42
=
s
IB
l
Sol. :
0 01 2 22 4
242
s
I IB
a a
= =
0
2c
IB
R
=
2 2 2
1s
c
B R
B a
=
PART - A (PHYSICS)
Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
3/12
Now, 4a = 2R 2
Ra
=
2
4 2 2 8 2S
C
B R
B R
= =
7. Answer (3)
Hint : AV = constant
Sol.: AV = constant
Since A2 < A1
V2 > V1
8. Answer (2)
Hint : cot2 = cot21 + cot22
Sol. : cot2 = cot21 + cot22
1 10
33 3
= + =
10
cot3
=
3
sin13
=
1 3sin
13
− =
9. Answer (2)
Hint : Use concept of effective length
Sol. :
3
0 12
2
R
x R
IdF I dx
x=
=
0 1 2 ln(3)2
I IF
=
10. Answer (2)
Hint and sol. :
VPQ = IPQ RPQ
Apply Kirchhoff law to get current in wire
PQ
|VQ – VP| = I(1)
I = 0.13 A from Q to P
11. Answer (4)
Hint : Baxis 0 0equatorial3 3
2,
4 4
= =
m mB
d d
Sol. : 01 3
2 ˆ4
mB i
a
=
( )02 3
ˆ4
mB i
a
= −
01 2 3
ˆ( )4
p
mB B B i
a
= + =
12. Answer (1)
Hint : At t = 0 all capacitors will act like a
conducting wire
Sol. : total
eq
VI
R=
2
V
R
=
1
3 62
VVI
RR
= =
13. Answer (3)
Hint : V = IR
Sol. : If S1 is closed
3
1
3
3 1 4
E EV
= =
+
Similarly 2
6(6)
7 7
E EV
= =
and 3
2
3
EV =
14. Answer (2)
Hint : Let resistance be X, then 2
12
= ++
XX
X
Sol. :
We have
2
12
XX
X= +
+
2X =
All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-A) (Hints & Solutions)
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15. Answer (4)
Hint : Use Kirchhoff law
Sol. : eq
15 304
15 30R
= +
+ = 14
30
2 A14 1
I = =+
16. Answer (1)
Hint : Use condition of balanced Wheatstone
bridge
Sol. : 3 2
PS SQ=
2
100 cm5
SQ = = 40 cm
17. Answer (2)
Hint : Deflection is directly proportional to current
Sol. : 3
max
510 A
4950 50i −= =
+
max imax
For 20 divisions,
3
max
2 210 A
3 3i i − = =
Let R be resistance in series
Then 35 210
50 3R
−= +
R = 7450
18. Answer (4)
Hint : Both fields are perpendicular to each
other
Sol. : 0 1 0 21 2,
2 2
I IB B
d d
= =
B1 ⊥ B2
2 2
net 1 2B B B = +
19. Answer (4)
Hint : Integrate magnetic moment due to an
elementary ring.
Sol. : N
N dxr
=
2
2( )r
x r
d N I x=
=
2
2
r
x r
Nd dx I x
r=
=
20. Answer (3)
Hint : B does not change K.E of particle
Sol. : The charge particle moves on a
cycloidal path when B E⊥
21. Answer (3)
Hint : Use Ampere’s circulation law
Sol. : 0 in( )B dl i =
= µo (+3 – 6) = –3µo
22. Answer (3)
Hint : q = qo (1 – e–t/)
= ReqC
Sol. : max
2
3q VC=
/22(1– )
3
VCq e−=
23. Answer (3)
Hint : V = IR
Sol. : 12 12 3
| |2 8
− = −C DV V = 1.5 V
24. Answer (2)
Hint : Redraw the circuit
Sol. :
eq
3
RR =
= 1
Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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25. Answer (4)
Hint : Electromagnet requires temporary
magnet
Sol. : Low retentivity and low coercivity
facilitates temporary magnetic property
in core material.
26. Answer (3)
Hint : At null point potential become same
Sol. : RAB = 10 Vend = 10 Volt
25 cm4
L= =
27. Answer (2)
Hint : Use U = ,i B− then dU
Fdx
−=
Sol. : 2
0 0
2 2 3/22( )
I RB
R x
=
+
µ = r2i
2
20 0
2 2 3/22( )
I RU r i
R x
=
+
2 2
0 0
2 2 5/2
3
2( )d
I R r iddU
R ddx
=
+
28. Answer (1)
Hint : Time spent =
Sol. : qB
m =
0
0
12sin
2
mV
d qB
mVr
qB
= = =
6
=
6
6
mt
qB qB
m
= =
29. Answer (1)
Hint : Force = iB
Sol. : 01
2
IF i a
a
=
02
2 (2 )
IF i a
a
=
0net 1 2
11
2 2
= − = −
IiF F F 0
4
=
Ii
30. Answer (2)
Hint : Magnetic field is uniform
Sol. :
By symmetry,
2 2B d B B = =
0 in( )B d I =
02 ( 1)B i =
0
2
iB
=
31. Answer (2)
Hint: By using gas law and kinetics law, it can
be assumed that at 2 5 2
0N O OV V
2 5 2
2 5 2 2
N O O0
N O O Ott
V VKt ln ln
V V V
= =
−
Sol. : 50
K 10 In50 10
=−
K = 0.02 min–1
Verifying the value of K
50
K 18 In50 15
=−
K = 0.02 min–1
PART - B (CHEMISTRY)
All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-A) (Hints & Solutions)
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32. Answer (1)
Hint : a2
1 1 2
Ek 1 1log
k 2.303 R T T
= −
3
a
4
E2 10 1 1log
2.303 R 400 42010
−
−
= −
aE 201.3
2.303 R 400 420=
Ea = 2.303 × 10920 × R
Sol. : a1
1
Elog k log A
RT 2.303= −
4 2.303 10920 Rlog10 log A
2.303 R 400
− = −
log A = 23.3
A 1023
33. Answer (3)
Hint : Chemisorption has high Ea that’s why
rate of chemisorption is enchanced by
increase in temp.
Sol. : But after a certain temperature, rate of
adsorption may decrease as the formed
bonds may dissociate at higher
temperature.
34. Answer (4)
Hint : B 1
C 2
n 2K
n 3K=
Sol. : –2
B 1–2
C 2
P 2K 2 3 101
P 3K 3 2 10
= = =
PB = PC
PB = 2 atm
35. Answer (3)
Hint : x = 1 ; y = 0, z = 2
Sol. : In Graph I , As rate increases linearly
with concentration reaction is of first
order
In Graph II, As concentration decreases
linearly, reaction is of zero order.
In Graph III, Product of half life and initial
concentration is constant, hence
reaction is of second order.
36. Answer (2)
Hint : r = K [R]x [H+]2
Sol. : r = K [R]x . [H+]2
2
A A2
BB
Hr4
r H
+
+
= =
a(HA)
a(HB)
K4
K=
6
a(HB)K 2.5 10−=
37. Answer (4)
Hint : Gases with greater critical temperature
are adsorbed to a greater extent.
Sol. :
SO CH H2 4 2
1 32
P P P
i.e P P P
38. Answer (4)
Hint : At high temperature physisorption may
change into chemisorptions.
Sol. : Physisorption is multilayered whereas
chemisorption is monolayered.
39. Answer (3)
Hint : Lead chamber process employs NO as
a catalyst
Sol. : ( ) ( ) ( ) ( )NO g
2 2 32SO g O g 2SO g+ ⎯⎯⎯⎯→
40. Answer (2)
Hint : Emulsions can be diluted with any
amount of dispersion medium
Sol. : The dispersed liquid on mixing with an
emulsion forms a separate layer
41. Answer (1)
Hint : Preferential adsorption of common ion
present in excess.
Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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Sol. : I. Fe2O3.xH2O/Fe3+
II. Fe2O3.xH2O/OH–
42. Answer (4)
Hint : (1), (2) and (3) represent adsorption
Sol. : Water vapour are absorbed on
anhydrous calcium chloride.
43. Answer (2)
Hint : Leaching of silver
( )–– –
2 2 24Ag 8CN 2H O O 4 Ag CN 4OH + + + → +
Sol. : Coordination number of Ag in [Ag(CN)2]–
is 2
44. Answer (1)
Hint : It induces the precipitation of hydrated
Al2O3
Sol. : 2Na[Al(OH)4] + CO2 → 2NaHCO3
+ Al2O3xH2O
45. Answer (4)
Hint : ( )
( )( )
2 3 2 2 3 2S x
Fe O .xH O Fe O xH O g
⎯⎯→ +
2 2(y)
ZnS O ZnO SO (g)
+ ⎯⎯→ +
Sol. : Oxidation number of O in H2O is –2 and
of S in SO2 is 4.
Difference = 6
46. Answer (3)
Hint : Composition of the slag formed is
FeSiO3.
Sol. : % of oxygen in FeSiO3
48
100132
= 36%
47. Answer (4)
Hint : Al from Al2O3 is not extracted by carbon
based reduction method.
Sol. : Oxides of Fe, Sn, Zn, Pb are reduced by
carbon
48. Answer (1)
Hint : That characteristic type of copper is
called as blister copper due to the
evolution of SO2
Sol. : Ag and Au impurities setteles as anode
mud.
49. Answer (1)
Hint : 4 2 2 5P SO Cl give PCl+
Sol. : 5 3 2PCl PCl Cl
⎯⎯→ +
4 2 3 2 2 2P 8SOCl 4PCl 4SO 2S Cl+ ⎯⎯→ + +
50. Answer (4)
Hint: NH4NO3 liberates NH3 gas with KOH
and KNO3 on heating dissociates to give
O2 which supports combustion.
Sol. :
4 3 3 2 3NH NO KOH NH H O KNO+ ⎯⎯→ + +
3 3 2 2 2KNO 4Zn 7KOH NH 4K ZnO 2H O+ + ⎯⎯→ + +
3 2 22KNO 2KNO O
⎯⎯→ +
51. Answer (2)
Hint : Red coloured solution show the
presence of Br2.
Sol. : Cl2 being stronger oxidising agent can
oxidise Br– to Br2.
52. Answer (1)
Hint : HI and HBr are strong reducing hydra
acids and hence they reduce H2SO4.
Sol. : HCI is quite stable and hence is oxidised
by strong oxidising agent like KMnO4.
HF is not a reducing agent as in F– ion, the
electron which is to be removed during
oxidation is very close to the nucleus.
53. Answer (3)
Hint : 3O2 → 2O3
( )H 298K 142 kJ / mol
= + process is
endothermic.
Sol. : O3 decompose at high temperature
All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-A) (Hints & Solutions)
8/12
54. Answer (3)
Hint : ( ) ( )
6 2 3BA
XeF 3H O XeO 6HF+ ⎯⎯→ +
Sol. : A is pyramidal
55. Answer (2)
Hint : SiO2 is polymeric solid
Sol. : SeO2, TeF4 are solid
56. Answer (1)
Hint : Greater the number of unpaired electrons
greater will be the paramagnetic character
Sol. : Mn2+ in MnSO4 .4H2O is 3d5
Cu2+ in CuSO4.5H2O is 3d9
Fe2+ in FeSO4.6H2O is 3d6
Ni2+ in NiSO4 .6H2O is 3d8
57. Answer (3)
Hint : 2MnO2 + 4KOH + O2 → 2K2MnO4 +
2H2O
Sol. : K2MnO4 is dark green in colour.
58. Answer (1)
Hint : Products are I2 and IO3–
Sol. :
( )
– 24 2 2
in acidicconditions
10 2MnO 16H 2Mn I 8H O+ ++ + ⎯⎯→ + +
( )
– – – –4 2 2 3
in neutralconditions
I 2MnO H O 2MnO 2OH IO+ + ⎯⎯→ + +
59. Answer (4)
Hint : Colour of d-block cations is due to d – d
transitions.
Sol. : Cu2+ (CuCl2) and VOCl2 (V+4) both show
blue colour in aq. medium.
60. Answer (2)
Hint : Lu+2 is 4f14 5d1
Sol. : Ce+2 4f2
Gd+2 4f7 5d1
Lu+2 4f14 5d1
Eu+2 4f7 5d0 6s0
61. Answer (2)
Hint : Use G.P. to get sum then divide by
51/n
Sol. : The given value
=
( )
1
1
1
2 12 2
5 1 55
lim
1 5 1 5
nn
n
nn n
nn
n
n n
+
→
− −
− −
= ( )
111
211
5 1 5 5lim
5 15 111
nn
n
nn
nn
+
→
− +
− −
= –4(log5e)2 + 5log5e
62. Answer (4)
Hint : Factorize then cancel the common term
Sol. :
1
2
4cos ( 1)
(tan 1)(tan tan 1)lim
(tan 1)(tan tan 1)x
x x x
x x x−→ −
+ − +
+ + −
2
4
( 1) ( 1) 13
( 1) 1 1
− − − += = −
− − −
63. Answer (2)
Hint : Properties of G.I.F. and modulus function
Sol. :
2
2( ) cos(| |) | | 3 | 2 |
2 1
xf x x x x x
x
= + + + −
+
Only non differentiable at x = 2
PART - C (MATHEMATICS)
Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
9/12
64. Answer (2)
Hint : Sum of infinite series then differentiation
Sol. : y = sinx y+
y2 = sinx + y
2 cosdy dy
y xdx dx
= +
cos
2 1
dy x
dx y=
−
2
cos
2 1 2sin
dy x
dx y x=
− −
65. Answer (4)
Hint : Algebra of limit then solve simultaneous
eq.
Sol. : p(x) = f(x) + g(x)
and q(x) = f(x) – g(x)
( ) ( )
lim2→
+
x a
p x q x and
( ) ( )lim
2→
−
x a
p x q x exist
( ) ( )
lim 42→
+=
x a
p x q x and
( ) ( )lim 1
2→
−=
x a
p x q x
( ) 4
lim 4( ) 1x a
f x
g x→= =
66. Answer (1)
Hint : is repeated root then f() = f () = 0
Sol. : A six degree polynomial satisfied by
exactly 5 real numbers then exactly one
of the f(0), f (–1), f (1),f (2) or f (–2) is
zero
67. Answer (4)
Hint :
dydy dt
dxdx
dt
= , then t = 1 to get coordinate
Sol. : 224
3 38
dy tt
dx t= = =
t = 1
Required coordinate = (7, 7)
68. Answer (2)
Hint : Differentiation of composite function
then interval
Sol. : Here g(x) = f(cos2x + 4cosx + 7)
On differentiating both sides w.r.t. x we
get :
g (x) = –2sinx(2 + cosx)f (cos2x + 4cosx + 7)
Here g(x) > 0 if sinx < 0
x(, 2)
69. Answer (3)
Hint : f (x) = 0 sin2x = 1
4. Then f(x) has
maxima
Sol. : f(x) = sin2x + cos4x
f (x) = 2cos2x – 4sin4x
= 2cos2x(1 – 4 sin2x)
Here f (x) = 0 if cos2x = 0 or sin2x = 1
4
f(x) has maximum value when sin2x = 1
4
f(x)maximum =
21 1
1 24 4
+ −
=
9
8
70. Answer (4)
Hint : Find the range by maximum and
minimum value
Sol. : We can see that f(x) cannot lie between
25 1and
2 2− −
71. Answer (2)
Hint : f(x) is discontinuous at x = 10
Sol. : The given function is only discontinuous
at x = 10.
10 10
lim ( ) lim ( )x x
f x f x− +→ →
All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-A) (Hints & Solutions)
10/12
72. Answer (3)
Hint :
( )
1 1
2 2
1
2x 2xcos sin
21 x 1 x
2tan x2
− −
−
= −
+ +
= − −
Sol. : f(x) = 1
2
2cos
1
xx
x
− −
+
= 1
2
2sin
2 1
xx
x
− − +
+
= 12tan2
x x−− + −
{ x = tan1}
= 12tan2
x x−+ −
f (x) = 2
2 2
2 11
1 1
x
x x
−− =
+ +
and g(x) = 1tan
42
xx
− −
= 2tan–1x +1 – x, as x = tan1
g (x) = 2
2 2
2 11
1 1
x
x x
−− =
+ +
2
2
2
2
1
( ( )) 1 1( ( )) 1
1
x
d f x xd g x x
x
−
+= = −−
+
73. Answer (2)
Hint : 3
2 , 154
dA daa a
dt dt= =
and 0.1=da
dt cm/s
Sol. : 0.1da
dt= cm/sec and
23
4A a=
3 3
0.1 152 2
dA daa
dt dt= = cm2/sec
= 3 3
4
74. Answer (3)
Hint : f(g(x)) = x if f and g are inverse of each
other
f (g(x)) = g (f(x)) = 1
Sol. : g(f(x)) = x
1
( ( ))( )
g f xf x
=
1
( (1))(1)
g ff
=
1 1
(3)6 4 1 11
g = =+ +
75. Answer (4)
Hint : 0 0
(0) lim ( ) lim ( )− +→ →
= =x x
f f x f x
Sol. :
0
lim ( )x
f x→
= tan
0
ln(sec tan )lim
tan
x x
x
e e x x x
x x→
− + + −
−
= tan 2
20
sec sec 1lim
sec 1
x x
x
e x e x
x→
− + −
−
=
( )
tan 2 tan
20
(sec 1) (sec 1)lim
sec 1
x x x
x
e x x e e
x→
− + − + −
−
= tan
20
11 lim
2 tan
x x
x
e e
x→
−+ +
= 3 3
02 2
+ =
76. Answer (4)
Hint : 1 form
Sol. :
1
11
lim2 1
21
lim lim2
n
n n
en
n
n n
ee e
→
−
→ →
+= =
77. Answer (2)
Hint : 2
2
2
2 4( ) ln 4
2 4
+ − = − − − −
xf x x
x
Let x = 2sin
Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
11/12
( ) 2ln cot 2cos2
= −
f x
24( ) ,
− − =
xf x
x
Sol. : 2
2
2
2 4( ) ln 4
2 4
xf x x
x
+ − = − − − −
Let x = 2sin then
24
( )− −
=x
f xx
and x(–2, 2) – {0}
f(x) is decreasing in (0, 1)
78. Answer (2)
Hint : f(a – h) < f(a) > f(a + h) then get b
Sol. : 2
2
2 | 5 6 |, 2( )
1, 2
x x xf x
b x
− − + =
+ =
b2 + 1 > 2 for local maxima
|b| 1
79. Answer (2)
Hint : Find domain
Sol. : –1 x 1 or x(–, –1] [1, )
x = ± 1
4
64
= −
80. Answer (1)
Hint : C1 → C1 + C2
Sol. : C1 → C1 + C2
{ } 1 1 { } 1
– {– } –1 –1 {– } –1 0
0 1 1 0 1
x x x
x x x x
x
= = =
81. Answer (2)
Hint : Coefficient of x in (x) = (0)
Sol. : Coefficient of x in (x) is (0)
2 1
1
21 2
2
1sin –1
1 sin cos –11
( ) cos 0 cos – sin 0 01
tan sec 11
tan 11
xx x x
x x x xx
x x
xx
−
−
−
−
= − + +−
+
0 1 11 0 1
(0) 1 1 0 0 0 (1 1) 22 2 2
1 0 10 1 1
−−
= − + = + + − − = −
82. Answer (1)
Hint : Multiplication of two determinants
Sol. :
2 2
2 2
1 1
( ) 1 1 0
1 1 1 0 0 0
x x x x
f x x x x x= =
( )2 1 0f − =
83. Answer (1)
Hint : Expansion
Sol. :
1 1 1 1 2 31 1
2 2 2 1 2 32 2
1 2 31 0 0
A B C a a aB C
A B C b b bB C
c c c
=
1 1 23
2 21 2 3
0 0
0 0B C
aB C
a a a
= =
1 1
32 2
B Ca
B C =
84. Answer (3)
Hint : = 0
Sol. : For non-trivial solution
1 2
2 1 3 0,
3 1
t t t
t t t
t t t
+ +
+ + + =
+ +
R2→R2 – R1, R3 → R3 –R1
All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-A) (Hints & Solutions)
12/12
1 2
1 1 3 0
1 1 1
t t t+ +
− =
−
(t + 1)(–1–3)–(t + 2)(1 + 3) + t(1 – 1) = 0
–4t – 4 – 4t –8 = 0
–8t = 12 3
2t = −
85. Answer (1)
Hint : Property of diagonal matrix
Sol. :
88
0
88
0 1 2 880
88
0
tan( 1) 0 0
... 0 cot( 1) 0
0 0 1
r
r
r
r
A A A A r
=
=
=
+
= +
88
0
tan( 1)r
r=
+ = tan1°tan2°..tan45°…tan88°.
tan89° = 1
88
0
cot( 1)r
r=
+ = cot1°cot2°…cot45°…cot88°
.cot89°= 1
88
0
1 1r =
=
86. Answer (4)
Hint : |A – I| = 0
Sol. :
1 2 3
| | 0, 2 3 1 0
3 1 2
A I
−
− = − =
−
3 – 6 2 – 3 + 18 = 0
A3 – 6A2 – 3A – 18I = 0
A2 – 6A – 18A–1 = 3I
87. Answer (1)
Hint : Property of adj. and inverse
Sol. : (adjA–1)–1(adjA)–1 = ((adjA)(adjA–1))–1
= (adj(A–1A))–1 = (adjI)–1 = (I)–1 = I
88. Answer (3)
Hint : Multiplication of matrices
Sol. :
2
0 0 1 0 0 1 1 0 0
0 1 0 0 1 0 0 1 0
1 0 0 1 0 0 0 0 1
A I
= − − = =
A is involutory matrix
|A| = 1 non-singular matrix
89. Answer (1)
Hint : Properties of adj. and inverse
Sol. :
|(2adj3A–1)–1| = |2adj(3A–1)|–1 = 2–3|adj(3A–1)|–1
= 2–3|32adj(A–1)|–1 = 2–3(32)–3|adj(A–1)|–1
= 18–3|(adjA)–1|–1 = 18–3|adjA| = 18–3|A|2
= 1 1
27 27 218 18 18 4
=
90. Answer (1)
Hint : Find
Sol. :
1 1 1
2 1 3 1( 3 3) 1(6 3) 1(2 1)
1 1 3
= − = − − − − + +
= –6 – 3 + 3 = – 6 0
System has a unique solution
Test - 2 (Code-B) (Answers) All India Aakash Test Series for JEE (Main)-2020
All India Aakash Test Series for JEE (Main)-2020
Test Date : 11/08/2019
ANSWERS
1/12
TEST - 2 - Code-B
PHYSICS CHEMISTRY MATHEMATICS
1. (2)
2. (1)
3. (1)
4. (2)
5. (3)
6. (4)
7. (2)
8. (3)
9. (3)
10. (3)
11. (3)
12. (4)
13. (4)
14. (2)
15. (1)
16. (4)
17. (2)
18. (3)
19. (1)
20. (4)
21. (2)
22. (2)
23. (2)
24. (3)
25. (3)
26. (3)
27. (3)
28. (3)
29. (3)
30. (4)
31. (2)
32. (4)
33. (1)
34. (3)
35. (1)
36. (2)
37. (3)
38. (3)
39. (1)
40. (2)
41. (4)
42. (1)
43. (1)
44. (4)
45. (3)
46. (4)
47. (1)
48. (2)
49. (4)
50. (1)
51. (2)
52. (3)
53. (4)
54. (4)
55. (2)
56. (3)
57. (4)
58. (3)
59. (1)
60. (2)
61. (1)
62. (1)
63. (3)
64. (1)
65. (4)
66. (1)
67. (3)
68. (1)
69. (1)
70. (2)
71. (1)
72. (2)
73. (2)
74. (2)
75. (4)
76. (4)
77. (3)
78. (2)
79. (3)
80. (2)
81. (4)
82. (3)
83. (2)
84. (4)
85. (1)
86. (4)
87. (2)
88. (2)
89. (4)
90. (2)
All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-B) (Hints & Solutions)
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1. Answer (2)
Hint : Magnetic field is uniform
Sol. :
By symmetry,
2 2 = = B d B B
0 in( ) = B d I
02 ( 1)B i =
0
2
iB
=
2. Answer (1)
Hint : Force = iB
Sol. : 01
2
IF i a
a
=
02
2 (2 )
IF i a
a
=
0net 1 2
11
2 2
= − = −
IiF F F 0
4
=
Ii
3. Answer (1)
Hint : Time spent =
Sol. : qB
m =
0
0
12sin
2
mV
d qB
mVr
qB
= = =
6
=
6
6
mt
qB qB
m
= =
4. Answer (2)
Hint : Use U = ,i B− then dU
Fdx
−=
Sol. : 2
0 0
2 2 3/22( )
I RB
R x
=
+
µ = r2i
2
20 0
2 2 3/22( )
I RU r i
R x
=
+
2 2
0 0
2 2 5/2
3
2( )d
I R r iddU
R ddx
=
+
5. Answer (3)
Hint : At null point potential become same
Sol. : RAB = 10 Vend = 10 Volt
25 cm4
L= =
6. Answer (4)
Hint : Electromagnet requires temporary
magnet
Sol. : Low retentivity and low coercivity
facilitates temporary magnetic property
in core material.
7. Answer (2)
Hint : Redraw the circuit
Sol. :
eq
3
RR =
= 1
8. Answer (3)
Hint : V = IR
Sol. : 12 12 3
| |2 8
− = −C DV V = 1.5 V
9. Answer (3)
Hint : q = qo (1 – e–t/)
= ReqC
PART - A (PHYSICS)
Test - 2 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
3/12
Sol. : max
2
3q VC=
/22(1– )
3
VCq e−=
10. Answer (3)
Hint : Use Ampere’s circulation law
Sol. : 0 in( )B dl i =
= µo (+3 – 6) = –3µo
11. Answer (3)
Hint : B does not change K.E of particle
Sol. : The charge particle moves on a
cycloidal path when B E⊥
12. Answer (4)
Hint : Integrate magnetic moment due to an
elementary ring.
Sol. : N
N dxr
=
2
2( )r
x r
d N I x=
=
2
2
r
x r
Nd dx I x
r=
=
13. Answer (4)
Hint : Both fields are perpendicular to each
other
Sol. : 0 1 0 21 2,
2 2
I IB B
d d
= =
B1 ⊥ B2
2 2
net 1 2B B B = +
14. Answer (2)
Hint : Deflection is directly proportional to current
Sol. : 3
max
510 A
4950 50i −= =
+
max imax
For 20 divisions,
3
max
2 210 A
3 3i i − = =
Let R be resistance in series
Then 35 210
50 3R
−= +
R = 7450
15. Answer (1)
Hint : Use condition of balanced Wheatstone
bridge
Sol. : 3 2
PS SQ=
2
100 cm5
SQ = = 40 cm
16. Answer (4)
Hint : Use Kirchhoff law
Sol. : eq
15 304
15 30R
= +
+ = 14
30
2 A14 1
I = =+
17. Answer (2)
Hint : Let resistance be X, then 2
12
= ++
XX
X
Sol. :
We have
2
12
XX
X= +
+
2X =
18. Answer (3)
Hint : V = IR
Sol. : If S1 is closed
3
1
3
3 1 4
E EV
= =
+
All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-B) (Hints & Solutions)
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Similarly 2
6(6)
7 7
E EV
= =
and 3
2
3
EV =
19. Answer (1)
Hint : At t = 0 all capacitors will act like a
conducting wire
Sol. : total
eq
VI
R=
2
V
R
=
1
3 62
VVI
RR
= =
20. Answer (4)
Hint : Baxis 0 0equatorial3 3
2,
4 4
= =
m mB
d d
Sol. : 01 3
2 ˆ4
mB i
a
=
( )02 3
ˆ4
mB i
a
= −
01 2 3
ˆ( )4
p
mB B B i
a
= + =
21. Answer (2)
Hint and Sol. :
VPQ = IPQ RPQ
Apply Kirchhoff law to get current in wire PQ
|VQ – VP| = I(1)
I = 0.13 A from Q to P
22. Answer (2)
Hint : Use concept of effective length
Sol. :
3
0 12
2
R
x R
IdF I dx
x=
=
0 1 2 ln(3)2
I IF
=
23. Answer (2)
Hint : cot2 = cot21 + cot22
Sol. : cot2 = cot21 + cot22
1 10
33 3
= + =
10
cot3
=
3
sin13
=
1 3sin
13
− =
24. Answer (3)
Hint : AV = constant
Sol.: AV = constant
Since A2 < A1
V2 > V1
25. Answer (3)
Hint :
0 (4 2)
42
=
s
IB
l
Sol. :
0 01 2 22 4
242
s
I IB
a a
= =
0
2c
IB
R
=
2 2 2
1s
c
B R
B a
=
Now, 4a = 2R 2
Ra
=
2
4 2 2 8 2S
C
B R
B R
= =
26. Answer (3)
Hint : , 2 ( )mv
r mv m KEqB
= =
Sol.: 2 ( )m KEmv
rqB qB
= =
Test - 2 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
5/12
2 ( . )
,ee
m K Er
exB =
2 ( . )p
p
m K Er
eB=
2 ( )2 ( )
(2 )
= =
pm KEm KEr
e B eB
Clearly, rp > re and rp = r
27. Answer (3)
Hint: Identify the equipotential points and
redraw the circuit.
Sol. : It can be reduced as
RAB = 7
12R by simple calculation
28. Answer (3)
Hint : Use Ampere’s circulation law
Sol. : 0 inB dl i = 1forB r r R
30 forB r R=
29. Answer (3)
Hint : 2= I
TMB
Sol. : 0
603 s
20T = =
Now, 01
0 2 02
BBT
T B B= =
0 3s
2 2
TT = =
30. Answer (4)
Hint : 0
1 2(cos cos )4
= +
IB
d
Sol. : 2 3
d =
= 30°
03 (cos30 cos30 )
42 3
IB
= +
( )0 2 33 3
4
I =
09
2
I=
31. Answer (2)
Hint : Lu+2 is 4f14 5d1
Sol. : Ce+2 4f2
Gd+2 4f7 5d1
Lu+2 4f14 5d1
Eu+2 4f7 5d0 6s0
32. Answer (4)
Hint : Colour of d-block cations is due to d – d
transitions.
Sol. : Cu2+ (CuCl2) and VOCl2 (V+4) both show
blue colour in aq. medium.
33. Answer (1)
Hint : Products are I2 and IO3–
Sol. :
( )
– 24 2 2
in acidicconditions
10 2MnO 16H 2Mn I 8H O+ ++ + ⎯⎯→ + +
( )
– – – –4 2 2 3
in neutralconditions
I 2MnO H O 2MnO 2OH IO+ + ⎯⎯→ + +
PART - B (CHEMISTRY)
All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-B) (Hints & Solutions)
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34. Answer (3)
Hint : 2MnO2 + 4KOH + O2 → 2K2MnO4 +
2H2O
Sol. : K2MnO4 is dark green in colour.
35. Answer (1)
Hint : Greater the number of unpaired electrons
greater will be the paramagnetic character
Sol. : Mn2+ in MnSO4 .4H2O is 3d5
Cu2+ in CuSO4.5H2O is 3d9
Fe2+ in FeSO4.6H2O is 3d6
Ni2+ in NiSO4 .6H2O is 3d8
36. Answer (2)
Hint : SiO2 is polymeric solid
Sol. : SeO2, TeF4 are solid
37. Answer (3)
Hint : ( ) ( )
6 2 3BA
XeF 3H O XeO 6HF+ ⎯⎯→ +
Sol. : A is pyramidal
38. Answer (3)
Hint : 3O2 → 2O3
( )H 298K 142 kJ / mol
= + process is
endothermic.
Sol. : O3 decompose at high temperature
39. Answer (1)
Hint : HI and HBr are strong reducing hydra
acids and hence they reduce H2SO4.
Sol. : HCI is quite stable and hence is oxidised
by strong oxidising agent like KMnO4.
HF is not a reducing agent as in F– ion, the
electron which is to be removed during
oxidation is very close to the nucleus.
40. Answer (2)
Hint : Red coloured solution show the
presence of Br2.
Sol. : Cl2 being stronger oxidising agent can
oxidise Br– to Br2.
41. Answer (4)
Hint: NH4NO3 liberates NH3 gas with KOH
and KNO3 on heating dissociates to give
O2 which supports combustion.
Sol. :
4 3 3 2 3NH NO KOH NH H O KNO+ ⎯⎯→ + +
3 3 2 2 2KNO 4Zn 7KOH NH 4K ZnO 2H O+ + ⎯⎯→ + +
3 2 22KNO 2KNO O
⎯⎯→ +
42. Answer (1)
Hint : 4 2 2 5P SO Cl give PCl+
Sol. : 5 3 2PCl PCl Cl
⎯⎯→ +
4 2 3 2 2 2P 8SOCl 4PCl 4SO 2S Cl+ ⎯⎯→ + +
43. Answer (1)
Hint : That characteristic type of copper is
called as blister copper due to the
evolution of SO2
Sol. : Ag and Au impurities setteles as anode
mud.
44. Answer (4)
Hint : Al from Al2O3 is not extracted by carbon
based reduction method.
Sol. : Oxides of Fe, Sn, Zn, Pb are reduced by
carbon
45. Answer (3)
Hint : Composition of the slag formed is
FeSiO3.
Sol. : % of oxygen in FeSiO3
48
100132
= 36%
46. Answer (4)
Hint : ( )
( )( )
2 3 2 2 3 2S x
Fe O .xH O Fe O xH O g
⎯⎯→ +
2 2
(y)
ZnS O ZnO SO (g)
+ ⎯⎯→ +
Test - 2 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
7/12
Sol. : Oxidation number of O in H2O is –2 and
of S in SO2 is 4.
Difference = 6
47. Answer (1)
Hint : It induces the precipitation of hydrated
Al2O3
Sol. : 2Na[Al(OH)4] + CO2 → 2NaHCO3
+ Al2O3xH2O
48. Answer (2)
Hint : Leaching of silver
( )–– –
2 2 24Ag 8CN 2H O O 4 Ag CN 4OH + + + → +
Sol. : Coordination number of Ag in [Ag(CN)2]–
is 2
49. Answer (4)
Hint : (1), (2) and (3) represent adsorption
Sol. : Water vapour are absorbed on
anhydrous calcium chloride.
50. Answer (1)
Hint : Preferential adsorption of common ion
present in excess.
Sol. : I. Fe2O3.xH2O/Fe3+
II. Fe2O3.xH2O/OH–
51. Answer (2)
Hint : Emulsions can be diluted with any
amount of dispersion medium
Sol. : The dispersed liquid on mixing with an
emulsion forms a separate layer
52. Answer (3)
Hint : Lead chamber process employs NO as
a catalyst
Sol. : ( ) ( ) ( ) ( )NO g
2 2 32SO g O g 2SO g+ ⎯⎯⎯⎯→
53. Answer (4)
Hint : At high temperature physisorption may
change into chemisorptions.
Sol. : Physisorption is multilayered whereas
chemisorption is monolayered.
54. Answer (4)
Hint : Gases with greater critical temperature
are adsorbed to a greater extent.
Sol. :
SO CH H2 4 2
1 32
P P P
i.e P P P
55. Answer (2)
Hint : r = K [R]x [H+]2
Sol. : r = K [R]x . [H+]2
2
A A2
BB
Hr4
r H
+
+
= =
a(HA)
a(HB)
K4
K=
6a(HB)K 2.5 10−=
56. Answer (3)
Hint : x = 1 ; y = 0, z = 2
Sol. : In Graph I , As rate increases linearly
with concentration reaction is of first
order
In Graph II, As concentration decreases
linearly, reaction is of zero order.
In Graph III, Product of half life and initial
concentration is constant, hence
reaction is of second order.
57. Answer (4)
Hint : B 1
C 2
n 2K
n 3K=
Sol. : –2
B 1–2
C 2
P 2K 2 3 101
P 3K 3 2 10
= = =
PB = PC
PB = 2 atm
58. Answer (3)
Hint : Chemisorption has high Ea that’s why
rate of chemisorption is enchanced by
increase in temp.
All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-B) (Hints & Solutions)
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Sol. : But after a certain temperature, rate of
adsorption may decrease as the formed
bonds may dissociate at higher
temperature.
59. Answer (1)
Hint : a2
1 1 2
Ek 1 1log
k 2.303 R T T
= −
3
a
4
E2 10 1 1log
2.303 R 400 42010
−
−
= −
aE 201.3
2.303 R 400 420=
Ea = 2.303 × 10920 × R
Sol. : a1
1
Elog k log A
RT 2.303= −
4 2.303 10920 Rlog10 log A
2.303 R 400
− = −
log A = 23.3
A 1023
60. Answer (2)
Hint: By using gas law and kinetics law, it can
be assumed that at 2 5 2
0N O OV V
2 5 2
2 5 2 2
N O O0
N O O Ott
V VKt ln ln
V V V
= =
−
Sol. : 50
K 10 In50 10
=−
K = 0.02 min–1
Verifying the value of K
50
K 18 In50 15
=−
K = 0.02 min–1
61. Answer (1)
Hint : Find
Sol. :
1 1 1
2 1 3 1( 3 3) 1(6 3) 1(2 1)
1 1 3
= − = − − − − + +
= –6 – 3 + 3 = – 6 0
System has a unique solution
62. Answer (1)
Hint : Properties of adj. and inverse
Sol. :
|(2adj3A–1)–1| = |2adj(3A–1)|–1 = 2–3|adj(3A–1)|–1
= 2–3|32adj(A–1)|–1 = 2–3(32)–3|adj(A–1)|–1
= 18–3|(adjA)–1|–1 = 18–3|adjA| = 18–3|A|2
= 1 1
27 27 218 18 18 4
=
63. Answer (3)
Hint : Multiplication of matrices
Sol. :
2
0 0 1 0 0 1 1 0 0
0 1 0 0 1 0 0 1 0
1 0 0 1 0 0 0 0 1
A I
= − − = =
A is involutory matrix
|A| = 1 non-singular matrix
64. Answer (1)
Hint : Property of adj. and inverse
Sol. : (adjA–1)–1(adjA)–1 = ((adjA)(adjA–1))–1
= (adj(A–1A))–1 = (adjI)–1 = (I)–1 = I
65. Answer (4)
Hint : |A – I| = 0
Sol. :
1 2 3
| | 0, 2 3 1 0
3 1 2
A I
−
− = − =
−
3 – 6 2 – 3 + 18 = 0
A3 – 6A2 – 3A – 18I = 0
A2 – 6A – 18A–1 = 3I
PART - C (MATHEMATICS)
Test - 2 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
9/12
66. Answer (1)
Hint : Property of diagonal matrix
Sol. :
88
0
88
0 1 2 880
88
0
tan( 1) 0 0
... 0 cot( 1) 0
0 0 1
r
r
r
r
A A A A r
=
=
=
+
= +
88
0
tan( 1)r
r=
+ = tan1°tan2°..tan45°…tan88°
tan89° = 1
88
0
cot( 1)r
r=
+ = cot1°cot2°…cot45°…cot88°
. cot89°= 1
88
0
1 1r =
=
67. Answer (3)
Hint : = 0
Sol. : For non-trivial solution
1 2
2 1 3 0,
3 1
t t t
t t t
t t t
+ +
+ + + =
+ +
R2→R2 – R1, R3 → R3 –R1
1 2
1 1 3 0
1 1 1
t t t+ +
− =
−
(t + 1)(–1–3)–(t + 2)(1 + 3) + t(1 – 1) = 0
–4t – 4 – 4t –8 = 0
–8t = 12 3
2t = −
68. Answer (1)
Hint : Expansion
Sol. :
1 1 1 1 2 31 1
2 2 2 1 2 32 2
1 2 31 0 0
A B C a a aB C
A B C b b bB C
c c c
=
1 1 23
2 21 2 3
0 0
0 0B C
aB C
a a a
= =
1 1
32 2
B Ca
B C =
69. Answer (1)
Hint : Multiplication of two determinants
Sol. :
2 2
2 2
1 1
( ) 1 1 0
1 1 1 0 0 0
x x x x
f x x x x x= =
( )2 1 0f − =
70. Answer (2)
Hint : Coefficient of x in (x) = (0)
Sol. : Coefficient of x in (x) is (0)
2 1
1
21 2
2
1sin –1
1 sin cos –11
( ) cos 0 cos – sin 0 01
tan sec 11
tan 11
xx x x
x x x xx
x x
xx
−
−
−
−
= − + +−
+
0 1 11 0 1
(0) 1 1 0 0 0 (1 1) 22 2 2
1 0 10 1 1
−−
= − + = + + − − = −
71. Answer (1)
Hint : C1 → C1 + C2
Sol. : C1 → C1 + C2
{ } 1 1 { } 1
– {– } –1 –1 {– } –1 0
0 1 1 0 1
x x x
x x x x
x
= = =
72. Answer (2)
Hint : Find domain
Sol. : –1 x 1 or x(–, –1] [1, )
x = ± 1
4
64
= −
All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-B) (Hints & Solutions)
10/12
73. Answer (2)
Hint : f(a – h) < f(a) > f(a + h) then get b
Sol. : 2
2
2 | 5 6 |, 2( )
1, 2
x x xf x
b x
− − + =
+ =
b2 + 1 > 2 for local maxima
|b| 1
74. Answer (2)
Hint : 2
2
2
2 4( ) ln 4
2 4
+ − = − − − −
xf x x
x
Let x = 2sin
( ) 2ln cot 2cos2
= −
f x
24( ) ,
− − =
xf x
x
Sol. : 2
2
2
2 4( ) ln 4
2 4
xf x x
x
+ − = − − − −
Let x = 2sin then
24( ) ,
− − =
xf x
x and x(–2, 2) – {0}
f(x) is decreasing in (0, 1)
75. Answer (4)
Hint : 1 form
Sol. :
1
11
lim2 1
21
lim lim2
n
n n
en
n
n n
ee e
→
−
→ →
+= =
76. Answer (4)
Hint : 0 0
(0) lim ( ) lim ( )− +→ →
= =x x
f f x f x
Sol. :
0
lim ( )x
f x→
= tan
0
ln(sec tan )lim
tan
x x
x
e e x x x
x x→
− + + −
−
= tan 2
20
sec sec 1lim
sec 1
x x
x
e x e x
x→
− + −
−
= ( )
tan 2 tan
20
(sec 1) (sec 1)lim
sec 1
x x x
x
e x x e e
x→
− + − + −
−
= tan
20
11 lim
2 tan
x x
x
e e
x→
−+ +
= 3 3
02 2
+ =
77. Answer (3)
Hint : f(g(x)) = x if f and g are inverse of each
other
f (g(x)) = g (f(x)) = 1
Sol. : g(f(x)) = x
1
( ( ))( )
g f xf x
=
1
( (1))(1)
g ff
=
1 1
(3)6 4 1 11
g = =+ +
78. Answer (2)
Hint : 3
2 , 154
dA daa a
dt dt= =
and 0.1=da
dt cm/s
Sol. : 0.1da
dt= cm/sec and
23
4A a=
3 3
0.1 152 2
dA daa
dt dt= = cm2/sec
= 3 3
4
Test - 2 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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79. Answer (3)
Hint :
( )
1 1
2 2
1
2x 2xcos sin
21 x 1 x
2tan x2
− −
−
= −
+ +
= − −
Sol. : f(x) = 1
2
2cos
1
xx
x
− −
+
= 1
2
2sin
2 1
xx
x
− − +
+
= 12tan2
x x−− + −
{ x = tan1}
= 12tan2
x x−+ −
f (x) = 2
2 2
2 11
1 1
x
x x
−− =
+ +
and g(x) = 1tan
42
xx
− −
= 2tan–1x +1 – x, as x = tan1
g (x) = 2
2 2
2 11
1 1
x
x x
−− =
+ +
2
2
2
2
1
( ( )) 1 1( ( )) 1
1
x
d f x xd g x x
x
−
+= = −−
+
80. Answer (2)
Hint : f(x) is discontinuous at x = 10
Sol. : The given function is only discontinuous
at x = 10.
10 10
lim ( ) lim ( )x x
f x f x− +→ →
81. Answer (4)
Hint : Find the range by maximum and
minimum value
Sol. : We can see that f(x) cannot lie between
25 1and
2 2− −
82. Answer (3)
Hint : f (x) = 0 sin2x = 1
4. Then f(x) has
maxima
Sol. : f(x) = sin2x + cos4x
f (x) = 2cos2x – 4sin4x
= 2cos2x(1 – 4 sin2x)
Here f (x) = 0 if cos2x = 0 or sin2x = 1
4
f(x) has maximum value when sin2x = 1
4
f(x)maximum = 2
1 11 2
4 4
+ −
=
9
8
83. Answer (2)
Hint : Differentiation of composite function
then interval
Sol. : Here g(x) = f(cos2x + 4cosx + 7)
On differentiating both sides w.r.t. x we
get :
g (x) = –2sinx(2 + cosx)f (cos2x + 4cosx + 7)
Here g(x) > 0 if sinx < 0
x(, 2)
84. Answer (4)
Hint :
dydy dt
dxdx
dt
= , then t = 1 to get coordinate
Sol. : 224
3 38
dy tt
dx t= = =
t = 1
Required coordinate = (7, 7)
85. Answer (1)
Hint : is repeated root then f() = f () = 0
All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-B) (Hints & Solutions)
12/12
Sol. : A six degree polynomial satisfied by
exactly 5 real numbers then exactly one
of the f(0), f (–1), f (1),f (2) or f (–2) is
zero
86. Answer (4)
Hint : Algebra of limit then solve simultaneous
eq.
Sol. : p(x) = f(x) + g(x)
and q(x) = f(x) – g(x)
( ) ( )
lim2→
+
x a
p x q x and
( ) ( )lim
2→
−
x a
p x q x exist
( ) ( )
lim 42→
+=
x a
p x q x and
( ) ( )lim 1
2→
−=
x a
p x q x
( ) 4
lim 4( ) 1x a
f x
g x→= =
87. Answer (2)
Hint : Sum of infinite series then differentiation
Sol. : y = sinx y+
y2 = sinx + y
2 cosdy dy
y xdx dx
= +
cos
2 1
dy x
dx y=
−
2
cos
2 1 2sin
dy x
dx y x=
− −
88. Answer (2)
Hint : Properties of G.I.F. and modulus function
Sol. :
2
2( ) cos(| |) | | 3 | 2 |
2 1
xf x x x x x
x
= + + + −
+
Only non differentiable at x = 2
89. Answer (4)
Hint : Factorize then cancel the common term
Sol. : 1
2
4cos ( 1)
(tan 1)(tan tan 1)lim
(tan 1)(tan tan 1)x
x x x
x x x−→ −
+ − +
+ + −
2
4
( 1) ( 1) 13
( 1) 1 1
− − − += = −
− − −
90. Answer (2)
Hint : Use G.P. to get sum then divide by
51/n
Sol. : The given value
=
( )
1
1
1
2 12 2
5 1 55
lim
1 5 1 5
nn
n
nn n
nn
n
n n
+
→
− −
− −
= ( )
111
211
5 1 5 5lim
5 15 111
nn
n
nn
nn
+
→
− +
− −
= –4(log5e)2 + 5log5e