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Test - 1 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2020
All India Aakash Test Series for JEE (Main)-2020
Test Date : 09/06/2019
ANSWERS
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TEST - 1 - Code-A
PHYSICS CHEMISTRY MATHEMATICS
1. (3)
2. (4)
3. (3)
4. (1)
5. (3)
6. (4)
7. (1)
8. (1)
9. (1)
10. (2)
11. (4)
12. (3)
13. (4)
14. (3)
15. (3)
16. (4)
17. (2)
18. (2)
19. (2)
20. (4)
21. (2)
22. (4)
23. (1)
24. (2)
25. (4)
26. (4)
27. (3)
28. (2)
29. (1)
30. (1)
31. (4)
32. (3)
33. (3)
34. (2)
35. (4)
36. (2)
37. (2)
38. (4)
39. (3)
40. (1)
41. (4)
42. (3)
43. (4)
44. (2)
45. (1)
46. (2)
47. (4)
48. (1)
49. (2)
50. (2)
51. (3)
52. (2)
53. (4)
54. (3)
55. (2)
56. (4)
57. (2)
58. (1)
59. (1)
60. (2)
61. (4)
62. (3)
63. (4)
64. (4)
65. (3)
66. (4)
67. (1)
68. (3)
69. (4)
70. (3)
71. (1)
72. (2)
73. (2)
74. (1)
75. (4)
76. (2)
77. (1)
78. (1)
79. (3)
80. (4)
81. (3)
82. (3)
83. (3)
84. (2)
85. (2)
86. (2)
87. (2)
88. (3)
89. (4)
90. (3)
All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-A) (Hints & Solutions)
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1. Answer (3)
Hint : EA = 2
KQ
r due to shell at outside point
Sol. : 2
(3 )A
A
K QE
r=
rA is distance of A from O
⟹ EA = constant
2. Answer (4)
Hint : VP = V1 + V2
Sol. : 0 0
2 (2 )2 2
P
RR
V
= +
0
2 R =
3. Answer (3)
Hint : Parallel arrangement
Sol. : Ceq = ...2 4 8
C C CC + + + + ∞
1
21
12
C C= =
−
4. Answer (1)
Hint : 0Q =
Sol. : 0Q =
⟹ Q + 2Q – Q + Q1 = 0
⟹ Q1 = –2Q
5. Answer (3)
Hint : 2
2
QU
C
=
Sol. : 2
0
(2 )
(2){4 2 }
QU
a=
2 2
0 0
4
16 4
Q QU
a a= =
6. Answer (4)
Hint : 1 2=kq q
Ur
Sol. : 2 2 2
2 ...2 3
kq kq kqU
a a a
= − + − +
2 2 1 1
1 ...2 3
kq
a
= − − + −
Note: 2 3 4
ln(1 ) ...2 3 4
x x xx x+ = − + − +
2 2
ln(2)kq
Ua
= −
2
0
ln(2)2
qU
a = −
7. Answer (1)
Hint : 0 =kQ
VR
Sol. : V at center is 00
3 3
4 2 2
QV
R =
30
0 00
3 1 1 4
2 2 2 4 3
VV V V R
R = − = =
2
06
RV
=
8. Answer (1)
Hint : Vinside = Vsurface
Sol. : Potential at inside point will be same as
potential at the surface of inside sphere.
PART - A (PHYSICS)
Test - 1 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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9. Answer (1)
Hint : 0 0 0QU V Q=
Sol. : 00 0
( )ln2 ln2
4 4
LV
L
= =
0
0
ln2
4
QU
=
10. Answer (2)
Hint : 0
inq =
Sol. : 11
0
q =
22
0
q = −
3 0 =
11. Answer (4)
Hint : 21
2U CV=
Sol. : 1 2
1 2
Q Q
C C=
21 48
(2)2 3
U
=
= 256 µJ
12. Answer (3)
Hint : Electric field near a point charge will
dominated by nearer charge.
Sol. : Electric field at x = ± ∞ will tend to zero.
13. Answer (4)
Hint : Use symmetry
Sol. : Field due to each spherical shell will be
along –y direction.
14. Answer (3)
Hint : Use super position principle
Sol. : sheet holePE E E= −
02 2
PE
=
15. Answer (3)
Hint : Work done in closed loop is zero due to
conservative field.
Sol. : Electric field lines cannot form a closed
loop.
16. Answer (4)
Hint : Use Gauss’ law
Sol. : Use concept of solid angle
Total flux = 0
2
10
Q
05
Q =
17. Answer (2)
Hint : VB = 0 after switch S is closed
Sol. :
VB = 0
2
A
kQ kQV
R R= −
2
kQ
R=
18. Answer (2)
Hint : Use combination of capacitor
Sol. : 2 8
23 3
PQ
CC C C= + =
1 1
8 31
3
ABV V
=
+
1
110 10 V11
= =
10 VABV =
All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-A) (Hints & Solutions)
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19. Answer (2)
Hint : 0
=Q
E ds
Sol. : 0
4Q
=
0
1
4
Q =
20. Answer (4)
Hint : 0
=
P
RV
Sol. : 0
P
RV
=
( )2
2Q
R =
2
0
2P
R QV
R
=
2
0
2Q
R
=
21. Answer (2)
Hint : 2ABeqC C=
Sol. : ABeqQ C V=
= (2C) V
= 2 × 1 × 12
= 24 µC
22. Answer (4)
Hint : Potential increases opposite to the
direction of electric field
Sol. : EA > EB
Since field lines are more dense at A and
VA > VB
23. Answer (1)
Hint : Use symmetry
Sol. : 1 1 1 1
...1 3 9C
= + + + ∞
1 3
1 21
3
= =
−
2
F3
C =
4
2 F3
ABC C = =
24. Answer (2)
Hint : Show charge distribution on ring
Sol. :
25. Answer (4)
Hint : 0
inqE ds =
Sol. : Total0
(2 )Q =
0 0
1 2
3 4
Q Q = −
= 0
7
12
Q
Test - 1 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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26. Answer (4)
Hint : Potential become same
Sol. : V1 = V2
⟹ Field will be zero in between them
27. Answer (3)
Hint : For maximum field
0dE
dz=
Sol. : Calculate electric field at z
Then 02
d RE z
dz= =
28. Answer (2)
Hint :
=
KQV
R
Sol. : 2
2 2 3B
KQ K Q KQV
a a a= + −
1 1
12 3
KQ
a
= + −
3 6 2
6
KQ
a
+ − =
7
6
KQ
a=
29. Answer (1)
Hint : | |P Q=
Sol. : ( 3)rP Q=
30. Answer (1)
Hint : 03
rE
=
Sol. : 03
rE
=
0
| |6
RE
=
31. Answer (4)
Hint : Body diagonal plane contains 2 Zn2+
ions.
Sol. : Body diagonal plane will consist of four
S2– ions at the corners 1
48
and two
S2– ions at the face centre 1
22
.
32. Answer (3)
Hint : Mole of H2 = 2 mole of e–
Sol. : 2
3
H
1.93 5 60n 3 10
2 96500
− = =
33. Answer (3)
Hint : LHE is anode.
RHE is cathode.
Sol. :
11
22
1 22 1
2M
P
2M
P
2 2M M
P P
A : 2Cl Cl 2e
C : Cl 2e 2Cl
2Cl Cl Cl 2Cl
− −
−−
−−
⎯→ +
+ ⎯→
+ ⎯→ +
34. Answer (2)
Hint : In the titration, AD is precipitated so A+ are replaced by C+.
Sol. : Since, conductance does not change so mobility of C+ is comparable to A+.
35. Answer (4)
Hint : The liquids may be immiscible or showing positive deviation from Raoults law.
Sol. : For solution with positive deviation, the more volatile component may have very low mole fraction in the liquid phase.
36. Answer (2)
Hint : 0
23 3
z(M )d
6 10 a=
PART - B (CHEMISTRY)
All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-A) (Hints & Solutions)
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M0 = 58
d = 2.48 gcm–3
Sol. : 3 22
23
4 58a 1.56 10
6 10 2.48
−= =
a = 5.38 × 10–8 cm = 538 pm
a
269 pm2
=
37. Answer (2)
Hint : Mole fraction of solute will be equal in both the beakers after a long time.
Sol. : Initially
Beaker A : n moles solute
4 moles of water
Beaker B : 2 moles of solute
3 moles of water
Finally
There is shifting of 1
2 mole of water to
beaker B.
n 2
1 1n 4 2 3
2 2
=
+ − + +
n = 2
28 g of X contains 2 moles
M = 14 g/mol
38. Answer (4)
Hint : o o ocell cathode AnodeE E E= −
Sol. : For (4), Eo = 2.126 V.
39. Answer (3)
Hint : Electrolyte is paste of KOH and ZnO
Sol. : It does not involve any ion whose conc. can change during its life time
40. Answer (1)
Hint : For positive deviation, Pactual > PRaoult
Sol. : ∆H > 0 and ∆V > 0
41. Answer (4)
Hint : van't Hoff factor = number of ions furnished by 1 mole.
Sol. : Ca3(PO4)2, i = 5
Na4[Fe(CN)6], i = 5
42. Answer (3)
Hint : 24 2
MnO 8H 5e Mn 4H O− + − ++ + ⎯→ +
Sol. : E = 8
0.06 1E log
5 [H ]+ −
= E° – 0.096 pH
= 1.51 – 0.096 (pH)
43. Answer (4)
Hint : meq of NaOH = meq of H2SO4
Sol. : Neutralisation
1.21000 M 20 2
40 =
M = 0.75
Now, wt. of H2SO4 in 1 lit solution = 0.75 × 98 = 73.5 and wt. of 1 lit solution = 1024.5
wt. of solvent = 951
Mass percent of solute
= 73.5
10001024.5
= 7.17%
0.75 1000
m 0.79951
= =
And mole fraction
= 0.75
0.014951
0.7518
=
+
44. Answer (2)
Hint : P = KH · X
Sol. : H
500K
0.01= = 5 × 10+4
= 50 k torr
As temperature increases, Henry’s
constant increases. As solvent solute
interactions become stronger, value of
Henry’s constant decreases.
45. Answer (1)
Hint : When the last trace of liquid disappears,
the vapor contains 6 moles of A and 4
moles of B.
Test - 1 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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Sol. : oA
P 30= , oB
P 50=
xA = ?, xB = ?
A
3y
5= ,
B
2y
5=
Now o
A A AA o o
T A A B B
P P xy
P P x P x= =
+
( )
o
A AA o o o
A B A B
P xy
P P x P=
− +
A
A
30x3
5 50 20x=
−
10 – 4xA = 10xA
A
10x
14=
∴ T
10 4 500P 30 50
14 14 14= + =
46. Answer (2)
Hint : meq of acid = meq of base
1.51 1000 25 0.2
M =
M = 300
Sol. : Molality = 2.5 1000 1
300 250 30
=
Tf = iKfm
10.1 i(1.86)
30=
i = 1.61
1 + = 1.61
= 0.61
47. Answer (4)
Hint : Tb (solution) = Tb (solvent) + Kb · m.
Sol. : Solution with the least boiling point will be decided by the value of ‘m’.
48. Answer (1)
Hint : Entropy of solid is lesser than liquid.
Sol. : As Ssolvent < Ssolution
So ( ) ( )fusion fusionsolvent solutionS S
49. Answer (2)
Hint : Reactions take place in basic medium.
Sol. : Cathode: O2(g) + H2O(l) + 4e– ⎯→
4OH–(aq)
Anode: 2H2(g) + 4OH–(aq) ⎯→
4H2O(l) + 4e–
50. Answer (2)
Hint : Charge in faradays passed = g eq of species produced
Sol. : Charge passed =1.93 2500 40
100
= 1930 C
Number of m moles of e– = 20
Now, A : H2O ⎯→ 1
2O2 + 2H+ + 2e–
C : Cu2+ + 2e– ⎯→ Cu
m moles of Cu deposited = 10
m moles of H+ produced = 20
3
220 10 2[H ] 10
3 3
−+ −
= =
pH = 2 + log3 – log2 = 2.18
51. Answer (3)
Hint : At Cathode, reduction takes place
Sol. : At Anode, oxidation takes place
52. Answer (2)
Hint : In conductivity cell, AC is used.
Sol. : DC changes the composition of solution
and KCl cannot be used as Cl– react with
Ag+ to form AgCl.
53. Answer (4)
Hint : o
cell cell
0.06E E logQ
2= −
Sol. :
3 2
2
3 2 2
C : B e B
A : A A 2e
A 2B A 2B
+ − +
+ −
+ + +
+ ⎯→
⎯→ +
+ ⎯→ +
2 2 2 2
3 2
[B ] [A ] xQ
0.1[B ]
+ +
+
= =
2o 0.06 x
E E log2 0.1
= −
2x2.03 2 0.03 log
0.1
= −
All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-A) (Hints & Solutions)
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2xlog 1
0.1
= −
x = 0.1
54. Answer (3)
Hint : m
K 1000
C
=
Sol. : K 1000
1000.1
0.82
=
K = 4 × 10–3
55. Answer (2)
Hint : Tetragonal system has all angles equal to 90°.
Sol. : Trigonal system has all angles not equal to 90°.
56. Answer (4)
Hint : Electrical neutrality must be maintained in ionic solids.
Sol. : Presence of F-centres impart colour.
57. Answer (2)
Hint : If edge length is a, then 3 a 4R= .
Sol. : 4R
a3
=
Now 2R + 2x = a
x = a 2R
2
−
=
4R2R
3
2
−
=
2 3R
3
−
58. Answer (1)
Hint : In AB type structure, the cation A+ can occupy either all octahedral voids or half the tetrahedral voids.
Sol. : In AB2 type structure, co-ordination number ratio of A2+ : B– is 2 : 1.
59. Answer (1)
Hint : In HCP, there are 6 atoms per unit cell.
Sol. : 8 tetrahedral voids are completely inside. There are tetrahedral voids on the vertical edges which are shared.
60. Answer (2)
Hint : Paramagnetic substances get magnetised in a magnetic field and lose their magnetism when the field is removed.
Sol. : Ferrimagnetic substances have domains oriented oppositely in unequal numbers.
61. Answer (4)
Hint : Draw graph
Sol. :
n(A ∩ B) = 3
⟹ n(P (A ∩ B)) = 23 = 8
62. Answer (3)
Hint : Venn diagram
Sol. :
63. Answer (4)
Hint : y = x
Sol. : ∵ sin–1x + cos–1y =
2 …(1)
PART - C (MATHEMATICS)
Test - 1 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
9/13
(x, y) ∈ R
∵ sin–1x + cos–1x =
2 ⟹ (x, x) ∈ R
⟹ R is reflexive
Let sin–1y + cos–1x = k …(2)
From (1) + (2)
− − = + =1 1sin cos
2 2k y x
⟹ (y, x) ∈ R ⟹ R is symmetric
Let (y, z) ∈ R ⟹ − − + =1 1sin cos
2y z …(3)
From (1) + (3),
− − + = 1 1sin cos ( , )
2x z x z R
⟹ R is transitive
⟹ R is equivalence relation
64. Answer (4)
Hint : − −= −1 1cos sin
2x x
Sol. :
− − − − − + − −
−
1 1 1 1
2
sin sin sin sin2 4 2
016
x x x x
− − − − + − −
2 21 1 2 1sin (sin ) sin 0
2 8 2 16x x x
− −
21 2(sin ) 0
16x
− − 1sin
4 4x
1 1
2 2x−
65. Answer (3)
Hint : + 1
2,tt
when t > 0
Sol. :
1 1| | 2, | | 2,
| | | |
1| | | | 2
| || |
x yx y
x yx y
+ +
+
+ + + + + 1 1 1
| | | | | || | 6| | | | | || |
x y x yx y x y
+ + + + + =1 1 1
| | | | | || | 6| | | | | || |
x y x yx y x y
= = = = = =1 1 1
| | | | | || | 1| | | | | || |
x y x yx y x y
⟹ |x| = |y| = 1 ⟹ x = ± 1, y = ± 1
Ordered pairs are (1, 1), (–1, 1), (1, –1), (–1, –1)
66. Answer (4)
Hint : Put 9x = t
Sol. : 34x + 9|x – 1| – 10 ≤ 0
81x + 9|x – 1| – 10 ≤ 0
Put 9x = t
If x – 1 ≤ 0 If x – 1 ≥ 0
+ − 2 910 0t
t + − 2 10 0
9
tt
− + 3 10 9 0t t 1 9 9xx
(t – 1)(t2 + t – 9) ≤ 0 9t
− + − −
21 1
( 1) 9 02 4
t t + 2 829
tt
− + +
+ −
1 37( 1)
2 2
1 370
2 2
t t
t
+ − 2 10 09
tt
Has no solution.
− − −
37 1( 1) 0
2t t
−
37 11
2t
All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-A) (Hints & Solutions)
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−
37 11 9
2
x
−
9
37 10 log
2x
67. Answer (1)
Hint : Draw graph
Sol. : 1
11 1
xy
x x= = −
+ +
(y – 1)(x + 1) = –1
[0,1)y
f(x) is one-one and into function
68. Answer (3)
Hint : sinx ∈ [–1, 1]
Sol. : 2sinx + 2y = 1
− sin11 sin 1 2 2
2
xx
− − −sin12 2
2
x
− − − − sin 111 2 1 1 2 2
2
x y
− −( , 1]y
69. Answer (4)
Hint : ln(1 + x) is an integer
Sol. : x + 1 > 0 1x −
ln(1 + x) + [(1+ x)2] – 3 is an integer
⇒ ln(1 + x) is an integer
⇒ [ln(1 + x)] = ln(1 + x)
⇒ [(1 + x)2] = 3
3 ≤ (1 + x)2 < 4
+ 3 |1 | 2x
ln 3 ln(1 x) ln2 +
ln(1 + x) is integer for no value of x
70. Answer (3)
Hint : Break G.I.F
Sol. :
3, [ 1, sin1)
2, [ sin1, 0)( )
0, [0, sin1)
1, [sin1,1]
− − −
− −=
x
xf x
x
x
71. Answer (1)
Hint : |sinx| + |cosx| ∈ [1, 2]
Sol. : + | sin | | cos | [1, 2]x x
+ =[| sin | | cos |] 1x x
⟹ domain of f(x) is ϕ (empty set)
72. Answer (2)
Hint : 0 ≤ {x} < 1
Sol. : ∵ sgn(sin–1x) = {–1, 0, 1} and {2x} ∈ [0, 1)
⟹ sgn(sin–1x) = {2x} = 0 ⟹ x = 0
73. Answer (2)
Hint : − −=−
1 1
2
22tan tan
1
xx
x
Sol. : − − =
1 11 14tan 2 2tan
5 5
− −
= = −
1 1
12
552 tan 2tan1 12
125
Test - 1 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
11/13
− −
= = −
1 1
52
12012tan tan25 119
1144
− − −
1 114tan cot (239)
5
− − = −
1 1120 1tan tan
119 239
− −
−
= = = +
1 1
120 1
119 239tan tan (1)120 1 4
1119 239
74. Answer (1)
Hint : Find domain
Sol. : − − 1 1
1 2 12 2
x x
Let f(x) = sin–1(2x) – cos–1x + tan–1(2x)
sin–12x is increasing
cos–1x is decreasing
and tan–1(2x) is increasing function
−
= − = − − − =
min
1 2 17( )
2 2 3 4 12f x f
= = − + =
max.
1 5( )
2 2 3 4 12f x f
=( )2
f x does not have any solution
75. Answer (4)
Hint : Draw graph of sin–1(sinx)
Sol. :
76. Answer (2)
Hint : − − −
=
1 1 1tan tan tan1
x yx y
xy
Sol. : tan cot 1 1x x = −
1 1 1 tan cottan (tan ) tan (cot ) tan
1 tan cot
− − − − − =
+
x xx x
x x
21 1tan 1
tan tan ( cot 2 )2 tan
− − −
= = −
xx
x
1 1tan (cot2 ) tan tan 22
− − = − = − −
x x
1tan tan 22
− = −
x
=
2T
77. Answer (1)
Hint : − − 1sin
2 2x
Sol. : − − 12tan
2 2a
− − − 1tan 1 1
4 4a a
78. Answer (1)
Hint : − −+= −
+ +
1 1( 1)tan tan
1 1n
n r nrT
n n
Sol. : − +
=+ + +
1
2 2
( 1)tan
( 1) ( 1)n
n nT
n n r r
−
+=
+ + + +
1 1tan( 1)
11 1
n
nn r nr
n n
1
( 1)
1 1tan( 1)
11 1
n r nr
n nn r nr
n n
−
+ − + +=
+ + + +
1 1( 1)tan tan
1 1
n r nr
n n
− −+= −
+ +
All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-A) (Hints & Solutions)
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− − = −
+
1 1tan ( ) tan1
nS n
n
→
= − =lim
2 4 4nS
79. Answer (3)
Hint : Break the function at –1, 0, 1, 2, 3
Sol. : [ ] 0 [0,1) x x
1, 1
2, 1 0
( ) 2, 1 2
1, 2 3
0, 3
− − −
− −
=
x
x
f x x
x
x
80. Answer (4)
Hint : − − + =1 1sin cos
2x x
Sol. : − −+1 2 1 2(sin ) (cos )x x
− − − − = + − −
1 1 2 1 1(sin cos ) 2sin sin2
x x x x
− −= + −
21 2 12(sin ) sin
4x x
− −
= + − + −
2 2 21 2 12 (sin ) sin
4 2 16 8x x
= −
+ −
2212 sin
8 4x
1 13sin sin
2 2 4 4 4
− − − − − x x
− −
2 21 9
0 sin4 16
x
− − +
2 21 2 1 2 10
(sin ) (cos )8 8
x x
81. Answer (3)
Hint : − − − −− =
+
1 1 1tan tan tan1
x yx y
xy
Sol. : − − −
+ =
1 1 11tan tan tan (3)x
y
− − − − − = − = +
1 1 1 11 3tan tan 3 tan tan
1 3
xx
y x
+= = − +
− −
1 3 103
3 3
xy
x x
For positive integer y, x = 1, 2
⟹ Solutions are (1, 2)(2, 7)
82. Answer (3)
Hint : 1 – sin2 = (cos1 – sin1)2
Sol. : 1 sin2 1
tancos2
− −
1 1 tan1tan
1 tan1
− − = −
+
1tan tan 1
4
− = − −
14
= −
83. Answer (3)
Hint : Solve graphically
Sol. : −
− [ ] 1
1 1, 0x
xx
If x > 0 If x < 0
–x ≤ [x] – 1 ≤ x –x ≥ [x] – 1 ≥ x
1 – x ≤ [x] ≤ x + 1 1 – x ≥ [x] ≥ 1+ x
1 – x ≤ x – {x} ≤ x + 1 1 – x ≥ x – {x} ≥ 1 + x
− − − −1 { } 1x x x x − − − −1 { } 1x x x x
− −2 1 { } 1x x
− −2 1 { } 1x x
1x = x
[1, )x
Test - 1 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
13/13
84. Answer (2)
Hint : f(x + T) = f(x)
Sol. : ( )2
f x f x
+ =
85. Answer (2)
Hint : Find domain
Sol. : –1 ≤ x ≤ 1, –1 ≤ x + 2 ≤ 1
⟹ x = –1
( 1) 0 02 2
f
− = − − + = −
86. Answer (2)
Hint : Onto function = 0
Sol. : Number of onto functions = 0
Number of functions = 43
⟹ Number of functions which are not onto = 64
87. Answer (2)
Hint : Reflexive relations = −2
2n n
Sol. : Number of relations = 2 252 2n =
Number of reflexive relations
= 2 25 5 202 2 2n n− −= =
88. Answer (3)
Hint : Draw graph
Sol. :
• y = sin–1(sinx)
• y = sin–1(cosx) = 1cos (cos )2
x−−
89. Answer (4)
Hint : R–1 is also an equivalence relation
Sol. : The inverse of an equivalence relation is
also an equivalence relation.
90. Answer (3)
Hint : Venn diagram
Sol. :
n(P ∪ C ∪ M) = 640
⟹ n(who did not opt)
= 800 – 640 = 160
Test - 1 (Code-B) (Answers) All India Aakash Test Series for JEE (Main)-2020
All India Aakash Test Series for JEE (Main)-2020
Test Date : 09/06/2019
ANSWERS
1/13
TEST - 1 - Code-B
PHYSICS CHEMISTRY MATHEMATICS
1. (1)
2. (1)
3. (2)
4. (3)
5. (4)
6. (4)
7. (2)
8. (1)
9. (4)
10. (2)
11. (4)
12. (2)
13. (2)
14. (2)
15. (4)
16. (3)
17. (3)
18. (4)
19. (3)
20. (4)
21. (2)
22. (1)
23. (1)
24. (1)
25. (4)
26. (3)
27. (1)
28. (3)
29. (4)
30. (3)
31. (2)
32. (1)
33. (1)
34. (2)
35. (4)
36. (2)
37. (3)
38. (4)
39. (2)
40. (3)
41. (2)
42. (2)
43. (1)
44. (4)
45. (2)
46. (1)
47. (2)
48. (4)
49. (3)
50. (4)
51. (1)
52. (3)
53. (4)
54. (2)
55. (2)
56. (4)
57. (2)
58. (3)
59. (3)
60. (4)
61. (3)
62. (4)
63. (3)
64. (2)
65. (2)
66. (2)
67. (2)
68. (3)
69. (3)
70. (3)
71. (4)
72. (3)
73. (1)
74. (1)
75. (2)
76. (4)
77. (1)
78. (2)
79. (2)
80. (1)
81. (3)
82. (4)
83. (3)
84. (1)
85. (4)
86. (3)
87. (4)
88. (4)
89. (3)
90. (4)
All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-B) (Hints & Solutions)
2/13
1. Answer (1)
Hint : 03
rE
=
Sol. : 03
rE
=
0
| |6
RE
=
2. Answer (1)
Hint : | |P Q=
Sol. : ( 3)rP Q=
3. Answer (2)
Hint :
=
KQV
R
Sol. : 2
2 2 3B
KQ K Q KQV
a a a= + −
1 1
12 3
KQ
a
= + −
3 6 2
6
KQ
a
+ − =
7
6
KQ
a=
4. Answer (3)
Hint : For maximum field
0dE
dz=
Sol. : Calculate electric field at z
Then 02
d RE z
dz= =
5. Answer (4)
Hint : Potential become same
Sol. : V1 = V2
⟹ Field will be zero in between them
6. Answer (4)
Hint : 0
inqE ds =
Sol. : Total0
(2 )Q =
0 0
1 2
3 4
Q Q = −
= 0
7
12
Q
7. Answer (2)
Hint : Show charge distribution on ring
Sol. :
8. Answer (1)
Hint : Use symmetry
Sol. : 1 1 1 1
...1 3 9C
= + + + ∞
1 3
1 21
3
= =
−
2
F3
C =
4
2 F3
ABC C = =
PART - A (PHYSICS)
Test - 1 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
3/13
9. Answer (4)
Hint : Potential increases opposite to the
direction of electric field
Sol. : EA > EB
Since field lines are more dense at A and
VA > VB
10. Answer (2)
Hint : 2ABeqC C=
Sol. : ABeqQ C V=
= (2C) V
= 2 × 1 × 12
= 24 µC
11. Answer (4)
Hint : 0
=
P
RV
Sol. : 0
P
RV
=
( )2
2Q
R =
2
0
2P
R QV
R
=
2
0
2Q
R
=
12. Answer (2)
Hint : 0
=Q
E ds
Sol. : 0
4Q
=
0
1
4
Q =
13. Answer (2)
Hint : Use combination of capacitor
Sol. : 2 8
23 3
PQ
CC C C= + =
1 1
8 31
3
ABV V
=
+
1
110 10 V11
= =
10 VABV =
14. Answer (2)
Hint : VB = 0 after switch S is closed
Sol. :
VB = 0
2
A
kQ kQV
R R= −
2
kQ
R=
15. Answer (4)
Hint : Use Gauss’ law
Sol. : Use concept of solid angle
Total flux = 0
2
10
Q
05
Q =
16. Answer (3)
Hint : Work done in closed loop is zero due to
conservative field.
Sol. : Electric field lines cannot form a closed
loop.
All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-B) (Hints & Solutions)
4/13
17. Answer (3)
Hint : Use super position principle
Sol. : sheet holePE E E= −
02 2
PE
=
18. Answer (4)
Hint : Use symmetry
Sol. : Field due to each spherical shell will be
along –y direction.
19. Answer (3)
Hint : Electric field near a point charge will
dominated by nearer charge.
Sol. : Electric field at x = ± ∞ will tend to zero.
20. Answer (4)
Hint : 21
2U CV=
Sol. : 1 2
1 2
Q Q
C C=
21 48
(2)2 3
U
=
= 256 µJ
21. Answer (2)
Hint : 0
inq =
Sol. : 11
0
q =
22
0
q = −
3 0 =
22. Answer (1)
Hint : 0 0 0QU V Q=
Sol. : 00 0
( )ln2 ln2
4 4
LV
L
= =
0
0
ln2
4
QU
=
23. Answer (1)
Hint : Vinside = Vsurface
Sol. : Potential at inside point will be same as
potential at the surface of inside sphere.
24. Answer (1)
Hint : 0 =kQ
VR
Sol. : V at center is 00
3 3
4 2 2
QV
R =
30
0 00
3 1 1 4
2 2 2 4 3
VV V V R
R = − = =
2
06
RV
=
25. Answer (4)
Hint : 1 2=kq q
Ur
Sol. : 2 2 2
2 ...2 3
kq kq kqU
a a a
= − + − +
2 2 1 11 ...
2 3
kq
a
= − − + −
Note: 2 3 4
ln(1 ) ...2 3 4
x x xx x+ = − + − +
2 2ln(2)
kqU
a
= −
2
0
ln(2)2
qU
a = −
Test - 1 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
5/13
26. Answer (3)
Hint : 2
2
QU
C
=
Sol. : 2
0
(2 )
(2){4 2 }
QU
a=
2 2
0 0
4
16 4
Q QU
a a= =
27. Answer (1)
Hint : 0Q =
Sol. : 0Q =
⟹ Q + 2Q – Q + Q1 = 0
⟹ Q1 = –2Q
28. Answer (3)
Hint : Parallel arrangement
Sol. : Ceq = ...2 4 8
C C CC + + + + ∞
1
21
12
C C= =
−
29. Answer (4)
Hint : VP = V1 + V2
Sol. : 0 0
2 (2 )2 2
P
RR
V
= +
0
2 R =
30. Answer (3)
Hint : EA = 2
KQ
r due to shell at outside point
Sol. : 2
(3 )A
A
K QE
r=
rA is distance of A from O
⟹ EA = constant
31. Answer (2)
Hint : Paramagnetic substances get magnetised in a magnetic field and lose their magnetism when the field is removed.
Sol. : Ferrimagnetic substances have domains oriented oppositely in unequal numbers.
32. Answer (1)
Hint : In HCP, there are 6 atoms per unit cell.
Sol. : 8 tetrahedral voids are completely inside. There are tetrahedral voids on the vertical edges which are shared.
33. Answer (1)
Hint : In AB type structure, the cation A+ can occupy either all octahedral voids or half the tetrahedral voids.
Sol. : In AB2 type structure, co-ordination number ratio of A2+ : B– is 2 : 1.
34. Answer (2)
Hint : If edge length is a, then 3 a 4R= .
Sol. : 4R
a3
=
Now 2R + 2x = a
x = a 2R
2
−
=
4R2R
3
2
−
=
2 3R
3
−
35. Answer (4)
Hint : Electrical neutrality must be maintained in ionic solids.
Sol. : Presence of F-centres impart colour.
36. Answer (2)
Hint : Tetragonal system has all angles equal to 90°.
Sol. : Trigonal system has all angles not equal to 90°.
PART - B (CHEMISTRY)
All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-B) (Hints & Solutions)
6/13
37. Answer (3)
Hint : m
K 1000
C
=
Sol. : K 1000
1000.1
0.82
=
K = 4 × 10–3
38. Answer (4)
Hint : o
cell cell
0.06E E logQ
2= −
Sol. :
3 2
2
3 2 2
C : B e B
A : A A 2e
A 2B A 2B
+ − +
+ −
+ + +
+ ⎯→
⎯→ +
+ ⎯→ +
2 2 2 2
3 2
[B ] [A ] xQ
0.1[B ]
+ +
+
= =
2o 0.06 x
E E log2 0.1
= −
2x2.03 2 0.03 log
0.1
= −
2xlog 1
0.1
= −
x = 0.1
39. Answer (2)
Hint : In conductivity cell, AC is used.
Sol. : DC changes the composition of solution
and KCl cannot be used as Cl– react with
Ag+ to form AgCl.
40. Answer (3)
Hint : At Cathode, reduction takes place
Sol. : At Anode, oxidation takes place
41. Answer (2)
Hint : Charge in faradays passed = g eq of species produced
Sol. : Charge passed =1.93 2500 40
100
= 1930 C
Number of m moles of e– = 20
Now, A : H2O ⎯→ 1
2O2 + 2H+ + 2e–
C : Cu2+ + 2e– ⎯→ Cu
m moles of Cu deposited = 10
m moles of H+ produced = 20
3220 10 2
[H ] 103 3
−+ −
= =
pH = 2 + log3 – log2 = 2.18
42. Answer (2)
Hint : Reactions take place in basic medium.
Sol. : Cathode: O2(g) + H2O(l) + 4e– ⎯→
4OH–(aq)
Anode: 2H2(g) + 4OH–(aq) ⎯→
4H2O(l) + 4e–
43. Answer (1)
Hint : Entropy of solid is lesser than liquid.
Sol. : As Ssolvent < Ssolution
So ( ) ( )fusion fusionsolvent solutionS S
44. Answer (4)
Hint : Tb (solution) = Tb (solvent) + Kb · m.
Sol. : Solution with the least boiling point will be decided by the value of ‘m’.
45. Answer (2)
Hint : meq of acid = meq of base
1.51 1000 25 0.2
M =
M = 300
Sol. : Molality = 2.5 1000 1
300 250 30
=
Tf = iKfm
10.1 i(1.86)
30=
i = 1.61
1 + = 1.61
= 0.61
46. Answer (1)
Hint : When the last trace of liquid disappears,
the vapor contains 6 moles of A and 4
moles of B.
Sol. : oA
P 30= , oB
P 50=
xA = ?, xB = ?
Test - 1 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
7/13
A
3y
5= ,
B
2y
5=
Now o
A A AA o o
T A A B B
P P xy
P P x P x= =
+
( )
o
A AA o o o
A B A B
P xy
P P x P=
− +
A
A
30x3
5 50 20x=
−
10 – 4xA = 10xA
A
10x
14=
∴ T
10 4 500P 30 50
14 14 14= + =
47. Answer (2)
Hint : P = KH · X
Sol. : H
500K
0.01= = 5 × 10+4
= 50 k torr
As temperature increases, Henry’s
constant increases. As solvent solute
interactions become stronger, value of
Henry’s constant decreases.
48. Answer (4)
Hint : meq of NaOH = meq of H2SO4
Sol. : Neutralisation
1.21000 M 20 2
40 =
M = 0.75
Now, wt. of H2SO4 in 1 lit solution = 0.75 × 98 = 73.5 and wt. of 1 lit solution = 1024.5
wt. of solvent = 951
Mass percent of solute
= 73.5
10001024.5
= 7.17%
0.75 1000
m 0.79951
= =
And mole fraction
= 0.75
0.014951
0.7518
=
+
49. Answer (3)
Hint : 24 2
MnO 8H 5e Mn 4H O− + − ++ + ⎯→ +
Sol. : E = 8
0.06 1E log
5 [H ]+ −
= E° – 0.096 pH
= 1.51 – 0.096 (pH)
50. Answer (4)
Hint : van't Hoff factor = number of ions furnished by 1 mole.
Sol. : Ca3(PO4)2, i = 5
Na4[Fe(CN)6], i = 5
51. Answer (1)
Hint : For positive deviation, Pactual > PRaoult
Sol. : ∆H > 0 and ∆V > 0
52. Answer (3)
Hint : Electrolyte is paste of KOH and ZnO
Sol. : It does not involve any ion whose conc. can change during its life time
53. Answer (4)
Hint : o o ocell cathode AnodeE E E= −
Sol. : For (4), Eo = 2.126 V.
54. Answer (2)
Hint : Mole fraction of solute will be equal in both the beakers after a long time.
Sol. : Initially
Beaker A : n moles solute
4 moles of water
Beaker B : 2 moles of solute
3 moles of water
Finally
There is shifting of 1
2 mole of water to
beaker B.
n 2
1 1n 4 2 3
2 2
=
+ − + +
All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-B) (Hints & Solutions)
8/13
n = 2
28 g of X contains 2 moles
M = 14 g/mol
55. Answer (2)
Hint : 0
23 3
z(M )d
6 10 a=
M0 = 58
d = 2.48 gcm–3
Sol. : 3 22
23
4 58a 1.56 10
6 10 2.48
−= =
a = 5.38 × 10–8 cm = 538 pm
a
269 pm2
=
56. Answer (4)
Hint : The liquids may be immiscible or showing positive deviation from Raoults law.
Sol. : For solution with positive deviation, the more volatile component may have very low mole fraction in the liquid phase.
57. Answer (2)
Hint : In the titration, AD is precipitated so A+ are replaced by C+.
Sol. : Since, conductance does not change so mobility of C+ is comparable to A+.
58. Answer (3)
Hint : LHE is anode.
RHE is cathode.
Sol. :
11
22
1 22 1
2M
P
2M
P
2 2M M
P P
A : 2Cl Cl 2e
C : Cl 2e 2Cl
2Cl Cl Cl 2Cl
− −
−−
−−
⎯→ +
+ ⎯→
+ ⎯→ +
59. Answer (3)
Hint : Mole of H2 = 2 mole of e–
Sol. : 2
3
H
1.93 5 60n 3 10
2 96500
− = =
60. Answer (4)
Hint : Body diagonal plane contains 2 Zn2+
ions.
Sol. : Body diagonal plane will consist of four
S2– ions at the corners 1
48
and two
S2– ions at the face centre 1
22
.
61. Answer (3)
Hint : Venn diagram
Sol. :
n(P ∪ C ∪ M) = 640
⟹ n(who did not opt)
= 800 – 640 = 160
62. Answer (4)
Hint : R–1 is also an equivalence relation
Sol. : The inverse of an equivalence relation is
also an equivalence relation.
63. Answer (3)
Hint : Draw graph
Sol. :
• y = sin–1(sinx)
• y = sin–1(cosx)
= 1cos (cos )2
x−−
PART - C (MATHEMATICS)
Test - 1 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
9/13
64. Answer (2)
Hint : Reflexive relations = −2
2n n
Sol. : Number of relations = 2 252 2n =
Number of reflexive relations
= 2 25 5 202 2 2n n− −= =
65. Answer (2)
Hint : Onto function = 0
Sol. : Number of onto functions = 0
Number of functions = 43
⟹ Number of functions which are not onto = 64
66. Answer (2)
Hint : Find domain
Sol. : –1 ≤ x ≤ 1, –1 ≤ x + 2 ≤ 1
⟹ x = –1
( 1) 0 02 2
f
− = − − + = −
67. Answer (2)
Hint : f(x + T) = f(x)
Sol. : ( )2
f x f x
+ =
68. Answer (3)
Hint : Solve graphically
Sol. : −
− [ ] 1
1 1, 0x
xx
If x > 0 If x < 0
–x ≤ [x] – 1 ≤ x –x ≥ [x] – 1 ≥ x
1 – x ≤ [x] ≤ x + 1 1 – x ≥ [x] ≥ 1+ x
1 – x ≤ x – {x} ≤ x + 1 1 – x ≥ x – {x} ≥ 1 + x
− − − −1 { } 1x x x x − − − −1 { } 1x x x x
− −2 1 { } 1x x
− −2 1 { } 1x x
1x = x
[1, )x
69. Answer (3)
Hint : 1 – sin2 = (cos1 – sin1)2
Sol. : 1 sin2 1
tancos2
− −
1 1 tan1tan
1 tan1
− − = −
+
1tan tan 1
4
− = − −
14
= −
70. Answer (3)
Hint : − − − −− =
+
1 1 1tan tan tan1
x yx y
xy
Sol. : − − −
+ =
1 1 11tan tan tan (3)x
y
− − − − − = − = +
1 1 1 11 3tan tan 3 tan tan
1 3
xx
y x
+= = − +
− −
1 3 103
3 3
xy
x x
All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-B) (Hints & Solutions)
10/13
For positive integer y, x = 1, 2
⟹ Solutions are (1, 2)(2, 7)
71. Answer (4)
Hint : − − + =1 1sin cos
2x x
Sol. : − −+1 2 1 2(sin ) (cos )x x
− − − − = + − −
1 1 2 1 1(sin cos ) 2sin sin2
x x x x
− −= + −
21 2 12(sin ) sin
4x x
− −
= + − + −
2 2 21 2 12 (sin ) sin
4 2 16 8x x
= −
+ −
2212 sin
8 4x
1 13sin sin
2 2 4 4 4
− − − − − x x
− −
2 21 9
0 sin4 16
x
− − +
2 21 2 1 2 10
(sin ) (cos )8 8
x x
72. Answer (3)
Hint : Break the function at –1, 0, 1, 2, 3
Sol. : [ ] 0 [0,1) x x
1, 1
2, 1 0
( ) 2, 1 2
1, 2 3
0, 3
− − −
− −
=
x
x
f x x
x
x
73. Answer (1)
Hint : − −+= −
+ +
1 1( 1)tan tan
1 1n
n r nrT
n n
Sol. : − +
=+ + +
1
2 2
( 1)tan
( 1) ( 1)n
n nT
n n r r
−
+=
+ + + +
1 1tan( 1)
11 1
n
nn r nr
n n
1
( 1)
1 1tan( 1)
11 1
n r nr
n nn r nr
n n
−
+ − + +=
+ + + +
1 1( 1)tan tan
1 1
n r nr
n n
− −+= −
+ +
− − = −
+
1 1tan ( ) tan1
nS n
n
→
= − =lim
2 4 4nS
74. Answer (1)
Hint : − − 1sin
2 2x
Sol. : − − 12tan
2 2a
− − − 1tan 1 1
4 4a a
75. Answer (2)
Hint : − − −
=
1 1 1tan tan tan1
x yx y
xy
Sol. : tan cot 1 1x x = −
1 1 1 tan cottan (tan ) tan (cot ) tan
1 tan cot
− − − − − =
+
x xx x
x x
21 1tan 1
tan tan ( cot 2 )2 tan
− − −
= = −
xx
x
1 1tan (cot2 ) tan tan 22
− − = − = − −
x x
1tan tan 22
− = −
x
=
2T
76. Answer (4)
Hint : Draw graph of sin–1(sinx)
Test - 1 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
11/13
Sol. :
77. Answer (1)
Hint : Find domain
Sol. : − − 1 1
1 2 12 2
x x
Let f(x) = sin–1(2x) – cos–1x + tan–1(2x)
sin–12x is increasing
cos–1x is decreasing
and tan–1(2x) is increasing function
−
= − = − − − =
min
1 2 17( )
2 2 3 4 12f x f
= = − + =
max.
1 5( )
2 2 3 4 12f x f
=( )2
f x does not have any solution
78. Answer (2)
Hint : − −=−
1 1
2
22tan tan
1
xx
x
Sol. : − − =
1 11 14tan 2 2tan
5 5
− −
= = −
1 1
12
552 tan 2tan1 12
125
− −
= = −
1 1
52
12012tan tan25 119
1144
− − −
1 114tan cot (239)
5
− − = −
1 1120 1tan tan
119 239
− −
−
= = = +
1 1
120 1
119 239tan tan (1)120 1 4
1119 239
79. Answer (2)
Hint : 0 ≤ {x} < 1
Sol. : ∵ sgn(sin–1x) = {–1, 0, 1} and {2x} ∈ [0, 1)
⟹ sgn(sin–1x) = {2x} = 0 ⟹ x = 0
80. Answer (1)
Hint : |sinx| + |cosx| ∈ [1, 2]
Sol. : + | sin | | cos | [1, 2]x x
+ =[| sin | | cos |] 1x x
⟹ domain of f(x) is ϕ (empty set)
81. Answer (3)
Hint : Break G.I.F
Sol. :
3, [ 1, sin1)
2, [ sin1, 0)( )
0, [0, sin1)
1, [sin1,1]
− − −
− −=
x
xf x
x
x
82. Answer (4)
Hint : ln(1 + x) is an integer
Sol. : x + 1 > 0 1x −
ln(1 + x) + [(1+ x)2] – 3 is an integer
⇒ ln(1 + x) is an integer
⇒ [ln(1 + x)] = ln(1 + x)
⇒ [(1 + x)2] = 3
3 ≤ (1 + x)2 < 4
+ 3 |1 | 2x
ln 3 ln(1 x) ln2 +
ln(1 + x) is integer for no value of x
All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-B) (Hints & Solutions)
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83. Answer (3)
Hint : sinx ∈ [–1, 1]
Sol. : 2sinx + 2y = 1
− sin11 sin 1 2 2
2
xx
− − −sin12 2
2
x
− − − − sin 111 2 1 1 2 2
2
x y
− −( , 1]y
84. Answer (1)
Hint : Draw graph
Sol. : 1
11 1
xy
x x= = −
+ +
(y – 1)(x + 1) = –1
[0,1)y
f(x) is one-one and into function
85. Answer (4)
Hint : Put 9x = t
Sol. : 34x + 9|x – 1| – 10 ≤ 0
81x + 9|x – 1| – 10 ≤ 0
Put 9x = t
If x – 1 ≤ 0 If x – 1 ≥ 0
+ − 2 910 0t
t + − 2 10 0
9
tt
− + 3 10 9 0t t 1 9 9xx
(t – 1)(t2 + t – 9) ≤ 0 9t
− + − −
21 1
( 1) 9 02 4
t t + 2 829
tt
− + +
+ −
1 37( 1)
2 2
1 370
2 2
t t
t
+ − 2 10 09
tt
Has no solution.
− − −
37 1( 1) 0
2t t
−
37 11
2t
−
37 11 9
2
x
−
9
37 10 log
2x
86. Answer (3)
Hint : + 1
2,tt
when t > 0
Sol. :
1 1| | 2, | | 2,
| | | |
1| | | | 2
| || |
x yx y
x yx y
+ +
+
+ + + + + 1 1 1
| | | | | || | 6| | | | | || |
x y x yx y x y
+ + + + + =1 1 1
| | | | | || | 6| | | | | || |
x y x yx y x y
= = = = = =1 1 1
| | | | | || | 1| | | | | || |
x y x yx y x y
⟹ |x| = |y| = 1 ⟹ x = ± 1, y = ± 1
Ordered pairs are (1, 1), (–1, 1), (1, –1), (–1, –1)
Test - 1 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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87. Answer (4)
Hint : − −= −1 1cos sin
2x x
Sol. :
− − − − − + − −
−
1 1 1 1
2
sin sin sin sin2 4 2
016
x x x x
− − − − + − −
2 21 1 2 1sin (sin ) sin 0
2 8 2 16x x x
− −
21 2(sin ) 0
16x
− − 1sin
4 4x
1 1
2 2x−
88. Answer (4)
Hint : y = x
Sol. : ∵ sin–1x + cos–1y =
2 …(1)
(x, y) ∈ R
∵ sin–1x + cos–1x =
2 ⟹ (x, x) ∈ R
⟹ R is reflexive
Let sin–1y + cos–1x = k …(2)
From (1) + (2)
− − = + =1 1sin cos
2 2k y x
⟹ (y, x) ∈ R ⟹ R is symmetric
Let (y, z) ∈ R ⟹ − − + =1 1sin cos
2y z …(3)
From (1) + (3),
− − + = 1 1sin cos ( , )
2x z x z R
⟹ R is transitive
⟹ R is equivalence relation
89. Answer (3)
Hint : Venn diagram
Sol. :
90. Answer (4)
Hint : Draw graph
Sol. :
n(A ∩ B) = 3
⟹ n(P (A ∩ B)) = 23 = 8