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Test - 6 (Code-C) (Answers) All India Aakash Test Series for JEE (Main)-2020
1/14
1. (4)
2. (3)
3. (4)
4. (2)
5. (4)
6. (1)
7. (3)
8. (3)
9. (1)
10. (1)
11. (2)
12. (3)
13. (4)
14. (2)
15. (1)
16. (1)
17. (4)
18. (4)
19. (3)
20. (1)
21. (2)
22. (1)
23. (2)
24. (1)
25. (4)
26. (2)
27. (1)
28. (2)
29. (4)
30. (1)
PHYSICS CHEMISTRY MATHEMATICS
31. (3)
32. (1)
33. (1)
34. (2)
35. (3)
36. (2)
37. (3)
38. (4)
39. (2)
40. (2)
41. (3)
42. (3)
43. (3)
44. (1)
45. (4)
46. (2)
47. (3)
48. (4)
49. (3)
50. (1)
51. (2)
52. (3)
53. (3)
54. (3)
55. (2)
56. (2)
57. (2)
58. (3)
59. (4)
60. (3)
61. (4)
62. (3)
63. (3)
64. (2)
65. (3)
66. (3)
67. (2)
68. (4)
69. (2)
70. (4)
71. (3)
72. (1)
73. (3)
74. (1)
75. (3)
76. (1)
77. (1)
78. (3)
79. (2)
80. (1)
81. (2)
82. (4)
83. (2)
84. (2)
85. (2)
86. (3)
87. (3)
88. (1)
89. (4)
90. (3)
Test Date : 17/02/2019
ANSWERS
TEST - 6 - Code-C
All India Aakash Test Series for JEE (Main)-2020
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-C) (Hints & Solutions)
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1. Answer (4)
Hint : x = A sin(t + )
Sol. : x = A sin(t + )
20 sin 2A
T
0 sin4
A
4
cos( )dx
A tdt
2 24 cos 4
16 16 4A
32cos sin
2 4 4A A
32 2mA
2. Answer (3)
Hint : 12
kf
m
Sol. : 12
kf
m
12 1kg
kf
eq12 8
kf
eq
1 12 4 2
k k kk k
k k
1Hz
4 f
3. Answer (4)
Hint : KeA = K1A1 = K2A2
Sol. : Let the amplitude of oscillation is A
3 63
3 6 e p
K KK A A K A
K K
23
pA A
23
pv v
PART - A (PHYSICS)
4. Answer (2)
Hint : dyv
dt
Sol. : y = A sint
v = Acost
2 2 21KE cos
2 mA t
2 21 1 cos22 2
tmA
2 T
T
5. Answer (4)
Hint : Use projection of particle on circle and findphase.
Sol. :A
Al22
/6
1
x1 = A cos(t)
2 sin6
x A t
x1 = x2
cos( ) sin6
t t
22 6
t
32
6 t
43 t
T
12T
t
6. Answer (1)
Hint : The phase difference must be 2 when theymeet again.
Test - 6 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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Sol. : 6( 1)
5 T
nT n
5n = 6n – 6
n = 6
Time period = 6T
Method 2
2 25 2
6t
T T
1 51
6t
T T
t = 6T
7. Answer (3)
Hint : 2
k
Sol. :4
0.5 sin 2011
y t x
2 411
11cm 5.5 cm
2
12011 55 cm s
4
v
k
8. Answer (3)
Hint : Draw standing wave in string.
Sol. :
9. Answer (1)
Hint : Use dimensional analysis
Sol. :1
2
T
fL
2
1 12
4
T Tf
L LDD
1f
LD
10. Answer (1)
Hint :2
5
Sol. :2
5
= 10 cm
5 cm2
11. Answer (2)
Hint : The frequency will be maximum whenapproaching the sound tangentially.
Sol. :4 m
5 m1
37°
4 m37°
3 m
3 m
7410
180 t
t = 0.13 s
12. Answer (3)
Hint : vp = A
Sol. : vp (maximum) = 2n x0
22wn
v n
2nx0 = (n)2
022
x
= x0
13. Answer (4)
Hint : v2 = 2(A2 – x2)
Sol. : v2 = 2(4a2 – a2)
22 23
34
aa
12
at t = 0, x = a
1sin
2
6
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-C) (Hints & Solutions)
4/14
x = 1
2 sin2 6
a t
= 1 1
2 sin cos cos sin2 6 2 6
a t t
= 3 1
2 sin cos2 2 2 2
t ta
= 3 sin cos2 2t t
a
14. Answer (2)
Hint : f0 = f1 – f2Sol. : 2f1 = 410
14102
f
24542
f
454 410 44Beat 7 7 Hz
2 2 2 22
15. Answer (1)
Hint : Wavelength does not change after reflectionfrom the wall.
Sol. : = 0 – V T
= 0 1s
VV
= 3320 20
1 104 320
= 330010
4
= 75 × 10–3 m
= 75 mm
16. Answer (1)
Hint :
12
Tf [Second harmonic of wire]
Sol. :1 2
12V TL L
320 1 1002 0.8 0.5
1100
1
0.5 100m
m = 5 g
17. Answer (4)
Hint : 2 1 cos(2 )sin
2
Sol. :2
1 cos2 22
a xy nt
41 cos 4
2
a xnt
18. Answer (4)
Hint :60°
x
x
Fnet = 2Kx + 2Kx cos60°
Sol. :
KK
2K
60° x
x
x = x cos60°
–ma = 2Kx + 2K x cos60°
–ma = 2Kx + 2Kx cos260°
=
21
2 22
Kx Kx
–ma = 2
24Kx
Kx
–ma = 22
KxKx
–ma = 5
2Kx
52
Ka x
m
22
5 m
TK
19. Answer (3)
Hint : Use formula B
Test - 6 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
5/14
Sol. :3
5
100 10
5 10
PB
VV
B = 20 × 108
8
3
20 10
10
B
v
v � 1400 ms–1
20. Answer (1)
Hint : a = –2A
Sol. : = –2x1 = –2x2
( + ) = –2(x1 + x2)
u2 = 2A2 – 2x12
v2 = 2A2 – 2x22
v2 – u2 = 2(x1 + x2)(x1 – x2)
= –( + )(x1 – x2)
2 2
1 2
u v
x x
21. Answer (2)
Hint : Total energy remain conserved.
Sol. :2
2 21 1 12 2 4 2
AkA k mV
22 21 1 1
' 42 2 4 2
AkA k mV
2 22 21 1 1 1
' 42 2 4 2 2 4
A AkA k kA k
A2 = 2
234
AA
A2 = 2134
A
A = 132
A
22. Answer (1)
Hint : amax = 2A
Sol. :2
5AA
= 5 rad s–1
5 sin6 A
5 = 12
A A = 10
a = (5)2 × 10 = 250 m/s2
23. Answer (2)
Hint : 2 2net 1 2 1 22 cosA A A A A
Sol. :
A2
A1
1 sin4
x a t
22
2 sin63
ax a t
22
sin63
x a t
1
2
32
AA
12
24. Answer (1)
Hint : 2max 1 2( )I I I
Sol. :2
max 1 2
min 1 2
I A A
I A A
1 2
1 2
31
A A
A A
3A1 – 3A2 = A1 + A2
2A1 = 4A2
1
2
2A
A
25. Answer (4)
Hint : 2 2v A x
Sol. : 2 213 3 A
2 212 4 A
(13)2 – (12)2 = 2(42 – 32)
25 = 2 (16 – 9)
2 125 5rad s
7 7
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-C) (Hints & Solutions)
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26. Answer (2)
Hint : TE = KE + PE
Sol. :2
21 1 3KE
2 2 4
kA k A
21 7KE
2 16
kA
780 KE
16
KE = 35 J
27. Answer (1)
Hint : Amplitude of oscillation = 2L
Sol. :
2L
m1
m2
2KL – KL = (m1 + m2)g
KL = (m1 + m2)g
1 2( )
m mK g
L
Amplitude of oscillation
2
2
m g LK
1 22
( )
2
m mm g g
2m2 = m1 + m2
m2 = m1
2
1
1m
m
28. Answer (2)
Hint : 2
T TV
r
Sol. :1
2T
fLr
11 1 1
2 1 1 1
12 92
rf L T
f L r T
1
2
3f
f
2450
Hz 150 Hz3
f
29. Answer (4)
Hint : 1 2
2m
Tk k
Sol. : 1 2 mt
k
2 21
1
4m
tk
21 2
1
4 mk
t
22 2
2
4 mk
t
1 2
2 m
Tk k
22 21 2
21 1
4
mT
mt t
2 2 21 2
1 1 1 T t t
2 2 21 2
T t t
30. Answer (1)
Hint : I1 = 0.8I0
Sol. : I = (0.8I0)2 = 0.64I0
I = 36%
Test - 6 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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PART - B (CHEMISTRY)
31. Answer (3)
Hint : Check all possible positions available andcheck stereochemistry for each isomer.
Sol. : Here are all the possible isomers.
Br
,
Br
,
CH Br2
Br
,
CH —Br2
Br
CH—Br
CH Br2
,
Br
,
CH Br2
Br
,
CH — Br2
Br
Br
CH — Br2
BrCH2
32. Answer (1)
Hint : — NH — C — CH 3
O
is ortho/para director.
Sol. : NHCOCH3 NHCOCH3
NO2
(Major product)
HNO /H SO3 2 4
33. Answer (1)
Hint : Alkyne having acidic H.
Sol. : White precipitates of HC C–Ag.
34. Answer (2)
Hint : Benzene is reactive toward ozonolysis butdoes not react with Br2.
Sol. : Benzene generally does not give additionreaction but it reacts with H2/Pt and showsaddition reaction.
35. Answer (3)
Hint : With alc. KOH both cis and trans butene isformed.
Sol. :
OH
OH
P4
CH3
OH
OH
CH3
,36. Answer (2)
Hint : FeCl3 is a Lewis acid.
Sol. : Cl — Cl + FeCl3L.B L.A
37. Answer (3)
Hint : Dehydrohalogenation
Sol. :
(ii) K/dry ether
CH Cl3
(i) NaNH2
Cl
Br
Br
Br
CH3
38. Answer (4)
Hint : Different carbides gives differenthydrocarbons upon hydrolysis.
Sol. : 2H O4 3 4Al C CH
CaC2 and Na2C2 on hydrolysis producesethyne.
Mg2C3 on hydrolysis gives propyne.
39. Answer (2)
Hint : 3AlCl6 6 2 6 5
(gas)BenzeneC H Cl C H Cl HCl
Sol. : The difference in the molar mass of productsis 76 g/mol.
40. Answer (2)
Hint : Alkenes on reaction with dil. H2SO4 formcarbocation, which has tendency ofrearrangement.
Sol. :(+)
(+)
dil. H SO2 4
OH H O2 :
Methylshift
41. Answer (3)
Hint : Alkynes undergo 2 times of addition reactionwith HCl.
Sol. : CH C CH CH — C — CH3 3 3
HCl
—Cl
—
Cl
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-C) (Hints & Solutions)
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42. Answer (3)
Hint : Alcohols with conc. H2SO4 undergodehydration.
Sol. :
(+)
H+
OHH+
43. Answer (3)
Hint : Decarboxylation of carboxylates reduces1 C atom from the parent chain.
Sol. : NaOH CaO3 2 2
3 2 3
CH CH CH COOH
CH CH CH
44. Answer (1)
Hint : Decarboxylation
Sol. : Trimethyl cyclohexane is formed when (R)undergo sodalime decarboxylation.
45. Answer (4)
Hint :H+
+ CH – CH = CH3 2
CH – CH – CH3 3
Sol. : + CH — CH = CH3 2
CH — CH — CH3 3
(A)
H+
(P)
Br /h2
Br
CH — C — CH3 3
——
(Q)
Alc. KOH
(R)dil. H SO2 4
(S)
CH — C = CH3 2 CH — C — CH3 3
OH
46. Answer (2)
Hint : 1,2-dihalo molecules give alkene on reactionwith zinc.
Sol. : A : CH3 – CH = CH2
B : 3 3CH – CH – CH|
Cl
47. Answer (3)
Hint : Tert. carbocation is most stable.
Sol. : 3° alcohols produces alkenes fastest.
48. Answer (4)
Hint : Ozonolysis reaction
Sol. :
CH2
O3
Zn/H O2
O
+ HCOH
49. Answer (3)
Hint : Warm KMnO4 in acidic medium oxidise thecompound.
Sol. : A CH — CH — CH — CH3 3—
OH
—
OHB CH — C — OH3 —
O
—
50. Answer (1)
Hint : CH — C — CH3 3
O
(P)
CH — C CH3 dil. H SO /HgSO2 4 4
— —
Sol. : Product P is a ketone and degree ofunsaturation of P is one.
51. Answer (2)
Hint: Both naphthalene and phenanthrene arearomatic rings and have e– participating inresonance and after alkylation e– remainsame.
Sol. : Naphthalene
Phenanthrene
52. Answer (3)
Hint : Correct formula of the given molecule accordingto the question is (CH3)3C — C(CH3)3.
Sol. : CH — C — C — CH3 3
CH3 CH3
CH3 CH3
Test - 6 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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53. Answer (3)
Hint : To produce formaldehyde molecule must have= CH2 group.
Sol. : , ,
54. Answer (3)
Hint : F– being a poor leaving group, cannot leaveeasily.
Br– is a good leaving group, so can produceSaytzeff product.
Sol. :
CH — CH — CH — CH3 2 2—
F
CH — CH — CH == CH3 2 2
CH — CH — CH — CH3 2 2
—H
NH2—
F
More stable anion
CH — CH == CH — CH3 3
—H
—
Br
B
CH — CH — CH — CH3 3
any strong base
More substitutedalkene is formed
which is more stable
55. Answer (2)
Hint : With Lindlar's catalyst syn addition occurs.
Sol. :
CH — C C — CH3 3Lindlar's catalyst
C == CH
CH3CH3
Hcis-2-butene
56. Answer (2)
Hint : HCl does not show peroxide effect.
Sol. : Br
HBr
HBr
HCl
Peroxide
Peroxide
CH — CH == CH2 2
Br
Cl
57. Answer (2)
Hint : Cl2/H2O gives Markovnikov addition of OHand Cl+.
Sol. : CH — CH = CH CH — CH — CH3 2 3 2
Cl /H O2 2
OH Cl(P)
58. Answer (3)
Hint : NBS produce Br.
which majorly attacksallylic and benzylic position.
Sol. :
CH3
NBS
CH — Br2 CH2
+ Br—
CH3 CH3 CH3
++
Br
Br
+
Br
59. Answer (4)
Hint : Grignard reagents first undergoes acid basereaction and produces a salt.
Sol. :
Br /h2
0.5 mol 0.5 mol
Mg/dry ether
Br
0.5 mol
Mg Br
(P)( )R
S — CH — C C — H2
—H
two acidic
(A)
2 moles of P are required for 1 mole of A
S — CH — C C2 0.25 mol
(Salt)(Q)
+
60. Answer (3)
Hint : Hydroboration reaction.
Sol. :
CH — CH == CH3 2
B H /THF2 6
D O , OD2 2– —
H
—
OD(A)
CH — CH — CH3 2
CH — CH == CH3 2
B H /THF2 6
H O /OH2 2
–
D(B)
CH — CH — CH — OH3 2
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-C) (Hints & Solutions)
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PART - C (MATHEMATICS)
61. Answer (4)
Hint : 5cos(2 ) cos2
Sol. : As cos(25) = cos 3264
= cos2
= 0
cos cos2 cos22 ... cos29 = 0
62. Answer (3)
Hint : cosine law
Sol. : Triangle DEF is called the pedal triangle ofthe triangle ABC.
Circumradius of pedal triangle = 12
(Circumradius of triangle ABC) = 2R
cosC = 2 2 2
2 a b c
ab
= 25(3 1 2 3) 50 100
25 2( 3 1)8
4
= 50( 3 1) 1
50 2( 3 1) 2
C = 45°, similarly B = 30°, A = 105°
5 512sin 222
cR
C
5 5 2
2 42 2 R
R
63. Answer (3)
Hint : Formula
Sol. : Apply direct formula
64. Answer (2)
Hint : Counting
Sol. : 12 11 11( )
52 51 221P E
65. Answer (3)
Hint : sin + sin( + ) + sin( + 2) + ... upton terms
Sol :13
1
(sin cos )
r
r r
13sin
2[sin7 cos7 ] 0
sin2
13
sin 02
or tan7 = 1 and sin 02
2 4 6, ,
13 13 13
5 9 13, , ,
28 28 28 28
66. Answer (3)
Hint : Sum of digits should be divisible by 3.
Sol. : If the sum of digits of a number is divisible by3, then the number is divisible by 3.
The number should be formed by using
1, 2, 3, 6, 9 or 2, 3, 4, 6, 9
6
5
2 5 2 1( )
6 35P E
C
67. Answer (2)
Hint : cosine law
Sol. :2 2 2
cos2
b c aA
bc
b2 – 2bc cosA + c2 – a2 = 0
2 4 2 8 0b b
2( 2 2) 0 2 2 b b
68. Answer (4)
Hint : Combinations
Sol. :18 17
2 235
2
9 17 17 8 17( )
35 17 35
C CP E
C
Test - 6 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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69. Answer (2)
Hint : Square both sides
Sol. : 1 + sin + cos = sincos
sin + cos = sincos– 1
Squaring on both sides,
1 + 2sincos
= 1 + sin2cos2 – 2sincos
212sin2 sin 2
4
sin2(sin2 – 8) = 0
3sin2 0 , ,
2 2
Only and 32
satisfy the original equation.
3 5Sum
2 2
70. Answer (4)
Hint : Adding a number in every observation doesnot change .
Sol. :2( )ix x
N
11 2 3 ... 10 11
10 2 x
25 6 7 ... 14 95 19 11
410 10 2 2
x
1 = 2
71. Answer (3)
Hint : cosine law
Sol. :2 2 26 (3 5) 3
cos2 6 3 5
36 45 9 2
36 5 5
1
tan 2 12
1 1tan
2 3
tan tan8 tan tan
6
8
6
72. Answer (1)
Hint : Venn diagram
Sol. : The probability of solving the question by
three students A, B and C are 2 1
,3 3
and 16
respectively.
2 1 1( ) , ( ) , ( )
3 3 6 P A P B P C
Probability that the question is solved byexactly two students is
P(E) = ( ) ( ) P A B C P A B C
( ) P A B C
= ( ) ( ) ( ) ( ) ( ) ( ) P A P B P C P A P B P C
( ) ( ) ( )P A P B P C
= 1 1 1 2 2 1 2 1 5 53 3 6 3 3 6 3 3 6 18
73. Answer (3)
Hint : Solve equation
Sol. :sin1 sin 22
sin cos2
sin2(1 sin ) 2
2sin cos cos2 2 2
22(1 sin ) 2sin2
2(1 + sin) = 1 – cos
2sin = –(1 + cos)
24 sin cos 2 cos2 2 2
2 sin cos2 2
tan 12
74. Answer (1)
Hint :2
2 2( )ixx
N
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-C) (Hints & Solutions)
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Sol. : x4 + x5 = 30 – (2 + 4 + 9) = 15 ...(i)
2 22 4 54 16 81
36 9.25
x x
x42 + x5
2 = 125 ...(ii)
From (i) and (ii) gives,
x4 = 5, x5 = 10
x4 : x5 = 1 : 2
75. Answer (3)
Hint : Venn diagram
Sol. : E
55
25
H
Percentage of population which readexactly one newspaper is
( ) ( )n E H n E H
= n(E) – n(E H) + n(H) – n(E H)
= 10 + 30 – 2 x 5 = 30%
76. Answer (1)
Hint : sec2 = 1 + tan2
Sol. : R = sec2Asec2B + tan2Atan2B +
2 secAsecBtanAtanB – sec2Atan2B –
sec2Btan2A – 2 secAsecBtanA tanB
= sec2A(sec2B – tan2B) +
tan2A(tan2B – sec2B)
= sec2A – tan2A = 1
77. Answer (1)
Hint : P(E) = 1 – P(X = 1) – P(X = 0)
Sol. : S = {HHH, HHT, HTH, THH, HTT, THT, TTH}
E = {HTT, THT, TTH}
( ) 3 4( ) 1 1
( ) 7 7n E
P En S
78. Answer (3)
Hint :( )( )
sin2
A s b s cbc
( )cos
2
A s s abc
Sol. : 22cos cos 4sin2 2 2
B C B C A
22sin cos 4sin2 2 2A B C A
cos 2sin2 2
B C A
cos cos sin sin 2sin2 2 2 2 2
B C B C A
( ) ( )s s b s s cac ab
( )( ) ( )( )s a s c s a s bac ab
( )( )
2s b s c
bc
2 2b c
b c aa
79. Answer (2)
Hint : 2 2 21( ) ix x
n
Sol. : 22
1122
xx
N N
2 21 1 1 22x x x x
21 100 80 20 x
2 20 1001
10 100
= 1
80. Answer (1)
Hint : sin2x = 1, cos2x = 0
Sol. : sin2x – cos2x = 1
sin2x – 1 + sin2x = 1
sin2x = 1 and cos2x = 0
2 2
1
(sin cos )
n
r r
r
x x
= (sin2x + sin4x + ... + sin2nx)
+ (cos2x + cos4x + ... + cos2nx)
= n + 0 = n
Test - 6 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
13/14
81. Answer (2)
Hint : Circular permutations
Sol. :4
45 4( )
1
nn CP E
n
5( 4)( 5)( 6)( 7)( 1)( 2)( 3) 4
n n n n n
n n n n
( 5)( 6)( 7)( )
( 1)( 2)( 3)n n n
P En n n
82. Answer (4)
Hint : Perpendiculartan
Base
Sol. :
10 m x
H
45° 60°
tan45 , tan6010
H Hx x
10 ,3
HH x x
103
HH
11 10
3
H
10 3 10 3( 3 1)
23 1
H
5(3 3)mH
83. Answer (2)
Hint : P(E) = 1 – P(A B)
Sol. : P(E) =
5 54
2 21
62 2
= 60 60 24
1180
= 96
1180
= 84 7
180 15
84. Answer (2)
Hint : Check domain
Sol. : tan3x = tanx
3 (2 1) , (2 1)2 2
x n x n
3x = n + x
3 5 7 11, , , , ,
2 2 6 6 6 6 x
2 n
x
3
0, , , , 22 2 x
x = 0, , 2
Sum = 3
85. Answer (2)
Hint : D = 0
Sol. : For exactly one solution,
D = 0 a2 = 4b
4 4( )
11 11 121P E
86. Answer (3)
Hint : sine law
Sol. : 2b2 = a2 + c2
2sin2B = sin2A + sin2C
2[1 – cos2B] = [1 – cos2A] + [1 – cos2C]
2cos2B = cos2A + cos2C
cos2B – cos2A = cos2C – cos2B
2sin(A + B) sin(A – B)
= 2sin(B + C) sin (B – C)
sinC [sinA cosB – cosAsinB]
= sinA[sinB cosC – cosBsinC]
cotB – cotA = cotC – cotB
2cotB = cotA + cotC
cotA, cotB, cotC are in A.P.
tanA, tanB, tanC are in H.P.
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-C) (Hints & Solutions)
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�����
87. Answer (3)
Hint : Venn diagram formula
Sol. :1 1 13 6 6
A
B
16
C
1 1 12 6 3
P(A B C)
= P(A) + P(B) + P(C) – P(A B) – P(B C)
– P(C A) + P(A B C)
1 1 1 3 2
( )3 2 6 6 3
P A B C
2 1 1 1 13 2 3 2 6
8 6 4 6 2 112 6
1 1 1( )
6 3 2P E
88. Answer (1)
Hint : 2sin cos 2 cos 12 2
Sol. : ( ) sin 2 cos4 8
f x x x
sin 2 cos8 2 8
x x
cos2 cos8 8
x x
22cos cos 18 8
x x
21 9
2 cos8 4 8
x
min9
( ( ))8
f x
89. Answer (4)
Hint : P(E1) + P(E2) + P(E3) = 1
and 0 P(E1), P(E2), P(E3) 1
Sol. :
–1 143
16
23
56 1 2
,6 3
P
1 2 3 1 62 6 6 P P P
3 3 2 3 1 61
6 P P P
10 1
2P 2 3
0 16
P 1 60 1
6P
0 P – 1 – 2 0 3P – 2 – 6 0 6P + 1 6
1 P – 1 2 3P – 4 –1 6P 5
2 43 3
P 1 56 6
P
90. Answer (3)
Hint : Factorization
Sol. : (sin2x – y2) – (sinx – y) = 0
(sinx – y)(sinx + y – 1) = 0
y = sinx OR sinx = 1 – y
–1 sinx 1 –1 sinx 1
y = –1, 0, 1 –1 1 – y 1
–2 –y 0
0 y 2
y = 0, 1, 2
y = {–1, 0, 1, 2}
Test - 6 (Code-D) (Answers) All India Aakash Test Series for JEE (Main)-2020
1/14
1. (1)
2. (4)
3. (2)
4. (1)
5. (2)
6. (4)
7. (1)
8. (2)
9. (1)
10. (2)
11. (1)
12. (3)
13. (4)
14. (4)
15. (1)
16. (1)
17. (2)
18. (4)
19. (3)
20. (2)
21. (1)
22. (1)
23. (3)
24. (3)
25. (1)
26. (4)
27. (2)
28. (4)
29. (3)
30. (4)
PHYSICS CHEMISTRY MATHEMATICS
31. (3)
32. (4)
33. (3)
34. (2)
35. (2)
36. (2)
37. (3)
38. (3)
39. (3)
40. (2)
41. (1)
42. (3)
43. (4)
44. (3)
45. (2)
46. (4)
47. (1)
48. (3)
49. (3)
50. (3)
51. (2)
52. (2)
53. (4)
54. (3)
55. (2)
56. (3)
57. (2)
58. (1)
59. (1)
60. (3)
61. (3)
62. (4)
63. (1)
64. (3)
65. (3)
66. (2)
67. (2)
68. (2)
69. (4)
70. (2)
71. (1)
72. (2)
73. (3)
74. (1)
75. (1)
76. (3)
77. (1)
78. (3)
79. (1)
80. (3)
81. (4)
82. (2)
83. (4)
84. (2)
85. (3)
86. (3)
87. (2)
88. (3)
89. (3)
90. (4)
Test Date : 17/02/2019
ANSWERS
TEST - 6 - Code-D
All India Aakash Test Series for JEE (Main)-2020
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-D) (Hints & Solutions)
2/14
1. Answer (1)
Hint : I1 = 0.8I0Sol. : I = (0.8I0)
2 = 0.64I0I = 36%
2. Answer (4)
Hint : 1 2
2m
Tk k
Sol. : 1 2 mt
k
2 21
1
4m
tk
21 2
1
4 mk
t
22 2
2
4 mk
t
1 2
2 m
Tk k
22 21 2
21 1
4
mT
mt t
2 2 21 2
1 1 1 T t t
2 2 21 2
T t t
3. Answer (2)
Hint : 2
T TV
r
Sol. :1
2T
fLr
11 1 1
2 1 1 1
12 92
rf L T
f L r T
1
2
3f
f
2450
Hz 150 Hz3
f
PART - A (PHYSICS)
4. Answer (1)
Hint : Amplitude of oscillation = 2L
Sol. :
2L
m1
m2
2KL – KL = (m1 + m2)g
KL = (m1 + m2)g
1 2( )
m mK g
L
Amplitude of oscillation
2
2
m g LK
1 22
( )
2
m mm g g
2m2 = m1 + m2
m2 = m1
2
1
1m
m
5. Answer (2)
Hint : TE = KE + PE
Sol. :2
21 1 3KE
2 2 4
kA k A
21 7KE
2 16
kA
780 KE
16
KE = 35 J
6. Answer (4)
Hint : 2 2v A x
Test - 6 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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Sol. : 2 213 3 A
2 212 4 A
(13)2 – (12)2 = 2(42 – 32)
25 = 2 (16 – 9)
2 125 5rad s
7 7
7. Answer (1)
Hint : 2max 1 2( )I I I
Sol. :2
max 1 2
min 1 2
I A A
I A A
1 2
1 2
31
A A
A A
3A1 – 3A2 = A1 + A2
2A1 = 4A2
1
2
2A
A
8. Answer (2)
Hint : 2 2net 1 2 1 22 cosA A A A A
Sol. :
A2
A1
1 sin4
x a t
22
2 sin63
ax a t
22
sin63
x a t
1
2
32
AA
12
9. Answer (1)
Hint : amax = 2A
Sol. :2
5AA
= 5 rad s–1
5 sin6 A
5 = 12
A A = 10
a = (5)2 × 10 = 250 m/s2
10. Answer (2)
Hint : Total energy remain conserved.
Sol. :2
2 21 1 12 2 4 2
AkA k mV
22 21 1 1
' 42 2 4 2
AkA k mV
2 22 21 1 1 1
' 42 2 4 2 2 4
A AkA k kA k
A2 = 2
234
AA
A2 = 2134
A
A = 132
A
11. Answer (1)
Hint : a = –2A
Sol. : = –2x1 = –2x2
( + ) = –2(x1 + x2)
u2 = 2A2 – 2x12
v2 = 2A2 – 2x22
v2 – u2 = 2(x1 + x2)(x1 – x2)
= –( + )(x1 – x2)
2 2
1 2
u v
x x
12. Answer (3)
Hint : Use formula B
Sol. :3
5
100 10
5 10
PB
VV
B = 20 × 108
8
3
20 10
10
B
v
v � 1400 ms–1
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-D) (Hints & Solutions)
4/14
13. Answer (4)
Hint :60°
x
x
Fnet = 2Kx + 2Kx cos60°
Sol. :
KK
2K
60° x
x
x = x cos60°
–ma = 2Kx + 2K x cos60°
–ma = 2Kx + 2Kx cos260°
=
21
2 22
Kx Kx
–ma = 2
24Kx
Kx
–ma = 22
KxKx
–ma = 5
2Kx
52
Ka x
m
22
5 m
TK
14. Answer (4)
Hint : 2 1 cos(2 )sin
2
Sol. :2
1 cos2 22
a xy nt
41 cos 4
2
a xnt
15. Answer (1)
Hint :
12
Tf [Second harmonic of wire]
Sol. :1 2
12V TL L
320 1 1002 0.8 0.5
1100
1
0.5 100m
m = 5 g
16. Answer (1)
Hint : Wavelength does not change after reflectionfrom the wall.
Sol. : = 0 – V T
= 0 1s
VV
= 3320 20
1 104 320
= 330010
4
= 75 × 10–3 m
= 75 mm
17. Answer (2)
Hint : f0 = f1 – f2Sol. : 2f1 = 410
14102
f
24542
f
454 410 44Beat 7 7 Hz
2 2 2 22
18. Answer (4)
Hint : v2 = 2(A2 – x2)
Sol. : v2 = 2(4a2 – a2)
22 23
34
aa
12
at t = 0, x = a
1sin
2
6
Test - 6 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
5/14
x = 1
2 sin2 6
a t
= 1 1
2 sin cos cos sin2 6 2 6
a t t
= 3 1
2 sin cos2 2 2 2
t ta
= 3 sin cos2 2t t
a
19. Answer (3)
Hint : vp = A
Sol. : vp (maximum) = 2n x0
22wn
v n
2nx0 = (n)2
022
x
= x0
20. Answer (2)
Hint : The frequency will be maximum whenapproaching the sound tangentially.
Sol. :4 m
5 m1
37°
4 m37°
3 m
3 m
7410
180 t
t = 0.13 s
21. Answer (1)
Hint :2
5
Sol. :2
5
= 10 cm
5 cm2
22. Answer (1)
Hint : Use dimensional analysis
Sol. :1
2
T
fL
2
1 12
4
T Tf
L LDD
1f
LD
23. Answer (3)
Hint : Draw standing wave in string.
Sol. :
24. Answer (3)
Hint : 2
k
Sol. :4
0.5 sin 2011
y t x
2 411
11cm 5.5 cm
2
12011 55 cm s
4
v
k
25. Answer (1)
Hint : The phase difference must be 2 when theymeet again.
Sol. : 6( 1)
5 T
nT n
5n = 6n – 6
n = 6
Time period = 6T
Method 2
2 25 2
6t
T T
1 51
6t
T T
t = 6T
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-D) (Hints & Solutions)
6/14
26. Answer (4)
Hint : Use projection of particle on circle and findphase.
Sol. :A
Al22
/6
1
x1 = A cos(t)
2 sin6
x A t
x1 = x2
cos( ) sin6
t t
22 6
t
32
6 t
43 t
T
12T
t
27. Answer (2)
Hint : dyv
dt
Sol. : y = A sint
v = Acost
2 2 21KE cos
2 mA t
2 21 1 cos22 2
tmA
2 T
T
28. Answer (4)
Hint : KeA = K1A1 = K2A2
Sol. : Let the amplitude of oscillation is A
3 63
3 6 e p
K KK A A K A
K K
23
pA A
23
pv v
29. Answer (3)
Hint : 12
kf
m
Sol. : 12
kf
m
12 1kg
kf
eq12 8
kf
eq
1 12 4 2
k k kk k
k k
1Hz
4 f
30. Answer (4)
Hint : x = A sin(t + )
Sol. : x = A sin(t + )
20 sin 2A
T
0 sin4
A
4
cos( )dx
A tdt
2 24 cos 4
16 16 4A
32cos sin
2 4 4A A
32 2mA
Test - 6 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
7/14
PART - B (CHEMISTRY)
31. Answer (3)
Hint : Hydroboration reaction.
Sol. :
CH — CH == CH3 2
B H /THF2 6
D O , OD2 2– —
H
—
OD(A)
CH — CH — CH3 2
CH — CH == CH3 2
B H /THF2 6
H O /OH2 2
–
D(B)
CH — CH — CH — OH3 2
32. Answer (4)
Hint : Grignard reagents first undergoes acid basereaction and produces a salt.
Sol. :
Br /h2
0.5 mol 0.5 mol
Mg/dry ether
Br
0.5 mol
Mg Br
(P)( )R
S — CH — C C — H2
—H
two acidic
(A)
2 moles of P are required for 1 mole of A
S — CH — C C2 0.25 mol
(Salt)(Q)
+
33. Answer (3)
Hint : NBS produce Br.
which majorly attacksallylic and benzylic position.
Sol. :
CH3
NBS
CH — Br2 CH2
+ Br—
CH3 CH3 CH3
++
Br
Br
+
Br
34. Answer (2)
Hint : Cl2/H2O gives Markovnikov addition of OHand Cl+.
Sol. : CH — CH = CH CH — CH — CH3 2 3 2
Cl /H O2 2
OH Cl(P)
35. Answer (2)
Hint : HCl does not show peroxide effect.
Sol. : Br
HBr
HBr
HCl
Peroxide
Peroxide
CH — CH == CH2 2
Br
Cl
36. Answer (2)
Hint : With Lindlar's catalyst syn addition occurs.
Sol. :
CH — C C — CH3 3Lindlar's catalyst
C == CH
CH3CH3
Hcis-2-butene
37. Answer (3)
Hint : F– being a poor leaving group, cannot leaveeasily.
Br– is a good leaving group, so can produceSaytzeff product.
Sol. :
CH — CH — CH — CH3 2 2—
F
CH — CH — CH == CH3 2 2
CH — CH — CH — CH3 2 2
—H
NH2—
F
More stable anion
CH — CH == CH — CH3 3
—H
—
Br
B
CH — CH — CH — CH3 3
any strong base
More substitutedalkene is formed
which is more stable
38. Answer (3)
Hint : To produce formaldehyde molecule must have= CH2 group.
Sol. : , ,
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-D) (Hints & Solutions)
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39. Answer (3)
Hint : Correct formula of the given molecule accordingto the question is (CH3)3C — C(CH3)3.
Sol. : CH — C — C — CH3 3
CH3 CH3
CH3 CH3
40. Answer (2)
Hint: Both naphthalene and phenanthrene arearomatic rings and have e– participating inresonance and after alkylation e– remainsame.
Sol. : Naphthalene
Phenanthrene
41. Answer (1)
Hint : CH — C — CH3 3
O
(P)
CH — C CH3 dil. H SO /HgSO2 4 4
— —
Sol. : Product P is a ketone and degree ofunsaturation of P is one.
42. Answer (3)
Hint : Warm KMnO4 in acidic medium oxidise thecompound.
Sol. : A CH — CH — CH — CH3 3—
OH
—
OHB CH — C — OH3 —
O
—
43. Answer (4)
Hint : Ozonolysis reaction
Sol. :
CH2
O3
Zn/H O2
O
+ HCOH
44. Answer (3)
Hint : Tert. carbocation is most stable.
Sol. : 3° alcohols produces alkenes fastest.
45. Answer (2)
Hint : 1,2-dihalo molecules give alkene on reactionwith zinc.
Sol. : A : CH3 – CH = CH2
B : 3 3CH – CH – CH|
Cl
46. Answer (4)
Hint :H+
+ CH – CH = CH3 2
CH – CH – CH3 3
Sol. : + CH — CH = CH3 2
CH — CH — CH3 3
(A)
H+
(P)
Br /h2
Br
CH — C — CH3 3
——
(Q)
Alc. KOH
(R)dil. H SO2 4
(S)
CH — C = CH3 2 CH — C — CH3 3
OH
47. Answer (1)
Hint : Decarboxylation
Sol. : Trimethyl cyclohexane is formed when (R)undergo sodalime decarboxylation.
48. Answer (3)
Hint : Decarboxylation of carboxylates reduces1 C atom from the parent chain.
Sol. : NaOH CaO3 2 2
3 2 3
CH CH CH COOH
CH CH CH
49. Answer (3)
Hint : Alcohols with conc. H2SO4 undergodehydration.
Test - 6 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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Sol. :
(+)
H+
OHH+
50. Answer (3)
Hint : Alkynes undergo 2 times of addition reactionwith HCl.
Sol. : CH C CH CH — C — CH3 3 3
HCl
—Cl
—
Cl
51. Answer (2)
Hint : Alkenes on reaction with dil. H2SO4 formcarbocation, which has tendency ofrearrangement.
Sol. :(+)
(+)
dil. H SO2 4
OH H O2 :
Methylshift
52. Answer (2)
Hint : 3AlCl6 6 2 6 5
(gas)BenzeneC H Cl C H Cl HCl
Sol. : The difference in the molar mass of productsis 76 g/mol.
53. Answer (4)
Hint : Different carbides gives differenthydrocarbons upon hydrolysis.
Sol. : 2H O4 3 4Al C CH
CaC2 and Na2C2 on hydrolysis producesethyne.
Mg2C3 on hydrolysis gives propyne.
54. Answer (3)
Hint : Dehydrohalogenation
Sol. :
(ii) K/dry ether
CH Cl3
(i) NaNH2
Cl
Br
Br
Br
CH3
55. Answer (2)
Hint : FeCl3 is a Lewis acid.
Sol. : Cl — Cl + FeCl3L.B L.A
56. Answer (3)
Hint : With alc. KOH both cis and trans butene isformed.
Sol. :
OH
OH
P4
CH3
OH
OH
CH3
,
57. Answer (2)
Hint : Benzene is reactive toward ozonolysis butdoes not react with Br2.
Sol. : Benzene generally does not give additionreaction but it reacts with H2/Pt and showsaddition reaction.
58. Answer (1)
Hint : Alkyne having acidic H.
Sol. : White precipitates of HC C–Ag.
59. Answer (1)
Hint : — NH — C — CH3
O
is ortho/para director.
Sol. : NHCOCH3 NHCOCH3
NO2
(Major product)
HNO /H SO3 2 4
60. Answer (3)
Hint : Check all possible positions available andcheck stereochemistry for each isomer.
Sol. : Here are all the possible isomers.
Br
,
Br
,
CH Br2
Br
,
CH —Br2
Br
CH—Br
CH Br2
,
Br
,
CH Br2
Br
,
CH — Br2
Br
Br
CH — Br2
BrCH2
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-D) (Hints & Solutions)
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PART - C (MATHEMATICS)
61. Answer (3)
Hint : Factorization
Sol. : (sin2x – y2) – (sinx – y) = 0
(sinx – y)(sinx + y – 1) = 0
y = sinx OR sinx = 1 – y
–1 sinx 1 –1 sinx 1
y = –1, 0, 1 –1 1 – y 1
–2 –y 0
0 y 2
y = 0, 1, 2
y = {–1, 0, 1, 2}
62. Answer (4)
Hint : P(E1) + P(E2) + P(E3) = 1
and 0 P(E1), P(E2), P(E3) 1
Sol. :
–1 143
16
23
56 1 2
,6 3
P
1 2 3 1 62 6 6 P P P
3 3 2 3 1 61
6 P P P
10 1
2P 2 3
0 16
P 1 60 1
6P
0 P – 1 – 2 0 3P – 2 – 6 0 6P + 1 6
1 P – 1 2 3P – 4 –1 6P 5
2 43 3
P 1 56 6
P
63. Answer (1)
Hint : 2sin cos 2 cos 12 2
Sol. : ( ) sin 2 cos4 8
f x x x
sin 2 cos8 2 8
x x
cos2 cos8 8
x x
22cos cos 18 8
x x
21 9
2 cos8 4 8
x
min9
( ( ))8
f x
64. Answer (3)
Hint : Venn diagram formula
Sol. :1 1 13 6 6
A
B
16
C
1 1 12 6 3
P(A B C)
= P(A) + P(B) + P(C) – P(A B) – P(B C)
– P(C A) + P(A B C)
1 1 1 3 2
( )3 2 6 6 3
P A B C
2 1 1 1 13 2 3 2 6
8 6 4 6 2 112 6
1 1 1( )
6 3 2P E
65. Answer (3)
Hint : sine law
Sol. : 2b2 = a2 + c2
2sin2B = sin2A + sin2C
2[1 – cos2B] = [1 – cos2A] + [1 – cos2C]
2cos2B = cos2A + cos2C
cos2B – cos2A = cos2C – cos2B
Test - 6 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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2sin(A + B) sin(A – B)
= 2sin(B + C) sin (B – C)
sinC [sinA cosB – cosAsinB]
= sinA[sinB cosC – cosBsinC]
cotB – cotA = cotC – cotB
2cotB = cotA + cotC
cotA, cotB, cotC are in A.P.
tanA, tanB, tanC are in H.P.
66. Answer (2)
Hint : D = 0
Sol. : For exactly one solution,
D = 0 a2 = 4b
4 4( )
11 11 121P E
67. Answer (2)
Hint : Check domain
Sol. : tan3x = tanx
3 (2 1) , (2 1)2 2
x n x n
3x = n + x
3 5 7 11, , , , ,
2 2 6 6 6 6 x
2 n
x
3
0, , , , 22 2 x
x = 0, , 2
Sum = 3
68. Answer (2)
Hint : P(E) = 1 – P(A B)
Sol. : P(E) =
5 54
2 21
62 2
= 60 60 24
1180
= 96
1180
= 84 7
180 15
69. Answer (4)
Hint : Perpendiculartan
Base
Sol. :
10 m x
H
45° 60°
tan45 , tan6010
H Hx x
10 ,3
HH x x
103
HH
11 10
3
H
10 3 10 3( 3 1)
23 1
H
5(3 3)mH
70. Answer (2)
Hint : Circular permutations
Sol. :4
45 4( )
1
nn CP E
n
5( 4)( 5)( 6)( 7)( 1)( 2)( 3) 4
n n n n n
n n n n
( 5)( 6)( 7)( )
( 1)( 2)( 3)n n n
P En n n
71. Answer (1)
Hint : sin2x = 1, cos2x = 0
Sol. : sin2x – cos2x = 1
sin2x – 1 + sin2x = 1
sin2x = 1 and cos2x = 0
2 2
1
(sin cos )
n
r r
r
x x
= (sin2x + sin4x + ... + sin2nx)
+ (cos2x + cos4x + ... + cos2nx)
= n + 0 = n
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-D) (Hints & Solutions)
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72. Answer (2)
Hint : 2 2 21( ) ix x
n
Sol. : 22
1122
xx
N N
2 21 1 1 22x x x x
21 100 80 20 x
2 20 1001
10 100
= 1
73. Answer (3)
Hint :( )( )
sin2
A s b s cbc
( )cos
2
A s s abc
Sol. : 22cos cos 4sin2 2 2
B C B C A
22sin cos 4sin2 2 2A B C A
cos 2sin2 2
B C A
cos cos sin sin 2sin2 2 2 2 2
B C B C A
( ) ( )s s b s s cac ab
( )( ) ( )( )s a s c s a s bac ab
( )( )
2s b s c
bc
2 2b c
b c aa
74. Answer (1)
Hint : P(E) = 1 – P(X = 1) – P(X = 0)
Sol. : S = {HHH, HHT, HTH, THH, HTT, THT, TTH}
E = {HTT, THT, TTH}
( ) 3 4( ) 1 1
( ) 7 7n E
P En S
75. Answer (1)
Hint : sec2 = 1 + tan2
Sol. : R = sec2Asec2B + tan2Atan2B +
2 secAsecBtanAtanB – sec2Atan2B –
sec2Btan2A – 2 secAsecBtanA tanB
= sec2A(sec2B – tan2B) +
tan2A(tan2B – sec2B)
= sec2A – tan2A = 1
76. Answer (3)
Hint : Venn diagram
Sol. : E
55
25
H
Percentage of population which readexactly one newspaper is
( ) ( )n E H n E H
= n(E) – n(E H) + n(H) – n(E H)
= 10 + 30 – 2 x 5 = 30%
77. Answer (1)
Hint :2
2 2( )ixx
N
Sol. : x4 + x5 = 30 – (2 + 4 + 9) = 15 ...(i)
2 22 4 54 16 81
36 9.25
x x
x42 + x5
2 = 125 ...(ii)
From (i) and (ii) gives,
x4 = 5, x5 = 10
x4 : x5 = 1 : 2
78. Answer (3)
Hint : Solve equation
Sol. :sin1 sin 22
sin cos2
sin2(1 sin ) 2
2sin cos cos2 2 2
Test - 6 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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22(1 sin ) 2sin2
2(1 + sin) = 1 – cos
2sin = –(1 + cos)
24 sin cos 2 cos2 2 2
2 sin cos2 2
tan 12
79. Answer (1)
Hint : Venn diagram
Sol. : The probability of solving the question by
three students A, B and C are 2 1
,3 3
and 16
respectively.
2 1 1( ) , ( ) , ( )
3 3 6 P A P B P C
Probability that the question is solved byexactly two students is
P(E) = ( ) ( ) P A B C P A B C
( ) P A B C
= ( ) ( ) ( ) ( ) ( ) ( ) P A P B P C P A P B P C
( ) ( ) ( )P A P B P C
= 1 1 1 2 2 1 2 1 5 53 3 6 3 3 6 3 3 6 18
80. Answer (3)
Hint : cosine law
Sol. :2 2 26 (3 5) 3
cos2 6 3 5
36 45 9 2
36 5 5
1
tan 2 12
1 1tan
2 3
tan tan8 tan tan
6
8
6
81. Answer (4)
Hint : Adding a number in every observation doesnot change .
Sol. :2( )ix x
N
11 2 3 ... 10 11
10 2 x
25 6 7 ... 14 95 19 11
410 10 2 2
x
1 = 2
82. Answer (2)
Hint : Square both sides
Sol. : 1 + sin + cos = sincos
sin + cos = sincos– 1
Squaring on both sides,
1 + 2sincos
= 1 + sin2cos2 – 2sincos
212sin2 sin 2
4
sin2(sin2 – 8) = 0
3sin2 0 , ,
2 2
Only and 32
satisfy the original equation.
3 5Sum
2 2
83. Answer (4)
Hint : Combinations
Sol. :18 17
2 235
2
9 17 17 8 17( )
35 17 35
C CP E
C
84. Answer (2)
Hint : cosine law
Sol. :2 2 2
cos2
b c aA
bc
b2 – 2bc cosA + c2 – a2 = 0
2 4 2 8 0b b
2( 2 2) 0 2 2 b b
All India Aakash Test Series for JEE (Main)-2020 Test - 6 (Code-D) (Hints & Solutions)
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85. Answer (3)
Hint : Sum of digits should be divisible by 3.
Sol. : If the sum of digits of a number is divisible by3, then the number is divisible by 3.
The number should be formed by using
1, 2, 3, 6, 9 or 2, 3, 4, 6, 9
6
5
2 5 2 1( )
6 35P E
C
86. Answer (3)
Hint : sin + sin( + ) + sin( + 2) + ... upton terms
Sol :13
1
(sin cos )
r
r r
13sin
2[sin7 cos7 ] 0
sin2
13
sin 02
or tan7 = 1 and sin 02
2 4 6, ,
13 13 13
5 9 13, , ,
28 28 28 28
87. Answer (2)
Hint : Counting
Sol. : 12 11 11( )
52 51 221P E
88. Answer (3)
Hint : Formula
Sol. : Apply direct formula
89. Answer (3)
Hint : cosine law
Sol. : Triangle DEF is called the pedal triangle ofthe triangle ABC.
Circumradius of pedal triangle = 12
(Circumradius of triangle ABC) = 2R
cosC = 2 2 2
2 a b c
ab
= 25(3 1 2 3) 50 100
25 2( 3 1)8
4
= 50( 3 1) 1
50 2( 3 1) 2
C = 45°, similarly B = 30°, A = 105°
5 512sin 222
cR
C
5 5 2
2 42 2 R
R
90. Answer (4)
Hint : 5cos(2 ) cos2
Sol. : As cos(25) = cos 3264
= cos2
= 0
cos cos2 cos22 ... cos29 = 0