Mock Test - 3 (Code-A) (Answers) All India Aakash …...All India Aakash Test Series for JEE (Main)-2019 Mock Test - 3 (Code-A) (Hints & Solutions) 4/11 20. Answer (4) Hint : It will

Mock Test - 3 (Code-A)(Answers) All India Aakash Test Series for JEE (Main)-2019

1/11

1. (1)

2. (2)

3. (1)

4. (2)

5. (3)

6. (2)

7. (3)

8. (2)

9. (2)

10. (3)

11. (4)

12. (4)

13. (1)

14. (4)

15. (3)

16. (2)

17. (2)

18. (2)

19. (4)

20. (4)

21. (2)

22. (1)

23. (2)

24. (4)

25. (2)

26. (1)

27. (3)

28. (4)

29. (1)

30. (1)

PHYSICS CHEMISTRY MATHEMATICS

31. (3)

32. (2)

33. (1)

34. (4)

35. (4)

36. (3)

37. (2)

38. (4)

39. (1)

40. (4)

41. (4)

42. (4)

43. (3)

44. (3)

45. (3)

46. (2)

47. (1)

48. (4)

49. (4)

50. (4)

51. (3)

52. (2)

53. (1)

54. (3)

55. (3)

56. (4)

57. (3)

58. (2)

59. (3)

60. (2)

61. (2)

62. (1)

63. (3)

64. (2)

65. (4)

66. (1)

67. (4)

68. (2)

69. (1)

70. (1)

71. (1)

72. (4)

73. (2)

74. (3)

75. (1)

76. (2)

77. (4)

78. (1)

79. (1)

80. (2)

81. (3)

82. (4)

83. (2)

84. (4)

85. (2)

86. (4)

87. (1)

88. (4)

89. (3)

90. (1)

Test Date : 24/03/2019

ANSWERS

MOCK TEST - 3 Code-A

All India Aakash Test Series for JEE (Main)-2019

All India Aakash Test Series for JEE (Main)-2019 Mock Test - 3 (Code-A) (Hints & Solutions)

2/11

1. Answer (1)

Hint : ∫A y dx

Sol. : A y dx ∫ , Also, 2R – BR2 = 0

2B

R

32

0

11

3 3

R

Bxx R ⇒

2. Answer (2)

Hint : 2

(1 sin )

uR

g

Sol. :2

max( )

(1 sin )

uR

g

(Rmax

) on ground 2

u

g

3. Answer (1)

Hint : Famous question of dog cat problem.

Sol. : Deduce the question like

u

v

d

O

2 2

13 24 13

144 6

vdt

v u

4. Answer (2)

Hint : Write down expression for displacement and

acceleration.

Sol. :sec

ratiol

xa

sec

sin cos

lx

g g

∵ 0dx

d ⇒

cos2 sin2 0

1

tan2 3

= 60°

PART - A (PHYSICS)

5. Answer (3)

Hint : Normal force will provide neccessary

contripetal force.

Sol. : N

mg

N cos = mg

N sin = ma

a = g tan6. Answer (2)

Hint : Final momentum is zero.

Sol. : Before hitting the surface the ball acquires the

momentum p�

because of gravitational force

only.

From thereafter

Igravitational

= 2 3 42 2 2 2 ...p ep e p e p e p

� � � � �

1 1 12 1 ....

2 4 8p ep

⎛ ⎞ ⎜ ⎟⎝ ⎠

�

12 2

2I p p

� �

3I p⇒ �

gravitation 3I p⇒ �

7. Answer (3)

Hint : Linear momentum and mechanical energy is

conserved.

Sol. :

2 2

21

2 6 2

p pkx

m m

2

24

3

pkx

m

33

2 2

x x kp mk v

m ⇒

8. Answer (2)

Hint : Apply rigid body dynamics.

Sol. : T – f = ma

F – f – T = 2ma

fR = I2a

R

2

2 3

Fa

m m m

Mock Test - 3 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019

3/11

9. Answer (2)

Hint : Angular momentum is conserved about point Q.

Sol. : Conservation of angular momentum about Q.

2 2

0

3 2 12

Mv lML Ml⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

0

2

v

l

10. Answer (3)

Hint : .dw F ds��

Sol. : Field is perpendicular to displacement.

11. Answer (4)

Hint : Weight of liquid displaced is equal to weight

of ice cube.

Sol. : When water is lighter than liquid, it forms a

layer over the liquid.

12. Answer (4)

Hint : Apply Bernoulli's equation.

Sol. :2

0 0

16 3

2P gh v

2v gh

13. Answer (1)

Hint :eq

k

m

Sol. : 1 2

1 2( )

k k

m k k

10 rad/sAmplitude of point A = 6 cm

So vmax

= A

vmax

= 6 10 cm/s

14. Answer (4)

Hint : Case of resonance column tube.

Sol. : 14

l e

2

3

4l e

e = 0.9 cm and 50 cm2

So, third resonance occurs at

l3

= 74.1 + 50

= 124.1 cm

15. Answer (3)

Hint : = 1 +

2 +

3

Sol. :

30°

30°

60°

30°

16. Answer (2)

Hint : Wb = U + H

Sol. : 2

0

1 2

2 3i

CU V

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

0

1

2f

U CV

20

battery3

CVW

2

2 2 0

0 0

2 1

3 2 6

CVH CV CV

⎛ ⎞ ⎜ ⎟⎝ ⎠

17. Answer (2)

Hint : Rext

= rint

, for maximum power consumption.

Sol. : For maximum power consumption, circuit

external resistance must be equal to internal

resistance.

22

3

rR

3

rR

18. Answer (2)

Hint : Compare with PVn = constant

Sol. : P2 V2 Vn = constant.

∵ 1 1

021

12

n

⎡ ⎤⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦

4

3n

19. Answer (4)

Hint :1 1 1 v u f

Sol. :

15 cm

P

Q

Q

P

10 cm

10 cm

Image of Q will be formed at Qbut P will be

formed P . 1 1 1

15 5v

v = +7.5 cm

length = 10 – 7.5 = 2.5 cm


4/11

20. Answer (4)

Hint : It will be equatorial plane.

Sol. : It will be equatorial plane and field given by

d

3

2 2 2

2

( )

KQaE

a d

3 2

2 2 2

2 2

8( )

KQa KQ

aa d

3

3 2 2 28 ( )a a d

1

2 2 22 ( )a a d

2 2 2

2 2 2

4

3

a a d

y z a

⇒

21. Answer (2)

Hint : Potential drop must balance

Sol. : 3VAC

V

1

3

AC

AB

22. Answer (1)

Hint : Typical case RC circuit

Sol. : /1

3 3

t RCCE CEQ CE e

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

/21

3 3

t RCCE CEQ e

23. Answer (2)

Hint : Draw pattern of field

Sol. : Field variation will be like

24. Answer (4)

Hint : Temperature gap will reduce exponentially.

Sol. : Ratio of specific heat capacity 25 5

15 3 .

( ) 50 50

( ) 50 30

t

A

t

B

T t e

T t e

ln2So

(1hr)

Temperature gap will reduce exponentially.

After 1 hr, temperature gap will be 20°C.

575 20

8A

T

= 62.5°C

25. Answer (2)

Hint : cosR

Z

Sol. : VL = 132 V

176cos

220

R

Z

= 0.8

26. Answer (1)

Hint :2

1 1 2 2

dAA A

dt

Sol. : 1 2 21 0

2 0

2 1

[ ]t t t tN

N e e N e e

For maximum, 2 ln20

dNt

dt ⇒

27. Answer (3)

Hint :o

e

fm

f

Sol. : First image will be formed at focal plane of

objective fe = 5 cm, f

o = 25 cm.

1 1 1

20 5u

1 1 1

5 20u

u = – 4 cm

Distance between lenses

l = 25 + 4 = 29 cm

28. Answer (4)

Hint : 2

0 2 2

1 2

1 1E E Z

n n

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎣ ⎦; E

0 = 13.6 eV


5/11

Sol. : 2

0 2 2

1 2

1 1E E Z

n n

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎣ ⎦

2 1 113.6 68

4 9Z

⎡ ⎤ ⎢ ⎥⎣ ⎦

6Z

2 1Now, 13.6 1

9K Z

⎡ ⎤ ⎢ ⎥⎣ ⎦

813.6 36 435.2 eV

9K⇒

29. Answer (1)

Hint : Velocity has to be along AC��

PART - B (CHEMISTRY)

31. Answer (3)

Hint : CO + C No reaction

CO2 + C 2CO

Sol. : Let volume of CO2 = x L

Volume of CO = (2 – x) L

CO2 + C 2CO

on heating, total volume

2x + 2 – x = 3

2 + x = 3

x = 1

Volume of CO2 = 1 L

Volume of CO = 1 L

32. Answer (2)

Hint : Correct order is

Tb > T

c

Sol. : b

aT

Rb

c

8aT

27Rb

33. Answer (1)

Hint : Mol of Hydrogen = PV

nRT

atom of H = A

PV2 N

RT

Sol. : No. of H-atom

232 6.023 10 1 2

0.0821 300

= 9.78 x 1022 atom.

Energy required to excite 1 atom in 3rd orbit is

19 1 113.6 1.6 10

1 9

⎛ ⎞ ⎜ ⎟⎝ ⎠

for 9.78 x 1022 atom is

22 199.78 10 13.6 1.6 10 8

1000 9

= 189.17 kJ

Bond dissociation energy of

H2 =

2 1436

0.082 300

= 35.45

Total energy = 224.62 kJ.

34. Answer (4)

Hint : Order is w > x > z > y

Sol. : Both Mg2+ and Ne has Noble gas configuration

bet Zeff

for Mg2+ is very high due to high

number of

proton into nucleus.

35. Answer (4)

Hint : Statement S1, S

3 and S

4 are incorrect.

Sol. :

O OO

H

Hy

O OO

F

Fx

Bond length : y > x

O2

is paramagnetic.

36. Answer (3)

Hint : On increasing P, reaction moves forward only

when ng = –ve

Sol. : For endothermic reaction, as T increases

reaction moves forward.

For 2 2 2

N (g) 2O (g) 2NO (g), ��⇀↽��

H = +ve,ng = –ve

Sol. : Velocity has to be along AC��

8 4

2k

k = 4 m/s

ˆ ˆ8 4 4 5 m/sv i j

30. Answer (1)

Hint : 2d RH

Sol. : 1620 = d2

2d Rh

h = 40 m


6/11

37. Answer (2)

Hint : 2

2S 2S

Ca(OH) Ca 2OH ��⇀↽��

pOH = –log10

[2S]

pH = 14 – pOH at 25°C.

Sol. : 4S3 = 1.08 × 10–4 = 108 × 10–6

S = 3 × 10–2

2S = 6 × 10–2

pOH = –log10

(6 × 10–2)

pOH = 1.22

pH = 12.78

38. Answer (4)

Hint : Cations as well as the electrons are

ammoniated.

Sol. : W + (x + 2y)NH3 [W(NH

3)x]2+ + 2[e–(NH

3)y]–

39. Answer (1)

Hint : Tert-butyl cation is very stable.

Sol. :+

+

+ CO + AlCl4

–AlCl

3

O

Cl

+

40. Answer (4)

Hint : 1eq

2

k 1For M N, K

k 3 ��⇀

↽��

At t = 0, Q = 3

so reaction moves backward

Sol. :1 x 3 x

M N ��⇀↽��

3 x 1

1 x 3

9 – 3x = 1 + x

8 = 4x

x = 2

[M]eq

= 3 [N]eq

= 1

41. Answer (4)

Hint : CH3CH

2CHO does not give positive Iodoform

test.

Sol. : x =

y =

z = HCHO

O

42. Answer (4)

Hint : CH – CH – 3 2 2

CH – CH – C – OH

OH O

*

show optical activity.

Sol. :

CH – CH – 3 2 2

CH – CH – C – Cl

CH – CH – 3 2 2


CH – CH – 3 2 2


Cl

OH

Cl

O

O

O

H O2

aq.KOH

43. Answer (3)

Hint : Brx :

BrCH3

Y :

CH CH CH2 2 3

CH3

Z :

Sol. :

Br2/Fe CH

3Cl/AlCl

3

Br Br

CH3

(X)

(Y)

(Z)

CH3

CH3

Br + Br – CH CH CH2 2 3

CH CH CH2 2 3

Na/ether

44. Answer (3)

Hint : HCOOH reduces the Tollen's test.

Sol. : CH3CHO, HCOOH and

C H

O

can reduce Tollen's reagent.

45. Answer (3)

Hint : X is sucrose.

Sol. : X Non reducing

X 1 mole of glucose + 1 mole fructose

invert sugar

1 mole

H O3

+


7/11

46. Answer (2)

Hint : X Buna - S

Sol. :

C H

6 5

1, 3 Butadiene + Styrene

(CH CH CH CH CH )2 2 2 n

Buna - S

Na

47. Answer (1)

Hint : Equanil is tranquilizer.

Sol. : Cimetidine and Ranitidine are examples of

antacids.

48. Answer (4)

Hint : O and P are positional isomers.

Sol. :

(M) (N)

(Meso) (Optically Active)

Me Me

Me Me

Br Br

Br HH Br

H H

(Q) (R)

CH3

CH3CH

3CH

3

CH CH

Et Et

Me Et

Me MeH H

Et Me

49. Answer (4)

Hint : X : (a) NaNH2

(b) C2H5Br

Y : (a) NaNH2

(b)

Br

V =

C C

H H

C C

H

H

W =

Sol. : Path A provide better yield because in A,

less hindered alkyl halide is used so chance

of elimination is less.

50. Answer (4)

Hint : In pyrrole, lone pair is involved in aromaticity.

Sol. :

CH3

CH3

CH3

CH3

N

Lone pair is not in conjugation with electron

of benzene so this is very strong base.

51. Answer (3)

Hint : [Zn+2] increases while [Cu2+] decreases

Sol. : As cell starts working

then, Zn Zn2+, so [Zn2+] increases

but Cu2+ +2e– Cu, so [Cu2+] decreases

52. Answer (2)

Hint :

2

100100

99

2100 100

ARate

(Rate) (A )

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

Sol. : Concentration of A after 100

s9

is

t

1 1001 0.09

A 9

t

11 1

A

At = 0.5

Concentration of A after 100 s

t

11 0.09 100

A

t

t

11 9, A 0.1

A

2100

(Rate)after0.59 25

(Rate)after100 0.1

⎛ ⎞ ⎜ ⎟⎝ ⎠

53. Answer (1)

Hint : P1 < P

2 < P

3 < P

4

Sol. : As temperature increases, vapor pressure

increases.


8/11

PART - C (MATHEMATICS)

54. Answer (3)

Hint : i = 2, it means structure may be NaCl type,

CsCl type and ZnS type.

Sol. : = i CRT

= 60 R

T = 300 K

9.6

96C 0.1

1L

60R = 0.1 x i x R x 300

i = 2

3

23 24

Z Md 8g / cm

6 10 8 10

80 6 0.1 8Z

96

Z = 4

If body centre is vacant, it means it is ZnS

type structure.

55. Answer (3)

Hint : From the above formula it is clear that x is 2.

Sol. : M+x is Ni2+ that form low spin complex with

CN– but high spin complex with H2O and Cl–.

56. Answer (4)

Hint : P = I2

Q = KIO3

61. Answer (2)

Hint : Substitution method

Sol. : (5x4 + 4x3 + 3x2)(x5 + x4 + x3)5dx

Let x5 + x4 + x3 = t

I = t5dx = 5 4 3 6( )

6

x x xc

62. Answer (1)

Hint : Make cases

Sol. :

Case 1: Non-zero integral solutions = 7C2222

= 168

Case 2: If exactly one of x, y, z is zero, number of

solutions = 3 2 2 7C1 = 84

Case 3: If exactly two of x, y, z is zero, number of

solutions = 3 2 = 6

Total solutions = 258

63. Answer (3)

Hint : |adjA| = |A|n–1

Sol. :

∵

2 2

2 2

2

adj 1

1 1

ax b x ab b xa b x

x b bx a a x x ab

x a x bx a ax b

⎡ ⎤ ⎛ ⎞⎡ ⎤⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥⎢ ⎥⎜ ⎟⎢ ⎥ ⎣ ⎦ ⎢ ⎥⎝ ⎠ ⎣ ⎦

D12 = D

2

64. Answer (2)

Hint : 1 + ei = 2cos2

2i

e

Sol. : Im(1 + ei)2n =

2

2cos sin2

n

n⎛ ⎞ ⎜ ⎟

⎝ ⎠

Sol. :

4 2 4 2 4 4 2 2(P)

KMnO H SO KI K SO MnSO H O I

4 2 3 2(Q)

KMnO H O KI KIO MnO KOH

57. Answer (3)

Hint : SO2 shows temporary bleaching while

Cl2 + H

2O show permanent bleaching action.

Sol. :4 6 2

2 42SO H O SO

(Strong reducing agent)

Cl2 + H

2O Cl– +(O) (Strong oxidising agent)

58. Answer (2)

Hint : Mond process is used for the refining of Ni.

Sol. : Ni + CO Ni(CO)4

Volatile compound

sp3, diamagnetic.

59. Answer (3)

Hint : Activity of enzyme increases in presence of

activators.

Sol. : Activators and co-enzymes both increase the

activity of enzymes.

60. Answer (2)

Hint : H3O+, CH

4 and H

3PO

4 has sp3 hybridised

central atom.

Sol. : H3O+ sp3

CH4 sp3

H3PO

4 sp3


9/11

65. Answer (4)

Hint : Telescopic method

Sol. : S = 2 + 4 + 7 + 11 + ...

S = 2 + 4 + 7 + ... + Tn

Tn = 2 + (2 + 3 + 4 + ...)

Tn = 1 + (1+ 2 + 3 + 4 + ...)

Tn =

21 21 ( 1)

2 2

n nn n

Sn = 2( 3 8)

6

nn n

S25

= 25

(708) 29506

66. Answer (1)

Hint : D 0

Sol. : x2 – 10x + (y2 + 15) = 0

D 0

100 – 4(y2 + 15) 0

y [ 10, 10]

Also, 10x – x2 – 15 0

x [5 10, 5 10]

n(A B) = 2

67. Answer (4)

Hint : Multiply terms

Sol. : 11 = (0, 11), (1, 10), (2, 9), (3, 8), (4, 7), (5, 6)

= 6 cases

Also, 11 = (1, 2, 8), (1, 3, 7), (1, 4, 6), (2, 3, 6),

(2, 4, 5), (1, 2, 3, 5)

= 6 cases

68. Answer (2)

Hint : Property of inverse matrices

Sol. : (B–1A–1)–1 = AB = 2 3

1 4

⎡ ⎤⎢ ⎥⎣ ⎦

69. Answer (1)

Hint : Find 1 1 1

x y z

Sol. :

7 111 sin sin

1 1 1 6 60

x y z k

70. Answer (1)

Hint : D < 0

Sol. : 4x2 + 4(a + 2)x – 3a – 2 > 0

D < 0

(a + 2)2 + 3a + 2 < 0

a (–6, –1)

Probability = 5 1

15 3

71. Answer (1)

Hint : Solve pairwise

Sol. : S = (–2)[4 + 12 + 20 + .... 60]

= (–2)(4)(1 + 3 + 5 + ... 15)

= –512

72. Answer (4)

Hint : Property of inverse

Sol. : 1 1

2

8sin 2 tan 4

1 16

xx

x

1 1

2

8sin 2 tan 4

1 16

xx

x

4x 1

x 1

4

73. Answer (2)

Hint : Parametric form of a line

Sol. : BC: 3x – 11y – 37 = 0

Slope of PD is m = 11

3

3 11 390cos , sin ,

2130 130PD

(x, y) 1 390 3 7 390 11

,2 2 2 2130 130

⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

1 3 3 7 11 3

,2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

60°

P x y( , )

A(1, 2)

C(5, –2)B(–6, –5) 1 7,

2 2D

⎛ ⎞⎜ ⎟⎝ ⎠

74. Answer (3)

Hint : S + L = 0

Sol. : Equation of circles is

x2 + (y – a)(y + a) + x = 0

2 2

2 2

2 4x y a

⎛ ⎞ ⎜ ⎟⎝ ⎠


10/11

22

2

| |

4 1

ca

m

2

2 2 2 2(1 ) 24

a m c c

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 + 8c + 20a2 – 4c2 = 0

1

2 = 20a2 – 4c2

As 2g1g

2 + 2f

1f2 = c

1 + c

2

5a2 – c2 = –a2

c2 = 6a2

75. Answer (1)

Hint : Exact differential equation

Sol. :2 4

4

( 1)

2 ( 1)

dy y y

dx xy y

dy(2xy5 – 2xy) = y2(y4 + 1)dx

2 2

2 2 4

2 2xydy y dx y dx xydy

x x y

2

2

1yd d

x xy

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

2

12

y

x xy

76. Answer (2)

Hint : Parabolas are symmetric about y = x

Sol. : Symmetric about y = x

1 1

1

1 1 171 ,

2 2 4

dyy x

dx y ⇒

Also, 2 2 2

1 172 1 ,

2 4

dyx x y

dx ⇒

21 1 17 15 2

22 2 4 8

r⎛ ⎞ ⎜ ⎟⎝ ⎠

Area = 225

32

77. Answer (4)

Hint : Graph

Sol. :

cos (sin )–1

x

sinx

78. Answer (1)

Hint : Dot product

Sol. : c

�

= (a�

b�

)

c

�

c�

= 2

3

2

Also, |c�

|2 = 2|a�

b�

|2

3 = 2 2 29( sin )

4a b

sin2 = 2

3

= 1 1

cos3

79. Answer (1)

Hint : Asymptotes are perpendicular.

Sol. : As asymptotes are perpendicular,

11

2

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

– = 2

Aslo,

3 1 0

1 2 0c

c

+ 3 = –7

= 14, = –7

80. Answer (2)

Hint : Factorization

Sol. :5

0

(1 cos ) tan (1 cos )limx

x x x x

x

2 30

(1 cos ) ( tan )lim

1

6

x

x x x

x x

81. Answer (3)

Hint : T = S1

Sol. : T = S1

22

4 4

xh hyk k

2

2

12 2 4

h k hk

3 1,

10 20h k

6

xy


11/11

82. Answer (4)

Hint : D 0

Sol. : (ysecx – sinx)2 = sin2x + 3

4

y2 tan2x – 2y tanx + y2 – 3

4 = 0

D 0

y2 – 3

4 – 1 0

y2 7

4

y 7 7,

4 4

⎡ ⎤⎢ ⎥⎣ ⎦

83. Answer (2)

Hint : P1 + P

2 = 0

Sol. : Let the plane be

ˆ ˆ ˆ ˆ ˆ ˆ[ (2 ) 3] [ ( 2 3 ) 2] 0r i j k r i j k � �

ˆ ˆ ˆ[( 2) (2 1) ( 3 1) ] (2 3) 0r i j k �

( + 2) – (2 + 1) + 3(–3 – 1) = 0

= 1

5

Plane is 9x + 3y – 2z – 17 = 0

84. Answer (4)

Hint :dy

dx

Sol. : 22 2

xdye x

dx

At (0, 1), 2dy

dx

y = 2x + 1

85. Answer (2)

Hint :1 2 3

( ) 0n n n � � �

Sol. :

ˆ ˆ ˆ

ˆ ˆ ˆ2 1 1 2 5

3 4 2

i j k

i j k

(2)(1) – 1(–) – 5(1) = 0

= 3

86. Answer (4)

Hint : Maxima/Minima

Sol. : Let minimum distance from centre = D

D2 = (t2)2 + (2t + 3)2

2

0dD

dt

t = –1

87. Answer (1)


Sol. :

/4

2

0

(sin cos )

9 4(sin cos )

x x dx

x x

∫

Let sinx – cosx = t

= 0

2

1

1log5

129 4

dt

t

∫

88. Answer (4)

Hint : Find tangent

Sol. : Tangent is 4x + 12y – 35 = 0

Area =

11

17

4

1 3 35 17(3 2)

2 2 4 4x dx

⎛ ⎞ ⎜ ⎟⎝ ⎠∫

= 9 27 9

2 8 8

89. Answer (3)

Hint : Definition

Sol. : Contrapositive of (p q) r is

~r ~p ~q

90. Answer (1)

Hint :2

2 2( )ix

xN

Sol. : 1, 3, 5 ... 39

2

2

20

400

i

i

xx

n

x

N

As 12 + 32 + ... + 392

= 40 41 81 4 20 21 41

6 6

= 20 41 13

2 = 133

� � �

Mock Test - 3 (Code-B)(Answers) All India Aakash Test Series for JEE (Main)-2019

1/11

1 (1)

2 (1)

3 (4)

4 (3)

5 (1)

6 (2)

7 (4)

8 (2)

9 (1)

10 (2)

11 (4)

12 (4)

13 (2)

14 (2)

15 (2)

16 (3)

17 (4)

18 (1)

19 (4)

20 (4)

21 (3)

22 (2)

23 (2)

24 (3)

25 (2)

26 (3)

27 (2)

28 (1)

29 (2)

30 (1)

PHYSICS CHEMISTRY MATHEMATICS

31 (2)

32 (3)

33 (2)

34 (3)

35 (4)

36 (3)

37 (3)

38 (1)

39 (2)

40 (3)

41 (4)

42 (4)

43 (4)

44 (1)

45 (2)

46 (3)

47 (3)

48 (3)

49 (4)

50 (4)

51 (4)

52 (1)

53 (4)

54 (2)

55 (3)

56 (4)

57 (4)

58 (1)

59 (2)

60 (3)

61 (1)

62 (3)

63 (4)

64 (1)

65 (4)

66 (2)

67 (4)

68 (2)

69 (4)

70 (3)

71 (2)

72 (1)

73 (1)

74 (4)

75 (2)

76 (1)

77 (3)

78 (2)

79 (4)

80 (1)

81 (1)

82 (1)

83 (2)

84 (4)

85 (1)

86 (4)

87 (2)

88 (3)

89 (1)

90 (2)

Test Date : 24/03/2019

ANSWERS

MOCK TEST - 3 Code-B

All India Aakash Test Series for JEE (Main)-2019

All India Aakash Test Series for JEE (Main)-2019 Mock Test - 3 (Code-B) (Hints & Solutions)

2/11

1. Answer (1)

Hint : 2d RH

Sol. : 1620 = d2

2d Rh

h = 40 m

2. Answer (1)

Hint : Velocity has to be along AC��

Sol. : Velocity has to be along AC��

8 4

2k

k = 4 m/s

ˆ ˆ8 4 4 5 m/sv i j

3. Answer (4)

Hint : 2

0 2 2

1 2

1 1E E Z

n n

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎣ ⎦; E

0 = 13.6 eV

Sol. : 2

0 2 2

1 2

1 1E E Z

n n

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎣ ⎦

2 1 113.6 68

4 9Z

⎡ ⎤ ⎢ ⎥⎣ ⎦

6Z

2 1Now, 13.6 1

9K Z

⎡ ⎤ ⎢ ⎥⎣ ⎦

813.6 36 435.2 eV

9K⇒

4. Answer (3)

Hint :o

e

fm

f

Sol. : First image will be formed at focal plane of

objective fe = 5 cm, f

o = 25 cm.

1 1 1

20 5u

1 1 1

5 20u

u = – 4 cm

Distance between lenses

l = 25 + 4 = 29 cm

PART - A (PHYSICS)

5. Answer (1)

Hint :2

1 1 2 2

dAA A

dt

Sol. : 1 2 21 0

2 0

2 1

[ ]t t t tN

N e e N e e

For maximum, 2 ln20

dNt

dt ⇒

6. Answer (2)

Hint : cosR

Z

Sol. : VL = 132 V

176cos

220

R

Z

= 0.8

7. Answer (4)

Hint : Temperature gap will reduce exponentially.

Sol. : Ratio of specific heat capacity 25 5

15 3 .

( ) 50 50

( ) 50 30

t

A

t

B

T t e

T t e

ln2So

(1hr)

Temperature gap will reduce exponentially.

After 1 hr, temperature gap will be 20°C.

575 20

8A

T

= 62.5°C

8. Answer (2)

Hint : Draw pattern of field

Sol. : Field variation will be like

Mock Test - 3 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019

3/11

9. Answer (1)

Hint : Typical case RC circuit

Sol. : /1

3 3

t RCCE CEQ CE e

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

/21

3 3

t RCCE CEQ e

10. Answer (2)

Hint : Potential drop must balance

Sol. : 3VAC

V

1

3

AC

AB

11. Answer (4)

Hint : It will be equatorial plane.

Sol. : It will be equatorial plane and field given by

d

3

2 2 2

2

( )

KQaE

a d

3 2

2 2 2

2 2

8( )

KQa KQ

aa d

3

3 2 2 28 ( )a a d

1

2 2 22 ( )a a d

2 2 2

2 2 2

4

3

a a d

y z a

⇒ 12. Answer (4)

Hint :1 1 1 v u f

Sol. :

15 cm

P

Q

Q

P

10 cm

10 cm

Image of Q will be formed at Qbut P will be

formed P . 1 1 1

15 5v

v = +7.5 cm

length = 10 – 7.5 = 2.5 cm

13. Answer (2)

Hint : Compare with PVn = constant

Sol. : P2 V2 Vn = constant.

∵ 1 1

021

12

n

⎡ ⎤⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦

4

3n

14. Answer (2)

Hint : Rext

= rint

, for maximum power consumption.

Sol. : For maximum power consumption, circuit

external resistance must be equal to internal

resistance.

22

3

rR

3

rR

15. Answer (2)

Hint : Wb = U + H

Sol. : 2

0

1 2

2 3i

CU V

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

0

1

2f

U CV

20

battery3

CVW

2

2 2 0

0 0

2 1

3 2 6

CVH CV CV

⎛ ⎞ ⎜ ⎟⎝ ⎠

16. Answer (3)

Hint : = 1 +

2 +

3

Sol. :

30°

30°

60°

30°

17. Answer (4)

Hint : Case of resonance column tube.

Sol. : 14

l e

2

3

4l e

e = 0.9 cm and 50 cm2

So, third resonance occurs at

l3

= 74.1 + 50

= 124.1 cm


4/11

18. Answer (1)

Hint :eq

k

m

Sol. : 1 2

1 2( )

k k

m k k

10 rad/sAmplitude of point A = 6 cm

So vmax

= A

vmax

= 6 10 cm/s

19. Answer (4)

Hint : Apply Bernoulli's equation.

Sol. :2

0 0

16 3

2P gh v

2v gh

20. Answer (4)

Hint : Weight of liquid displaced is equal to weight

of ice cube.

Sol. : When water is lighter than liquid, it forms a

layer over the liquid.

21. Answer (3)

Hint : .dw F ds��

Sol. : Field is perpendicular to displacement.

22. Answer (2)

Hint : Angular momentum is conserved about point Q.

Sol. : Conservation of angular momentum about Q.

2 2

0

3 2 12

Mv lML Ml⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

0

2

v

l

23. Answer (2)

Hint : Apply rigid body dynamics.

Sol. : T – f = ma

F – f – T = 2ma

fR = I2a

R

2

2 3

Fa

m m m

24. Answer (3)

Hint : Linear momentum and mechanical energy is

conserved.

Sol. :

2 2

21

2 6 2

p pkx

m m

2

24

3

pkx

m

33

2 2

x x kp mk v

m ⇒

25. Answer (2)

Hint : Final momentum is zero.

Sol. : Before hitting the surface the ball acquires the

momentum p�

because of gravitational force only.

From thereafter

Igravitational

= 2 3 42 2 2 2 ...p ep e p e p e p

� � � � �

1 1 12 1 ....

2 4 8p ep

⎛ ⎞ ⎜ ⎟⎝ ⎠

�

12 2

2I p p

� �

3I p⇒ �

gravitation 3I p⇒ �

26. Answer (3)

Hint : Normal force will provide neccessary

contripetal force.

Sol. : N

mg

N cos = mg

N sin = ma

a = g tan27. Answer (2)

Hint : Write down expression for displacement and

acceleration.

Sol. :sec

ratiol

xa

sec

sin cos

lx

g g

∵ 0dx

d ⇒

cos2 sin2 0

1

tan2 3

= 60°

28. Answer (1)

Hint : Famous question of dog cat problem.


5/11

PART - B (CHEMISTRY)

31. Answer (2)

Hint : H3O+, CH

4 and H

3PO

4 has sp3 hybridised

central atom.

Sol. : H3O+ sp3

CH4 sp3

H3PO

4 sp3

32. Answer (3)

Hint : Activity of enzyme increases in presence of

activators.

Sol. : Activators and co-enzymes both increase the

activity of enzymes.

33. Answer (2)

Hint : Mond process is used for the refining of Ni.

Sol. : Ni + CO Ni(CO)4

Volatile compound

sp3, diamagnetic.

34. Answer (3)

Hint : SO2 shows temporary bleaching while

Cl2 + H

2O show permanent bleaching action.

Sol. :4 6 2

2 42SO H O SO

(Strong reducing agent)

Cl2 + H

2O Cl– +(O) (Strong oxidising agent)

35. Answer (4)

Hint : P = I2

Q = KIO3

Sol. :

4 2 4 2 4 4 2 2(P)

KMnO H SO KI K SO MnSO H O I

4 2 3 2(Q)

KMnO H O KI KIO MnO KOH

Sol. : Deduce the question like

u

v

d

O

2 2

13 24 13

144 6

vdt

v u

29. Answer (2)

Hint : 2

(1 sin )

uR

g

Sol. :2

max( )

(1 sin )

uR

g

(Rmax

) on ground 2

u

g

30. Answer (1)

Hint : ∫A y dx

Sol. : A y dx ∫ , Also, 2R – BR2 = 0

2B

R

32

0

11

3 3

R

Bxx R ⇒

36. Answer (3)

Hint : From the above formula it is clear that x is 2.

Sol. : M+x is Ni2+ that form low spin complex with

CN– but high spin complex with H2O and Cl–.

37. Answer (3)

Hint : i = 2, it means structure may be NaCl type,

CsCl type and ZnS type.

Sol. : = i CRT

= 60 R

T = 300 K

9.6

96C 0.1

1L

60R = 0.1 x i x R x 300

i = 2

3

23 24

Z Md 8g / cm

6 10 8 10

80 6 0.1 8

Z96

Z = 4

If body centre is vacant, it means it is ZnS

type structure.

38. Answer (1)

Hint : P1 < P

2 < P

3 < P

4

Sol. : As temperature increases, vapor pressure

increases.

39. Answer (2)

Hint :

2

100100

99

2100 100

ARate

(Rate) (A )

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠


6/11

Sol. : Concentration of A after 100

s9

is

t

1 1001 0.09

A 9

t

11 1

A

At = 0.5

Concentration of A after 100 s

t

11 0.09 100

A

t

t

11 9, A 0.1

A

2100

(Rate)after0.59 25

(Rate)after100 0.1

⎛ ⎞ ⎜ ⎟⎝ ⎠

40. Answer (3)

Hint : [Zn+2] increases while [Cu2+] decreases

Sol. : As cell starts working

then, Zn Zn2+, so [Zn2+] increases

but Cu2+ +2e– Cu, so [Cu2+] decreases

41. Answer (4)

Hint : In pyrrole, lone pair is involved in aromaticity.

Sol. :

CH3

CH3

CH3

CH3

N

Lone pair is not in conjugation with electron

of benzene so this is very strong base.

42. Answer (4)

Hint : X : (a) NaNH2

(b) C2H5Br

Y : (a) NaNH2

(b)

Br

V =

C C

H H

C C

H

H

W =

Sol. : Path A provide better yield because in A,

less hindered alkyl halide is used so chance

of elimination is less.

43. Answer (4)

Hint : O and P are positional isomers.

Sol. :

(M) (N)

(Meso) (Optically Active)

Me Me

Me Me

Br Br

Br HH Br

H H

(Q) (R)

CH3

CH3CH

3CH

3

CH CH

Et Et

Me Et

Me MeH H

Et Me

44. Answer (1)

Hint : Equanil is tranquilizer.

Sol. : Cimetidine and Ranitidine are examples of

antacids.

45. Answer (2)

Hint : X Buna - S

Sol. :

C H

6 5

1, 3 Butadiene + Styrene

(CH CH CH CH CH )2 2 2 n

Buna - S

Na

46. Answer (3)

Hint : X is sucrose.

Sol. : X Non reducing

X 1 mole of glucose + 1 mole fructose

invert sugar

1 mole

H O3

+

47. Answer (3)

Hint : HCOOH reduces the Tollen's test.

Sol. : CH3CHO, HCOOH and

C H

O

can reduce Tollen's reagent.


7/11

48. Answer (3)

Hint : Brx :

BrCH3

Y :

CH CH CH2 2 3

CH3

Z :

Sol. :

Br2/Fe CH

3Cl/AlCl

3

Br Br

CH3

(X)

(Y)

(Z)

CH3

CH3

Br + Br – CH CH CH2 2 3

CH CH CH2 2 3

Na/ether

49. Answer (4)

Hint : CH – CH – 3 2 2


OH O

*

show optical activity.

Sol. :

CH – CH – 3 2 2

CH – CH – C – Cl

CH – CH – 3 2 2


CH – CH – 3 2 2


Cl

OH

Cl

O

O

O

H O2

aq.KOH

50. Answer (4)

Hint : CH3CH

2CHO does not give positive Iodoform

test.

Sol. : x =

y =

z = HCHO

O

51. Answer (4)

Hint : 1eq

2

k 1For M N, K

k 3 ��⇀

↽��

At t = 0, Q = 3

so reaction moves backward

Sol. :1 x 3 x

M N ��⇀↽��

3 x 1

1 x 3

9 – 3x = 1 + x

8 = 4x

x = 2

[M]eq

= 3 [N]eq

= 1

52. Answer (1)

Hint : Tert-butyl cation is very stable.

Sol. :+

+

+ CO + AlCl4

–AlCl

3

O

Cl

+

53. Answer (4)

Hint : Cations as well as the electrons are

ammoniated.

Sol. : W + (x + 2y)NH3 [W(NH

3)x]2+ + 2[e–(NH

3)y]–

54. Answer (2)

Hint : 2

2S 2S

Ca(OH) Ca 2OH ��⇀↽��

pOH = –log10

[2S]

pH = 14 – pOH at 25°C.

Sol. : 4S3 = 1.08 × 10–4 = 108 × 10–6

S = 3 × 10–2

2S = 6 × 10–2

pOH = –log10

(6 × 10–2)

pOH = 1.22

pH = 12.78

55. Answer (3)

Hint : On increasing P, reaction moves forward only

when ng = –ve

Sol. : For endothermic reaction, as T increases

reaction moves forward.

For 2 2 2

N (g) 2O (g) 2NO (g), ��⇀↽��

H = +ve,ng = –ve

56. Answer (4)

Hint : Statement S1, S

3 and S

4 are incorrect.

Sol. : O OO

H

Hy

O OO

F

Fx

Bond length : y > x

O2

is paramagnetic.


8/11

PART - C (MATHEMATICS)

61. Answer (1)

Hint :2

2 2( )ix

xN

Sol. : 1, 3, 5 ... 39

2

2

20

400

i

i

xx

n

x

N

As 12 + 32 + ... + 392

= 40 41 81 4 20 21 41

6 6

= 20 41 13

2 = 133

62. Answer (3)

Hint : Definition

Sol. : Contrapositive of (p q) r is

~r ~p ~q

63. Answer (4)

Hint : Find tangent

57. Answer (4)

Hint : Order is w > x > z > y

Sol. : Both Mg2+ and Ne has Noble gas configuration

bet Zeff

for Mg2+ is very high due to high

number of proton into nucleus.

58. Answer (1)

Hint : Mol of Hydrogen = PV

nRT

atom of H = A

PV2 N

RT

Sol. : No. of H-atom

232 6.023 10 1 2

0.0821 300

= 9.78 x 1022 atom.

Energy required to excite 1 atom in 3rd orbit is

19 1 113.6 1.6 10

1 9

⎛ ⎞ ⎜ ⎟⎝ ⎠

for 9.78 x 1022 atom is

22 199.78 10 13.6 1.6 10 8

1000 9

= 189.17 kJ

Bond dissociation energy of

H2 =

2 1436

0.082 300

= 35.45

Total energy = 224.62 kJ.

59. Answer (2)

Hint : Correct order is

Tb > T

c

Sol. : b

aT

Rb

c

8aT

27Rb

60. Answer (3)

Hint : CO + C No reaction

CO2 + C 2CO

Sol. : Let volume of CO2 = x L

Volume of CO = (2 – x) L

CO2 + C 2CO

on heating, total volume

2x + 2 – x = 3

2 + x = 3

x = 1

Volume of CO2 = 1 L

Volume of CO = 1 L

Sol. : Tangent is 4x + 12y – 35 = 0

Area =

11

17

4

1 3 35 17(3 2)

2 2 4 4x dx

⎛ ⎞ ⎜ ⎟⎝ ⎠∫

= 9 27 9

2 8 8

64. Answer (1)


Sol. :

/4

2

0

(sin cos )

9 4(sin cos )

x x dx

x x

∫

Let sinx – cosx = t

= 0

2

1

1log5

129 4

dt

t

∫

65. Answer (4)

Hint : Maxima/Minima

Sol. : Let minimum distance from centre = D

D2 = (t2)2 + (2t + 3)2


9/11

2

0dD

dt

t = –1

66. Answer (2)

Hint :1 2 3

( ) 0n n n � � �

Sol. :

ˆ ˆ ˆ

ˆ ˆ ˆ2 1 1 2 5

3 4 2

i j k

i j k

(2)(1) – 1(–) – 5(1) = 0

= 3

67. Answer (4)

Hint :dy

dx

Sol. : 22 2

xdye x

dx

At (0, 1), 2dy

dx

y = 2x + 1

68. Answer (2)

Hint : P1 + P

2 = 0

Sol. : Let the plane be

ˆ ˆ ˆ ˆ ˆ ˆ[ (2 ) 3] [ ( 2 3 ) 2] 0r i j k r i j k � �

ˆ ˆ ˆ[( 2) (2 1) ( 3 1) ] (2 3) 0r i j k �

( + 2) – (2 + 1) + 3(–3 – 1) = 0

= 1

5

Plane is 9x + 3y – 2z – 17 = 0

69. Answer (4)

Hint : D 0

Sol. : (ysecx – sinx)2 = sin2x + 3

4

y2 tan2x – 2y tanx + y2 – 3

4 = 0

D 0

y2 – 3

4 – 1 0

y2 7

4

y 7 7,

4 4

⎡ ⎤⎢ ⎥⎣ ⎦

70. Answer (3)

Hint : T = S1

Sol. : T = S1

22

4 4

xh hyk k

2

2

12 2 4

h k hk

3 1,

10 20h k

6

xy

71. Answer (2)

Hint : Factorization

Sol. :5

0

(1 cos ) tan (1 cos )limx

x x x x

x

2 30

(1 cos ) ( tan )lim

1

6

x

x x x

x x

72. Answer (1)

Hint : Asymptotes are perpendicular.

Sol. : As asymptotes are perpendicular,

11

2

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

– = 2

Aslo,

3 1 0

1 2 0c

c

+ 3 = –7

= 14, = –7

73. Answer (1)

Hint : Dot product

Sol. : c

�

= (a�

b�

)

c

�

c�

= 2

3

2

Also, |c�

|2 = 2|a�

b�

|2

3 = 2 2 29( sin )

4a b

sin2 = 2

3

= 1 1

cos3


10/11

74. Answer (4)

Hint : Graph

Sol. :

cos (sin )–1

x

sinx

75. Answer (2)

Hint : Parabolas are symmetric about y = x

Sol. : Symmetric about y = x

1 1

1

1 1 171 ,

2 2 4

dyy x

dx y ⇒

Also, 2 2 2

1 172 1 ,

2 4

dyx x y

dx ⇒

21 1 17 15 2

22 2 4 8

r⎛ ⎞ ⎜ ⎟⎝ ⎠

Area = 225

32

76. Answer (1)

Hint : Exact differential equation

Sol. :2 4

4

( 1)

2 ( 1)

dy y y

dx xy y

dy(2xy5 – 2xy) = y2(y4 + 1)dx

2 2

2 2 4

2 2xydy y dx y dx xydy

x x y

2

2

1yd d

x xy

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

2

12

y

x xy

77. Answer (3)

Hint : S + L = 0

Sol. : Equation of circles is

x2 + (y – a)(y + a) + x = 0

2 2

2 2

2 4x y a

⎛ ⎞ ⎜ ⎟⎝ ⎠

22

2

| |

4 1

ca

m

2

2 2 2 2(1 ) 24

a m c c

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 + 8c + 20a2 – 4c2 = 0

1

2 = 20a2 – 4c2

As 2g1g

2 + 2f

1f2 = c

1 + c

2

5a2 – c2 = –a2

c2 = 6a2

78. Answer (2)

Hint : Parametric form of a line

Sol. : BC: 3x – 11y – 37 = 0

Slope of PD is m = 11

3

3 11 390cos , sin ,

2130 130PD

(x, y) 1 390 3 7 390 11

,2 2 2 2130 130

⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

1 3 3 7 11 3

,2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

60°

P x y( , )

A(1, 2)

C(5, –2)B(–6, –5) 1 7,

2 2D

⎛ ⎞⎜ ⎟⎝ ⎠

79. Answer (4)

Hint : Property of inverse

Sol. : 1 1

2

8sin 2 tan 4

1 16

xx

x

1 1

2

8sin 2 tan 4

1 16

xx

x

4x 1

x 1

4

80. Answer (1)

Hint : Solve pairwise

Sol. : S = (–2)[4 + 12 + 20 + .... 60]

= (–2)(4)(1 + 3 + 5 + ... 15)

= –512

81. Answer (1)

Hint : D < 0


11/11

� � �

Sol. : 4x2 + 4(a + 2)x – 3a – 2 > 0

D < 0

(a + 2)2 + 3a + 2 < 0

a (–6, –1)

Probability = 5 1

15 3

82. Answer (1)

Hint : Find 1 1 1

x y z

Sol. :

7 111 sin sin

1 1 1 6 60

x y z k

83. Answer (2)

Hint : Property of inverse matrices

Sol. : (B–1A–1)–1 = AB = 2 3

1 4

⎡ ⎤⎢ ⎥⎣ ⎦

84. Answer (4)

Hint : Multiply terms

Sol. : 11 = (0, 11), (1, 10), (2, 9), (3, 8), (4, 7), (5, 6)

= 6 cases

Also, 11 = (1, 2, 8), (1, 3, 7), (1, 4, 6), (2, 3, 6),

(2, 4, 5), (1, 2, 3, 5)

= 6 cases

85. Answer (1)

Hint : D 0

Sol. : x2 – 10x + (y2 + 15) = 0

D 0

100 – 4(y2 + 15) 0

y [ 10, 10]

Also, 10x – x2 – 15 0

x [5 10, 5 10]

n(A B) = 2

86. Answer (4)

Hint : Telescopic method

Sol. : S = 2 + 4 + 7 + 11 + ...

S = 2 + 4 + 7 + ... + Tn

Tn = 2 + (2 + 3 + 4 + ...)

Tn = 1 + (1+ 2 + 3 + 4 + ...)

Tn =

21 21 ( 1)

2 2

n nn n

Sn = 2( 3 8)

6

nn n

S25

= 25

(708) 29506

87. Answer (2)

Hint : 1 + ei = 2cos2

2i

e

Sol. : Im(1 + ei)2n =

2

2cos sin2

n

n⎛ ⎞ ⎜ ⎟

⎝ ⎠

88. Answer (3)

Hint : |adjA| = |A|n–1

Sol. :

∵

2 2

2 2

2

adj 1

1 1

ax b x ab b xa b x

x b bx a a x x ab

x a x bx a ax b

⎡ ⎤ ⎛ ⎞⎡ ⎤⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥⎢ ⎥⎜ ⎟⎢ ⎥ ⎣ ⎦ ⎢ ⎥⎝ ⎠ ⎣ ⎦

D12 = D

2

89. Answer (1)

Hint : Make cases

Sol. :

Case 1: Non-zero integral solutions = 7C2222

= 168

Case 2: If exactly one of x, y, z is zero, number of

solutions = 3 2 2 7C1 = 84

Case 3: If exactly two of x, y, z is zero, number of

solutions = 3 2 = 6

Total solutions = 258

90. Answer (2)


Sol. : (5x4 + 4x3 + 3x2)(x5 + x4 + x3)5dx

Let x5 + x4 + x3 = t

I = t5dx = 5 4 3 6( )

6

x x xc

Documents

Mock Test - 3 (Code-A) (Answers) All India Aakash …...All India Aakash Test Series for JEE (Main)-2019 Mock Test - 3 (Code-A) (Hints & Solutions) 4/11 20. Answer (4) Hint : It will