Mock Test - 2 (Code-E) (Answers) All India Aakash Test ... · All India Aakash Test Series for JEE...

Mock Test - 2 (Code-E) (Answers) All India Aakash Test Series for JEE (Main)-2019

1/14

1. (3)

2. (1)

3. (3)

4. (4)

5. (3)

6. (2)

7. (3)

8. (2)

9. (2)

10. (1)

11. (2)

12. (1)

13. (2)

14. (3)

15. (3)

16. (3)

17. (3)

18. (3)

19. (3)

20. (4)

21. (3)

22. (1)

23. (2)

24. (3)

25. (3)

26. (4)

27. (4)

28. (2)

29. (2)

30. (2)

PHYSICS CHEMISTRY MATHEMATICS

31. (4)

32. (2)

33. (4)

34. (3)

35. (2)

36. (1)

37. (2)

38. (1)

39. (1)

40. (3)

41. (3)

42. (2)

43. (2)

44. (2)

45. (1)

46. (4)

47. (4)

48. (1)

49. (4)

50. (3)

51. (4)

52. (4)

53. (3)

54. (1)

55. (3)

56. (2)

57. (1)

58. (3)

59. (4)

60. (3)

61. (4)

62. (1)

63. (3)

64. (2)

65. (3)

66. (1)

67. (2)

68. (1)

69. (1)

70. (3)

71. (2)

72. (2)

73. (2)

74. (2)

75. (4)

76. (3)

77. (3)

78. (3)

79. (1)

80. (2)

81. (3)

82. (3)

83. (1)

84. (2)

85. (2)

86. (3)

87. (2)

88. (3)

89. (4)

90. (4)

Test Date : 03/03/2019

ANSWERS

MOCK TEST - 2 - Code-E

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

All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-E) (Hints & Solutions)

2/14

1. Answer (3)

Hint :dx

vdt

Sol. : cosv A t

cosv

tA

sinx

tA

2 22 2

2 2cos sin

v xt t

A A

2 2 2v x A

2. Answer (1)

Hint : There will be static friction between 2 kg and

3 kg blocks.

Sol. : (2 3)G

F f a 0.1 50 5 NGf

10 – 5 = 5a a = 1 m/s2

2 kgf

2 1 2 Nf

3. Answer (3)

Hint : Thermal stress = Y T

Sol. : Stressmg

Y TA

5 11 5

100 10

10 2 10 10

mgT

AY

50 CT T = 27 – 50 = –23°C

4. Answer (4)

Hint : sin( )y A t kx

Sol. : 2 2 200 400f

100100 m/s

0.01

Tv

100 1m

200 2

v

f

12

2 24k

2 mm sin(400 4 )y t x

at t = 0, at x = 0 y = –2 mm

2 mm 2 mmsin 3

2

PART - A (PHYSICS)

5. Answer (3)

Hint :1 1

ML T ⎡ ⎤ ⎣ ⎦

Sol. :1 1

ML T ⎡ ⎤ ⎣ ⎦

1

1 12 3

4

⎡ ⎤ ⎢ ⎥⎣ ⎦

8

3

6. Answer (2)

Hint : inputBE

B

VR

I

Sol. :4

6

(1.4 1.1) 0.310

70 A 40 A 30 10

VR

7. Answer (3)

Hint : Particle should reach a point where forces on

it are balanced.

Sol. :2 2

44

(12 )

GMm G Mmx R

x R x ⇒

For v0 minimum

2

0

4 1 4

11 2 4 8

GMm G Mm GMm G Mmmv

R R R R

0

27

22

GMv

R

8. Answer (2)

Hint : Mechanical Energy Conservation

Sol. : Major axismin max

2a r r

2 2 3a r r r

2

max

1

2 2

GMm GMmmv

a r

2

3 2 2

GM v GM

r r

3

GMv

r

Mock Test - 2 (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019

3/14

9. Answer (2)

Hint :primary

AB

VV r

R R

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

Sol. : dxdR

A

1

0

0

9xdx

dRA

∫ ∫

09

2A

2 2010 22 9

(6 9) 2 3V x x

A

⎛ ⎞ ⇒ ⎜ ⎟ ⎝ ⎠

1

3x

10. Answer (1)

Hint :1 1 1

f v u

0

ih v

h u

Sol. : For M1, u = –30 cm, f = –20 cm

1 1 1

20 30v

60 cmv

( 60)

1mm 30

ih

2mmi

h

For M2

, u = –40 cm so v = –40 cm

03 mmh so 3 mm

ih

11. Answer (2)

Hint : E dlt

∫

� �

�

Sol. : 2

0B r

2

0 02

2

B r B rE r E

t t

⇒

0( 2 )2

B rr F rqE r r

t

��

�

Angular impulse = 3

0t B r

12. Answer (1)

Hint :2 m

TqB

Sol. :

R

90

2

R

3

2

R

12sin

2

R

R 30

Total angle 2

90 303

0

2

3

mt

eB

13. Answer (2)

Hint and Sol. :

Hinge HingeI

22

4 12 4

L mL Lmg m

⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

12

7

g

L

CM CM

12 3

4 7 7

L g ga R a

L ⇒

14. Answer (3)

Hint :max c m

min c m

Sol. :max

10100 c = 10000

min9900 f

c = 5000 Hz

m100

50 Hz2

m

mf

15. Answer (3)

Hint :rms rms

cosP V I


4/14

Sol. :rms

13V

2V rms

rms

VI

Z

rms2 2

13 1

22 12 (15 10)I A

2 2

12 12

1312 (15 10)cos

13 1 126 W

132 2P

16. Answer (3)

Hint : A charge moving with constant velocity

produces magnetic field that does not change

with time. So it does not produce time varying

electric field.

Sol. : A charge moving with constant velocity



electric field.

17. Answer (3)

Hint : string CM2

Rv v

Sol. :

CM

2

Rv

CMv R 2

R

string CM

3

2 2 2

R Rv v R R

CMv R

CM

string

2

3 3

2

v R

Rv

CM string

2

3v v

18. Answer (3)

Hint : 2xxT

mgd

�

Sol. : For physical pendulum

2xxT

mgd

�22

12 2xx

mm⎛ ⎞ ⎜ ⎟⎝ ⎠

� ��

27

12

m �

27

2

122

mT

mg

�

�

7 22

12T

g �

19. Answer (3)

Hint : 2

2 2

1 2

1 113.6E Z

n n

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

Sol. : 2 2

1 113.6

1 3H

E⎛ ⎞ ⎜ ⎟⎝ ⎠

2

2 2

1 113.6( )

3E Z

n

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

2 2 2

1 1 1 113.6 13.6

13 3Z

n

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

2 2

13.6 1 1 1 113.6

9 1 1 3

3

Z

n

⎛ ⎞⎜ ⎟

⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Comparing Z = 3, n = 9

20. Answer (4)

Hint : Intensitynh

At

Sol. :nh

IAt

double

So becomes halfn

So saturation current becomes half.

21. Answer (3)

Hint : Take a semicircular ring as an element and

integrate it


5/14

Sol. :

dr

r

0 0

CM

0 0

2( ) ( )

R R

R R

rdm y dA

y

dm dA

∫ ∫

∫ ∫

0

0

0

0

2( )

R

R

rr rdr

r rdr

⎛ ⎞ ⎜ ⎟⎝ ⎠

∫

∫

CM

3

2

Ry

22. Answer (1)

Hint : Minimum for pure rolling

min 2

CM

tan

1MR

I

Sol. : min 2

2

tan 2tan

51

2

3

MR

MR

min

So there is pure rolling.

So work done by friction is zero.

23. Answer (2)

Hint : 0B C

Q

Sol. : ( ) 3 ( )1

A B p B A B A

n RQ nC T T T nR T T

0 03

A BQ PV

1 10,

B C B CB CQ T V T V

1.5 1 1.5 1

0(2 )(2 )

A A CT V T V

08

CV V

0

0 0

0

ln ln8

A

C A A

C

VVQ nRT P V

V V

⎛ ⎞ ⎜ ⎟

⎝ ⎠

0 03 ln2P V

0 02.08PV

0 0 0 0 0 03 0 2.08 0.92

ABCAQ P V P V P V

24. Answer (3)

Hint : 1 2

0 02 2

E

Sol. :

+

+

+

vd

+

+

+

+

+

+

+

+

+

1

2

e

F = eE

1 2

0

( )

2

eF ma

1 2

0

( )

2

ea

m

21

2y y y

S u t a t

2

1 2

0

( )1

2 2 2

e td

m

0

1 2

2

( )

mdt

e

x xS u t

0

1 2

2

( )

mdl v

e

25. Answer (3)

Hint : Net 1 2

0

q

Sol. : Through both the cones

q

x


6/14

Net 1 2

0

q

1

0

3

7

q

So 2

0

4

7

q

26. Answer (4)

Hint : N = No

e–t

Sol. : 60% decay 40% remaining

90% decay 10% remaining

So remaining portion becomes 1th

4

So 2 1

2t t T

27. Answer (4)

Hint : ( 1)x t n

Sol. : (1.525 1)10 m 5000Ån 6 7

0.525 10 10 5 10n

52.510.5

5n

11th dark at path difference of 10.5 .

28. Answer (2)

Hint :/R M R M

V V V � � �

Sol. : Take ˆi along east, ˆj along north, ˆk along

upwards

Case 1:/R M R M

V V V � � �

ˆ ˆ10 5R

k V i �

ˆ ˆ5 10R

V i k �

Case 2: ˆ5M

V j�

/ˆ ˆ ˆ5 10 5

R MV i k j �

Angle with ˆk 2 2 2

10 10cos

1505 5 10

2cos

3

29. Answer (2)

Hint :

22 ( )

9T

r gV

Sol. : VT

4

3

2 10 (1000 700)10

9 10

200m/s 66.7 m/s

3

30. Answer (2)

Hint :sep

app

1v

e

v

APP2 2v gR

Sol. :2 1

2 2M gR m gR mV MV ....(1)

2 12 2M gR MV V M ....(2)

MV1

mV2

2

2 35

gR M mV gR

M m

( 5 2)

(3 2 5)

mM

PART - B (CHEMISTRY)

31. Answer (4)

Hint : Magnetic nature of complex compound.

Sol. : Only IV has unpaired electron.

32. Answer (2)

Hint : Hg22+ ions disproportionate in presence of

cyanide.

Sol. : 2

2 2Hg 2CN Hg Hg(CN)

33. Answer (4)

Hint : Alkali metal hydrides are denser than parent

metal.

Sol. : Metallic hydrides have a lattice which is

different from that of the parent metal.


7/14

34. Answer (3)

Hint : Oxymercuration of alkene.

Sol. :

OHOH

HgOAc+

+

Hg

CH2 CH2

HgOAc

NaBH4

+OAc O

+

O

H

35. Answer (2)

Hint : Formation of hydrocarbons.

Sol. :

CHC OH

O

Reaction I: +CH

2

CH2

OHC

CH

O

Reaction II: +C

H

C

C O CH3

O

CH

C O CH3

C

O

36. Answer (1)

Hint : Combustion of hydrocarbon.

Sol. : In Reaction-I, highest heat per gram of the

hydrocarbon is liberated.

37. Answer (2)

Hint : Reaction of Alkyne

Sol. :

C H2 5

C C H3 7

CO

3

Zn, H O2

C H2 5

C C

O O

C H3 7

38. Answer (1)

Hint : Elimination reaction; addition of H2O.

Sol. :

CH OH3

heat, KOH

Major

Br

CH3

CH3

BH –THF3

H O ; OH2 2

–

CH3

CH3

O Et

CH3

(i) Na

(ii) CH – 3 CH – Br2

OH

39. Answer (1)

Hint : Preparation of phenol from cumene.

Sol. :

(i) O2

(ii) H3O

+

C

C

C OH

O C

CH3

CH3

A B

+

CH3

H3C H

HCl

CH3

CH3

O C

CH3

CH3

2 OH

HO C OH

+

40. Answer (3)

Hint : I and III are mirror images

Sol. : (I, III) enantiomers

(I, IV), (III, IV) diasteromers

II is a positional isomer of the others.

41. Answer (3)

Hint : Alkylation reaction of Amine.


8/14

Sol. :

CH I3

N N+

N+

N

CH3

Ag2O

CH3

CH3

I–

OH–

42. Answer (2)

Hint : Oxidation of 1, 4-Dihydroxy benzene.

Sol. :

O O

OO

H

H

::

:

:

Na Cr O2 2 7

H SO /H O2 4 2

43. Answer (2)

Hint : Reaction of carbonyl compound.

Sol. :

C CH CH3

C C CH3

Ph

CH3

CH2

O

Ph

CH3

O

(i) LDA

(ii) Ph CH2 Br

44. Answer (2)

Hint : Reaction due to –H

Sol. :

CH2

O OEt

C C

O O

C O CH2

CH2

CH3

CH3

OCH2

CH2

CH3

CH3

CH

C

O

O

CO

O

OEt– –

O

45. Answer (1)

Hint : Limiting Reagent

Sol. :4 8 2 2 2

C H 6O 4CO 4H O

moles of O2 available = 6

⇒ Volume = 6 × 22.4 = 134.4 litre

46. Answer (4)

Hint : HCP is ABAB type of packing

Sol. :

T

T

O

T

T

O

T

T

O

T

T

O

T

T

T T T T T

T T T T T T

O

T

O

O O O O O O

47. Answer (4)

Hint : Elevation in boiling point.

Sol. : Elevation in boiling point is proportional to the

no. of solute particle.

NaCl Na Cl

2

2CaCl Ca 2Cl

48. Answer (1)

Hint : Order of reactivity on the basis of (ECS).

Sol. : E°red

K < Mg < Zn < Au

49. Answer (4)

Hint : Ln+3 ions have unpaired e– in f orbital.

Sol. : Most of the Ln+3 ions are coloured both in the

solid state and in the aqueous solutions.

50. Answer (3)

Hint : Normality of 1 litre solution = 0.99

Sol. :0

10xd 10 70 1.54M

M 98

= 11

Now, meq of H3PO

4 in V ml = meq of H

3PO

4 in 1 litre

11 × V × 3 = 0.99 × 1000

V = 30 ml.


9/14

51. Answer (4)

Hint : Calculation of pressure in equilibrium mixture.

Sol. : Since pressure of the gases are same in both

the containers. Therefore the final pressure will

not be changed.

52. Answer (4)

Hint : G = H – TS

Sol. : As S increases, the value of G becomes

more negative.

53. Answer (3)

Hint : Equilibrium constant and Gibb's free energy

Sol. : G°= –RT ln Keq

.

54. Answer (1)

Hint : Conc. of reactant and eq. const. (Le-Chatelier's

principle).

Sol. :

2

2 2

HI 0.4 0.4Q 8 ; Q K

H I 0.1 0.2

55. Answer (3)

Hint : Slaked lime and formation of bleaching powder.

Sol. : MnO2 + 4HCl MnCl

2 + Cl

2 + 2H

2O

Cl2 + Ca(OH)

2 CaOCl

2 + H

2O

(Y)

CaOCl2

+ 2HCl CaCl2

+ Cl2

+ H2O

56. Answer (2)

Hint : Chemical properties (epoxidation) of alkene.

Sol. :

CH2

CH2

LiAlH4

O

C

OH

CC

O

O

Cl

O H

57. Answer (1)

Hint : Factual

Sol. : Insulin is an hormone.

58. Answer (3)

Hint : Fact Based

Sol. : Soluble fluoride is added to drinking water to

bring the concentration to 1 ppm.

59. Answer (4)

Hint : Lassaigne solution (test of nitrogen)

Sol. : Cyanide ions are formed on treatment with Na

metal which from a complex with Fe+2 ions.

Some Fe2+ ion also get oxidised to Fe+3 and

thus prussian blue is formed.

60. Answer (3)

Hint : Cellulose – Hydrolysis

Sol. : Cellulose is a polymer of -D-Glucose.

PART - C (MATHEMATICS)

61. Answer (4)

Hint :3

2

0

2( )

3 ∫

bb

f x dx b

Sol. :3

2

0

2( )

3

b bf x dx b ∫

d. w.r. t. b

f(b) = 2b – 2b2 f(x) = 2x – 2x2

f(x) = 2 1 2

24 4

x x⎡ ⎤ ⎢ ⎥⎣ ⎦

=

21 1

22 2

x⎛ ⎞ ⎜ ⎟⎝ ⎠

f(x)max.

= 1

2 when x =

1

2

62. Answer (1)

Hint : Form : 1

Sol. : l =

1

1 cot

0

sinlim

x x

x

x

x

⎛ ⎞⎜ ⎟⎝ ⎠

form : 1

∵0

sinlimx

x

x

⎛ ⎞⎜ ⎟⎝ ⎠

= 1

0

lim ( cot )x

x x

= 0

limtanx

x

x = 1

l =

sin 1lim 1

1 cot0

x

x x xxe

⎛ ⎞ ⎜ ⎟ ⎝ ⎠


10/14

=

sin tanlim

tan0

x x x

x x xxe

⎛ ⎞⎜ ⎟⎝ ⎠

=

sin tanlim

tan0

x x x

x x xxe

⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

=

11

2e

=

1

e

63. Answer (3)

Hint : Take log on both sides then convert in definite

integral

Sol. :

logl =1 1 2

lim log log log ... logn

n n n n n

n n n n n

⎛ ⎞⎡ ⎤ ⎛ ⎞ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎝ ⎠

=

21lim log

n

nr n

r

n n

⎛ ⎞⎜ ⎟⎝ ⎠

∑ = 2

1logxdx∫

= [xlogx – x]12 = 2log2 – 2 + 1 = 2log2 – 1

= 4

loge

⎛ ⎞⎜ ⎟⎝ ⎠

l = 4

e

64. Answer (2)

Hint : Draw graph

Sol. :

x

y

AB

C

x = 1

y a =

(1, 1)

O

y = 2x – x2 (x – 1)2 = –(y – 1)

If y = a, then x = 1 1 a

A1 =

2 2

0(2 )x x dx∫ =

23

2

03

xx

⎡ ⎤⎢ ⎥

⎣ ⎦

= 8 4

43 3

A(OABC) = 1

3

1 1 2

0

1(2 ) 1

3

a

x x dx a a

∫

2 31 11 1 1 1 1

3 3a a a a

Put 1 – a = t2

2t3 = 1 t6 = 1

4 = 2–2

t2 =

2

32

a = 1 – t2 = 2

31 2

65. Answer (3)

Hint : 3sinx + 2cosx

= 3cos 2sin (3cos 2sin ) dx x x x

dx

Sol. :

I = 3sin 2cos

3cos 2sin

∫

x xdx

x x

=

(3cos 2sin )3cos 2sin

3cos 2sin 3cos 2sin

∫ ∫

dx x

x x dxA dx B dx

x x x x

3sinx + 2cosx = A(3cosx + 2sinx)

+ B(–3sinx + 2cosx)

Comparing coefficients of sinx and cosx on both

sides

A = 12

13, B =

5

13

121

1 13

5

13

⎛ ⎞

⎜ ⎟⎝ ⎠

A

B = –5

66. Answer (1)

Hint : Parabolic curve

Sol. : 0dy

dx

3ax2 + 6 0

y ax = 3 + 62

3a > 0, D < 0


11/14

a > 0

0 – 4(3a)6 < 0

a > 0

at a = 0, y = 6x, an increasing function

67. Answer (2)

Hint : Plane contains points A, B, C

Sol. : Plane containing all medians is the plane

passing through the vertices A, B and C

Equation of the plane is

PA BA CA

⎡ ⎤⎢ ⎥⎣ ⎦

= 0 where P(x, y, z)

1 1 1

1 0 1

x y z

= 0

x – 2y + z = 0

68. Answer (1)

Hint : Multiply and divide by

1 1

2 2(sec ) (tan )n n

x x

Sol. :

R =

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

11

1 1 11

2 2 22(sec ) (tan ) (sec ) tan

...(sec tan )

n n n

nx x x x

x x

R =

2 2

22

1 1

2 2

11

22

1 1

2 2

(sec ) (tan )

(sec ) tan ...(sec tan )

(sec ) (tan )

n n

n

n

n n

x x

x x x x

x x

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟ ⎝ ⎠

= 1 1

2 2

1

(sec ) (tan )n n

x x

=

1

2

1

2

(cos )

1 (sin )

n

n

x

x

69. Answer (1)

Hint :�

a is perpendicular to c�

Sol. : a c� �

= a c� �

a c� �

ˆ ˆ ˆ( )c i j k �

a c� �

= b�

ˆ ˆ ˆ

ˆ ˆ1 1 0

i j k

i j

= 0, = 1

c�

= ˆk

70. Answer (3)

Hint : AA–1 = I

Sol. : A–1A = I, then compare corresponding

elements.

71. Answer (2)

Hint : (d) = r

Sol. : Equation of tangent of x2 = – 4y is

tx = –y + t2 ...(i)

(i) is also tangent to x2 + y2 = 4

2

2

| |

1

t

t = 2

t4 = 4 + 4t2 t2 = 2 2 2

m = – t m2 = 2 2 2 [m2] = 4

72. Answer (2)

Hint : S1 – S

2 = 0

Sol. : c1(–g, 0), (r

1 = |g|, c

2(0, – f), r

2 = |f |, equation

of common chord is

A

B

MC

1

(– , 0)g

r1

gx fy – = 0

gx fy – = 0

C1M =

2

2 2

g

g f

AM2 = r12 – (C

1M)2 =

42

2 2

gg

g f

=

2 2

2 2

g f

g f

AM = 2 2

| |gf

g f

AB = 2 2

2 | |gf

g f


12/14

73. Answer (2)

Hint : y = x is axis of both hyperbolas

Sol. :

x

y y x =

xy = 4

xy = 1 A(1, 1)

(2, 2)B

AB = 2

74. Answer (2)

Hint : 2 2 2 y mx a m b

Sol. : Equations of common tangents to both

hyperbolas are

y = 2 2

x a b

For 4 common tangents, a > b

75. Answer (4)

Hint :2

a b c

⎡ ⎤⎢ ⎥⎣ ⎦

=

a a a b a c

b a b b b c

c a c b c c

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥

⎣ ⎦

Sol. :2

a b c

⎡ ⎤⎢ ⎥⎣ ⎦

=

a a a b a c

b a b b b c

c a c b c c

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥

⎣ ⎦

=

1 3

1 2

3 2

1

1

1

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦

= 1 + 21

2

3 –

12 –

22 –

32

a b c

⎡ ⎤⎢ ⎥⎣ ⎦ =

2 2 21 2 3 1 2 31 2

76. Answer (3)

Hint : = 0

Sol. : = 0

5 31

2 2

50

2

30 4

2

k

= 0

5 3 3

4 (10)2 2 2

k k⎛ ⎞ ⎜ ⎟⎝ ⎠

= 0

94

4k k = 25

25

4k = 25 k = 4

tan = 2

2 h ab

a b

= 3

5

= 1 3

tan5

⎛ ⎞⎜ ⎟⎝ ⎠

77. Answer (3)

Hint : = 0

Sol. : = 0

1 1

1 2

1 1 0

= 0

2 – 2 + (–– 1) = 0

2 + = 0

= 0 or –1

78. Answer (3)

Hint : If n(A) = n, then number of reflexive relations

on the set A = 2n(n–1)

Sol. : If n(A) = n, then number of reflexive relations


79. Answer (1)

Hint : Make all cases

Sol. : I, I, I, I, M, P, P, S, S, S, S

1(I)

1(M)

+

1(I)

1(P)

+

1(I)

1(S)


13/14

Number of words = 9! 9! 9!

3!2!4! 3! 4! 3! 2!3!

= 9!

74!3!2!

= 1260 × 7

80. Answer (2)

Hint : A.M. G.M.

Sol. :

1

3(2 ) (4 ) (8 )

((2 )·(4 ) (8 ))3

xy yz zxxy yz zx

1

2 2 2 327

(64 )3

x y z

64x2y2z2 93

2

2 2 2 27

8x y z

⎛ ⎞ ⎜ ⎟⎝ ⎠

27

8xyz Greatest possible value of xyz

is 3.375

81. Answer (3)

Hint : Sum of G.P.

Sol. :

S = 2 3

12 12 121 ...i i i

e e e

upto 25 terms

=

2525

12 12

1212

1 1

11

i i

ii

e e

ee

=

212

12

1

1

i

i

e

e

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 12

12

1

1

i

i

e

e

= 1

82. Answer (3)

Hint : 2 3 2

2 3 2 2 3 3(32) | | | | z z z z z z

Sol. : |(32)2z1 + (16)2z

2 + (8)2z

3|

= |(|z2| |z

3|)2z

1 + (|z

1| |z

3|)2z

2 + (|z

1||z

2|)2z

3|

= 2 2 3 3 1 1 1 3 3 2 1 1 2 2 3z z z z z z z z z z z z z z z

= 1 2 3 2 3 3 1 1 2z z z z z z z z z

= 1 2 3 2 3 3 1 1 2| | | | | |z z z z z z z z z

= 2 × 4 × 8 |z1z

2 + z

2z

3 + z

3z

1|

= 64

83. Answer (1)

Hint : (A + B)(A – B) = A2 – B2

Sol. : (1 + x + x2)35 (1 – x + x2)35

= ((x2 + 1)2 – x2)35

= (x4 + x2 + 1)35

Coefficient of x15 is zero

84. Answer (2)

Hint : Factorization

Sol. : 22 3 3 2 1z z i z z = 0

22 2 3 3z zi z i i z = 0

2z(z – i) + 1(z – i) – 3 ( )i z i = 0

( ) 2 1 3z i z i = 0

z = i, z = 1 3

2 2i =

18 + 18 + 1 = i18 + 18 + 1

= –1 + 1 + 1 = 1

85. Answer (2)

Hint : Find n, L.D. eq

dx

dy

Sol. :1

y

y

y

dx xee x

dy e

dx

dy + x = e–y

I.F = 1dy

e∫ = ey

Solution of differential equation is

x.ey = .

y ye e dy c ∫xey = y + c passes then (0, 0)

c = 0

Equation of curve is y = xey

86. Answer (3)

Hint :

2

2 2( )ix

xN

∑

Sol. :1 2 3 ... 10 10(11) 11

10 2 10 2x


14/14

Variance =

22

2 ix

xN

∑

=

21 10(10 1) (2 10 1) 11

10 6 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 11 21 11 11

6 4

= 11 7 11 11 1114 11

2 4 4

2 33

4 =

33

2

87. Answer (2)

Hint : Required probability

= P(odd, odd) + P(even, even)

Sol. : Let p be the probability of showing an odd

number

Then 3p = 1 p = 1

3

Required probability


= p2 + 4p2

= 5p2

= 5

9

88. Answer (3)

Hint : Line is parallel to the vector 1 2n n��

Sol. : Line is parallel to the vector 1 2n n��

Where � �

1 22 2 , 3 4 6n i j k n i j k

��

1 2 3. 0n n n

��

(where �

34 5 8n i j k

�� )

Line is perpendicular to normal to the plane

Line is parallel to the plane

89. Answer (4)

Hint : Draw graph

Sol. :

y = 2

x

y =

x320

y =

x

y = x

y =

x +

2 y

x

=

2

y x = cos (cos )1

y

Number of real solution(s) of the equation

cos–1(cosx) + x = 2 is intersection point of

y = cos–1(cosx) and y = 2 – x

y = cos–1(cosx) and y = 2 – x intersect at

infinitely many points.

90. Answer (4)

Hint : Truth table

Sol. : Truth values of r and s are T and T respectively.

� � �

Mock Test - 2 (Code-F) (Answers) All India Aakash Test Series for JEE (Main)-2019

1/14

1. (2)

2. (2)

3. (2)

4. (4)

5. (4)

6. (3)

7. (3)

8. (2)

9. (1)

10. (3)

11. (4)

12. (3)

13. (3)

14. (3)

15. (3)

16. (3)

17. (3)

18. (2)

19. (1)

20. (2)

21. (1)

22. (2)

23. (2)

24. (3)

25. (2)

26. (3)

27. (4)

28. (3)

29. (1)

30. (3)

PHYSICS CHEMISTRY MATHEMATICS

31. (3)

32. (4)

33. (3)

34. (1)

35. (2)

36. (3)

37. (1)

38. (3)

39. (4)

40. (4)

41. (3)

42. (4)

43. (1)

44. (4)

45. (4)

46. (1)

47. (2)

48. (2)

49. (2)

50. (3)

51. (3)

52. (1)

53. (1)

54. (2)

55. (1)

56. (2)

57. (3)

58. (4)

59. (2)

60. (4)

61. (4)

62. (4)

63. (3)

64. (2)

65. (3)

66. (2)

67. (2)

68. (1)

69. (3)

70. (3)

71. (2)

72. (1)

73. (3)

74. (3)

75. (3)

76. (4)

77. (2)

78. (2)

79. (2)

80. (2)

81. (3)

82. (1)

83. (1)

84. (2)

85. (1)

86. (3)

87. (2)

88. (3)

89. (1)

90. (4)

Test Date : 03/03/2019

ANSWERS

MOCK TEST - 2 - Code-F

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

All India Aakash Test Series for JEE (Main)-2019 Mock Test - 2 (Code-F) (Hints & Solutions)

2/14

1. Answer (2)

Hint :sep

app

1v

e

v

APP2 2v gR

Sol. :2 1

2 2M gR m gR mV MV ....(1)

2 12 2M gR MV V M ....(2)

MV1

mV2

2

2 35

gR M mV gR

M m

( 5 2)

(3 2 5)

mM

2. Answer (2)

Hint :

22 ( )

9T

r gV

Sol. : VT

4

3

2 10 (1000 700)10

9 10

200m/s 66.7 m/s

3

3. Answer (2)

Hint :/R M R M

V V V � � �

Sol. : Take ˆi along east, ˆj along north, ˆk along

upwards

Case 1:/R M R M

V V V � � �

ˆ ˆ10 5R

k V i �

ˆ ˆ5 10R

V i k �

Case 2: ˆ5M

V j�

/ˆ ˆ ˆ5 10 5

R MV i k j �

Angle with ˆk 2 2 2

10 10cos

1505 5 10

2cos

3

PART - A (PHYSICS)

4. Answer (4)

Hint : ( 1)x t n

Sol. : (1.525 1)10 m 5000Ån

6 70.525 10 10 5 10n

52.510.5

5n

11th dark at path difference of 10.5 .

5. Answer (4)

Hint : N = No

e–t

Sol. : 60% decay 40% remaining

90% decay 10% remaining

So remaining portion becomes 1th

4

So 2 1

2t t T

6. Answer (3)

Hint : Net 1 2

0

q

Sol. : Through both the cones

q

x

Net 1 2

0

q

1

0

3

7

q

So 2

0

4

7

q

7. Answer (3)

Hint : 1 2

0 02 2

E

Mock Test - 2 (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019

3/14

Sol. :

+

+

+

vd

+

+

+

+

+

+

+

+

+

1

2

e

F = eE

1 2

0

( )

2

eF ma

1 2

0

( )

2

ea

m

21

2y y y

S u t a t

2

1 2

0

( )1

2 2 2

e td

m

0

1 2

2

( )

mdt

e

x xS u t

0

1 2

2

( )

mdl v

e

8. Answer (2)

Hint : 0B C

Q

Sol. : ( ) 3 ( )1

A B p B A B A

n RQ nC T T T nR T T

0 03

A BQ PV

1 10,

B C B CB CQ T V T V

1.5 1 1.5 1

0(2 )(2 )

A A CT V T V

08

CV V

0

0 0

0

ln ln8

A

C A A

C

VVQ nRT P V

V V

⎛ ⎞ ⎜ ⎟

⎝ ⎠

0 03 ln2P V

0 02.08PV

0 0 0 0 0 03 0 2.08 0.92

ABCAQ P V P V P V

9. Answer (1)

Hint : Minimum for pure rolling

min 2

CM

tan

1MR

I

Sol. : min 2

2

tan 2tan

51

2

3

MR

MR

min

So there is pure rolling.

So work done by friction is zero.

10. Answer (3)

Hint : Take a semicircular ring as an element and

integrate it

Sol. :

dr

r

0 0

CM

0 0

2( ) ( )

R R

R R

rdm y dA

y

dm dA

∫ ∫

∫ ∫

0

0

0

0

2( )

R

R

rr rdr

r rdr

⎛ ⎞ ⎜ ⎟⎝ ⎠

∫

∫

CM

3

2

Ry

11. Answer (4)

Hint : Intensitynh

At

Sol. :nh

IAt

double

So becomes halfn

So saturation current becomes half.

12. Answer (3)

Hint : 2

2 2

1 2

1 113.6E Z

n n

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠


4/14

Sol. : 2 2

1 113.6

1 3H

E⎛ ⎞ ⎜ ⎟⎝ ⎠

2

2 2

1 113.6( )

3E Z

n

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

2 2 2

1 1 1 113.6 13.6

13 3Z

n

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

2 2

13.6 1 1 1 113.6

9 1 1 3

3

Z

n

⎛ ⎞⎜ ⎟

⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Comparing Z = 3, n = 9

13. Answer (3)

Hint : 2xxT

mgd

�

Sol. : For physical pendulum

2xxT

mgd

�22

12 2xx

mm⎛ ⎞ ⎜ ⎟⎝ ⎠

� ��

27

12

m �

27

2

122

mT

mg

�

�

7 22

12T

g �

14. Answer (3)

Hint : string CM2

Rv v

Sol. :

CM

2

Rv

CMv R 2

R

string CM

3

2 2 2

R Rv v R R

CMv R

CM

string

2

3 3

2

v R

Rv

CM string

2

3v v

15. Answer (3)

Hint : A charge moving with constant velocity



electric field.

Sol. : A charge moving with constant velocity



electric field.

16. Answer (3)

Hint :rms rms

cosP V I

Sol. :rms

13V

2V rms

rms

VI

Z

rms2 2

13 1

22 12 (15 10)I A

2 2

12 12

1312 (15 10)cos

13 1 126 W

132 2P

17. Answer (3)

Hint :max c m

min c m

Sol. :max

10100 c = 10000

min9900 f

c = 5000 Hz

m100

50 Hz2

m

mf

18. Answer (2)

Hint and Sol. :

Hinge HingeI

22

4 12 4

L mL Lmg m

⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦


5/14

12

7

g

L

CM CM

12 3

4 7 7

L g ga R a

L ⇒

19. Answer (1)

Hint :2 m

TqB

Sol. :

R

90

2

R

3

2

R

12sin

2

R

R 30

Total angle 2

90 303

0

2

3

mt

eB

20. Answer (2)

Hint : E dlt

∫

� �

�

Sol. : 2

0B r

2

0 02

2

B r B rE r E

t t

⇒

0( 2 )2

B rr F rqE r r

t

��

�

Angular impulse = 3

0t B r

21. Answer (1)

Hint :1 1 1

f v u

0

ih v

h u

Sol. : For M1, u = –30 cm, f = –20 cm

1 1 1

20 30v

60 cmv

( 60)

1mm 30

ih

2mmi

h

For M2

, u = –40 cm so v = –40 cm

03 mmh so 3 mm

ih

22. Answer (2)

Hint :primary

AB

VV r

R R

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

Sol. : dxdR

A

1

0

0

9xdx

dRA

∫ ∫

09

2A

2 2010 22 9

(6 9) 2 3V x x

A

⎛ ⎞ ⇒ ⎜ ⎟ ⎝ ⎠

1

3x

23. Answer (2)

Hint : Mechanical Energy Conservation

Sol. : Major axismin max

2a r r

2 2 3a r r r

2

max

1

2 2

GMm GMmmv

a r

2

3 2 2

GM v GM

r r

3

GMv

r

24. Answer (3)

Hint : Particle should reach a point where forces on

it are balanced.

Sol. :2 2

44

(12 )

GMm G Mmx R

x R x ⇒


6/14

PART - B (CHEMISTRY)

31. Answer (3)

Hint : Cellulose – Hydrolysis

Sol. : Cellulose is a polymer of -D-Glucose.

32. Answer (4)

Hint : Lassaigne solution (test of nitrogen)

Sol. : Cyanide ions are formed on treatment with Na

metal which from a complex with Fe+2 ions.

Some Fe2+ ion also get oxidised to Fe+3 and

thus prussian blue is formed.

For v0 minimum

2

0

4 1 4

11 2 4 8

GMm G Mm GMm G Mmmv

R R R R

0

27

22

GMv

R

25. Answer (2)

Hint : inputBE

B

VR

I

Sol. :4

6

(1.4 1.1) 0.310

70 A 40 A 30 10

VR

26. Answer (3)

Hint :1 1

ML T ⎡ ⎤ ⎣ ⎦

Sol. :1 1

ML T ⎡ ⎤ ⎣ ⎦

1

1 12 3

4

⎡ ⎤ ⎢ ⎥⎣ ⎦

8

3

27. Answer (4)

Hint : sin( )y A t kx

Sol. : 2 2 200 400f

100100 m/s

0.01

Tv

100 1m

200 2

v

f

12

2 24k

2 mm sin(400 4 )y t x

at t = 0, at x = 0 y = –2 mm

2 mm 2 mmsin 3

2

28. Answer (3)

Hint : Thermal stress = Y T

Sol. : Stressmg

Y TA

5 11 5

100 10

10 2 10 10

mgT

AY

50 CT T = 27 – 50 = –23°C

29. Answer (1)

Hint : There will be static friction between 2 kg and

3 kg blocks.

Sol. : (2 3)G

F f a 0.1 50 5 NGf

10 – 5 = 5a a = 1 m/s2

2 kgf

2 1 2 Nf

30. Answer (3)

Hint :dx

vdt

Sol. : cosv A t

cosv

tA

sinx

tA

2 22 2

2 2cos sin

v xt t

A A

2 2 2v x A

33. Answer (3)

Hint : Fact Based

Sol. : Soluble fluoride is added to drinking water to

bring the concentration to 1 ppm.

34. Answer (1)

Hint : Factual

Sol. : Insulin is an hormone.

35. Answer (2)

Hint : Chemical properties (epoxidation) of alkene.


7/14

Sol. :

CH2

CH2

LiAlH4

O

C

OH

CC

O

O

Cl

O H

36. Answer (3)

Hint : Slaked lime and formation of bleaching powder.

Sol. : MnO2 + 4HCl MnCl

2 + Cl

2 + 2H

2O

Cl2 + Ca(OH)

2 CaOCl

2 + H

2O

(Y)

CaOCl2

+ 2HCl CaCl2

+ Cl2

+ H2O

37. Answer (1)

Hint : Conc. of reactant and eq. const. (Le-Chatelier's

principle).

Sol. :

2

2 2

HI 0.4 0.4Q 8 ; Q K

H I 0.1 0.2

38. Answer (3)

Hint : Equilibrium constant and Gibb's free energy

Sol. : G°= –RT ln Keq

.

39. Answer (4)

Hint : G = H – TS

Sol. : As S increases, the value of G becomes

more negative.

40. Answer (4)

Hint : Calculation of pressure in equilibrium mixture.

Sol. : Since pressure of the gases are same in both

the containers. Therefore the final pressure will

not be changed.

41. Answer (3)

Hint : Normality of 1 litre solution = 0.99

Sol. :0

10xd 10 70 1.54M

M 98

= 11

Now, meq of H3PO

4 in V ml = meq of H

3PO

4 in 1 litre

11 × V × 3 = 0.99 × 1000

V = 30 ml.

42. Answer (4)

Hint : Ln+3 ions have unpaired e– in f orbital.

Sol. : Most of the Ln+3 ions are coloured both in the

solid state and in the aqueous solutions.

43. Answer (1)

Hint : Order of reactivity on the basis of (ECS).

Sol. : E°red

K < Mg < Zn < Au

44. Answer (4)

Hint : Elevation in boiling point.

Sol. : Elevation in boiling point is proportional to the

no. of solute particle.

NaCl Na Cl 2

2CaCl Ca 2Cl

45. Answer (4)

Hint : HCP is ABAB type of packing

Sol. :

T

T

O

T

T

O

T

T

O

T

T

O

T

T

T T T T T

T T T T T T

O

T

O

O O O O O O

46. Answer (1)

Hint : Limiting Reagent

Sol. :4 8 2 2 2

C H 6O 4CO 4H O

moles of O2 available = 6

⇒ Volume = 6 × 22.4 = 134.4 litre

47. Answer (2)

Hint : Reaction due to –H

Sol. :

CH2

O OEt

C C

O O

C O CH2

CH2

CH3

CH3

OCH2

CH2

CH3

CH3

CH

C

O

O

CO

O

OEt– –

O


8/14

48. Answer (2)

Hint : Reaction of carbonyl compound.

Sol. :

C CH CH3

C C CH3

Ph

CH3

CH2

O

Ph

CH3

O

(i) LDA

(ii) Ph CH2 Br

49. Answer (2)

Hint : Oxidation of 1, 4-Dihydroxy benzene.

Sol. :

O O

OO

H

H

::

:

:

Na Cr O2 2 7

H SO /H O2 4 2

50. Answer (3)

Hint : Alkylation reaction of Amine.

Sol. :

CH I3

N N+

N+

N

CH3

Ag2O

CH3

CH3

I–

OH–

51. Answer (3)

Hint : I and III are mirror images

Sol. : (I, III) enantiomers

(I, IV), (III, IV) diasteromers

II is a positional isomer of the others.

52. Answer (1)

Hint : Preparation of phenol from cumene.

Sol. :

(i) O2

(ii) H3O

+

C

C

C OH

O C

CH3

CH3

A B

+

CH3

H3C H

HCl

CH3

CH3

O C

CH3

CH3

2 OH

HO C OH

+

53. Answer (1)

Hint : Elimination reaction; addition of H2O.

Sol. :

CH OH3

heat, KOH

Major

Br

CH3

CH3

BH –THF3

H O ; OH2 2

–

CH3

CH3

O Et

CH3

(i) Na

(ii) CH – 3 CH – Br2

OH

54. Answer (2)

Hint : Reaction of Alkyne

Sol. :

C H2 5

C C H3 7

CO

3

Zn, H O2

C H2 5

C C

O O

C H3 7

55. Answer (1)

Hint : Combustion of hydrocarbon.

Sol. : In Reaction-I, highest heat per gram of the

hydrocarbon is liberated.


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PART - C (MATHEMATICS)

61. Answer (4)

Hint : Truth table

Sol. : Truth values of r and s are T and T

respectively.

62. Answer (4)

Hint : Draw graph

Sol. :

y = 2

x

y =

x320

y =

x

y = x

y =

x +

2 y

x

=

2

y x= cos (cos )1

y

56. Answer (2)

Hint : Formation of hydrocarbons.

Sol. :

CHC OH

O

Reaction I: +CH

2

CH2

OHC

CH

O

Reaction II: +C

H

C

C O CH3

O

CH

C O CH3

C

O

57. Answer (3)

Hint : Oxymercuration of alkene.

Sol. :

OHOH

HgOAc+

+

Hg

CH2 CH2

HgOAc

NaBH4

+OAc O

+

O

H

58. Answer (4)

Hint : Alkali metal hydrides are denser than parent

metal.

Sol. : Metallic hydrides have a lattice which is

different from that of the parent metal.

59. Answer (2)

Hint : Hg22+ ions disproportionate in presence of

cyanide.

Sol. : 2

2 2Hg 2CN Hg Hg(CN)

60. Answer (4)

Hint : Magnetic nature of complex compound.

Sol. : Only IV has unpaired electron.

Number of real solution(s) of the equation

cos–1(cosx) + x = 2 is intersection point of

y = cos–1(cosx) and y = 2 – x

y = cos–1(cosx) and y = 2 – x intersect at

infinitely many points.

63. Answer (3)

Hint : Line is parallel to the vector 1 2n n��

Sol. : Line is parallel to the vector 1 2n n��

Where � �

1 22 2 , 3 4 6n i j k n i j k

��

1 2 3. 0n n n

��

(where �

34 5 8n i j k

�� )

Line is perpendicular to normal to the plane

Line is parallel to the plane


10/14

64. Answer (2)

Hint : Required probability


Sol. : Let p be the probability of showing an odd

number

Then 3p = 1 p = 1

3

Required probability


= p2 + 4p2

= 5p2

= 5

9

65. Answer (3)

Hint :

2

2 2( )ix

xN

∑

Sol. :1 2 3 ... 10 10(11) 11

10 2 10 2x

Variance =

22

2 ix

xN

∑

=

21 10(10 1) (2 10 1) 11

10 6 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 11 21 11 11

6 4

= 11 7 11 11 1114 11

2 4 4

2 33

4 =

33

2

66. Answer (2)

Hint : Find n, L.D. eq

dx

dy

Sol. :1

y

y

y

dx xee x

dy e

dx

dy + x = e–y

I.F = 1dy

e∫ = ey

Solution of differential equation is

x.ey = .

y ye e dy c ∫xey = y + c passes then (0, 0)

c = 0

Equation of curve is y = xey

67. Answer (2)

Hint : Factorization

Sol. : 22 3 3 2 1z z i z z = 0

22 2 3 3z zi z i i z = 0

2z(z – i) + 1(z – i) – 3 ( )i z i = 0

( ) 2 1 3z i z i = 0

z = i, z = 1 3

2 2i =

18 + 18 + 1 = i18 + 18 + 1

= –1 + 1 + 1 = 1

68. Answer (1)

Hint : (A + B)(A – B) = A2 – B2

Sol. : (1 + x + x2)35 (1 – x + x2)35

= ((x2 + 1)2 – x2)35

= (x4 + x2 + 1)35

Coefficient of x15 is zero

69. Answer (3)

Hint : 2 3 2

2 3 2 2 3 3(32) | | | | z z z z z z

Sol. : |(32)2z1 + (16)2z

2 + (8)2z

3|

= |(|z2| |z

3|)2z

1 + (|z

1| |z

3|)2z

2 + (|z

1||z

2|)2z

3|

= 2 2 3 3 1 1 1 3 3 2 1 1 2 2 3z z z z z z z z z z z z z z z

= 1 2 3 2 3 3 1 1 2z z z z z z z z z

= 1 2 3 2 3 3 1 1 2| | | | | |z z z z z z z z z

= 2 × 4 × 8 |z1z

2 + z

2z

3 + z

3z

1|

= 64

70. Answer (3)

Hint : Sum of G.P.


11/14

Sol. :

S = 2 3

12 12 121 ...i i i

e e e

upto 25 terms

=

2525

12 12

1212

1 1

11

i i

ii

e e

ee

=

212

12

1

1

i

i

e

e

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 12

12

1

1

i

i

e

e

= 1

71. Answer (2)

Hint : A.M. G.M.

Sol. :

1

3(2 ) (4 ) (8 )

((2 )·(4 ) (8 ))3

xy yz zxxy yz zx

1

2 2 2 327

(64 )3

x y z

64x2y2z2 93

2

2 2 2 27

8x y z

⎛ ⎞ ⎜ ⎟⎝ ⎠

27

8xyz Greatest possible value of xyz

is 3.375

72. Answer (1)

Hint : Make all cases

Sol. : I, I, I, I, M, P, P, S, S, S, S

1(I)

1(M)

+

1(I)

1(P)

+

1(I)

1(S)

Number of words = 9! 9! 9!

3!2!4! 3! 4! 3! 2!3!

= 9!

74!3!2!

= 1260 × 7

73. Answer (3)

Hint : If n(A) = n, then number of reflexive relations


Sol. : If n(A) = n, then number of reflexive relations


74. Answer (3)

Hint : = 0

Sol. : = 0

1 1

1 2

1 1 0

= 0

2 – 2 + (–– 1) = 0

2 + = 0

= 0 or –1

75. Answer (3)

Hint : = 0

Sol. : = 0

5 31

2 2

50

2

30 4

2

k

= 0

5 3 3

4 (10)2 2 2

k k⎛ ⎞ ⎜ ⎟⎝ ⎠

= 0

94

4k k = 25

25

4k = 25 k = 4

tan = 2

2 h ab

a b

= 3

5

= 1 3

tan5

⎛ ⎞⎜ ⎟⎝ ⎠

76. Answer (4)

Hint :2

a b c

⎡ ⎤⎢ ⎥⎣ ⎦

=

a a a b a c

b a b b b c

c a c b c c

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥

⎣ ⎦


12/14

Sol. :2

a b c

⎡ ⎤⎢ ⎥⎣ ⎦

=

a a a b a c

b a b b b c

c a c b c c

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥

⎣ ⎦

=

1 3

1 2

3 2

1

1

1

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦

= 1 + 21

2

3 –

12 –

22 –

32

a b c

⎡ ⎤⎢ ⎥⎣ ⎦ =

2 2 21 2 3 1 2 31 2

77. Answer (2)

Hint : 2 2 2 y mx a m b

Sol. : Equations of common tangents to both

hyperbolas are

y = 2 2

x a b

For 4 common tangents, a > b

78. Answer (2)

Hint : y = x is axis of both hyperbolas

Sol. :

x

y y x =

xy = 4

xy = 1 A(1, 1)

(2, 2)B

AB = 2

79. Answer (2)

Hint : S1 – S

2 = 0

Sol. : c1(–g, 0), (r

1 = |g|, c

2(0, – f), r

2 = |f |, equation

of common chord is

A

B

MC

1

(– , 0)g

r1

gx fy – = 0

gx fy – = 0

C1M =

2

2 2

g

g f

AM2 = r12 – (C

1M)2 =

42

2 2

gg

g f

=

2 2

2 2

g f

g f

AM = 2 2

| |gf

g f

AB = 2 2

2 | |gf

g f

80. Answer (2)

Hint : (d) = r

Sol. : Equation of tangent of x2 = – 4y is

tx = –y + t2 ...(i)

(i) is also tangent to x2 + y2 = 4

2

2

| |

1

t

t = 2

t4 = 4 + 4t2 t2 = 2 2 2

m = – t m2 = 2 2 2 [m2] = 4

81. Answer (3)

Hint : AA–1 = I

Sol. : A–1A = I, then compare corresponding

elements.

82. Answer (1)

Hint :�

a is perpendicular to c�

Sol. : a c� �

= a c� �

a c� �

ˆ ˆ ˆ( )c i j k �

a c� �

= b�

ˆ ˆ ˆ

ˆ ˆ1 1 0

i j k

i j

= 0, = 1

c�

= ˆk

83. Answer (1)

Hint : Multiply and divide by

1 1

2 2(sec ) (tan )n n

x x


13/14

Sol. :

R =

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

11

1 1 11

2 2 22(sec ) (tan ) (sec ) tan

...(sec tan )

n n n

nx x x x

x x

R =

2 2

22

1 1

2 2

11

22

1 1

2 2

(sec ) (tan )

(sec ) tan ...(sec tan )

(sec ) (tan )

n n

n

n

n n

x x

x x x x

x x

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟ ⎝ ⎠

= 1 1

2 2

1

(sec ) (tan )n n

x x

=

1

2

1

2

(cos )

1 (sin )

n

n

x

x

84. Answer (2)

Hint : Plane contains points A, B, C

Sol. : Plane containing all medians is the plane

passing through the vertices A, B and C

Equation of the plane is

PA BA CA

⎡ ⎤⎢ ⎥⎣ ⎦

= 0 where P(x, y, z)

1 1 1

1 0 1

x y z

= 0

x – 2y + z = 0

85. Answer (1)

Hint : Parabolic curve

Sol. : 0dy

dx

3ax2 + 6 0

y ax = 3 + 62

3a > 0, D < 0

a > 0

0 – 4(3a)6 < 0

a > 0

at a = 0, y = 6x, an increasing function

86. Answer (3)

Hint : 3sinx + 2cosx

= 3cos 2sin (3cos 2sin ) dx x x x

dx

Sol. :

I = 3sin 2cos

3cos 2sin

∫

x xdx

x x

=

(3cos 2sin )3cos 2sin

3cos 2sin 3cos 2sin

∫ ∫

dx x

x x dxA dx B dx

x x x x

3sinx + 2cosx = A(3cosx + 2sinx)

+ B(–3sinx + 2cosx)

Comparing coefficients of sinx and cosx on both

sides

A = 12

13, B =

5

13

121

1 13

5

13

⎛ ⎞

⎜ ⎟⎝ ⎠

A

B = –5

87. Answer (2)

Hint : Draw graph

Sol. :

x

y

AB

C

x = 1

y a =

(1, 1)

O

y = 2x – x2 (x – 1)2 = –(y – 1)

If y = a, then x = 1 1 a

A1 =

2 2

0(2 )x x dx∫ =

23

2

03

xx

⎡ ⎤⎢ ⎥

⎣ ⎦

= 8 4

43 3

A(OABC) = 1

3

1 1 2

0

1(2 ) 1

3

a

x x dx a a

∫

2 31 11 1 1 1 1

3 3a a a a


14/14

� � �

Put 1 – a = t2

2t3 = 1 t6 = 1

4 = 2–2

t2 =

2

32

a = 1 – t2 = 2

31 2

88. Answer (3)

Hint : Take log on both sides then convert in definite

integral

Sol. :

logl =1 1 2

lim log log log ... logn

n n n n n

n n n n n

⎛ ⎞⎡ ⎤ ⎛ ⎞ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎝ ⎠

=

21lim log

n

nr n

r

n n

⎛ ⎞⎜ ⎟⎝ ⎠

∑ = 2

1logxdx∫

= [xlogx – x]12 = 2log2 – 2 + 1 = 2log2 – 1

= 4

loge

⎛ ⎞⎜ ⎟⎝ ⎠

l = 4

e

89. Answer (1)

Hint : Form : 1

Sol. : l =

1

1 cot

0

sinlim

x x

x

x

x

⎛ ⎞⎜ ⎟⎝ ⎠

form : 1

∵0

sinlimx

x

x

⎛ ⎞⎜ ⎟⎝ ⎠

= 1

0

lim ( cot )x

x x

= 0

limtanx

x

x = 1

l =

sin 1lim 1

1 cot0

x

x x xxe

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

=

sin tanlim

tan0

x x x

x x xxe

⎛ ⎞⎜ ⎟⎝ ⎠

=

sin tanlim

tan0

x x x

x x xxe

⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

=

11

2e

=

1

e

90. Answer (4)

Hint :3

2

0

2( )

3 ∫

bb

f x dx b

Sol. :3

2

0

2( )

3

b bf x dx b ∫

d. w.r. t. b

f(b) = 2b – 2b2 f(x) = 2x – 2x2

f(x) = 2 1 2

24 4

x x⎡ ⎤ ⎢ ⎥⎣ ⎦

=

21 1

22 2

x⎛ ⎞ ⎜ ⎟⎝ ⎠

f(x)max.

= 1

2 when x =

1

2

Mock Test - 2 (Code-E) (Answers) All India Aakash Test ... · All India Aakash Test Series for JEE...

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Transcript of Mock Test - 2 (Code-E) (Answers) All India Aakash Test ... · All India Aakash Test Series for JEE...