mock test - 1 (code-a) (answers) all india aakash test series for … · 2019-06-04 · mock test -...

Mock Test - 1 (Code-A)(Answers) All India Aakash Test Series for JEE (Main)-2019

1/14

1. (3)

2. (4)

3. (1)

4. (2)

5. (3)

6. (2)

7. (2)

8. (3)

9. (3)

10. (1)

11. (3)

12. (3)

13. (2)

14. (2)

15. (3)

16. (3)

17. (1)

18. (1)

19. (4)

20. (3)

21. (1)

22. (1)

23. (3)

24. (3)

25. (4)

26. (4)

27. (1)

28. (1)

29. (1)

30. (4)

PHYSICS CHEMISTRY MATHEMATICS

31. (1)

32. (1)

33. (1)

34. (4)

35. (1)

36. (1)

37. (2)

38. (3)

39. (2)

40. (4)

41. (2)

42. (2)

43. (3)

44. (2)

45. (2)

46. (4)

47. (4)

48. (4)

49. (4)

50. (2)

51. (3)

52. (2)

53. (1)

54. (2)

55. (3)

56. (2)

57. (3)

58. (3)

59. (3)

60. (4)

61. (3)

62. (1)

63. (3)

64. (4)

65. (1)

66. (4)

67. (3)

68. (2)

69. (4)

70. (3)

71. (4)

72. (2)

73. (4)

74. (2)

75. (1)

76. (2)

77. (2)

78. (1)

79. (2)

80. (4)

81. (3)

82. (3)

83. (3)

84. (3)

85. (2)

86. (3)

87. (2)

88. (1)

89. (4)

90. (2)

Test Date : 10/02/2019

ANSWERS

MOCK TEST - 1 - Code-A

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

All India Aakash Test Series for JEE (Main)-2019 Mock Test - 1 (Code-A) (Hints & Solutions)

2/14

1. Answer (3)

Hint : Time taken should be least.

Sol. : tmin

= 350

47

= 54 s

v

t

7

3

4O

2. Answer (4)

Hint :/

� � �

C B C BV V V .

Sol. : AB = BC

x cos30° = 1x cos30°

= 1

3. Answer (1)

Hint : Buoyant force will not change.

Sol. : B = 3

4

Mg

4. Answer (2)

Hint : P = Fv

Sol. : mv2dv

dx = P

3 3(2 ) ( )3

mu u = Px

P = 3

7

3

mu

x

5. Answer (3)

Hint : 2Constant

V

T

P2V = Constant

Sol. : Q = U + W

W = 2R(T)

U = 5

2R T

6. Answer (2)

Hint : Velocities along LOI will be interchanged.

Sol. : LOI makes an angle of 30° with initial

direction of motion.

PART - A (PHYSICS)

v2 = v

0 cos 30°

v1 = v

0 sin 30°

f = cos230° = 3

4

7. Answer (2)

Hint : Acceleration is vector sum of normal and

tangential acceleration

Sol. :2 21

6 �m =

2mg

�

2� = 3g

2mg

� =

2

3

m �

= 3

2

g

�

� = 3

2

g

anet

= 3 5

2

g

8. Answer (3)

Hint : Field due to straight section PQ, PR is zero

Sol. :0 0 (cos37 cos53 )

4P

IB

d

53°

37°d

0 0 7

4 (4.8) 5

IB

0 07

96

IB

9. Answer (3)

Hint : e = Velocity of separation

Velocity of approach

Sol. : v = 0 0

6 6

mv v

m

= 0

0

2

22

26

12

mvv

m

�

�� =

0v

� m v, 0

Mock Test - 1 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019

3/14

e = 0

2

�v

v =

0 0

0

26 2

3

v v

v

10. Answer (1)

Hint : Loss in PE = gain in KE

Sol. :21

2 2

GMm GMmmv

R R

v gR

11. Answer (3)

Hint :21

2

rv U

Sol. :

22

rel

1 2

2 3

mv

m = (2m)gh

vrel

= 6gh

12. Answer (3)

Hint : Level will get stabilised.

Sol. : 2a gh = Q

h =

2 2 2

2 3 2

(10 )5 m

2 2 10 (10 )

Q

ga

13. Answer (2)

Hint : Linear momentum remains conserved.

Sol. :1

1

hp

2

2

hp

2 1

h h h

14. Answer (2)

Hint : Temperature should be minimum.

Sol. : x = pv

= 0 2p v

v

dx

dv = 0

p0 –

20

v

v = 0

p

Tmin

= 0

0

2p

R p

= 0

2p

R

Umin

= 0

32

2p =

03 p

15. Answer (3)

Hint : Draw the corresponding phase diagram.

Sol. : Asin1 = 3

2

A

60°30°

23 sin( )A =

3

2A

( – 2) =

6

1 = 60°,

2 = 150°

2 –

1 = 90°

16. Answer (3)

Hint :1

3

4

v

� =

2

4

2

v

�

Sol. :1

3

4

v

� =

2

4

2

v

�

�1 = 2

3

8�

17. Answer (1)

Hint : F = dE

pdx

Sol. : E = 2

04

Q

x

dE

dx = 3

02

Q

x

F = 3 3

0 0

(2 )

2

Qq a Qqa

x x

18. Answer (1)

Hint : Field due to dipole at equatorial point.

Sol. : Net dipole moment is 4Qa along –ve x axis. At

a point in equatorial plane field 3

04

pE

x

�

�


4/14

19. Answer (4)

Hint : Equivalent time constant is 2

CR

Sol. : Q = CV0(1 – e–2t/RC)

V0

R/2

C

Q =

2ln3

20

1

CR

RCCV e

= 0

11

3

CV

= 0

2

3

V C

20. Answer (3)

Hint : Particle will emerge out radially.

Sol. : Deviation = 2

Change in momentum = 0

2 mv

R

O

RR

O

21. Answer (1)

Hint :2

v �

Sol. : 02

v �

1( 2 )

v � �

2( 2 )

v � �

(2 –

1) =

2

48

v

� �

��

22. Answer (1)

Hint : Ray should fall normally on the mirror.

Sol. : Image formed by 1st lens should be in focal

plane of 2nd.

23. Answer (3)

Hint : = 1 +

2

Sol. : = 2(i – r)

i

r r i

r = 30°

sin

sin

i

r = 3

i = 60°

= 60°

24. Answer (3)

Hint : Path difference is equal to 9

2

at position of

5th dark fringe

Sol. : = d

D

d

D

2d

(2d) = 9

2

22d

D =

9

2

d =3

2D

25. Answer (4)

Hint :2

D x b

r x

Sol. : D = 2

( )rx b

x

dD

dt = 2

2 br dx

dtx =

1m/s

2

x

r

D


5/14

26. Answer (4)

Hint : K 2E

Sol. : Let kinetic energy be K

2

K = 13.6

K = 27.2 eV

27. Answer (1)

Hint : Total number of nuclei will be constant.

Sol. : After long time, number will become

constant. Rate of formation = rate of decay

28. Answer (1)

Hint : Its a combination of OR, AND gate.

Sol. : X = A(A + B)

PART - B (CHEMISTRY)

31. Answer (1)

Hint : Down the group stability of group-1 hydrides

decreases

Sol. : According to VSEPR, its structure is

Xe

••

FF

FF

O

Lone pair is trans to oxygen

32. Answer (1)

Hint : Basic strength depends on availability of lone

pair

Sol. : H N2 C NH2• •

• •

NH

Because of resonance, doubly bonded N

atom in this structure has most negative

charge density.

33. Answer (1)

Hint : Ozonolysis followed by Aldol condensation

Sol. :O

3

Zn

X

O

O

OH (dil.)–

O

Y

H

34. Answer (4)

Hint : At constant T, Kp remians constant

Sol. : (g) (g)2R A��⇀

↽��

At P1

P0

11

3

P0

1

6

Kp =

1

5

16 P

At P2

P0

21

3

P0

1

3

Kp =

2

2

P

∵ Kp remain constant

1

5

16 P = 2

2

P

5

32 =

1

2

P

P

35. Answer (1)

Hint : NaCl is a neutral salt

Sol. : At pH = 3

[H+] = 10–3 molar

[H+] after adding equal volume of NaCl

solution

[H+] = 3

10

2

molar

pH = 4 – log 5

= 3.3

29. Answer (1)

Hint : = C

B

I

I

Sol. : 51C

B

I

I

IE = I

C + I

B

30. Answer (4)

Hint : To overcome the long range repulsive effect

of electrostatic forces of protons

Sol. : To overcome the long range repulsive effect



6/14

36. Answer (1)

Hint : = n(n 2)

Sol. : = 5.91 B.M

Number of unpaired e– in M+2 is 5

Mn+2 has 5 unpaired e– (3d5)

37. Answer (2)

Hint : moles = V(ml)

22400

Sol. : 1120 ml of N2O = 0.05 mol = 2.2 g

∵ d = mass

volume

V = 2.2 110

ml1.18 59

38. Answer (3)

Hint : Neutralisation of weak acid and strong base

Sol. : Phenol will behave as a weak acid with strong

base KOH

39. Answer (2)

Hint : Preparation of ketone

Sol. : A = CH3

CH CH3

Cl

B = CH3

CH CH3

MgCl

C = CH3CH CH

3

CN

D = CH3

CH CH3

C NMgCl

CH

CH3

CH3

E = C CH(CH )3 2

O

CH(CH 3 2)

40. Answer (4)

Hint : -keto acid undergo decarboxylation on

heating

Sol. : A

O

COOH

B

O

41. Answer (2)

Hint : Solution boils at higher temperature

Sol. : One unit cell contains

Four formula units

42. Answer (2)

Hint : meq of Fe2(C

2O

4)3 = meq of K

2Cr

2O

7

Sol. : 6e 3

2 2 4 3 2Fe (C O ) 2Fe 6CO

(nF = 6)

6e2 3

2 7Cr O 2Cr

(nF = 6)

43. Answer (3)

Hint : S atoms have O.S. of +6 and –2

Sol. : Structure of 2

2 3S O

S

S

O

–

O

44. Answer (2)

Hint : Probability is zero at nucleus

Sol. : 3p will have one radial node

45. Answer (2)

Hint : Stability of C+ increases the rate of SN1

Sol. :

+

N

H

aromatic,

+

anti-aromatic

46. Answer (4)

Hint : Boiling point of cis butene is greater than

trans butene

Sol. : A is butane

B is cis-but-2-ene

C, D are trans-but-2-ene

47. Answer (4)

Hint : Rearrangement of C+


7/14

Sol. :

Cl anhyd.FeCl

3

+

Rearrangement

+

Product

48. Answer (4)

Hint : Periodic acid is used for oxidation

Sol. :C OH

C OH

+ HIO4

C O

C O

I

O

O–

to form this

complex both

the –OH groups

should be in

syn conformation.

49. Answer (4)

Hint : An acid is added to strong base, pH will

decrease sharply.

Sol. :

pH

HCl(ml)

50. Answer (2)

Hint : Bond length in N2O is least

Sol. : Bond length in N2O

3 is 186 pm and in N

2O

4

is 175 pm

51. Answer (3)

Hint : Zn causes dehalogenation

Sol. : P1 is

NH2

P2 is

+

Aromatic

P3 is

OH

52. Answer (2)

Hint : 0

0

a2kt ln

a 2x

Sol. : 2A 3B

t = 0 a 0

t = tf a – 2x 3x

a 2x

3x

=

4

3

3a – 6x = 12x

3a = 18x

x = a

6

0

0

a2kt ln

a 2x

t = 3

2.303 a2 log

a2.303 10 a3

= 85 min

53. Answer (1)

Hint : Charge passed in faradays is equal to g meq

of H2

Sol. : Volume at STP will be equal to 8V ml

H2O H

2 + 2

1O

2

2 1

8V ml 8V ml3 3

= 16V

3

Number of moles of H2 produced

= 16V

mole3 22400

to produce 1 mole of H2 = 2F C charge is

required.

So, to produce 16V

ml3 22400

of H2

= 16V VF

2F C3 22400 2100

54. Answer (2)

Hint : Preferential adsorption theory


8/14

PART - C (MATHEMATICS)

Sol. : Generally, common ion present in excess is

adsorbed.

55. Answer (3)

Hint : E is a state function

Sol. : E1 = E

2 = E

3

56. Answer (2)

Hint : Cu2+ belongs to group II

Sol. : Mn2+, Ni2+, Co2+ belong to group IV

57. Answer (3)

Hint : N2O

4 is a dimer of NO

2

Sol. : N2O

4 is an acidic gas

58. Answer (3)

Hint : PMMA is polymethyl methacrylate

61. Answer (3)

Hint : I = 2sec tan

2 2 2

x x xdx dx

(Integrate 1st by parts)

Sol. :2

sin

2cos2

x xdx

x

=

21sec

2 2

xx dx + tan

2

xdx

= 2

tan1 2

sec 112 2

2

x

xx dx dx

tan2

xdx

= tan

1 122 tan tan

12 2 2 2

2

x

x xx dx dx

= x tan2

xc

62. Answer (1)

Hint : A matrix A is symmetric if AT = A and skew-

symmetric if AT = –A and its all diagonal

elements are zeros and zero matrix is both

symmetric and skew- symmetric

Sol. : Probability = 6 3

9 9 9

5 5 1

5 5 5 = 3 6 9

1 1 1

5 5 5

where probability (that matrix is symmetric)

= 9

total no. of ways to fill up ' ' places

5

x

x x x

x x

x

�

�

=

6

9

5

5

Probability (that matrix is skew-symmetric)

= 9


5

x

0

0

0

x x

x

�

� �

=

3

9

5

5

Probability (that matrix is both symmetric

and skew-symmetric) = 9

1

5

63. Answer (3)

Hint : |adj(adj(adj A))| = |A|(3–1)3 = |A|8

Sol. : |A| = x + y + z

We have, |adj(adj(adj A))| = |A|(3–1)3 = |A|8 = 78

(x + y + z) = 7 ...(1)

Number of such matrices = No. of +ve integral

solutions of (1)

= 7–1C3–1

= 6C2 = 15

64. Answer (4)

Hint :

1 1

2

1 1

1 1

4 2 2tan tan

1 14 31

2 2

r r

r r

rr r

Sol. : CH2

CH Ph CH2CH

Ph

n

Polystyrene

Styrene

59. Answer (3)

Hint : Fact based.

Sol. : For preparation of XeF6, Xe:F

2 is 1:20

60. Answer (4)

Hint : Glycogen is known as animal starch

Sol. : It is a branched chain polysaccharide of

glucose and is found in liver, muscles and

brain.


9/14

Sol. :1

21

1tan

11

4

r r

= 1

1

1tan

1 11

2 2

rr r

= 1 1

1

1 1tan tan

2 2r

r r

=1 1 1 13 1 5 3

lim tan tan tan tan2 2 2 2

n

1 11 1... tan tan

2 2

n n

= 1 11 1

lim tan tan2 2

n

n

= 1 1

tan2 2

= 1 11

cot tan 22

65. Answer (1)

Hint : Here equation of chord of two circles and the

equation of chord of contact made by the

tangents from P(x1, y

1) to first circle will

represent same equation comparing we get

values of x1 and y

1.

Sol. : Equation of common chord AB of circles is

S1 – S

2 = 0

6x + 8y – 31 = 0 ...(1)

P x y( , )1 1

A

B

Also equation of chord of contact AB made

by the tangents from P(x1, y

1) to the circle

S1 = 0 is

T = 0

xx1 + yy

1 – 18 = 0 ...(2)

comparing (1) and (2)

1 118

6 8 31

x y

(x1, y

1) =

108 144,

31 31

66. Answer (4)

Hint : g(x) = f(|x|) – |f(x)|

= 6, 2 0

2( 3), 0 2

x x

x x

Sol. : g(x) = 6, 2 0

2( 3), 0 2

x x

x x

Required area = area of trapezium OPQR +

area of trapezium ORST

= 1 1(10)(2) (8)(2)

2 2 = 18 sq. units

xP

R (0, –6)

S(2, –2)(–2, –4)

Q

–2O T

x = 2x = – 2

y x = 2( – 3)y–x

= – 6

y

2

67. Answer (3)

Hint : Integrate by parts

Sol. :

2

1

1log 1x dx

x

=

22 2 22

1

1

1

1log 1

12 21

x x xdx

x

x

=

22

2

11

1 1log 1

2 2 1

x xdx

x x

=

2

1

3 1 1 12 log log 2 log (1 )

2 2 2 2

x x

= 9 1 1log log 2 (2 1) log 3 log 2

4 2 2 e e

= 3 3

log4

e

= 27

log log 274 4

e kk e


10/14

68. Answer (2)

Hint : (C0 + C

1 + C

2 + ... + C

n)2

= (C02 + C

12 + ... + C

n

2) + 20

i ji j n

CC

Sol. :0

i ji j n

CC

= (C

0C

1 + C

0C

2 + ...C

0C

n) +

(C1C

2 + C

1C

3 + ... C

1C

n) + (C

2C

3 + C

2C

4 +

...C2C

n) + ... = R (say)

As (C0 + C

1 + C

2 + ... C

n)2

= 2 2 2

0 1... 2

nC C C R

R = 2

2 12

2

n

n nC

,

since C0 + C

1 + C

2 + ... + C

n = 2n

ad 2 2 2 2

0 1 2...

nC C C C

= coefficient of xn in (1 + x)n (x + 1)n

= 2nCn

69. Answer (4)

Hint : Check for reflexive, symmetric and transitive

relation

Sol. : Reflexivity: We have sin2p + cos2p = 1

aR

pRp pR R is reflexive

Symmetry: Let pRq

sin2p + cos2q = 1

1 – cos2p + 1 – sin2q = 1

sin2q + cos2p = 1

qRp R is symmetric

Transitivity:

Let pRq and qRr

sin2p + cos2q = 1 ...(1)

and sin2q + cos2r = 1 ...(2)

By [(1) + (2)]

sin2p + (cos2q + sin2q) + cos2r = 2

sin2p + cos2r = 1

pRr R is transitive also.

Hence R is equivalence relation.

70. Answer (3)

Hint : Draw the figure and then use tan =

p

h for

two triangles

Sol. :

2 3

A B C Q

P

a

Let PQ be the tower.

From APQ, tan = PQ

AQ

AQ = tan

PQ

...(1)

From BPQ, tan2 = PQ

BQ

BQ = tan2PQ

...(2)

Here a = AB – AQ – BQ = tan tan2

PQ PQ

a =

2

1 1

2tantan

1 tan

PQ

21 1 tan

tan 2tanPQ

22 1 tan

2tanPQ a

PQ2

2tansin2

1 tan

a a

71. Answer (4)

Hint : Equation of tangent to the parabola is

y = mx + a

m

if it passes through (h, k)

k = mh + a

m

Sol. : Equation of tangent to the parabola y2 = 4ax

is


11/14

y = a

mx

m

If it passes through (h, k), then

m2h – mk + a = 0

m1 + 4m

1 =

k

h,

2

14

am

h

2

4

25

k

h = a locus of (h, k) is

4y2 = 25ax

72. Answer (2)

Hint : Slope of tangent is m = dy

dx

For maxima or minima of m, 0dm

dx and find

value of x and if 2

20d m

dx, then m is minima

at that point

Sol. : Here, y = 2ex sin4 2

x

cos4 2

x

= ex sin 24 2

x

= ex sin (/2 – x) = ex cosx

m = dy

dx= slope of tangent

= ex(–sinx) + ex cosx

= ex(–sinx + cosx)

dy

dx = ex(–cosx – sinx) + (–sinx + cosx) ex

= ex(–2 sinx)

2

22 cos sin

x xd me x x e

dx

= –2ex(cosx + sinx)


dx

–2ex sinx = 0 sinx = 0

x = 0, , 2

When

x = 0,

20

22 (1 0) 0

d me

dx

When

x = ,

2

22 (cos sin ) 2 0 d me e

dx

When

x = 2,

22

22 (cos2 sin2 ) d me

dx

= –2e2 (1 + 0) < 0

m is minimum at x =

73. Answer (4)

Hint :

12

1

2 1

2tan , 01cos

1 2tan , 0

x xx

x x x

Sol. :

12

1

2 1

2tan , 01cos

1 2tan , 0

x xx

x x x

Here

L.H.L. = 1

0

tanlim 2x

x

x

= –2 × 1 = –2

R.H.L. = 1

0

tanlim 2x

x

x

= 2 × 1 = 2

L.H.L R.H.L. given limit is non-existent

74. Answer (2)

Hint : Asymptotes are x2 – 2y2 = 0 x ± 2 0y

Any point on hyperbola are (2sec, 2 tan )

Sol. : Asymptotes are 2 0x y and any point

on hyperbola are (2sec, 2 tan)

Product of perpendiculars from

2 sec , 2 tan to the asymptotes =

2 sec 2 2 tan 2 sec 2 2 tan

3 3

= 2 2

4 sec tan4

3 3

75. Answer (1)

Hint : f(x) is defined if

log|x|–3

(x2 + 2x + 1) 0 ...(1)

x2 + 2x + 1 > 0, |x| – 3 1

Sol. : Here, f(x) is defined if

log|x|–3

(x2 + 2x + 1) 0 ...(1)


12/14

Case I:

When |x| – 3 > 1

|x| > 4

x < –4 or x > 4

x(–, –4) (4, ) ...(2)

and then from (1)

(x2 + 2x + 1) (|x| –3)0

x(x + 2) 0

x –2 or x 0 ...(3)

From (2) and (3),

x(–, –4) (4, ) ...(4)

Case II:

When 0 < |x| – 3 < 1

3 < |x| < 4

–4 < x < –3 or 3 < x < 4 ...(5)

Then from (1),

(x2 + 2x + 1) (|x| – 3)0

x(x + 2) 0

–2 x 0 ...(6)

From (5) and (6),

x ...(7)

x2 + 2x + 1 > 0

xR ...(8)

|x| – 3 1

x ±4 ...(9)

76. Answer (2)

Hint : f(x) =

( 2), if 2

( 2), if 2

x xx

x

x xx

x

=

2 2, if 2

2, if 2

xx

x

x

x

Sol. : f(x) =

( 2), if 2

( 2) 2, if 2

x xx

x

x xx

x x

f (x) = 2

2

(2 1 0) (2 2) 1, if 2

2, if 2

x xx

x

x

x

f (2–) = 2

4 (4 2) 1,

22

f (2+) = 2 1

4 2

f(x) is non-differentiable at x = 2

77. Answer (2)

Hint : Slope of tangent is

f(x) = 2 2

( 2)(1) ( 3)(1) 5

( 2) ( 2)

x x

x x

Sol. : f(x) = 2

5

( 2)x

= –5 at x = 3

Equation of tangent at (3, 6) is

y – 6 = –5(x – 3) 5x + y = 21

Hence, given line is a tangent on f(x) at (3, 6)

78. Answer (1)

Hint : Given equation is

2 2

2 2

(1 )log

( 1) ( 1)

e

dy x xy x

dx x x x

and IF =

2

2

( 1)

( 1)

xdx

x xe

= 2

1x

x

Sol. : From given differential equation

2 3

2 2

( 1)log

( 1) ( 1)e

dy x xy x

dx x x x x

...(1)

I.F. =

2

2

( 1)

( 1)

xdx

x xe

= 2

2 1

1

xdx

xxe

= 2log 1 log e e

x x

e

= 2

1x

x

Solution of given equation is 21y x

x

=

23

2

1log

1

e

x xx dx

x x x

= 2 1

log log2 2

e e

x xx x dx x dx

x

= 2

21log

2 4

e

xx x C


13/14

C = 0

( ) 0y e

79. Answer (2)

Hint : Image of A(2, 1) due to line x = 3 is P(4, 1)

and so equation of side BC is the line joining

points C(5, 3) and P(4, 1)

Sol. : Image of A(2, 1) due to the line x = 3 is

(4, 1) which lies on BC.

Equation of side BC is

y – 3 = 3 1

( 5)5 4

x

2x – y = 7

Which is satisfied by the point (3, –1)

B = (3, –1)

80. Answer (4)

Hint : 2 2

( 3) 3 (3 1)(1 0) 80

( 3) ( 3)

dy x x

dx x x

Sol. :2

8( ) 0

( 3)f x

x

Which is true for all x R

Also let 3 1

3

xy

x

13 1 3 1( )

3 3

y y

x f yy y

f –1(x) = 3 1

3

x

x

f(x) is its own inverse Also,

f(x) = 8

33x

which is unbounded.

81. Answer (3)

Hint : | a�

b�

| + |2( a�

b�

)| = || a�

|| b�

| cos| +

|2|a�

||b�

| sin ˆn |

Sol. : |(a�

b�

)| + |2(a�

b�

)| = 12 cos + 24 sin

2 212 24

Maximum value is 2 212 24 = 12 5

82. Answer (3)

Hint : Put x2 + x = y

Given equation becomes (y – 1)(y – 2) – 6 = 0

Sol. : Let x2 + x = y

(y – 1)(y – 2) – 6 = 0

y = 4, –1

x2 + x = 4 or x2 + x = –1

x2 + x – 4 = 0

Which has real roots

Sum of real (Non- real roots)

roots = –1

83. Answer (3)

Hint : From given equation (z – 2i) (iz2 – 4) = 0

Sol. : (z – 2i) (iz2) – 4(z – 2i) = 0

(z – 2i) (iz2 – 4) = 0

z = 2i or z2 = 4

4ii

|z| = |2i| = 2 |z2| = |–4i|

|z| = 2 |z|2 = 4 |z| = 2

84. Answer (3)

Hint : Use condition for coplanarity of planes

Sol. : If planes are coplanar, then

2 1 3

1 2 0

2 1

p

q

2(–1 + 4) –1(p + 2q) + (–3) (2p + q) = 0

6 – p – 2q – 6p – 3q = 0

7p + 5q = 6

85. Answer (2)

Hint : Find positive integral solutions of x1 + x

2 + x

3

= 12 or 13 or 14 or 15 and then add all of

them

Sol. : Here,

x1 + x

2 + x

3 = 12 or

x1 + x

2 + x

3 = 13 or

x1 + x

2 + x

3 = 14 or

x1 + x

2 + x

3 = 15


14/14

Number of solutions

= 12–1C3–1

+ 13–1C3–1

+ 14–1C3–1

+ 15–1C3–1

= 11C2 + 12C

2 + 13C

2 + 14C

2

= 55 + 66 + 78 + 91

= 290

86. Answer (3)

Hint : Length of projection = AB sin

Sol. : cos = | 8 1 4 |

33 6

=

11

18

sin = 7

18

Length of projection = AB sin

= 7 77

3318 6

87. Answer (2)

Hint :3 11,

36 12p q

Sol. : For single throw favourable cases are (4, 6),

(5, 5) and (5, 4)

p = 3 1

36 12

q = 1 11

1 112 12

p

Probability =

2

3

2

1 11 11

12 12 576C

88. Answer (1)

Hint : 1

1

2

2 ( 1)

n

n r

r

rS

r r

= 1

1

1 1

2 2 ( 1)

n

r r

rr r

Sol. : S = 1

1

2

2 ( 1)

n

r

r

r

r r

= 1

1

2( 1)

2 ( 1)

n

r

r

r r

r r

= 1

1

1 1

2 2 ( 1)

n

r r

rr r

Sn

= 1

1 1

2 2 ( 1)nn

= 1

2 ( 1) 1

2 ( 1)

n

n

n

n

7

6

S

S=

10 7

8 6

2 1 2 7

2 8 (2 .7 1)

= (1024 1) 7

2 8 447

=

1023 7

16 447

=

2387

2384

89. Answer (4)

Hint : Make truth table and then find equivalent

statement

Sol. : Making truth table we can find equivalent

statement as p q

Truth table

( ) ( ) p q p q q p p q q p p q

T T T T T T

T F F T F F

F T T F F F

F F T T T T

90. Answer (2)

Hint : = 2 21( )

i i i iN f x f x

N

Sol. : = 2 21( )

i i i iN f x f x

N

= 2

40 (12 36 112 180 490 1100)1

40 (6 12 28 30 70 110)

= 1

77200 6553040

= 1

108.03 2.740

� � �

Mock Test - 1 (Code-B)(Answers) All India Aakash Test Series for JEE (Main)-2019

1/14

1. (4)

2. (1)

3. (1)

4. (1)

5. (4)

6. (4)

7. (3)

8. (3)

9. (1)

10. (1)

11. (3)

12. (4)

13. (1)

14. (1)

15. (3)

16. (3)

17. (2)

18. (2)

19. (3)

20. (3)

21. (1)

22. (3)

23. (3)

24. (2)

25. (2)

26. (3)

27. (2)

28. (1)

29. (4)

30. (3)

PHYSICS CHEMISTRY MATHEMATICS

31. (4)

32. (3)

33. (3)

34. (3)

35. (2)

36. (3)

37. (2)

38. (1)

39. (2)

40. (3)

41. (2)

42. (4)

43. (4)

44. (4)

45. (4)

46. (2)

47. (2)

48. (3)

49. (2)

50. (2)

51. (4)

52. (2)

53. (3)

54. (2)

55. (1)

56. (1)

57. (4)

58. (1)

59. (1)

60. (1)

61. (2)

62. (4)

63. (1)

64. (2)

65. (3)

66. (2)

67. (3)

68. (3)

69. (3)

70. (3)

71. (4)

72. (2)

73. (1)

74. (2)

75. (2)

76. (1)

77. (2)

78. (4)

79. (2)

80. (4)

81. (3)

82. (4)

83. (2)

84. (3)

85. (4)

86. (1)

87. (4)

88. (3)

89. (1)

90. (3)

Test Date : 10/02/2019

ANSWERS

MOCK TEST - 1 - Code-B

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

All India Aakash Test Series for JEE (Main)-2019 Mock Test - 1 (Code-B) (Hints & Solutions)

2/14

1. Answer (4)

Hint : To overcome the long range repulsive effect


Sol. : To overcome the long range repulsive effect


2. Answer (1)

Hint : = C

B

I

I

Sol. : 51C

B

I

I

IE = I

C + I

B

3. Answer (1)

Hint : Its a combination of OR, AND gate.

Sol. : X = A(A + B)

4. Answer (1)

Hint : Total number of nuclei will be constant.

Sol. : After long time, number will become

constant. Rate of formation = rate of decay

5. Answer (4)

Hint : K 2E

Sol. : Let kinetic energy be K

2

K = 13.6

K = 27.2 eV

6. Answer (4)

Hint :2

D x b

r x

Sol. : D = 2

( )rx b

x

dD

dt = 2

2 br dx

dtx =

1m/s

2

x

r

D

PART - A (PHYSICS)

7. Answer (3)

Hint : Path difference is equal to 9

2

at position of

5th dark fringe

Sol. : = d

D

d

D

2d

(2d) = 9

2

22d

D =

9

2

d =3

2D

8. Answer (3)

Hint : = 1 +

2

Sol. : = 2(i – r)

i

r r i

r = 30°

sin

sin

i

r = 3

i = 60°

= 60°

9. Answer (1)

Hint : Ray should fall normally on the mirror.

Sol. : Image formed by 1st lens should be in focal

plane of 2nd.

10. Answer (1)

Hint :2

v �

Mock Test - 1 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019

3/14

Sol. : 02

v �

1( 2 )

v � �

2( 2 )

v � �

(2 –

1) =

2

48

v

� �

��

11. Answer (3)

Hint : Particle will emerge out radially.

Sol. : Deviation = 2

Change in momentum = 0

2 mv

R

O

RR

O

12. Answer (4)

Hint : Equivalent time constant is 2

CR

Sol. : Q = CV0(1 – e–2t/RC)

V0

R/2

C

Q =

2ln3

20

1

CR

RCCV e

= 0

11

3

CV

= 0

2

3

V C

13. Answer (1)

Hint : Field due to dipole at equatorial point.

Sol. : Net dipole moment is 4Qa along –ve x axis. At

a point in equatorial plane field 3

04

pE

x

�

�

14. Answer (1)

Hint : F = dE

pdx

Sol. : E = 2

04

Q

x

dE

dx = 3

02

Q

x

F = 3 3

0 0

(2 )

2

Qq a Qqa

x x

15. Answer (3)

Hint :1

3

4

v

� =

2

4

2

v

�

Sol. :1

3

4

v

� =

2

4

2

v

�

�1 = 2

3

8�

16. Answer (3)

Hint : Draw the corresponding phase diagram.

Sol. : Asin1 = 3

2

A

60°30°

23 sin( )A =

3

2A

( – 2) =

6

1 = 60°,

2 = 150°

2 –

1 = 90°

17. Answer (2)

Hint : Temperature should be minimum.

Sol. : x = pv

= 0 2p v

v

dx

dv = 0

p0 –

20

v

v = 0

p


4/14

Tmin

= 0

0

2p

R p

= 0

2p

R

Umin

= 0

32

2p =

03 p

18. Answer (2)

Hint : Linear momentum remains conserved.

Sol. :1

1

hp

2

2

hp

2 1

h h h

19. Answer (3)

Hint : Level will get stabilised.

Sol. : 2a gh = Q

h =

2 2 2

2 3 2

(10 )5 m

2 2 10 (10 )

Q

ga

20. Answer (3)

Hint :21

2

rv U

Sol. :

22

rel

1 2

2 3

mv

m = (2m)gh

vrel

= 6gh

21. Answer (1)

Hint : Loss in PE = gain in KE

Sol. :21

2 2

GMm GMmmv

R R

v gR

22. Answer (3)

Hint : e = Velocity of separation

Velocity of approach

Sol. : v = 0 0

6 6

mv v

m

= 0

0

2

22

26

12

mvv

m

�

�� =

0v

� m v, 0

e = 0

2

�v

v =

0 0

0

26 2

3

v v

v

23. Answer (3)

Hint : Field due to straight section PQ, PR is zero

Sol. :0 0 (cos37 cos53 )

4P

IB

d

53°

37°d

0 0 7

4 (4.8) 5

IB

0 07

96

IB

24. Answer (2)

Hint : Acceleration is vector sum of normal and

tangential acceleration

Sol. :2 21

6 �m =

2mg

�

2� = 3g

2mg

� =

2

3

m �

= 3

2

g

�

� = 3

2

g

anet

= 3 5

2

g

25. Answer (2)

Hint : Velocities along LOI will be interchanged.

Sol. : LOI makes an angle of 30° with initial

direction of motion.

v2 = v

0 cos 30°

v1 = v

0 sin 30°

f = cos230° = 3

4

26. Answer (3)

Hint : 2Constant

V

T

P2V = Constant


5/14

PART - B (CHEMISTRY)

31. Answer (4)

Hint : Glycogen is known as animal starch

Sol. : It is a branched chain polysaccharide of

glucose and is found in liver, muscles and

brain.

32. Answer (3)

Hint : Fact based.

Sol. : For preparation of XeF6, Xe:F

2 is 1:20

33. Answer (3)

Hint : PMMA is polymethyl methacrylate

Sol. : CH2

CH Ph CH2CH

Ph

n

Polystyrene

Styrene

34. Answer (3)

Hint : N2O

4 is a dimer of NO

2

Sol. : N2O

4 is an acidic gas

35. Answer (2)

Hint : Cu2+ belongs to group II

Sol. : Mn2+, Ni2+, Co2+ belong to group IV

36. Answer (3)

Hint : E is a state function

Sol. : E1 = E

2 = E

3

37. Answer (2)

Hint : Preferential adsorption theory

Sol. : Q = U + W

W = 2R(T)

U = 5

2R T

27. Answer (2)

Hint : P = Fv

Sol. : mv2dv

dx = P

3 3(2 ) ( )3

mu u = Px

P = 3

7

3

mu

x

28. Answer (1)

Hint : Buoyant force will not change.

Sol. : B = 3

4

Mg

29. Answer (4)

Hint :/

� � �

C B C BV V V

Sol. : AB = BC

x cos30° = 1x cos30°

= 1

30. Answer (3)

Hint : Time taken should be least.

Sol. : tmin

= 350

47

= 54 s

v

t

7

3

4O

Sol. : Generally, common ion present in excess is

adsorbed.

38. Answer (1)

Hint : Charge passed in faradays is equal to g meq

of H2

Sol. : Volume at STP will be equal to 8V ml

H2O H

2 + 2

1O

2

2 1

8V ml 8V ml3 3

= 16V

3

Number of moles of H2 produced

= 16V

mole3 22400

to produce 1 mole of H2 = 2F C charge is

required.

So, to produce 16V

ml3 22400

of H2

= 16V VF

2F C3 22400 2100

39. Answer (2)

Hint : 0

0

a2kt ln

a 2x


6/14

Sol. : 2A 3B

t = 0 a 0

t = tf a – 2x 3x

a 2x

3x

=

4

3

3a – 6x = 12x

3a = 18x

x = a

6

0

0

a2kt ln

a 2x

t = 3

2.303 a2 log

a2.303 10 a3

= 85 min

40. Answer (3)

Hint : Zn causes dehalogenation

Sol. : P1 is

NH2

P2 is

+

Aromatic

P3 is

OH

41. Answer (2)

Hint : Bond length in N2O is least

Sol. : Bond length in N2O

3 is 186 pm and in N

2O

4

is 175 pm

42. Answer (4)

Hint : An acid is added to strong base, pH willdecrease sharply.

Sol. :

pH

HCl(ml)

43. Answer (4)

Hint : Periodic acid is used for oxidation

Sol. :C OH

C OH

+ HIO4

C O

C O

I

O

O–

to form this

complex both

the –OH groups

should be in

syn conformation.

44. Answer (4)

Hint : Rearrangement of C+

Sol. :

Cl anhyd.FeCl

3

+

Rearrangement

+

Product

45. Answer (4)

Hint : Boiling point of cis butene is greater than

trans butene

Sol. : A is butane

B is cis-but-2-ene

C, D are trans-but-2-ene

46. Answer (2)

Hint : Stability of C+ increases the rate of SN1

Sol. :

+

N

H

aromatic,

+

anti-aromatic

47. Answer (2)

Hint : Probability is zero at nucleus

Sol. : 3p will have one radial node

48. Answer (3)

Hint : S atoms have O.S. of +6 and –2


7/14

Sol. : Structure of 2

2 3S O

S

S

O

–

O

49. Answer (2)

Hint : meq of Fe2(C

2O

4)3 = meq of K

2Cr

2O

7

Sol. : 6e 3

2 2 4 3 2Fe (C O ) 2Fe 6CO

(nF = 6)

6e2 3

2 7Cr O 2Cr

(nF = 6)

50. Answer (2)

Hint : Solution boils at higher temperature

Sol. : One unit cell contains

Four formula units

51. Answer (4)

Hint : -keto acid undergo decarboxylation on

heating

Sol. : A

O

COOH

B

O

52. Answer (2)

Hint : Preparation of ketone

Sol. : A = CH3

CH CH3

Cl

B = CH3

CH CH3

MgCl

C = CH3CH CH

3

CN

D = CH3

CH CH3

C NMgCl

CH

CH3

CH3

E = C CH(CH )3 2

O

CH(CH 3 2)

53. Answer (3)

Hint : Neutralisation of weak acid and strong base

Sol. : Phenol will behave as a weak acid with strong

base KOH

54. Answer (2)

Hint : moles = V(ml)

22400

Sol. : 1120 ml of N2O = 0.05 mol = 2.2 g

∵ d = mass

volume

V = 2.2 110

ml1.18 59

55. Answer (1)

Hint : = n(n 2)

Sol. : = 5.91 B.M

Number of unpaired e– in M+2 is 5

Mn+2 has 5 unpaired e– (3d5)

56. Answer (1)

Hint : NaCl is a neutral salt

Sol. : At pH = 3

[H+] = 10–3 molar

[H+] after adding equal volume of NaCl

solution

[H+] = 3

10

2

molar

pH = 4 – log 5

= 3.3

57. Answer (4)

Hint : At constant T, Kp remians constant

Sol. : (g) (g)2R A��⇀

↽��

At P1

P0

11

3

P0

1

6

Kp =

1

5

16 P

At P2

P0

21

3

P0

1

3

Kp =

2

2

P

∵ Kp remain constant

1

5

16 P = 2

2

P

5

32 =

1

2

P

P


8/14

PART - C (MATHEMATICS)

61. Answer (2)

Hint : = 2 21( )

i i i iN f x f x

N

Sol. : = 2 21( )

i i i iN f x f x

N

= 2

40 (12 36 112 180 490 1100)1

40 (6 12 28 30 70 110)

= 1

77200 6553040

= 1

108.03 2.740

62. Answer (4)

Hint : Make truth table and then find equivalent

statement

Sol. : Making truth table we can find equivalent

statement as p q

Truth table

( ) ( ) p q p q q p p q q p p q

T T T T T T

T F F T F F

F T T F F F

F F T T T T

58. Answer (1)

Hint : Ozonolysis followed by Aldol

condensation

Sol. :O

3

Zn

X

O

O

OH (dil.)–

O

Y

H

59. Answer (1)

Hint : Basic strength depends on availability of lone

pair

Sol. :H N2 C NH2

• •

• •

NH

Because of resonance, doubly bonded N

atom in this structure has most negative

charge density.

60. Answer (1)

Hint : Down the group stability of group-1 hydrides

decreases

Sol. : According to VSEPR, its structure is

Xe

••

FF

FF

O

Lone pair is trans to oxygen

63. Answer (1)

Hint : 1

1

2

2 ( 1)

n

n r

r

rS

r r

= 1

1

1 1

2 2 ( 1)

n

r r

rr r

Sol. : S = 1

1

2

2 ( 1)

n

r

r

r

r r

= 1

1

2( 1)

2 ( 1)

n

r

r

r r

r r

= 1

1

1 1

2 2 ( 1)

n

r r

rr r

Sn

= 1

1 1

2 2 ( 1)nn

= 1

2 ( 1) 1

2 ( 1)

n

n

n

n

7

6

S

S=

10 7

8 6

2 1 2 7

2 8 (2 .7 1)

= (1024 1) 7

2 8 447

=

1023 7

16 447

=

2387

2384


9/14

64. Answer (2)

Hint :3 11,

36 12p q

Sol. : For single throw favourable cases are (4, 6),

(5, 5) and (5, 4)

p = 3 1

36 12

q = 1 11

1 112 12

p

Probability =

2

3

2

1 11 11

12 12 576C

65. Answer (3)

Hint : Length of projection = AB sin

Sol. : cos = | 8 1 4 |

33 6

=

11

18

sin = 7

18

Length of projection = AB sin

= 7 77

3318 6

66. Answer (2)

Hint : Find positive integral solutions of x1 + x

2 + x

3

= 12 or 13 or 14 or 15 and then add all of

them

Sol. : Here,

x1 + x

2 + x

3 = 12 or

x1 + x

2 + x

3 = 13 or

x1 + x

2 + x

3 = 14 or

x1 + x

2 + x

3 = 15

Number of solutions

= 12–1C3–1

+ 13–1C3–1

+ 14–1C3–1

+ 15–1C3–1

= 11C2 + 12C

2 + 13C

2 + 14C

2

= 55 + 66 + 78 + 91

= 290

67. Answer (3)

Hint : Use condition for coplanarity of planes

Sol. : If planes are coplanar, then

2 1 3

1 2 0

2 1

p

q

2(–1 + 4) –1(p + 2q) + (–3) (2p + q) = 0

6 – p – 2q – 6p – 3q = 0

7p + 5q = 6

68. Answer (3)

Hint : From given equation (z – 2i) (iz2 – 4) = 0

Sol. : (z – 2i) (iz2) – 4(z – 2i) = 0

(z – 2i) (iz2 – 4) = 0

z = 2i or z2 = 4

4ii

|z| = |2i| = 2 |z2| = |–4i|

|z| = 2 |z|2 = 4 |z| = 2

69. Answer (3)

Hint : Put x2 + x = y

Given equation becomes (y – 1)(y – 2) – 6 = 0

Sol. : Let x2 + x = y

(y – 1)(y – 2) – 6 = 0

y = 4, –1

x2 + x = 4 or x2 + x = –1

x2 + x – 4 = 0

Which has real roots

Sum of real (Non- real roots)

roots = –1

70. Answer (3)

Hint : | a�

b�

| + |2( a�

b�

)| = || a�

|| b�

| cos| +

|2|a�

||b�

| sin ˆn |

Sol. : |(a�

b�

)| + |2(a�

b�

)| = 12 cos + 24 sin

2 212 24

Maximum value is 2 212 24 = 12 5

71. Answer (4)

Hint : 2 2

( 3) 3 (3 1)(1 0) 80

( 3) ( 3)

dy x x

dx x x

Sol. :2

8( ) 0

( 3)f x

x

Which is true for all x R

Also let 3 1

3

xy

x


10/14

13 1 3 1( )

3 3

y y

x f yy y

f –1(x) = 3 1

3

x

x

f(x) is its own inverse Also,

f(x) = 8

33x

which is unbounded.

72. Answer (2)

Hint : Image of A(2, 1) due to line x = 3 is P(4, 1)

and so equation of side BC is the line joining

points C(5, 3) and P(4, 1)

Sol. : Image of A(2, 1) due to the line x = 3 is

(4, 1) which lies on BC.

Equation of side BC is

y – 3 = 3 1

( 5)5 4

x

2x – y = 7

Which is satisfied by the point (3, –1)

B = (3, –1)

73. Answer (1)

Hint : Given equation is

2 2

2 2

(1 )log

( 1) ( 1)

e

dy x xy x

dx x x x

and IF =

2

2

( 1)

( 1)

xdx

x xe

= 2

1x

x

Sol. : From given differential equation

2 3

2 2

( 1)log

( 1) ( 1)e

dy x xy x

dx x x x x

...(1)

I.F. =

2

2

( 1)

( 1)

xdx

x xe

= 2

2 1

1

xdx

xxe

= 2log 1 log e e

x x

e

= 2

1x

x

Solution of given equation is 21y x

x

=

23

2

1log

1

e

x xx dx

x x x

= 2 1

log log2 2

e e

x xx x dx x dx

x

= 2

21log

2 4

e

xx x C

C = 0

( ) 0y e

74. Answer (2)

Hint : Slope of tangent is

f(x) = 2 2

( 2)(1) ( 3)(1) 5

( 2) ( 2)

x x

x x

Sol. : f(x) = 2

5

( 2)x

= –5 at x = 3

Equation of tangent at (3, 6) is

y – 6 = –5(x – 3) 5x + y = 21

Hence, given line is a tangent on f(x) at (3, 6)

75. Answer (2)

Hint : f(x) =

( 2), if 2

( 2), if 2

x xx

x

x xx

x

=

2 2, if 2

2, if 2

xx

x

x

x

Sol. : f(x) =

( 2), if 2

( 2) 2, if 2

x xx

x

x xx

x x

f (x) = 2

2

(2 1 0) (2 2) 1, if 2

2, if 2

x xx

x

x

x

f (2–) = 2

4 (4 2) 1,

22

f (2+) = 2 1

4 2

f(x) is non-differentiable at x = 2


11/14

76. Answer (1)

Hint : f(x) is defined if

log|x|–3

(x2 + 2x + 1) 0 ...(1)

x2 + 2x + 1 > 0, |x| – 3 1

Sol. : Here, f(x) is defined if

log|x|–3

(x2 + 2x + 1) 0 ...(1)

Case I:

When |x| – 3 > 1

|x| > 4

x < –4 or x > 4

x(–, –4) (4, ) ...(2)

and then from (1)

(x2 + 2x + 1) (|x| –3)0

x(x + 2) 0

x –2 or x 0 ...(3)

From (2) and (3),

x(–, –4) (4, ) ...(4)

Case II:

When 0 < |x| – 3 < 1

3 < |x| < 4

–4 < x < –3 or 3 < x < 4 ...(5)

Then from (1),

(x2 + 2x + 1) (|x| – 3)0

x(x + 2) 0

–2 x 0 ...(6)

From (5) and (6),

x ...(7)

x2 + 2x + 1 > 0

xR ...(8)

|x| – 3 1

x ±4 ...(9)

77. Answer (2)

Hint : Asymptotes are x2 – 2y2 = 0 x ± 2 0y

Any point on hyperbola are (2sec, 2 tan )

Sol. : Asymptotes are 2 0x y and any point

on hyperbola are (2sec, 2 tan)

Product of perpendiculars from

2 sec , 2 tan to the asymptotes =

2 sec 2 2 tan 2 sec 2 2 tan

3 3

= 2 2

4 sec tan4

3 3

78. Answer (4)

Hint :

121

2 1

2tan , 01cos

1 2tan , 0

x xx

x x x

Sol. :

121

2 1

2tan , 01cos

1 2tan , 0

x xx

x x x

Here

L.H.L. = 1

0

tanlim 2x

x

x

= –2 × 1 = –2

R.H.L. = 1

0

tanlim 2x

x

x

= 2 × 1 = 2

L.H.L R.H.L. given limit is non-existent

79. Answer (2)

Hint : Slope of tangent is m = dy

dx


dx and find

value of x and if 2

20d m

dx, then m is minima

at that point

Sol. : Here, y = 2ex sin4 2

x

cos4 2

x

= ex sin 24 2

x

= ex sin (/2 – x) = ex cosx

m = dy

dx= slope of tangent

= ex(–sinx) + ex cosx

= ex(–sinx + cosx)

dy

dx = ex(–cosx – sinx) + (–sinx + cosx) ex

= ex(–2 sinx)


12/14

2

22 cos sin

x xd me x x e

dx

= –2ex(cosx + sinx)


dx

–2ex sinx = 0 sinx = 0

x = 0, , 2

When

x = 0,

20

22 (1 0) 0

d me

dx

When

x = ,

2

22 (cos sin ) 2 0 d me e

dx

When

x = 2,

22

22 (cos2 sin2 ) d me

dx

= –2e2 (1 + 0) < 0

m is minimum at x =

80. Answer (4)

Hint : Equation of tangent to the parabola is

y = mx + a

m

if it passes through (h, k)

k = mh + a

m

Sol. : Equation of tangent to the parabola y2 = 4ax

is

y = a

mx

m

If it passes through (h, k), then

m2h – mk + a = 0

m1 + 4m

1 =

k

h,

2

14

am

h

2

4

25

k

h = a locus of (h, k) is

4y2 = 25ax

81. Answer (3)

Hint : Draw the figure and then use tan =

p

h for

two triangles

Sol. :

2 3

A B C Q

P

a

Let PQ be the tower.

From APQ, tan = PQ

AQ

AQ = tan

PQ

...(1)

From BPQ, tan2 = PQ

BQ

BQ = tan2PQ

...(2)

Here a = AB – AQ – BQ = tan tan2

PQ PQ

a =

2

1 1

2tantan

1 tan

PQ

21 1 tan

tan 2tanPQ

22 1 tan

2tanPQ a

PQ2

2tansin2

1 tan

a a

82. Answer (4)

Hint : Check for reflexive, symmetric and transitive

relation

Sol. : Reflexivity: We have sin2p + cos2p = 1

aR

pRp pR R is reflexive

Symmetry: Let pRq

sin2p + cos2q = 1

1 – cos2p + 1 – sin2q = 1

sin2q + cos2p = 1

qRp R is symmetric


13/14

Transitivity:

Let pRq and qRr

sin2p + cos2q = 1 ...(1)

and sin2q + cos2r = 1 ...(2)

By [(1) + (2)]

sin2p + (cos2q + sin2q) + cos2r = 2

sin2p + cos2r = 1

pRr R is transitive also.

Hence R is equivalence relation.

83. Answer (2)

Hint : (C0 + C

1 + C

2 + ... + C

n)2

= (C02 + C

12 + ... + C

n

2) + 20

i ji j n

CC

Sol. :0

i ji j n

CC

= (C

0C

1 + C

0C

2 + ...C

0C

n) +

(C1C

2 + C

1C

3 + ... C

1C

n) + (C

2C

3 + C

2C

4 +

...C2C

n) + ... = R (say)

As (C0 + C

1 + C

2 + ... C

n)2

= 2 2 2

0 1... 2

nC C C R

R = 2

2 12

2

n

n nC

,

since C0 + C

1 + C

2 + ... + C

n = 2n

ad 2 2 2 2

0 1 2...

nC C C C

= coefficient of xn in (1 + x)n (x + 1)n

= 2nCn

84. Answer (3)

Hint : Integrate by parts

Sol. :

2

1

1log 1x dx

x

=

22 2 22

1

1

1

1log 1

12 21

x x xdx

x

x

=

22

2

11

1 1log 1

2 2 1

x xdx

x x

=

2

1

3 1 1 12 log log 2 log (1 )

2 2 2 2

x x

= 9 1 1log log 2 (2 1) log 3 log 2

4 2 2 e e

= 3 3

log4

e

= 27

log log 274 4

e kk e

85. Answer (4)

Hint : g(x) = f(|x|) – |f(x)|

= 6, 2 0

2( 3), 0 2

x x

x x

Sol. : g(x) = 6, 2 0

2( 3), 0 2

x x

x x

Required area = area of trapezium OPQR +

area of trapezium ORST

= 1 1(10)(2) (8)(2)

2 2 = 18 sq. units

xP

R (0, –6)

S(2, –2)(–2, –4)

Q

–2O T

x = 2x = – 2

y x = 2( – 3)y–x

= – 6

y

2

86. Answer (1)

Hint : Here equation of chord of two circles and the

equation of chord of contact made by the

tangents from P(x1, y

1) to first circle will

represent same equation comparing we get

values of x1 and y

1.

Sol. : Equation of common chord AB of circles is

S1 – S

2 = 0

6x + 8y – 31 = 0 ...(1)

P x y( , )1 1

A

B


14/14

� � �

Also equation of chord of contact AB made

by the tangents from P(x1, y

1) to the circle

S1 = 0 is

T = 0

xx1 + yy

1 – 18 = 0 ...(2)

comparing (1) and (2)

1 118

6 8 31

x y

(x1, y

1) =

108 144,

31 31

87. Answer (4)

Hint :

1 1

2

1 1

1 1

4 2 2tan tan

1 14 31

2 2

r r

r r

rr r

Sol. :1

21

1tan

11

4

r r

= 1

1

1tan

1 11

2 2

rr r

= 1 1

1

1 1tan tan

2 2r

r r

=1 1 1 13 1 5 3

lim tan tan tan tan2 2 2 2

n

1 11 1... tan tan

2 2

n n

= 1 11 1

lim tan tan2 2

n

n

= 1 1

tan2 2

= 1 11

cot tan 22

88. Answer (3)

Hint : |adj(adj(adj A))| = |A|(3–1)3 = |A|8

Sol. : |A| = x + y + z

We have, |adj(adj(adj A))| = |A|(3–1)3 = |A|8 = 78

(x + y + z) = 7 ...(1)

Number of such matrices = No. of +ve integral

solutions of (1)

= 7–1C3–1

= 6C2 = 15

89. Answer (1)

Hint : A matrix A is symmetric if AT = A and skew-

symmetric if AT = –A and its all diagonal

elements are zeros and zero matrix is both

symmetric and skew- symmetric

Sol. : Probability = 6 3

9 9 9

5 5 1

5 5 5 = 3 6 9

1 1 1

5 5 5

where probability (that matrix is symmetric)

= 9


5

x

x x x

x x

x

�

�

=

6

9

5

5

Probability (that matrix is skew-symmetric)

= 9


5

x

0

0

0

x x

x

�

� �

=

3

9

5

5

Probability (that matrix is both symmetric

and skew-symmetric) = 9

1

5

90. Answer (3)

Hint : I = 2sec tan

2 2 2

x x xdx dx

(Integrate 1st by parts)

Sol. :2

sin

2cos2

x xdx

x

=

21sec

2 2

xx dx + tan

2

xdx

= 2

tan1 2

sec 112 2

2

x

xx dx dx

tan2

xdx

= tan

1 122 tan tan

12 2 2 2

2

x

x xx dx dx

= x tan2

xc

mock test - 1 (code-a) (answers) all india aakash test series for … · 2019-06-04 · mock test -...

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