mock test - 2 (paper - 2) (code-e) (answers) all india ......mock test - 2 (paper - 2) (code-e)...

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

1/14

1. (5)

2. (8)

3. (5)

4. (4)

5. (8)

6. (6)

7. (6)

8. (8)

9. (B, C)

10. (B, C)

11. (A, B, C, D)

12. (B, D)

13. (A, C, D)

14. (B, C)

15. (C, D)

16. (B, C)

17. (A, C)

18. (A, D)

19. (A, C)

20. (A, B)

21. (8)

22. (4)

23. (2)

24. (5)

25. (3)

26. (3)

27. (5)

28. (2)

29. (A, B, C, D)

30. (A, B)

31. (A, B, C)

32. (A, B, C, D)

33. (A, B, C, D)

34. (A, B, D)

35. (B)

36. (A, B, D)

37. (B, C, D)

38. (B, C, D)

39. (A, D)

40. (A, C)

41. (2)

42. (0)

43. (7)

44. (7)

45. (1)

46. (8)

47. (2)

48. (3)

49. (A, B, D)

50. (B, C)

51. (B, C)

52. (B, C)

53. (B, C)

54. (A, B)

55. (A, B, C)

56. (C, D)

57. (A, B, D)

58. (A, C, D)

59. (B, C)

60. (A, C, D)

ANSWERS

Test Date: 10/03/2019

PHYSICS CHEMISTRY MATHEMATICS

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

MOCK TEST - 2 (Paper-2) - Code-E

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

2/14

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (5)

Hint:

2 21 22

A tE t

E A t

Solution :

2 22 t

E A e

lnE = 2lnA – 2t2

2 21 22

A tE t

E A t

2 2% 2(2) (0.2) 2 (2.5) 2E

E

% 4 1 5%E

E

2. Answer (8)

Hint :

Total energy of system is conserved and speed will be

max. at centre of sphere.

Solution :

Speed will be maximum when point mass is at the

centre of sphere.

2

rel

1PE

2v

2rel

3 3 .3 1 .3

4 2 2 4

Gm m Gm m m mv

R R m

2

2

rel

15 1 3

4 2 4

Gm mv

R

rel

10 4

3 3

Gm vv v v

R

v = 45

8

Gm

R

3. Answer (5)

Hint :

Use rotational kinetic energy.

Solution :

When rod becomes vertical

2

21

2 2 3 L ML

Mg

23 g

L

T = 2

/2.

2 ∫

L

L

Mgdm x

=

22

/22 2

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎣ ⎦

L

L

Mg M x

L

= 9

2 8Mg

Mg

= 13

8Mg

131 5

8 8

nn ⇒

4. Answer (4)

Hint :

Kinetic energy = 2

2

P

m

Solution :

22

LL ⇒

KE =

221

2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

P h

m m

=

2

2

1

2 h

m

=

2

2

4

2

h

m L

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

3/14

=

2

2

2h

mL

2n

n = 4

5. Answer (8)

Hint :

Use KVL and KCL to solve circuit.

Solution :

1

402 A

20I

4 4

2 A2 A

S

R

100 V

4 A

20

DC

B

2 A

A

16 + 2R – 100 + 40 = 0

2R = 100 – 40 – 16

2R = 44

R = 22

P = 88 watt

6. Answer (6)

Hint :

0

0

s

v vf f

v v

⎛ ⎞ ⎜ ⎟⎝ ⎠

Solution :

Frequency received by approaching car

1 0

21

330f f

⎡ ⎤ ⎢ ⎥⎣ ⎦

Frequency received by source again

1

22

1330

ff

⎛ ⎞⎜ ⎟⎝ ⎠

So, beat frequency fB = f2 – f0 = 6 Hz

7. Answer (6)

Hint :

P = 2(P1 + P2) + Pm

Solution :

Peq = 2(P1 + P2) + Pm

–eq 1 2

1 1 12

f f f

⎛ ⎞ ⎜ ⎟

⎝ ⎠

2

51

1 1 14cm

1 30 120f

⎛ ⎞⎜ ⎟⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟

⎝ ⎠

1

31

1 –1 –12cm

1 30 60f

⎛ ⎞⎜ ⎟⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟

⎝ ⎠

8. Answer (8)

Hint :

The bottom surface of the container

remains covered with liquid during rotation (no dry spots).

Solution :

A vertical cylindrical container partially filled with a liquid

is rotated at constant speed. The drop in the liquid

level at the center of the cylinder is to be determined.

Free

surface

zs

z

g

R = 20 cm

r

h =0

60 cm

Taking the center of the bottom surface of

the rotating vertical cylinder as the origin (r = 0, z = 0),

the equation for the free surface of the liquid is given as

22 2

0( ) ( 2 )

4sz r h R r

g

where h0 = 0.6 m is the original height of the liquid

before rotation, and = 2n = 2(120 rev/min)1min

60s

⎛ ⎞⎜ ⎟⎝ ⎠

= 4 rad/s


4/14

Then the vertical height of the liquid at the center of the

container where r = 0 becomes

2 2 2 2

0 2 2

(4 ) (0.20m)(0) (0.60m)

4 4( m /s )s

Rz h

g

= 0.44 m

Therefore, the drop in the liquid level at the center of

the cylinder is

drop, center 0 (0) 0.60 0.44 0.16 m 16 cms

h h z

9. Answer (B, C)

Hint :

Radial component of spring force will balance the

centripetal force.

Solution :

A

BCO

(A) Elongation of spring =

2

2 8(2 )

5

⎛ ⎞ ⎜ ⎟⎝ ⎠

RR R

= 2

2 644

25 R

R R

= 6

5 5 R

R R

Radial component of the spring force

Fr = cos5

⎛ ⎞ ⎜ ⎟⎝ ⎠

Rk

8 4sin

5.2 5 R

R

3cos

5

fr = 3 3

.5 5 25 R kR

k

Normal force to be zero.

23

25

mv kR

R

2

2 3 3 10 16 2

25 25 1

kRv

M

v2 = 12 16

5

16.2 msv

�

sin5

t

RF k

sin5

dv kRm

dt

4 410 2

5 5 dv

dt

64

5

= 12.8 m/s2

10. Answer (B, C)

Hint :

Assume Q at the centre of cube of size 2L.

Solution :

The flux will same through the far faces and is same

through all near faces.

31 + 32 = 0

Q

2 = flux through far face

2 = 0 0

1

4 6 24

Q Q

31 = 0 0

8

Q Q

31 = 0

7

8

Q

11. Answer (A, B, C, D)

Hint :

Use dimensional analysis.


5/14

Solution : L hxcyGz ...(1)

M hpcqGr ...(2)

From (1)

0 0 2 1 1 1 3 2M L T ML T LT M L T

x y z ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

1 3 1

2 2 2x y z

From equation (2),

1 1 1, ,

2 2 2p q r

12. Answer (B, D)

Hint :

dNN

dt

Solution :

Initial activity of 24Na

A = 1.0 CdN

dt

= 1.0 × 10–6 × 3.71 × 1010

= 3.71 × 104 disintegration s–1

Half life = T = 15 h = 15 × 3600 s

Initial Activity

A = N0

3.71 × 104 = 0

0.7

15 3600N

4

0

3.71 10 15 3600

0.7N

N0 = 2862 × 106

= 2.86 × 109

N = Number of Nuclei present after 7.5 h in blood

sample per cc

' dNN

dt

294 0.7'

60 15 3600

N

N = 3.78 × 105 cm–3

N0 = Number of nuclei present in blood sample

Total activity now

43.71 10

2

294amt 1cc

60

4

60 3.71 105.35

294 2

13. Answer (A, C, D)

Hint :

H = W – u

Solution :

Qin = CE = 3 × 3 = 9 C

Uin = 1 27(3 9) J

2 2

Cf = 2C.6C 3

C2C 6C 2

9F

2f

C

Ufinal = 1 9 9 2

9 J2 9

Heat dissipated = 27

9 4.5 J2

QP = 1

1 Qk

⎛ ⎞⎜ ⎟⎝ ⎠

= 1 2

1 9 9 6 C3 3

⎛ ⎞ ⎜ ⎟⎝ ⎠

0 .

2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

A B

E d E dV V

k

= . 1

12

⎡ ⎤⎢ ⎥⎣ ⎦

E d

k

= 2

3Ed

23 2

3A B

V V V


6/14

14. Answer (B, C)

Hint :

The satellite will move in elliptical orbit and earth as at

a focus.

Solution :

TE = 4 8

GMm GMm

R R

TE = 8

GMm

R

2 8 GMm GMm

a R

a = 4R

21

8 2 6

GMm GMmmv

R R

21

8 6 2 GMm GMm

mvR R

2–6 8 1

48 2

GMm GMmmv

R

2

12

GMv

R

v = 12

GM

R

15. Answer (C, D)

Hint :

vnC T V

Solution :

vnC T V

3

2n RT V

T = 2

3

V

nR

PV = n R T

2

3

nR VPV

nR

2

3PV V

1

22

3PV

Molar specific heat

C = 1

v

RC

x

= 3

121

2

⎛ ⎞ ⎜ ⎟⎝ ⎠

RR

= 3

22

R R

C = 7

2R

U = n Cv T

3

2U nR T

w = 2R.nT

2 4

3 3

2

w R

UR

4 904 120J

3 3w U

16. Answer (B, C)

Hint :

Energy = 1

2Stress × Strain × Volume

Solution :

22 4

mg gmg m a a ⇒

3

2 4 4

mg gT m T mg

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

Stress = 3

/4mg A

⎛ ⎞⎜ ⎟⎝ ⎠

21 3

2 4

mg ALU

A Y

⎡ ⎤ ⎢ ⎥⎣ ⎦

17. Answer (A, C)

Hint :

ma = mg – Bid

Solution :

LdiB dv

dt

.Ldi dxB d

dt dt


7/14

. . B d xI

L

ma = mg – Bid

ma = 2 2

B dmg x

L

a = 2 2

B dg x

Lm

18. Answer (A, D)

Hint :

ma = mg – Bid

Solution :

2 2

dv B dg x

dt mL

v = v0 sin t

v = 0sin

Bdv t

mL

A = 0

V

at t = 0

Acceleration a = g

v0 = g

A = 2g

A = 2 2 2 2gmL mgL

B d B d

19. Answer (A, C)

Hint : The path difference at the slit is zero.

Solution :

In first case

sin (2 1)2

d n and 2

d

2 1sin 1

4

n

n 2.5 So, n = 1, 2

When

1 1

1 11, sin tan

4 15 ⇒ n

1

1tan

15y D

When

2 2

3 32, sin tan

4 7 ⇒ n

2

3m

7y

20. Answer (A, B)

Hint :

The additional path difference of 2

d exist at the slits.

Solution :

sin (2 1)2 2

dn

and 2

d

sin 1

n 1.5

n = 1

Position of centre maxima

= = 30°

tan = 0

y

D

y0 = 1

3 3

D

Path difference for p1, 3

2 2x

sin = 3 3

2 4d

tan1 = 13

7

y

D

y1 = 3

7

path difference for p2,

x = 2 2

tan2 = 2

1 1m

15 15y⇒


8/14

21. Answer (8)

Hint :

Root mean square velocity and most probable velocity

are 3RT 2RT

andM M

respectively.

Solution :

1 2

1 2

3RT 2RT

M M

2

400 3 2 60

80 M

15 = 2

120

M

M2 = 8

22. Answer (4)

Hint :

Due to addition of CH3Li to carbonyl group produce

alcohol.

Solution :

CH3

C

O CH3

(i) CH3Li

(ii) H O2

CH3

C

HO CH3

CH3

BCl3

Me

C+

Me

Me

Me(i) O

3

(ii) Zn; H O2

Me

O

Me

OH–

H C3

O

CH3

C

H

O

CCH

3

23. Answer (2)

Hint :

2 4 3 2NH CONH (s) 2NH CO��⇀

↽��

Partial pressure of NH3= 2(Partial pressure of CO

2)

Solution :

2 4 3 2NH CONH (s) 2NH (g) CO (g)

0 0

2x x

��⇀↽��

Ptotal

= P

2CO

x PP P 1

2x x 3

3NH

2x 2P P 3

2x x 3

= 2

KP =

3 2

2

NH COP P = 4

24. Answer (5)

Hint :

cis form is optically active. x = 3, y = 8

Solution : C.N of Cs = 8 = y

H3N

NH3

Co

ox

H3N

NH3

(d & �)Cis

ox

ox

ox

Co

Co

NH3

ox ox

NH3

(Trans)

25. Answer (3)

Hint : -hydroxy acid on heating form unsaturated

compound.

PART - II (CHEMISTRY)


9/14

Hydrolysis of ester and dehydration

Solution :

CH –C–OCH2 3 H

3O

+

O

CH –C–OH2

O

CH–C–OH

O

OH OH

26. Answer (3)

Hint :

During formation of osazone, 1 mol consume 3 mol of

PhNHNH2.

Solution : p = 3, q = 3

O||C–H

CHOH

(CHOH)3

CH OH2

CH=N–N–C H6 5

H C=N–N–C H

6 5

(CHOH)3

CH OH

Phenyl osazone2

H

+ 3C H NH–NH2

6 5

+ Ph–NH2

+NH

3

+H

2O

27. Answer (5)

Hint :

Lindlar reduction.

Solution :

H C–H C–C C–CH –C–H23 2

NaBH4

O

H C–H C–C C–CH – –OH23 2 2

CHPBr

3

H C–H C–C C–CH – –Br23 2 2

CH

H C–CH –C C–CH – –MgBr23 2 2

CH

Mg/ether

CO ; H O22

H C–CH –C C–CH – –C–OH23 2 2

CHsp

PSOCl2

H C–H C–C C–CH –C23 2 2–CH

H2

O

H C–C –C C–CH – –C–Cl23 2 2

H CH

Pd/BaSO4

O

H

sp2

sp

sp2

sp2

H H

O

28. Answer (2)

Hint :

[Ni(CO)4] are diamagnetic

Solution :

sp3

Ni 3 d8

4 s2

dsp2

[Ni(CN) 4

2–

]3d

8

Ni2+

CN is strong ligand–

Ni(CO)4

CO is strong ligand


Hint :

Arrhenius equation.

Solution :

K = Ae–Ea/RT

30. Answer (A, B)

Hint :

Mg and Al are highly electro positive metal.

Solution :

Fe2O3 and SnO

2 is reduced by carbon.

31. Answer (A, B, C)

Hint :

Diel-Alder reaction and ozonolysis.

Solution :

H

H

O

O

C

C

O(i) O

3

(ii) Zn; H O2

O

C

O

C

O

O

C

O

C

O

C

O

H

O

CH

CHO–H2

(i) LiAIH4

excess

(ii) H O

3

CHO–H2

CH2–OH

CH2–OH


10/14


Hint :

Oxidation of side chain and diazotisation.

Solution :

CH3

CH3

KMnO4

H+

NH3

COOH

COOH

C

C

N–H

(i) NaOH, Br2

(ii) H+

COOH

NH2

(i) NaNO +HCl2

0–5°C

C–O–

N N

O

+

–N

2

–CO2

(Q)(R)

(S)

(ii) OH–

O

O


Hint :

n = 3, formula is XY3 and molar mass of XY

3 is 108 g/mol.

Solution :

If (Tf) XY

n = 4 (T

f)urea

at same concentration.

It means n = 3

During electrolysis of molten XY3

Cathode X3+ + 3e– X

Anode : 3Y– 3Y + 3e–

34. Answer (A, B, D)

Hint :

S4 and S

6 are incorrect.

Solution :

Cysteine are non-essential -amino acid.

35. Answer (B)

Hint :

Frequency () of revolution of e– in orbit is Z2/n3

Solution :

2

2

3

Zv

n

nr

Z

Z

n

⎤ ⎥⎥⎥ ⎥⎥⎥

⎥⎦

5

He

2H

x 52;y 33


Hint : P Ca3P2, Q = MgO

R = PH3, S = Cu

3P2

Solution :

3 4 2 3 2(Q)(P)

Ca (PO ) + 8Mg Ca P + 8MgO

3 2 2 2 3(P) (R)

Ca P + 8H O 3Ca(OH) 2PH

3 4 3 2 2 42PH 3CuSO Cu P 3H SO

37. Answer (B, C, D)

Hint :

Perkin’s reaction.

Solution :

(CH –CH –C–) O3 2 2

OCH –CH –C–O

3 2

–

O

CH –CH–C–O–C–CH –CH3 2 3

O O

Ph C

H

O

+ CH–C–O–C

O

CH3

O

CH2

CH3

Ph–CH–CH–C–O–C–CH2

O–

CH3

O

CH3

H O2

Ph–CH C–C–OH

CH3

O

+ HO–C–CH –2

CH3

O

O


Hint :



11/14

Solution :

Ph

C

O

H CH2

C

O

O

O

C

Ph

CH

O

O

O

C O

OPh

O O

Ph

O O–

O

C OCH

HH C–C–O

3

–

O

PhCH

CH

CO

O–

–

O

O

O–

39. Answer (A, D)

Hint :

B2H6 on reaction with NH

3 form inorganic benzene.

Solution :

Y = B2H6, X = B

2O3

40. Answer (A, C)

Hint :

Boron and inorganic benzene.

Solution :

Organic benzene is more symmetrical molecules so

has higher mp(6°C) than inorganic benzene (–58ºC).

PART - III (MATHEMATICS)

41. Answer (2)

Hint : Taking log both side.

Solution :

xx xlog –

2log x = 0

xx x– log 0

2

⎛ ⎞ ⎜ ⎟⎝ ⎠

xx 1–

2

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

logx = 0

x = 1, 4 (∵x > 0)

42. Answer (0)

Hint : a b. 0� �

Solution :

a b. 0� �

x x2

2 2– log 6 4 log 0

–y2 + 4y + 6 < 0 , –< 0 , D < 0

162 + 4 × × 6 < 0

8[2 + 3] < 0

∵

0–3

2

–3

,02

⎛ ⎞ ⎜ ⎟⎝ ⎠

43. Answer (7)

Hint : Fav.cases = number of non-negative integral

solutions of x1 + x

2 + ..... + x

20 = 10

Solution :

The number non-negative integral solutions of x1 + x

2 +

..... x20

= 10 is C29

19

Required probability = C

29

19

1020

n = 19

n – 12 = 7

44. Answer (7)

Hint : Use transformation of roots.

Solution :

The equation where roots are y = f(i) i.e. y = x2 – 2

x y 2

y y5 2

2 2 1 0

y5 + 10y4 + 40y3 + 74y2 + 74y + 23 = 0

Product of roots = –23

Also, 1

2

3

4

5 = –1

i i

i

f f

5 5

1 1

– 30 7

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠


12/14

45. Answer (1)

Hint : Substitute all angles in half angle.

Solution :

i

i

r

r r r xx x x

f xr r

x x1

2 1 2 1 6 32sin cos 2sin .sin

2 2 2 2

6 3 2 12sin . cos

2 2

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑

i

r

rx r x

1

cos – cos 1

∑

fi(x) = cosx – cos (i + 1)x

f20

21cos – cos 2

4 4 4

⎛ ⎞ ⎜ ⎟⎝ ⎠

46. Answer (8)

Hint : Given family passes through (1, –2).

Solution :

(3x + 7y + 11) sec + (5x – 3y – 11) cosec = 0

(5x – 3y –11) + tan (3x + 7y –11) = 0

These family of lines always pass through (1, –2).

Let A(x, y) be a point on x – y + 3 = 0

Now, PB2 = PA2 + AB2 – 2PA. AB cos.

PA2 + AB2 – 2PA . AB

= (PA – PB)2

So maximum of A BP P– is PB.

PB2

40 40 ⇒

2

85

⇒

47. Answer (2)

Hint : R = 8r

Solution :

R = 8r

∵

A B C 1sin . sin . sin

2 2 2 32

∵ A B C9

cos cos cos8

B B2 9sin 1– 2sin

2 2 8

B4sin 1

2

48. Answer (3)

Hint : 7 tails and 3 heads.

Solution :

7 tails, 3 heads

Probability = C C

10 3

7 3

10

.

2

120 15

1024 128

= 3


Hint : For interior point, S1 < 0

Solution :

Point P lies inside the circle.

so, S1 < 0

(p + 1)2 + (2p –1)2 –4 (p + 1) –4(2p – 1) – 4 < 0

p7 – 59 7 59

,5 5

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

.....(1)

And for line x + 2y – 2 = 0 point P lies on the same side

origin lies

So, (p + 1) + 2(2p – 1) – 2 < 0

p3

5 ......(2)

From (1) and (2),

p7 – 59 3

,5 5

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

50. Answer (B, C)

Hint : Equation of common chord.

Solution :

Clearly AB is chord of contact with respect to (1, 1).

so equation of AB is 3x – y + 2 = 0 .......(1)

The tangent y2 = 4ax is a

y mxm

.......(2)

Comparing (1) and (2) a 6⇒

1. Length of L.R. = 24

2. y = 6x + 1 is tangent to curve y2 = 24x

3. Common tangent for x2 = 4ay and y2 = 4ax is

x + y + a = 0

4. Normal y = mx – 2am – am3


13/14

put m = 1 and a = 6

y = x – 12 – 6

y = x – 18

51. Answer (B, C)

Hint : i

P A P Ei P A Ei

12

0

. /

∑

Solution :

Let Ei denote the event that the bag contain i black

and (12 – i) white balls (i = 0, 1, 2, ....., 12 ). Let A

denote the event that the 4 balls drawn are black. Then

P Ei1,

13 i = 0, 1, 2, ...., 12

4

12

4

for 0, 1, 2, 3

/for 4, 5, ...., 12

e

O i

P A Ei Ci

C

⎧⎪ ⎨ ⎪⎩

(i) i

P A P Ei P A Ei

12

0

1 1/

13 12 ∑

4 5 12

4 4 4....C C C⎡ ⎤ ⎣ ⎦

13

5

12

4

1

513

C

C

(ii) By Baye’s Theorem,

c

P E P A EE cP

A P A

10

4

1210 1010 4

1

13. / 70

1 429

5

⎛ ⎞ ⎜ ⎟⎝ ⎠

52. Answer (B, C)

Hint : b b

a af x dx f a b x dx– ∫ ∫

Solution :

nx

nxI dx

x–

sin

sin 1

∫ ........(1)

n

x

n xI dx

x–

–

sin –

sin – 1

⎡ ⎤ ⎣ ⎦∫ .......(2)

(1) + (2),

n

nxI dx

x0

sin

sin

∫

I1 = I

3 = I

5 = ........ =

I2 = I

4 = I

6 = ........ = 0

53. Answer (B, C)

Hint : Multiplication of two matrices.

Solution :

a b c

2 2

2 2 3 3

2 2 3 3 4 4

3 1 1

, , 1 1 1

1 1 1

2 2 2 2

1 1 1 1 1 1

1 1

1 1

= ( – )2 ( – ( +) + 1)2

= (( + )2 – 4) ( – ( +) + 1)2

= b c c b

a a a a

2 2–

– 4 1

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

= a b c b ac

a

2 2

4

– 4

54. Answer (A, B)

Hint : Summation of series would be 90.

Solution :

r

r r

90

1

2 sin2 tan1

∑

= 2sin2 4sin4 6sin6 ......

tan1178sin178 180sin180

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

But = 1°

= tan1° × 90 cot1°

= 90

Therefore, (5 + 2 –6) (5+ 90 – 6) > 0

(–, –6) (1, )


Hint : AM G.M. and range sec–1x & cosec–1x


14/14

Solution :

xy x yx y

1 10 2, –2 ⇒

or, x yx y

1 1–2, 2

x x

x x

–11 12 sec ,

3 2

⎛ ⎞ ⎡ ⎞ ⇒ ⎜ ⎟ ⎟⎢⎝ ⎠ ⎣ ⎠

y yy y

–11 1 ––2 cosec ,0

6

⎛ ⎞ ⎡ ⎞ ⇒ ⎜ ⎟ ⎟⎢⎣ ⎠⎝ ⎠

So, sec–1 x yx y

–11 1cosec ,

6 2

⎛ ⎞ ⎛ ⎞ ⎡ ⎞ ⎜ ⎟⎜ ⎟ ⎟⎢⎝ ⎠ ⎣ ⎠⎝ ⎠

or x x

x x

–11 1 2–2 sec ,

2 3

⎛ ⎞ ⎛ ⎤ ⇒ ⎜ ⎟ ⎜ ⎥⎝ ⎠ ⎝ ⎦

y yy y

–11 12 cosec 0,

6

⎛ ⎞ ⎛ ⎤ ⇒ ⎜ ⎟ ⎜ ⎥⎝ ⎦⎝ ⎠

So, x yx y

–1 –1 1 5sec cosec ,

2 6

⎛ ⎞ ⎛ ⎞ ⎛ ⎤ ⎜ ⎟⎜ ⎟ ⎜ ⎥⎝ ⎠ ⎝ ⎦⎝ ⎠

56. Answer (C, D)


Solution :

cos sin 0 4 1 2

– sin cos 0 3 –2 5 0

1 – sin 0 0 0 0

Solution : Q. 57 and Q.58

Let P (x, y)

x ysin cosec sin – cosec

,2 2

x2 – y2 = 1

C x2 – y2 = 1

Normal of ‘C’ at (sec, tan) is

x tan + y sec= 2 tan. sec

Now, A (2 sec, 0)

B (0, 2 tan)

Let mid point of AB is m = (sec, tan) (x, y)

Again curve S is x2 – y2 = 1


Hint : x2 – y2 = 1

The equation of common tangents of ‘S’ is x2 + y2 = 1 is

x = + 1

Distance between common tangent = 2


Hint : x2 – y2 = 1, e 2

Solution :

Curve S x2 – y2 = 1

e 2 , does not intersect y-axis

Tangent 2, 3 is x y2 – 3 1

And curve C and S are identical

59. Answer (B, C)

Hint : Draw graph of x x1

sin , cos ,4

⎧ ⎫⎨ ⎬⎩ ⎭



sin , cos ,4

⎧ ⎫⎨ ⎬⎩ ⎭

Solution : Q. 59 and Q. 60

1

4

1

4

2

3

2

2

By observing graph,

1 = 1

2 = 5

� � �

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

1/14

1. (8)

2. (6)

3. (6)

4. (8)

5. (4)

6. (5)

7. (8)

8. (5)

9. (B, C)

10. (C, D)

11. (B, C)

12. (A, C, D)

13. (B, D)

14. (A, B, C, D)

15. (B, C)

16. (B, C)

17. (A, C)

18. (A, D)

19. (A, C)

20. (A, B)

21. (2)

22. (5)

23. (3)

24. (3)

25. (5)

26. (2)

27. (4)

28. (8)

29. (A, B, D)

30. (B)

31. (A, B, D)

32. (A, B, C, D)

33. (A, B, C, D)

34. (A, B, C)

35. (A, B)

36. (A, B, C, D)

37. (B, C, D)

38. (B, C, D)

39. (A, D)

40. (A, C)

41. (3)

42. (2)

43. (8)

44. (1)

45. (7)

46. (7)

47. (0)

48. (2)

49. (C, D)

50. (A, B, C)

51. (A, B)

52. (B, C)

53. (B, C)

54. (B, C)

55. (B, C)

56. (A, B, D)

57. (A, B, D)

58. (A, C, D)

59. (B, C)

60. (A, C, D)

ANSWERS

Test Date: 10/03/2019

PHYSICS CHEMISTRY MATHEMATICS

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

MOCK TEST - 2 (Paper-2) - Code-F

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

2/14

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (8)

Hint :

The bottom surface of the container

remains covered with liquid during rotation (no dry spots).

Solution :

A vertical cylindrical container partially filled with a liquid

is rotated at constant speed. The drop in the liquid

level at the center of the cylinder is to be determined.

Free

surface

zs

z

g

R = 20 cm

r

h =0

60 cm

Taking the center of the bottom surface of

the rotating vertical cylinder as the origin (r = 0, z = 0),

the equation for the free surface of the liquid is given as

22 2

0( ) ( 2 )

4sz r h R r

g

where h0 = 0.6 m is the original height of the liquid

before rotation, and = 2n = 2(120 rev/min)1min

60s

⎛ ⎞⎜ ⎟⎝ ⎠

= 4 rad/s

Then the vertical height of the liquid at the center of the

container where r = 0 becomes

2 2 2 2

0 2 2

(4 ) (0.20m)(0) (0.60m)

4 4( m /s )s

Rz h

g

= 0.44 m

Therefore, the drop in the liquid level at the center of

the cylinder is

drop, center 0 (0) 0.60 0.44 0.16 m 16 cms

h h z

2. Answer (6)

Hint :

P = 2(P1 + P2) + Pm

Solution :

Peq = 2(P1 + P2) + Pm

–eq 1 2

1 1 12

f f f

⎛ ⎞ ⎜ ⎟

⎝ ⎠

2

51

1 1 14cm

1 30 120f

⎛ ⎞⎜ ⎟⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟

⎝ ⎠

1

31

1 –1 –12cm

1 30 60f

⎛ ⎞⎜ ⎟⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟

⎝ ⎠

3. Answer (6)

Hint :

0

0

s

v vf f

v v

⎛ ⎞ ⎜ ⎟⎝ ⎠

Solution :

Frequency received by approaching car

1 0

21

330f f

⎡ ⎤ ⎢ ⎥⎣ ⎦

Frequency received by source again

1

22

1330

ff

⎛ ⎞⎜ ⎟⎝ ⎠

So, beat frequency fB = f2 – f0 = 6 Hz

4. Answer (8)

Hint :

Use KVL and KCL to solve circuit.

Solution :

1

402 A

20I

4 4

2 A2 A

S

R

100 V

4 A

20

DC

B

2 A

A

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

3/14

16 + 2R – 100 + 40 = 0

2R = 100 – 40 – 16

2R = 44

R = 22

P = 88 watt

5. Answer (4)

Hint :

Kinetic energy = 2

2

P

m

Solution :

22

LL ⇒

KE =

221

2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

P h

m m

=

2

2

1

2 h

m

=

2

2

4

2

h

m L

=

2

2

2h

mL

2n

n = 4

6. Answer (5)

Hint :

Use rotational kinetic energy.

Solution :

When rod becomes vertical

2

21

2 2 3 L ML

Mg

23 g

L

T = 2

/2.

2 ∫

L

L

Mgdm x

=

22

/22 2

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎣ ⎦

L

L

Mg M x

L

= 9

2 8Mg

Mg

= 13

8Mg

131 5

8 8

nn ⇒

7. Answer (8)

Hint :

Total energy of system is conserved and speed will be

max. at centre of sphere.

Solution :

Speed will be maximum when point mass is at the

centre of sphere.

2

rel

1PE

2v

2rel

3 3 .3 1 .3

4 2 2 4

Gm m Gm m m mv

R R m

2

2

rel

15 1 3

4 2 4

Gm mv

R

rel

10 4

3 3

Gm vv v v

R

v = 45

8

Gm

R

8. Answer (5)

Hint:

2 21 22

A tE t

E A t

Solution :

2 22 t

E A e

lnE = 2lnA – 2t2

2 21 22

A tE t

E A t

2 2% 2(2) (0.2) 2 (2.5) 2E

E

% 4 1 5%E

E


4/14

9. Answer (B, C)

Hint :

Energy = 1

2Stress × Strain × Volume

Solution :

22 4

mg gmg m a a ⇒

3

2 4 4

mg gT m T mg

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

Stress = 3

/4mg A

⎛ ⎞⎜ ⎟⎝ ⎠

21 3

2 4

mg ALU

A Y

⎡ ⎤ ⎢ ⎥⎣ ⎦

10. Answer (C, D)

Hint :

vnC T V

Solution :

vnC T V

3

2n RT V

T = 2

3

V

nR

PV = n R T

2

3

nR VPV

nR

2

3PV V

1

22

3PV

Molar specific heat

C = 1

v

RC

x

= 3

121

2

⎛ ⎞ ⎜ ⎟⎝ ⎠

RR

= 3

22

R R

C = 7

2R

U = n Cv T

3

2U nR T

w = 2R.nT

2 4

3 3

2

w R

UR

4 904 120J

3 3w U

11. Answer (B, C)

Hint :

The satellite will move in elliptical orbit and earth as at

a focus.

Solution :

TE = 4 8

GMm GMm

R R

TE = 8

GMm

R

2 8 GMm GMm

a R

a = 4R

21

8 2 6

GMm GMmmv

R R

21

8 6 2 GMm GMm

mvR R

2–6 8 1

48 2

GMm GMmmv

R

2

12

GMv

R

v = 12

GM

R


Hint :

H = W – u

Solution :

Qin = CE = 3 × 3 = 9 C

Uin = 1 27(3 9) J

2 2


5/14

Cf = 2C.6C 3

C2C 6C 2

9F

2f

C

Ufinal = 1 9 9 2

9 J2 9

Heat dissipated = 27

9 4.5 J2

QP = 1

1 Qk

⎛ ⎞⎜ ⎟⎝ ⎠

= 1 2

1 9 9 6 C3 3

⎛ ⎞ ⎜ ⎟⎝ ⎠

0 .

2 2

⎛ ⎞ ⎜ ⎟⎝ ⎠

A B

E d E dV V

k

= . 1

12

⎡ ⎤⎢ ⎥⎣ ⎦

E d

k

= 2

3Ed

23 2

3A B

V V V

13. Answer (B, D)

Hint :

dNN

dt

Solution :

Initial activity of 24Na

A = 1.0 CdN

dt

= 1.0 × 10–6 × 3.71 × 1010

= 3.71 × 104 disintegration s–1

Half life = T = 15 h = 15 × 3600 s

Initial Activity

A = N0

3.71 × 104 = 0

0.7

15 3600N

4

0

3.71 10 15 3600

0.7N

N0 = 2862 × 106

= 2.86 × 109

N = Number of Nuclei present after 7.5 h in blood

sample per cc

' dNN

dt

294 0.7'

60 15 3600

N

N = 3.78 × 105 cm–3

N0 = Number of nuclei present in blood sample

Total activity now

43.71 10

2

294amt 1cc

60

4

60 3.71 105.35

294 2


Hint :

Use dimensional analysis.

Solution : L hxcyGz ...(1)

M hpcqGr ...(2)

From (1)

0 0 2 1 1 1 3 2M L T ML T LT M L T

x y z ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

1 3 1

2 2 2x y z

From equation (2),

1 1 1, ,

2 2 2p q r

15. Answer (B, C)

Hint :

Assume Q at the centre of cube of size 2L.


6/14

Solution :

The flux will same through the far faces and is same

through all near faces.

31 + 32 = 0

Q

2 = flux through far face

2 = 0 0

1

4 6 24

Q Q

31 = 0 0

8

Q Q

31 = 0

7

8

Q

16. Answer (B, C)

Hint :

Radial component of spring force will balance the

centripetal force.

Solution :

A

BCO

(A) Elongation of spring =

2

2 8(2 )

5

⎛ ⎞ ⎜ ⎟⎝ ⎠

RR R

= 2

2 644

25 R

R R

= 6

5 5 R

R R

Radial component of the spring force

Fr = cos5

⎛ ⎞ ⎜ ⎟⎝ ⎠

Rk

8 4sin

5.2 5 R

R

3cos

5

fr = 3 3

.5 5 25 R kR

k

Normal force to be zero.

23

25

mv kR

R

2

2 3 3 10 16 2

25 25 1

kRv

M

v2 = 12 16

5

16.2 msv

�

sin5

t

RF k

sin5

dv kRm

dt

4 410 2

5 5 dv

dt

64

5

= 12.8 m/s2

17. Answer (A, C)

Hint :

ma = mg – Bid

Solution :

LdiB dv

dt

.Ldi dxB d

dt dt

. . B d xI

L

ma = mg – Bid

ma = 2 2

B dmg x

L

a = 2 2

B dg x

Lm


7/14

18. Answer (A, D)

Hint :

ma = mg – Bid

Solution :

2 2

dv B dg x

dt mL

v = v0 sin t

v = 0sin

Bdv t

mL

A = 0

V

at t = 0

Acceleration a = g

v0 = g

A = 2g

A = 2 2 2 2gmL mgL

B d B d

19. Answer (A, C)

Hint : The path difference at the slit is zero.

Solution :

In first case

sin (2 1)2

d n and 2

d

2 1sin 1

4

n

n 2.5 So, n = 1, 2

When

1 1

1 11, sin tan

4 15 ⇒ n

1

1tan

15y D

When

2 2

3 32, sin tan

4 7 ⇒ n

2

3m

7y

20. Answer (A, B)

Hint :

The additional path difference of 2

d exist at the slits.

Solution :

sin (2 1)2 2

dn

and 2

d

sin 1

n 1.5

n = 1

Position of centre maxima

= = 30°

tan = 0

y

D

y0 = 1

3 3

D

Path difference for p1, 3

2 2x

sin = 3 3

2 4d

tan1 = 13

7

y

D

y1 = 3

7

path difference for p2,

x = 2 2

tan2 = 2

1 1m

15 15y⇒


8/14

21. Answer (2)

Hint :

[Ni(CO)4] are diamagnetic

Solution :

sp3

Ni 3 d8

4 s2

dsp2

[Ni(CN) 4

2–

]3d

8

Ni2+

CN is strong ligand–

Ni(CO)4

CO is strong ligand

22. Answer (5)

Hint :

Lindlar reduction.

Solution :

H C–H C–C C–CH –C–H23 2

NaBH4

O

H C–H C–C C–CH – –OH23 2 2

CHPBr

3

H C–H C–C C–CH – –Br23 2 2

CH

H C–CH –C C–CH – –MgBr23 2 2

CH

Mg/ether

CO ; H O22

H C–CH –C C–CH – –C–OH23 2 2

CHsp

PSOCl2

H C–H C–C C–CH –C23 2 2–CH

H2

O

H C–C –C C–CH – –C–Cl23 2 2

H CH

Pd/BaSO4

O

H

sp2

sp

sp2

sp2

H H

O

23. Answer (3)

Hint :

During formation of osazone, 1 mol consume 3 mol of

PhNHNH2.

Solution : p = 3, q = 3

O||C–H

CHOH

(CHOH)3

CH OH2

CH=N–N–C H6 5

H C=N–N–C H

6 5

(CHOH)3

CH OH

Phenyl osazone2

H

+ 3C H NH–NH2

6 5

+ Ph–NH2

+NH

3

+H

2O

24. Answer (3)

Hint : -hydroxy acid on heating form unsaturated

compound.

Hydrolysis of ester and dehydration

Solution :

CH –C–OCH2 3 H

3O

+

O

CH –C–OH2

O

CH–C–OH

O

OH OH

25. Answer (5)

Hint :

cis form is optically active. x = 3, y = 8

Solution : C.N of Cs = 8 = y

H3N

NH3

Co

ox

H3N

NH3

(d & �)Cis

ox

ox

ox

Co

Co

NH3

ox ox

NH3

(Trans)

PART - II (CHEMISTRY)


9/14

26. Answer (2)

Hint :

2 4 3 2NH CONH (s) 2NH CO��⇀

↽��

Partial pressure of NH3= 2(Partial pressure of CO

2)

Solution :

2 4 3 2NH CONH (s) 2NH (g) CO (g)

0 0

2x x

��⇀↽��

Ptotal

= P

2CO

x PP P 1

2x x 3

3NH

2x 2P P 3

2x x 3

= 2

KP =

3 2

2

NH COP P = 4

27. Answer (4)

Hint :

Due to addition of CH3Li to carbonyl group produce

alcohol.

Solution :

CH3

C

O CH3

(i) CH3Li

(ii) H O2

CH3

C

HO CH3

CH3

BCl3

Me

C+

Me

Me

Me(i) O

3

(ii) Zn; H O2

Me

O

Me

OH–

H C3

O

CH3

C

H

O

CCH

3

28. Answer (8)

Hint :

Root mean square velocity and most probable velocity

are 3RT 2RT

andM M

respectively.

Solution :

1 2

1 2

3RT 2RT

M M

2

400 3 2 60

80 M

15 = 2

120

M

M2 = 8


Hint : P Ca3P2, Q = MgO

R = PH3, S = Cu

3P2

Solution :

3 4 2 3 2(Q)(P)

Ca (PO ) + 8Mg Ca P + 8MgO

3 2 2 2 3(P) (R)

Ca P + 8H O 3Ca(OH) 2PH

3 4 3 2 2 42PH 3CuSO Cu P 3H SO

30. Answer (B)

Hint :

Frequency () of revolution of e– in orbit is Z2/n3

Solution :

2

2

3

Zv

n

nr

Z

Z

n

⎤ ⎥⎥⎥ ⎥⎥⎥

⎥⎦

5

He

2H

x 52;y 33


Hint :

S4 and S

6 are incorrect.

Solution :

Cysteine are non-essential -amino acid.


Hint :

n = 3, formula is XY3 and molar mass of XY

3 is 108 g/mol.

Solution :

If (Tf) XY

n = 4 (T

f)urea

at same concentration.

It means n = 3


10/14

During electrolysis of molten XY3

Cathode X3+ + 3e– X

Anode : 3Y– 3Y + 3e–


Hint :

Oxidation of side chain and diazotisation.

Solution :

CH3

CH3

KMnO4

H+

NH3

COOH

COOH

C

C

N–H

(i) NaOH, Br2

(ii) H+

COOH

NH2

(i) NaNO +HCl2

0–5°C

C–O–

N N

O

+

–N

2

–CO2

(Q)(R)

(S)

(ii) OH–

O

O


Hint :

Diel-Alder reaction and ozonolysis.

Solution :

H

H

O

O

C

C

O(i) O

3

(ii) Zn; H O2

O

C

O

C

O

O

C

O

C

O

C

O

H

O

CH

CHO–H2

(i) LiAIH4

excess

(ii) H O

3

CHO–H2

CH2–OH

CH2–OH

35. Answer (A, B)

Hint :

Mg and Al are highly electro positive metal.

Solution :

Fe2O3 and SnO

2 is reduced by carbon.


Hint :

Arrhenius equation.

Solution :

K = Ae–Ea/RT


Hint :


Solution :

(CH –CH –C–) O3 2 2

OCH –CH –C–O

3 2

–

O

CH –CH–C–O–C–CH –CH3 2 3

O O

Ph C

H

O

+ CH–C–O–C

O

CH3

O

CH2

CH3

Ph–CH–CH–C–O–C–CH2

O–

CH3

O

CH3

H O2

Ph–CH C–C–OH

CH3

O

+ HO–C–CH –2

CH3

O

O


Hint :



11/14

Solution :

Ph

C

O

H CH2

C

O

O

O

C

Ph

CH

O

O

O

C O

OPh

O O

Ph

O O–

O

C OCH

HH C–C–O

3

–

O

PhCH

CH

CO

O–

–

O

O

O–

PART - III (MATHEMATICS)

41. Answer (3)

Hint : 7 tails and 3 heads.

Solution :

7 tails, 3 heads

Probability = C C

10 3

7 3

10

.

2

120 15

1024 128

= 3

42. Answer (2)

Hint : R = 8r

Solution :

R = 8r

∵

A B C 1sin . sin . sin

2 2 2 32

∵ A B C9

cos cos cos8

B B2 9sin 1– 2sin

2 2 8

B4sin 1

2

43. Answer (8)

Hint : Given family passes through (1, –2).

Solution :

(3x + 7y + 11) sec + (5x – 3y – 11) cosec = 0

(5x – 3y –11) + tan (3x + 7y –11) = 0

These family of lines always pass through (1, –2).

Let A(x, y) be a point on x – y + 3 = 0

Now, PB2 = PA2 + AB2 – 2PA. AB cos.

PA2 + AB2 – 2PA . AB

= (PA – PB)2

So maximum of A BP P– is PB.

PB2

40 40 ⇒ 2

85

⇒

44. Answer (1)

Hint : Substitute all angles in half angle.

Solution :

i

i

r

r r r xx x x

f xr r

x x1

2 1 2 1 6 32sin cos 2sin .sin

2 2 2 2

6 3 2 12sin . cos

2 2

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑

i

r

rx r x

1

cos – cos 1

∑

fi(x) = cosx – cos (i + 1)x

f20

21cos – cos 2

4 4 4

⎛ ⎞ ⎜ ⎟⎝ ⎠

39. Answer (A, D)

Hint :

B2H6 on reaction with NH

3 form inorganic benzene.

Solution :

Y = B2H6, X = B

2O3

40. Answer (A, C)

Hint :

Boron and inorganic benzene.

Solution :

Organic benzene is more symmetrical molecules so

has higher mp(6°C) than inorganic benzene (–58ºC).


12/14

45. Answer (7)

Hint : Use transformation of roots.

Solution :

The equation where roots are y = f(i) i.e. y = x2 – 2

x y 2

y y5 2

2 2 1 0

y5 + 10y4 + 40y3 + 74y2 + 74y + 23 = 0

Product of roots = –23

Also, 1

2

3

4

5 = –1

i i

i

f f

5 5

1 1

– 30 7

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

46. Answer (7)

Hint : Fav.cases = number of non-negative integral

solutions of x1 + x

2 + ..... + x

20 = 10

Solution :

The number non-negative integral solutions of x1 + x

2 +

..... x20

= 10 is C29

19

Required probability = C

29

19

1020

n = 19

n – 12 = 7

47. Answer (0)

Hint : a b. 0� �

Solution :

a b. 0� �

x x2

2 2– log 6 4 log 0

–y2 + 4y + 6 < 0 , –< 0 , D < 0

162 + 4 × × 6 < 0

8[2 + 3] < 0

∵

0–3

2

–3

,02

⎛ ⎞ ⎜ ⎟⎝ ⎠

48. Answer (2)

Hint : Taking log both side.

Solution :

xx xlog –

2log x = 0

xx x– log 0

2

⎛ ⎞ ⎜ ⎟⎝ ⎠

xx 1–

2

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

logx = 0

x = 1, 4 (∵x > 0)

49. Answer (C, D)


Solution :

cos sin 0 4 1 2

– sin cos 0 3 –2 5 0

1 – sin 0 0 0 0


Hint : AM G.M. and range sec–1x & cosec–1x

Solution :

xy x yx y

1 10 2, –2 ⇒

or, x yx y

1 1–2, 2

x x

x x

–11 12 sec ,

3 2

⎛ ⎞ ⎡ ⎞ ⇒ ⎜ ⎟ ⎟⎢⎝ ⎠ ⎣ ⎠

y yy y

–11 1 ––2 cosec ,0

6

⎛ ⎞ ⎡ ⎞ ⇒ ⎜ ⎟ ⎟⎢⎣ ⎠⎝ ⎠

So, sec–1 x yx y

–11 1cosec ,

6 2

⎛ ⎞ ⎛ ⎞ ⎡ ⎞ ⎜ ⎟⎜ ⎟ ⎟⎢⎝ ⎠ ⎣ ⎠⎝ ⎠

or x x

x x

–11 1 2–2 sec ,

2 3

⎛ ⎞ ⎛ ⎤ ⇒ ⎜ ⎟ ⎜ ⎥⎝ ⎠ ⎝ ⎦

y yy y

–11 12 cosec 0,

6

⎛ ⎞ ⎛ ⎤ ⇒ ⎜ ⎟ ⎜ ⎥⎝ ⎦⎝ ⎠

So, x yx y

–1 –1 1 5sec cosec ,

2 6

⎛ ⎞ ⎛ ⎞ ⎛ ⎤ ⎜ ⎟⎜ ⎟ ⎜ ⎥⎝ ⎠ ⎝ ⎦⎝ ⎠

51. Answer (A, B)

Hint : Summation of series would be 90.


13/14

Solution :

r

r r

90

1

2 sin2 tan1

∑

= 2sin2 4sin4 6sin6 ......

tan1178sin178 180sin180

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

But = 1°

= tan1° × 90 cot1°

= 90

Therefore, (5 + 2 –6) (5+ 90 – 6) > 0

(–, –6) (1, )

52. Answer (B, C)


Solution :

a b c

2 2

2 2 3 3

2 2 3 3 4 4

3 1 1

, , 1 1 1

1 1 1

2 2 2 2

1 1 1 1 1 1

1 1

1 1

= ( – )2 ( – ( +) + 1)2

= (( + )2 – 4) ( – ( +) + 1)2

= b c c b

a a a a

2 2–

– 4 1

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

= a b c b ac

a

2 2

4

– 4

53. Answer (B, C)

Hint : b b

a af x dx f a b x dx– ∫ ∫

Solution :

nx

nxI dx

x–

sin

sin 1

∫ ........(1)

n

x

n xI dx

x–

–

sin –

sin – 1

⎡ ⎤ ⎣ ⎦∫ .......(2)

(1) + (2),

n

nxI dx

x0

sin

sin

∫

I1 = I

3 = I

5 = ........ =

I2 = I

4 = I

6 = ........ = 0

54. Answer (B, C)

Hint : i

P A P Ei P A Ei

12

0

. /

∑

Solution :

Let Ei denote the event that the bag contain i black

and (12 – i) white balls (i = 0, 1, 2, ....., 12 ). Let A

denote the event that the 4 balls drawn are black. Then

P Ei1,

13 i = 0, 1, 2, ...., 12

4

12

4

for 0, 1, 2, 3

/for 4, 5, ...., 12

e

O i

P A Ei Ci

C

⎧⎪ ⎨ ⎪⎩

(i) i

P A P Ei P A Ei

12

0

1 1/

13 12 ∑

4 5 12

4 4 4....C C C⎡ ⎤ ⎣ ⎦

13

5

12

4

1

513

C

C

(ii) By Baye’s Theorem,

c

P E P A EE cP

A P A

10

4

1210 1010 4

1

13. / 70

1 429

5

⎛ ⎞ ⎜ ⎟⎝ ⎠

55. Answer (B, C)

Hint : Equation of common chord.

Solution :

Clearly AB is chord of contact with respect to (1, 1).

so equation of AB is 3x – y + 2 = 0 .......(1)

The tangent y2 = 4ax is a

y mxm

.......(2)


14/14

Comparing (1) and (2) a 6⇒

1. Length of L.R. = 24

2. y = 6x + 1 is tangent to curve y2 = 24x

3. Common tangent for x2 = 4ay and y2 = 4ax is

x + y + a = 0

4. Normal y = mx – 2am – am3

put m = 1 and a = 6

y = x – 12 – 6

y = x – 18


Hint : For interior point, S1 < 0

Solution :

Point P lies inside the circle.

so, S1 < 0

(p + 1)2 + (2p –1)2 –4 (p + 1) –4(2p – 1) – 4 < 0

p7 – 59 7 59

,5 5

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

.....(1)

And for line x + 2y – 2 = 0 point P lies on the same side

origin lies

So, (p + 1) + 2(2p – 1) – 2 < 0

p3

5 ......(2)

From (1) and (2),

p7 – 59 3

,5 5

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

Solution : Q. 57 and Q.58

Let P (x, y)

x ysin cosec sin – cosec

,2 2

x2 – y2 = 1

C x2 – y2 = 1

Normal of ‘C’ at (sec, tan) is

x tan + y sec= 2 tan. sec

Now, A (2 sec, 0)

B (0, 2 tan)

Let mid point of AB is m = (sec, tan) (x, y)

Again curve S is x2 – y2 = 1


Hint : x2 – y2 = 1

The equation of common tangents of ‘S’ is x2 + y2 = 1 is

x = + 1

Distance between common tangent = 2


Hint : x2 – y2 = 1, e 2

Solution :

Curve S x2 – y2 = 1

e 2 , does not intersect y-axis

Tangent 2, 3 is x y2 – 3 1

And curve C and S are identical

59. Answer (B, C)


sin , cos ,4

⎧ ⎫⎨ ⎬⎩ ⎭



sin , cos ,4

⎧ ⎫⎨ ⎬⎩ ⎭

Solution : Q. 59 and Q. 60

1

4

1

4

2

3

2

2

By observing graph,

1 = 1

2 = 5

� � �

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