Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions) All ... · 6/14 (B) H Cl 2 4 3 1 HO C H 25...

Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019

1/14

1. (2)

2. (1)

3. (2)

4. (3)

5. (3)

6. (5)

7. (7)

8. (6)

9. (A, D)

10. (A, C)

11. (A, C)

12. (A, B, C)

13. (A, C)

14. (B, D)

15. (B, C)

16. (C, D)

17. (A, B, D)

18. (A, B, D)

19. (A, C)

20. (A, D)

21. (3)

22. (6)

23. (6)

24. (4)

25. (4)

26. (2)

27. (6)

28. (7)

29. (A, C)

30. (A, B, D)

31. (A)

32. (A, C)

33. (B, D)

34. (A, B, D)

35. (A, C, D)

36. (A, B, C)

37. (D)

38. (D)

39. (A, D)

40. (A, B)

41. (7)

42. (4)

43. (3)

44. (8)

45. (1)

46. (5)

47. (2)

48. (6)

49. (A, B, C)

50. (A, B)

51. (A, B, C, D)

52. (A, D)

53. (A, C, D)

54. (C, D)

55. (A, C, D)

56. (A, B, C, D)

57. (A, B, D)

58. (C)

59. (A, D)

60. (B)

ANSWERS

Test Date: 17/02/2019

PHYSICS CHEMISTRY MATHEMATICS

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

MOCK TEST - 1 (Paper-2) - Code-C

All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions)

2/14

PART - I (PHYSICS)

ANSWERS & HINTS

1. Answer (2)

Hint : O B A BT T

1

2

Solution :

mgx = fR mg

f xR

g ga x

R R

R T R

T tg g

2 ,4 2

2. Answer (1)

Hint : E A2 2

0

1

4 2

Solution :

x1

7cm

8 x

2

17cm

8

2 212.5

4 2U E A

2 2

2

12.5

4 2

TU E A

v

2 2

2

5 1 1 4 100100 4

2 4 2 100 100U E

25U E 2

15

U

3. Answer (2)

L

H

Hint : 2dpx

dx

Solution :

22

LAL ghA

Lh

g

2 2

2

4. Answer (3)

Hint : dP

B VdV

–

Solution :

3 24, 4

3V a dV a da

–

–/

dP dPB V

dV V dV

MgdP

A

3

2

4–

3 4

a MgB

A a da

da Mg

a BA–3

5. Answer (3)

Hint : pr v

RC C

n–

–1

Solution :

T2 P3 = C

2

3 5 2

PVP C P V C

nR


3/14

2/5pV C

3 19–

22 6–1

5

pr

R R RC

Q = n Cpr T R

Q19

2 1006

RQ

1900

3

32 100

2

Ru

u = 300 R

1900 1000– – 300

3 3

R RW Q u R

6. Answer (5)

Hint : hc

Emin

max

Solution :

1 hC

ev

If V increase 1 decreases

7. Answer (7)

Hint : Length of string will remain constant

Solution :

V0 cos 37 = V

0

4

5 V V

VV0

5

4

8. Answer (6)

Hint :

Fmax

= 2 mAg

Solution :

Fmax

= 2 mAg

9. Answer (A, D)

Hint :

Apply Ampere’s law

Solution :

Amplere’s law

0 enclosed. .B dl I

��

�

0–

ˆ –2

��

B dxi

10. Answer (A, C)

Hint : Ea

02

Solution :

( , )a a

( )p

3

1

x

y2

31 2

0 0 0

1 1

2 . 2 . 2 . 2 2 2pE j i i j

a a a

��

3 3

2 1

0 0

1 1

2 . 2 2 . 2

��

pE i ja a

11. Answer (A, C)

Hint : Apply energy conservation and momentum

conservation

Solution :

T

2ma0

Ring

T cos 37° = ma

For 2 m

T cos 37° + 2 ma0 = (2 m) g sin 37°

We get the result

12. Answer (A, B, C)

Hint : Use R – C circuit analysis.

Solution :

K Q KQV

d a

1–

0 0

1

QaQ

d

t

RcQ Q e–

1 t = Rc ln3 = 4

0 .aR ln3

1

3

QQ

2 2 2

22 200

8 .2 4 .

VQ a Q a

dd d


4/14

13. Answer (A, C)

Hint : Apply I ��

Solution :

(2 2mR + mg2

)N

fO

0 = I

0

RmR mg mR mR2 2 2

2 2 2 22

2

4

g R

R

14. Answer (B, D)

Hint : ax

VI

Z

max

m

Solution :

15

I2

60

30 v

I1

I

I t1

100sin 50

2 3

I t2

100sin 50 –

2 6

I t50 2 sin 5012

Pam

= Vrm

Irm

cos

100 5 2cos15

2 2

15. Answer (B, C)

Hint : Apply superposition principle

Solution :

00

1

4

R

C

x dxV

x

0

4

R

02

P

RV

16. Answer (C, D)

Hint : After reflection from mirror there will be phase

change of Solution :

D 2D

y

O

Screen

d

D

y

P

Q

OP dOP d

D D2 2

d OROR d

D D3

3

y = 2d = 1 mm

also = yd

nD

1–

2 2

or y d n 5.5, n 15.5

17. Answer (A, B, D )

18. Answer (A, B, D)

Hint : f Nmax

1 1 1

Solution for Q. Nos. 17 and 18

4 kg

2 kg

1 kg f1

f2

f3

14 N


5/14

ca

14 – 7

7

= 1 m/s2

max

37 Nf

max

12 Nf

max

23 Nf

max

1a = 2 m/s2

max

2a = 1 m/s2

19. Answer (A, C)

20. Answer (A, D)

Hint :

Rt

LI I e–

max1–

in R - L circuit

Solution for Q. Nos. 19 and Q. Nos. 20

1

2

1 V

1

2

I

4H

3

Circuit can be reduced as

t

LI I e

R–

max1–

4R

3

4 2 4

4 2 3

R

t

I e

–4

31 4

1–4 3

Imax

1A

4

I = 0.25 (1 – e–t) A

also L1I1 = L

2I2

I t I I t I1 2

2 1,

3 3

tI t e–

1

11–

6 , tI t e

–

2

11–

12

Total

1 4 1 1J

2 3 16 24

U

PART - II (CHEMISTRY)

21. Answer (3)

Hint: Kp remain constant for a reaction, even in

simultaneous equilibria.

Solution :

a 2b a

PQ(s) P(g) Q(g)

��⇀↽�� K

p = 4.8 × 10–6

2

2b a

RQ (s) R(s) 2Q(g)

��⇀↽�� K

p = 5.76 × 10–6

(2b + a)2 = 5.76 × 10–6

(2b + a) = 2.4 × 10–3

(2.4 × 10–3) × a = 4.8 × 10–6

a = 2 × 10–3 = 2 × 10–x

x = 3

22. Answer (6)

Hint: To obtain aromatic character R can lose a

hydride H–, and become sp2 hybridised.

Solution :

C C

H

+

+

H

+ C(Ph)3

+ H – C(Ph)3

(R)

–

BF4

–

BF4H

a a

Number of electrons in ring R is 6.

23. Answer (6)

Hint: First convert molecules tetrahedral geometry into

Fisher form.

Solution :

(A) CH3

H

OH

2

4

3

1

(R)

(1)C H2 5


6/14

(B)

HH

Cl

24

3

1

HO C H2 5

(R)

(1)

(C)

HH

Cl2

4

3

Br (S)(2)

CH3

1

(D)

HH

C H2 5

Br

4

3

2

(S)(2)

CH3

1

24. Answer (4)

Hint: Check the chirality and elements of symmetry in

product.

Solution :

P1

Optically active

P2

Optically active

P3

Optically active

P4

Optically active

P5

Optically inactive

P6

Optically inactive

P7

Optically inactive

P8

Optically inactive

25. Answer (4)

Hint :

Number of moles CaC2O

4 = Number of moles of CaCI

2

= Number of moles of organic acid

Solution :

(Oxalic acid) N1V

1 = N

2V

2 (KMnO

4)

2 × no. of moles = 80

5 0.051000

= 0.01 moles

Number of moles Oxalic acid = number of moles

of CaC2O

4 = number of moles of CaCI

2

Mass of CaCI2 = 0.01 × 111 = 1.11 g

% of CaCI2 in mixture =

1.11100 60%

1.8

26. Answer (2)

Hint :

2

2Cd H S CdS 2H

��⇀↽��

Solution :

0.5 × 10–3 moles of Cd2+, 0.5 × 10–3 moles of S–2 ions

are required to precipitate.

0.5 ×10–3 moles of S2– 0.5 × 10–3 moles H2S

1×10–3 moles of H+

Number of moles of H+ from HCl = 50 × 0.16

= 8 ×10–3 moles

Total moles of H+ = 9 × 10–3 moles

–3–29 10

H 1000 10 molar900

27. Answer (6)

Hint: Diborane has two 3c –2e

H

H

H H

H HB B

Solution : Each 3c – 2e– bond involve 3 orbital, one

from each B and H.

28. Answer (7)

Hint: Consider the changes which occurs inthermodynamic properties during a spontaneousprocess like vapourisation.

Solution : Except (f) all statements are correct

statement (f) is false because when the temperature is

smaller than equilibrium temperature, the H term

dominates and G becomes +ve for the vapourisation

process.


7/14

29. Answer (A, C)

Hint: According to Nernst equation cell

RTE E lnQ

nf

cell

0.059E E logQ

n

y = c + mx

so, 1

mn

x = 0.059 logQ

Solution : if n = 1

tan = 1

= 45°

E°cell

= x (magnitude wise)

for hydrogen electrode

cell

2

0.059 [H ]E E log

1 H

0.770.059 log ( 0.6 0.77 and E 0)

0.6 ∵

= 0


Hint: AlCl3 in C

6H

6 present as dimer.

Solution : Borax loses water of crystallisation on

heating.

31. Answer (A)

Hint: Consider the factors affecting hydration energy.

Solution :

As size of cation increases the size of hydrated ions

in water decreases and thus ionic mobility increases.

Solubilities of bromides and hydroxides increases down

the group because their lattice energy change is more

than hydration energy on moving down the group.

32. Answer (A, C)

Hint: General properties of group-16 hydrides.

Solution : Correct order of M.P. is

H2S < H

2Se < H

2Te

as increase in mass of the molecule will cause

increase in intermolecular binding energy.

And bond angle of hydrides does not differ much on

moving down the group, from S to Te.

33. Answer (B, D)

Hint: Check the electron donating and withdrawing

extent of the group.

Solution :

Activating group activates the ring towards E.A.S. while

deactivating group deactivates it.


Hint: F2 is produced from anhyd. HF.

Solution : Water should not be present during

electrolysis otherwise O2 is liberated.

35. Answer (A, C, D)

Hint: Check all the possibilities of structures which

can be obtained after the reaction.

Solution :

H , Ni2 Cl , h2 3 products

H , Ni2 Cl , h

2

3 products


Hint: Apply concepts of gram equivalents.

Solution :

meq of KMnO4 = 25 × 0.2 = 5

meq of FeSO4 = 40 × 0.2 = 8

meq of H3ASO

3 = 30 × 0.1 × 2 = 6

meq of H2O

2 = 15 × 0.1 × 2 = 3

meq of SnCl2 = 25 × 0.1 × 2 = 5

37. Answer (D)

Hint: ∵ B can be taken as an intermediate of the

reaction.

d[B]

0dt

Solution : 1K

A B

2K

B F

1

d[A]k dt

dt

1 2

d[B]k [A] K [B]

dt

∵ 1K t

t 0[A] [A] e


8/14

PART - III (MATHEMATICS)

(∵ it has been assumed that all the reactions are

first order)

Now,

1K t

1 0 2

d[B]K [A] e K [B] 0

dt

1K t

02

1

[A] eK

K [B]

38. Answer (D)

Hint: We have to consider the given situation

i.e. K1 > K

2. So more of A is consumed as compared to

B.

Solution : ∵ K1 > K

2, the amount B becomes fairly

large before droping eventually to zero.

39. Answer (A, D)

Hint: As the solution is a good source of e–,

The solution will behave as a good reducing agent.

Solution : NaNH2 will be produced after a very long

time.

40. Answer (A, B)

Hint: Reaction of Na to H2O is highly exothermic due

to high polarity of H2O. So blue colour solution is not

obtained, as free electrons are absent.

Solution : Concentrated solutions are copper bronze

in colour.

Alkali metal solution in ammonia on evaporation yield

metals.

41. Answer (7)

Hint : Make cases

Solution : Here 3 vowels A, I, A can be placed in the

following five available places:

× N × G × N × N ×

OR

× N × G × G × G ×

OR

× N × G × G × N ×

Required number of ways =

3

3! 4! 4! 4!5

2! 3! 3! 2! 2!

C

= 5 4 3! 3 2! 4 3! 4 3! 4 3 2!

3! 2 1 2! 3! 3! 2! 2

= 30(4 + 4 + 6) = 420

4207

60 60

42. Answer (4)

Hint : A.M G.M.

Solution : Let , , , be the roots of given equation.

+ + + = 4 and = 1

Now A.M. of roots = 4

14 4

and G.M. of roots =

1 1

4 4( ) (1) 1

A.M. = G.M.

Which is possible only if

= = = = 1

Given equation will be of the form (x – 1)4 = 0

x4 – 4x3 + 4C2x2 – 4C

3x + 1 = 0

x4 – 4x3 + 6x2 – 4x + 1 = 0

Comparing, we get a = 6, b = – 4

a + b + 2 = 6 – 4 + 2 = 4

43. Answer (3)

Hint : Draw graph of y = x3 + 2x2 + x and y = –6

Solution : Given equation is x3 + 2x2 + x + 6 = 0

x3 + 2x2 + x = – 6

Let f(x) = x3 + 2x2 + x and g(x) = – 6

Here, f (x) = 3x2 + 4x + 1

and f (x) = 6x + 4

When f (x) = 0 3x2 + 3x + x + 1 = 0


9/14

3x (x + 1) + (x + 1) = 0

(x + 1) + (3x + 1) = 0

1

1,3

x

f(–1) = – 6 + 4 = – 2 < 0

and 1 1

6 4 2 03 3

f

f(x) has local maxima at x = – 1 and local minima

at x = 1

3

Also, max. f(x) = f(–1) = –1 + 2 – 1 = 0

and min. 1 1 2 1

( )3 27 9 3

f x f

1 6 9 4

27 27

( 2) 8 8 2 2 (1) 1 2 1 4

( 3) 27 18 3 12

f f

f

0

–3 –2 –1

x

y

y = g x( ) = – 6

(–3, –12)

(–2, –2)

y

x

1

3

From graph it is clear that one real ‘’ will lie between

(–3) and (–2) –3 < < –2

[ ] 3 3

44. Answer (8)

Hint : SP + SP = 2A

Solution : Locus of Z is ellipse

2,0 0, 2 ,2 4 s s A

1

2 22

ss Ae e

distance between directrices 2 4

81/ 2

A

e

45. Answer (1)

Hint : Solve Limit

Solution : From given equation

1

2

lim tan sin cos63 1

k

kx

kx kx k x

1

2

3lim tan sin

213 1

k

k x

xk x x

k k

= tan 33

2

3

3 1

x

x x

3x2 – 9x + 3 = x

3x2 – 10x + 3 = 0 = 1

46. Answer (5)

Hint : 1 2Area ( ) ( )

b

a

f x f x dx

Solution : y = x2 + x – 2, y = 2x

x2 + x – 2 = 2x x2 – x – 2 = 0

x2 – 2x + x – 2 = 0 x (x – 2) +1(x – 2) = 0

x = –1, 2

intersection points of y = x2 + x – 2 and y = 2x, are

(–1, –2) and (2, 4)

(–1, –2)

(–2, 0) xx

(2, 4)

y x = 2

y x + x – = 22

0

–1

(1, 0)

Here y = x2 + x – 2 is a parabolic curve with vertex

1 9,

2 4

which cuts x-axis at (–2, 0) and (1, 0) and

y-axis at (0, –2)


10/14

= Required area

2 2

2 2

1 1

(2x x x 2)dx ( x x 2)dx

23 2

1

x x2x

3 2

8 1 12 4 2

3 3 2

1 95

2 2

91 1 5

2

47. Answer (2)

Hint : f (x) < 0 at point P

Solution : At any point of maxima,

f(x) = 0 and f(x) < 0

f(x) = x2 – f2(x) –9 < 0

Let P(, )

2 – 2 – 9 < 0 and 2 – 2 = a2

a2 – 9 < 0 a (–3, 3)

48. Answer (6)

Hint : Centroid of a triangle divides the line joining

orthocentre and circumcentre in the ratio 2 : 1 internally.

Solution : Since centroid divides the line joining

orthocentre (2, 4) and circumcentre 7 5,

2 2

in the ratio

2 : 1

7 52 1 2 2 1 4

2 2, (3,3)2 1 2 1

A

2 2(3 0) (3 0) 3 2OA

A(3, 3)

y

xO

BP

L M N

45°

45°

45°

3 2

3

2AB

3

2 AP PB

3 33 ,3

2 2

B

+ = 6


Hint : Solve integration, then limit

Solution : We have 2 2

1a

ndx

n x

2

2

1

1( )a

dxn

xn

11 1tan

11

a

x

n

nn

11tan ( )

a

n nx

n

1 1 1tan tan tan ( )2

na an

1

2 2lim lim tan

21n n

a

ndxL an

n x

, if 02 2

0, if 02

, if 02 2

a

a

a

, if 0

, if 02

0, if 0

a

a

a


11/14

50. Answer (A, B)

Hint : Find AB

Solution :

2 2 23 3 3

6 6 6

ax bx cx

AB a b c

ax bx cx

Now, tr(AB) = tr(C)

3ax2 + b + 6 cx = (x + 2)2 + 2x + 5x2 x R

3ax2 + 6 cx + b = 6x2 + 6x + 4

Comparing, we get a = 2, b = 4, c = 1

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

Hint : Use properties of determinants

Solution : Given equation is

2 2

2 2

2 2

1 sin cos 2sin4

sin 1 cos 2sin4 0

sin 1 2sin4

A A

A A

A cos A

By R1 R

1 – R

2 and R

2 R

2 – R

3 we have

2 2

1 1 0

0 1 1 0

sin cos 1 2sin4A A

By C2 C

2 + C

3, we have

2 2

1 1 0

0 0 1 0

sin cos 1 2sin4 1 2sin4A A

– 0 + 0 – (–1) [cos2A + 1 + 2sin4 + sin2A ] = 0

2(1 + sin4) = 0

sin4 1 sin2

4 ( 1)2

nn

( 1) ( 1)4 8 4 8

n nn n

Which is independent of A.

When n = 0, = 0 ( 1)8 8

31,

4 8 8n

When n = –1, 2

4 8 8 8

Options (A), (B), (C), (D) are correct.

52. Answer (A, D)

Hint : Find dy

dxfrom quadratic equation.

Solution : Given differential equation is

2

( ) 0dy dy

y x y xdx dx

2( ) ( ) 4 ( )

2

x y x y y xdy

dx y

2( ) ( )

2

y x x y

y

( )

2

y x x y

y

or2 2

y x x y y x x y

y y

1ordy x

dx y

when 1dy

dx

0dx dy

x – y + c = 0 ....(1)

and when dy x

dx y

ydy + xdx = 0

2 2 2

2 2 2

x y k

x2 + y2 = k2 ....(2)

But curves passes through the point (1, 2)

1 – 2 + c = 0 c = 1

1st curve is the straight line

x – y + 1 = 0

and from (2), 1 + 4 = k2

2nd curve is x2 + y2 = 5


12/14


Hint : Take dot product of both sides with

, , andu v a w

��

.

Solution : Given u v w a ��

....(1)

Taking dot product of both sides of (1) with

, , andu v a w

��

, we get

2 3

2u u v u w a u

��

31

2u v u w � � � �

1

2u v u w � � � �

....(2)

2 7

4u v v v w a v

��

71

4u v v w � � � �

7 31

4 4u v v w � � � �

....(3)

2u a a v a w a a a

� � � � � ��

3 7

42 4

a w � �

13 34

4 4a w � �

....(4)

and 2

��

u w v w w a w

31

4

� � � �

u w v w

3 11

4 4

� � � �

u w v w ....(5)

By [(2) + (3) + (5)], we get

u v u w u v v w u v v w � � � � � � � � � � � �

1 3 1 2 3 11

2 4 4 4

1

2u v v w u w � � � � � �

....(6)

By [(6) – (2)] 0v w � �

By [(6) – (3)], 1 3 2 3 1

2 4 4 4w u

� �

From (6), 1 1

04 2

u v � �

1 1 2 1 3

2 4 4 4u v

� �

54. Answer (C, D)

Hint : 2 cosAcosB = cos(A + B) + cos(A – B)

Solution : Here 0

1( ) 2cos cos( )

4f x t x t dx

0

1cos ( ) cos ( )

4t t x t t x t dt

0

0

cos 1[ ] cos(2 )

4 4

xt t x dt

0

cos 1 sin(2 )

4 4 2

x t x

cos 1sin 2 sin 0

4 8

xx x

cos 1sin sin

4 8

xx x

cos( )

4

xf x

Which is clearly continuous and

differentiable is (0, 2)

( ) sin4

f x x

( ) cos4

f x x

When ( ) 0 sin 04

f x x

sin x = 0

x = 0, , 2,

At x = 0, 2, ( ) 1 04

f x

x = 0, 2 are the points of local maxima


13/14

and at x = , ( ) ( 1) 04

f x

x = is the point of local minima

At x = , ( ) sin 04

f x

atleast one c = (0, 2) such that f(c) = 0

and maximum value of ( ) cos4 4

f x

Max. ( ) at ( 0, 2 )4

f x x


Hint : Draw diagram.

Solution :

A

B

P(3, 4)C

(0, 1)

Power of point P(3, 4) w.r.t. circle

S x2 + y2 – 2y – 3 = 0 is defined as

PA PB = S1

= 9 + 16 – 8 – 30 = 14 option (A) is correct

2 2(3 0) (4 1) 9 9 3 2PC

CA = radius of given circle 2 2(0) ( 1) ( 3) = 2

From CAP,

2 2sin

33 2

CA

CP

1 2sin

3

Angle between tangents

1 22 2sin

3

Option (B) is wrong.

Also CP is a diameter of circumcircle of PAB

Equation of circumcircle is

(x – 0) (x – 3) + (y – 1)(y – 4) = 0

x2 – 3x + y2 – 4y – y + 4 = 0

x2 + y2 – 3x – 5y + 4 = 0

Option (C) is correct.

Also, area of quadrilateral PACB

12 2 14 2 14

2

Option (D) is correct.


Hint : Sun of two perfect square = 0

Solution : Given: 2a2 + 4b2 + c2 – 4ab – 2ac = 0

2 2 2 22 2 (2 ) ( 2 ) 0a a b b a c ac

(a – 2b)2 + (a – c)2 = 0

a = 2b = c ,2

ab c a

22 2

2 2 2

4cos

2 2

aa a

c a bB

ca a a

2

2

12

74

82

a

a

Option (A) is correct.

Again, a = c 2R sinA = 2R sinC, by sine rule

A = C A – C = 0

cos(A – C) = cos = 1

Option (B) is correct

Again 1

1r s a as

r s s

s a

1

2

a

a b c

21

2

a

aa a


14/14

4 5 4 11

5 5 5

a

a


Again, By sine rule

sin sin sinA B C

a b c

sin sin sin

2

A B C

aa a

sin sin sin

11 1

2

A B C

1sin : sin : sin 1: : 1 2 :1: 2

2A B C

Solution for Q. No. 57 and 58:

Given, 2

0

( ) ( )

x

tf x x e f x t dt

2 ( )

0

( )

x

x tf x x e f x x t dt

2

0

( ) ( )

x

x tf x x e e f t dt

....(1)

Differentiating both sides of (1) w.r.t. x, we get

0

( ) 2 ( ) 1 ( )

x

x x x tf x x e e f x e e f t dt

....(2)

By [(1) + (2)

f(x) + f (x) = x2 + 2x + f(x)

Integrating both sides we get

3 2

( ) 23 2

x xf x c

3

2( )3

xf x x c ....(1)

When x = 0, f(0) = 0 + 0 + c c = f(0)

But 0

2

0

(0) (0) (0 ) 0tf e f t dt

c = f(0) = 0

Hence from (1)

32( )

3

xf x x


Solution : 2

3( )

( 3)

g x

x x

g(x) is discontinuous at x = 0 and –3.

58. Answer (C)

Solution : 2

1 1 1( )

3( 3) 3

g x dx dx

x xx

1 1 3ln

3

xC

x x

59. Answer (A, D)

Solution : ( , , ) , ,2 2 2

a b c

a = 2, b = 2, c = 2 8 = ± 32 × 6 = ± 24

60. Answer (B)

Solution : Let P = (, , ).Now PA PB ( – a) + (– b) + r = 0

a + b = 2 + 2 + 2

PBPC b + c = 2 + 2 + 2

1 1 1

a b c

2 2 2 2 2 2

,2 2

a b and

2 2 2

2

c

abc = ± 6 × 32

(2 + 2 + 2)3 = ± 192 × 8 × (x2 + y2 + z2) = ± 1536 xyz

��

Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019

1/14

1. (6)

2. (7)

3. (5)

4. (3)

5. (3)

6. (2)

7. (1)

8. (2)

9. (C, D)

10. (B, C)

11. (B, D)

12. (A, C)

13. (A, B, C)

14. (A, C)

15. (A, C)

16. (A, D)

17. (A, B, D)

18. (A, B, D)

19. (A, C)

20. (A, D)

21. (7)

22. (6)

23. (2)

24. (4)

25. (4)

26. (6)

27. (6)

28. (3)

29. (A, B, C)

30. (A, C, D)

31. (A, B, D)

32. (B, D)

33. (A, C)

34. (A)

35. (A, B, D)

36. (A, C)

37. (D)

38. (D)

39. (A, D)

40. (A, B)

41. (6)

42. (2)

43. (5)

44. (1)

45. (8)

46. (3)

47. (4)

48. (7)

49. (A, B, C, D)

50. (A, C, D)

51. (C, D)

52. (A, C, D)

53. (A, D)

54. (A, B, C, D)

55. (A, B)

56. (A, B, C)

57. (A, B, D)

58. (C)

59. (A, D)

60. (B)

ANSWERS

Test Date: 17/02/2019

PHYSICS CHEMISTRY MATHEMATICS

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

MOCK TEST - 1 (Paper-2) - Code-D

All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 1 (Paper - 2) (Code-D) (Hints & Solutions)

2/14

PART - I (PHYSICS)

ANSWERS & HINTS

1. Answer (6)

Hint :

Fmax

= 2 mAg

Solution :

Fmax

= 2 mAg

2. Answer (7)

Hint : Length of string will remain constant

Solution :

V0 cos 37 = V

0

4

5 V V

VV0

5

4

3. Answer (5)

Hint : hc

Emin

max

Solution :

1 hC

ev

If V increase 1 decreases

4. Answer (3)

Hint : pr v

RC C

n–

–1

Solution :

T2 P3 = C

2

3 5 2

PVP C P V C

nR

2/5pV C

3 19–

22 6–1

5

pr

R R RC

Q = n Cpr T R

Q19

2 1006

RQ

1900

3

32 100

2

Ru

u = 300 R

1900 1000– – 300

3 3

R RW Q u R

5. Answer (3)

Hint : dP

B VdV

–

Solution :

3 24, 4

3V a dV a da

–

–/

dP dPB V

dV V dV

MgdP

A

3

2

4–

3 4

a MgB

A a da

da Mg

a BA–3

6. Answer (2)

L

H

Hint : 2dpx

dx

Solution :

22

LAL ghA

Lh

g

2 2

2

7. Answer (1)

Hint : E A2 2

0

1

4 2

Solution :

x1

7cm

8 x

2

17cm

8


3/14

2 212.5

4 2U E A

2 2

2

12.5

4 2

TU E A

v

2 2

2

5 1 1 4 100100 4

2 4 2 100 100U E

25U E 2

15

U

8. Answer (2)

Hint : O B A BT T

1

2

Solution :

mgx = fR mg

f xR

g ga x

R R

R T R

T tg g

2 ,4 2

9. Answer (C, D)

Hint : After reflection from mirror there will be phase

change of Solution :

D 2D

y

O

Screen

d

D

y

P

Q

OP dOP d

D D2 2

d OROR d

D D3

3

y = 2d = 1 mm

also = yd

nD

1–

2 2

or y d n 5.5, n 15.5

10. Answer (B, C)

Hint : Apply superposition principle

Solution :

00

1

4

R

C

x dxV

x

0

4

R

02

P

RV

11. Answer (B, D)

Hint : ax

VI

Z

max

m

Solution :

15

I2

60

30 v

I1

I

I t1

100sin 50

2 3

I t2

100sin 50 –

2 6

I t50 2 sin 5012

Pam

= Vrm

Irm

cos

100 5 2cos15

2 2


4/14

12. Answer (A, C)

Hint : Apply I ��

Solution :

(2 2mR + mg2

)N

fO

0 = I

0

RmR mg mR mR2 2 2

2 2 2 22

2

4

g R

R


Hint : Use R – C circuit analysis.

Solution :

K Q KQV

d a

1–

0 0

1

QaQ

d

t

RcQ Q e–

1 t = Rc ln3 = 4

0 .aR ln3

1

3

QQ

2 2 2

22 200

8 .2 4 .

VQ a Q a

dd d

14. Answer (A, C)

Hint : Apply energy conservation and momentum

conservation

Solution :

T

2ma0

Ring

T cos 37° = ma

For 2 m

T cos 37° + 2 ma0 = (2 m) g sin 37°

We get the result

15. Answer (A, C)

Hint : Ea

02

Solution :

( , )a a

( )p

3

1

x

y2

31 2

0 0 0

1 1

2 . 2 . 2 . 2 2 2pE j i i j

a a a

��

3 3

2 1

0 0

1 1

2 . 2 2 . 2

��

pE i ja a

16. Answer (A, D)

Hint :

Apply Ampere’s law

Solution :

Amplere’s law

0 enclosed. .B dl I

��

�

0–

ˆ –2

��

B dxi

17. Answer (A, B, D )


Hint : f Nmax

1 1 1


4 kg

2 kg

1 kg f1

f2

f3

14 N

ca

14 – 7

7

= 1 m/s2

max

37 Nf

max

12 Nf

max

23 Nf

max

1a = 2 m/s2

max

2a = 1 m/s2


5/14

19. Answer (A, C)

20. Answer (A, D)

Hints :

Rt

LI I e–

max1–

in R - L circuit


1

2

1 V

1

2

I

4H

3

Circuit can be reduced as

t

LI I e

R–

max1–

PART - II (CHEMISTRY)

21. Answer (7)

Hint: Consider the changes which occurs inthermodynamic properties during a spontaneousprocess like vapourisation.

Solution : Except (f) all statements are correct

statement (f) is false because when the temperature is

smaller than equilibrium temperature, the H term

dominates and G becomes +ve for the vapourisation

process.

22. Answer (6)

Hint: Diborane has two 3c –2e

H

H

H H

H HB B

Solution : Each 3c – 2e– bond involve 3 orbital, one

from each B and H.

23. Answer (2)

Hint :

2

2Cd H S CdS 2H

��⇀↽��

Solution :

0.5 × 10–3 moles of Cd2+, 0.5 × 10–3 moles of S–2 ions

are required to precipitate.

0.5 ×10–3 moles of S2– 0.5 × 10–3 moles H2S

1×10–3 moles of H+

Number of moles of H+ from HCl = 50 × 0.16

= 8 ×10–3 moles

Total moles of H+ = 9 × 10–3 moles

–3–29 10

H 1000 10 molar900

24. Answer (4)

Hint :

Number of moles CaC2O

4 = Number of moles of CaCI

2

= Number of moles of organic acid

Solution :

(Oxalic acid) N1V

1 = N

2V

2 (KMnO

4)

2 × no. of moles = 80

5 0.051000

= 0.01 moles

Number of moles Oxalic acid = number of moles

of CaC2O

4 = number of moles of CaCI

2

Mass of CaCI2 = 0.01 × 111 = 1.11 g

% of CaCI2 in mixture =

1.11100 60%

1.8

4

3R

4 2 4

4 2 3

R

t

I e

–4

31 4

1–4 3

Imax

1A

4

I = 0.25 (1 – e–t) A

also L1I1 = L

2I2

I t I I t I1 2

2 1,

3 3

tI t e–

1

11–

6 , tI t e

–

2

11–

12

Total

1 4 1 1J

2 3 16 24

U


6/14

25. Answer (4)

Hint: Check the chirality and elements of symmetry in

product.

Solution :

P1

Optically active

P2

Optically active

P3

Optically active

P4

Optically active

P5

Optically inactive

P6

Optically inactive

P7

Optically inactive

P8

Optically inactive

26. Answer (6)

Hint: First convert molecules tetrahedral geometry into

Fisher form.

Solution :

(A) CH3

H

OH

2

4

3

1

(R)

(1)C H2 5

(B)

HH

Cl

24

3

1

HO C H2 5

(R)

(1)

(C)

HH

Cl2

4

3

Br (S)(2)

CH3

1

(D)

HH

C H2 5

Br

4

3

2

(S)(2)

CH3

1

27. Answer (6)

Hint: To obtain aromatic character R can lose a

hydride H–, and become sp2 hybridised.

Solution :

C C

H

+

+

H

+ C(Ph)3

+ H – C(Ph)3

(R)

–

BF4

–

BF4H

a a

Number of electrons in ring R is 6.

28. Answer (3)

Hint: Kp remain constant for a reaction, even in

simultaneous equilibria.

Solution :

a 2b a

PQ(s) P(g) Q(g)

��⇀↽�� K

p = 4.8 × 10–6

2

2b a

RQ (s) R(s) 2Q(g)

��⇀↽�� K

p = 5.76 × 10–6

(2b + a)2 = 5.76 × 10–6

(2b + a) = 2.4 × 10–3

(2.4 × 10–3) × a = 4.8 × 10–6

a = 2 × 10–3 = 2 × 10–x

x = 3


Hint: Apply concepts of gram equivalents.

Solution :

meq of KMnO4 = 25 × 0.2 = 5

meq of FeSO4 = 40 × 0.2 = 8

meq of H3ASO

3 = 30 × 0.1 × 2 = 6

meq of H2O

2 = 15 × 0.1 × 2 = 3

meq of SnCl2 = 25 × 0.1 × 2 = 5


7/14


Hint: Check all the possibilities of structures which

can be obtained after the reaction.

Solution :

H , Ni2 Cl , h2 3 products

H , Ni2 Cl , h

2

3 products


Hint: F2 is produced from anhyd. HF.

Solution : Water should not be present during

electrolysis otherwise O2 is liberated.

32. Answer (B, D)

Hint: Check the electron donating and withdrawing

extent of the group.

Solution :

Activating group activates the ring towards E.A.S. while

deactivating group deactivates it.

33. Answer (A, C)

Hint: General properties of group-16 hydrides.

Solution : Correct order of M.P. is

H2S < H

2Se < H

2Te

as increase in mass of the molecule will cause

increase in intermolecular binding energy.

And bond angle of hydrides does not differ much on

moving down the group, from S to Te.

34. Answer (A)

Hint: Consider the factors affecting hydration energy.

Solution :

As size of cation increases the size of hydrated ions

in water decreases and thus ionic mobility increases.

Solubilities of bromides and hydroxides increases down

the group because their lattice energy change is more

than hydration energy on moving down the group.


Hint: AlCl3 in C

6H

6 present as dimer.

Solution : Borax loses water of crystallisation on

heating.

36. Answer (A, C)

Hint: According to Nernst equation cell

RTE E lnQ

nf

cell

0.059E E logQ

n

y = c + mx

so, 1

mn

x = 0.059 logQ

Solution : if n = 1

tan = 1

= 45°

E°cell

= x (magnitude wise)

for hydrogen electrode

cell

2

0.059 [H ]E E log

1 H

0.770.059 log ( 0.6 0.77 and E 0)

0.6 ∵

= 0

37. Answer (D)

Hint: ∵ B can be taken as an intermediate of the

reaction.

d[B]

0dt

Solution : 1K

A B

2K

B F

1

d[A]k dt

dt

1 2

d[B]k [A] K [B]

dt

∵ 1K t

t 0[A] [A] e

(∵ it has been assumed that all the reactions are

first order)

Now,

1K t

1 0 2

d[B]K [A] e K [B] 0

dt

1K t

02

1

[A] eK

K [B]


8/14

PART - III (MATHEMATICS)

41. Answer (6)

Hint : Centroid of a triangle divides the line joining

orthocentre and circumcentre in the ratio 2 : 1 internally.

Solution : Since centroid divides the line joining

orthocentre (2, 4) and circumcentre 7 5,

2 2

in the ratio

2 : 1

7 52 1 2 2 1 4

2 2, (3,3)2 1 2 1

A

2 2(3 0) (3 0) 3 2OA

A(3, 3)

y

xO

BP

L M N

45°

45°

45°

3 2

3

2AB

3

2 AP PB

3 33 ,3

2 2

B

+ = 6

42. Answer (2)

Hint : f (x) < 0 at point P

Solution : At any point of maxima,

f(x) = 0 and f(x) < 0

f(x) = x2 – f2(x) –9 < 0

Let P(, )

2 – 2 – 9 < 0 and 2 – 2 = a2

a2 – 9 < 0 a (–3, 3)

43. Answer (5)

Hint : 1 2Area ( ) ( )

b

a

f x f x dx

Solution : y = x2 + x – 2, y = 2x

x2 + x – 2 = 2x x2 – x – 2 = 0

x2 – 2x + x – 2 = 0 x (x – 2) +1(x – 2) = 0

x = –1, 2

intersection points of y = x2 + x – 2 and y = 2x, are

(–1, –2) and (2, 4)

(–1, –2)

(–2, 0) xx

(2, 4)

y x = 2

y x + x – = 22

0

–1

(1, 0)

Here y = x2 + x – 2 is a parabolic curve with vertex

38. Answer (D)

Hint: We have to consider the given situation

i.e. K1 > K

2. So more of A is consumed as compared to

B.

Solution : ∵ K1 > K

2, the amount B becomes fairly

large before droping eventually to zero.

39. Answer (A, D)

Hint: As the solution is a good source of e–,

The solution will behave as a good reducing agent.

Solution : NaNH2 will be produced after a very long

time.

40. Answer (A, B)

Hint: Reaction of Na to H2O is highly exothermic due

to high polarity of H2O. So blue colour solution is not

obtained, as free electrons are absent.

Solution : Concentrated solutions are copper bronze

in colour.

Alkali metal solution in ammonia on evaporation yield

metals.


9/14

1 9,

2 4

which cuts x-axis at (–2, 0) and (1, 0) and

y-axis at (0, –2)

= Required area

2 2

2 2

1 1

(2x x x 2)dx ( x x 2)dx

23 2

1

x x2x

3 2

8 1 12 4 2

3 3 2

1 95

2 2

91 1 5

2

44. Answer (1)

Hint : Solve Limit

Solution : From given equation

1

2

lim tan sin cos63 1

k

kx

kx kx k x

1

2

3lim tan sin

213 1

k

k x

xk x x

k k

= tan 33

2

3

3 1

x

x x

3x2 – 9x + 3 = x

3x2 – 10x + 3 = 0 = 1

45. Answer (8)

Hint : SP + SP = 2A

Solution : Locus of Z is ellipse

2,0 0, 2 ,2 4 s s A

1

2 22

ss Ae e

distance between directrices 2 4

81/ 2

A

e

46. Answer (3)

Hint : Draw graph of y = x3 + 2x2 + x and y = –6

Solution : Given equation is x3 + 2x2 + x + 6 = 0

x3 + 2x2 + x = – 6

Let f(x) = x3 + 2x2 + x and g(x) = – 6

Here, f (x) = 3x2 + 4x + 1

and f (x) = 6x + 4

When f (x) = 0 3x2 + 3x + x + 1 = 0

3x (x + 1) + (x + 1) = 0

(x + 1) + (3x + 1) = 0

1

1,3

x

f(–1) = – 6 + 4 = – 2 < 0

and 1 1

6 4 2 03 3

f

f(x) has local maxima at x = – 1 and local minima

at x = 1

3

Also, max. f(x) = f(–1) = –1 + 2 – 1 = 0

and min. 1 1 2 1

( )3 27 9 3

f x f

1 6 9 4

27 27

( 2) 8 8 2 2 (1) 1 2 1 4

( 3) 27 18 3 12

f f

f

0

–3 –2 –1

x

y

y = g x( ) = – 6

(–3, –12)

(–2, –2)

y

x

1

3

From graph it is clear that one real ‘’ will lie between

(–3) and (–2) –3 < < –2

[ ] 3 3


10/14

47. Answer (4)

Hint : A.M G.M.

Solution : Let , , , be the roots of given equation.

+ + + = 4 and = 1

Now A.M. of roots = 4

14 4

and G.M. of roots =

1 1

4 4( ) (1) 1

A.M. = G.M.

Which is possible only if

= = = = 1

Given equation will be of the form (x – 1)4 = 0

x4 – 4x3 + 4C2x2 – 4C

3x + 1 = 0

x4 – 4x3 + 6x2 – 4x + 1 = 0

Comparing, we get a = 6, b = – 4

a + b + 2 = 6 – 4 + 2 = 4

48. Answer (7)

Hint : Make cases

Solution : Here 3 vowels A, I, A can be placed in the

following five available places:

× N × G × N × N ×

OR

× N × G × G × G ×

OR

× N × G × G × N ×

Required number of ways =

3

3! 4! 4! 4!5

2! 3! 3! 2! 2!

C

= 5 4 3! 3 2! 4 3! 4 3! 4 3 2!

3! 2 1 2! 3! 3! 2! 2

= 30(4 + 4 + 6) = 420

4207

60 60


Hint : Sun of two perfect square = 0

Solution : Given: 2a2 + 4b2 + c2 – 4ab – 2ac = 0

2 2 2 22 2 (2 ) ( 2 ) 0a a b b a c ac

(a – 2b)2 + (a – c)2 = 0

a = 2b = c ,2

ab c a

22 2

2 2 2

4cos

2 2

aa a

c a bB

ca a a

2

2

12

74

82

a

a

Option (A) is correct.

Again, a = c 2R sinA = 2R sinC, by sine rule

A = C A – C = 0

cos(A – C) = cos = 1

Option (B) is correct

Again 1

1r s a as

r s s

s a

1

2

a

a b c

21

2

a

aa a

4 5 4 11

5 5 5

a

a


Again, By sine rule

sin sin sinA B C

a b c

sin sin sin

2

A B C

aa a

sin sin sin

11 1

2

A B C

1sin : sin : sin 1: : 1 2 :1: 2

2A B C


11/14


Hint : Draw diagram.

Solution :

A

B

P(3, 4)C

(0, 1)

Power of point P(3, 4) w.r.t. circle

S x2 + y2 – 2y – 3 = 0 is defined as

PA PB = S1

= 9 + 16 – 8 – 30 = 14 option (A) is correct

2 2(3 0) (4 1) 9 9 3 2PC

CA = radius of given circle 2 2(0) ( 1) ( 3) = 2

From CAP,

2 2sin

33 2

CA

CP

1 2sin

3

Angle between tangents

1 22 2sin

3

Option (B) is wrong.

Also CP is a diameter of circumcircle of PAB

Equation of circumcircle is

(x – 0) (x – 3) + (y – 1)(y – 4) = 0

x2 – 3x + y2 – 4y – y + 4 = 0

x2 + y2 – 3x – 5y + 4 = 0


Also, area of quadrilateral PACB

12 2 14 2 14

2

Option (D) is correct.

51. Answer (C, D)

Hint : 2 cosAcosB = cos(A + B) + cos(A – B)

Solution : Here 0

1( ) 2cos cos( )

4f x t x t dx

0

1cos ( ) cos ( )

4t t x t t x t dt

0

0

cos 1[ ] cos(2 )

4 4

xt t x dt

0

cos 1 sin(2 )

4 4 2

x t x

cos 1sin 2 sin 0

4 8

xx x

cos 1sin sin

4 8

xx x

cos( )

4

xf x

Which is clearly continuous and

differentiable is (0, 2)

( ) sin4

f x x

( ) cos4

f x x

When ( ) 0 sin 04

f x x

sin x = 0

x = 0, , 2,

At x = 0, 2, ( ) 1 04

f x

x = 0, 2 are the points of local maxima

and at x = , ( ) ( 1) 04

f x

x = is the point of local minima

At x = , ( ) sin 04

f x

atleast one c = (0, 2) such that f(c) = 0

and maximum value of ( ) cos4 4

f x

Max. ( ) at ( 0, 2 )4

f x x


12/14


Hint : Take dot product of both sides with

, , andu v a w

��

.

Solution : Given u v w a ��

....(1)

Taking dot product of both sides of (1) with

, , andu v a w

��

, we get

2 3

2u u v u w a u

��

31

2u v u w � � � �

1

2u v u w � � � �

....(2)

2 7

4u v v v w a v

��

71

4u v v w � � � �

7 31

4 4u v v w � � � �

....(3)

2u a a v a w a a a

� � � � � ��

3 7

42 4

a w � �

13 34

4 4a w � �

....(4)

and 2

��

u w v w w a w

31

4

� � � �

u w v w

3 11

4 4

� � � �

u w v w ....(5)

By [(2) + (3) + (5)], we get

u v u w u v v w u v v w � � � � � � � � � � � �

1 3 1 2 3 11

2 4 4 4

1

2u v v w u w � � � � � �

....(6)

By [(6) – (2)] 0v w � �

By [(6) – (3)], 1 3 2 3 1

2 4 4 4w u

� �

From (6), 1 1

04 2

u v � �

1 1 2 1 3

2 4 4 4u v

� �

53. Answer (A, D)

Hint : Find dy

dxfrom quadratic equation.

Solution : Given differential equation is

2

( ) 0dy dy

y x y xdx dx

2( ) ( ) 4 ( )

2

x y x y y xdy

dx y

2( ) ( )

2

y x x y

y

( )

2

y x x y

y

or2 2

y x x y y x x y

y y

1ordy x

dx y

when 1dy

dx

0dx dy

x – y + c = 0 ....(1)

and when dy x

dx y

ydy + xdx = 0

2 2 2

2 2 2

x y k

x2 + y2 = k2 ....(2)


13/14

But curves passes through the point (1, 2)

1 – 2 + c = 0 c = 1

1st curve is the straight line

x – y + 1 = 0

and from (2), 1 + 4 = k2

2nd curve is x2 + y2 = 5


Hint : Use properties of determinants

Solution : Given equation is

2 2

2 2

2 2

1 sin cos 2sin4

sin 1 cos 2sin4 0

sin 1 2sin4

A A

A A

A cos A

By R1 R

1 – R

2 and R

2 R

2 – R

3 we have

2 2

1 1 0

0 1 1 0

sin cos 1 2sin4A A

By C2 C

2 + C

3, we have

2 2

1 1 0

0 0 1 0

sin cos 1 2sin4 1 2sin4A A

– 0 + 0 – (–1) [cos2A + 1 + 2sin4 + sin2A ] = 0

2(1 + sin4) = 0

sin4 1 sin2

4 ( 1)2

nn

( 1) ( 1)4 8 4 8

n nn n

Which is independent of A.

When n = 0, = 0 ( 1)8 8

31,

4 8 8n

When n = –1, 2

4 8 8 8

Options (A), (B), (C), (D) are correct.

55. Answer (A, B)

Hint : Find AB

Solution :

2 2 23 3 3

6 6 6

ax bx cx

AB a b c

ax bx cx

Now, tr(AB) = tr(C)

3ax2 + b + 6 cx = (x + 2)2 + 2x + 5x2 x R

3ax2 + 6 cx + b = 6x2 + 6x + 4

Comparing, we get a = 2, b = 4, c = 1


Hint : Solve integration, then limit

Solution : We have 2 2

1a

ndx

n x

2

2

1

1( )a

dxn

xn

11 1tan

11

a

x

n

nn

11tan ( )

a

n nx

n

1 1 1tan tan tan ( )2

na an

1

2 2lim lim tan

21n n

a

ndxL an

n x

, if 02 2

0, if 02

, if 02 2

a

a

a

, if 0

, if 02

0, if 0

a

a

a


14/14

��

Solution for Q. Nos. 57 and 58:

Given, 2

0

( ) ( )

x

tf x x e f x t dt

2 ( )

0

( )

x

x tf x x e f x x t dt

2

0

( ) ( )

x

x tf x x e e f t dt

....(1)

Differentiating both sides of (1) w.r.t. x, we get

0

( ) 2 ( ) 1 ( )

x

x x x tf x x e e f x e e f t dt

....(2)

By [(1) + (2)

f(x) + f (x) = x2 + 2x + f(x)

Integrating both sides we get

3 2

( ) 23 2

x xf x c

3

2( )3

xf x x c ....(1)

When x = 0, f(0) = 0 + 0 + c c = f(0)

But 0

2

0

(0) (0) (0 ) 0tf e f t dt

c = f(0) = 0

Hence from (1)

32( )

3

xf x x


Solution : 2

3( )

( 3)

g x

x x

g(x) is discontinuous at x = 0 and –3.

58. Answer (C)

Solution : 2

1 1 1( )

3( 3) 3

g x dx dx

x xx

1 1 3ln

3

xC

x x

59. Answer (A, D)

Solution : ( , , ) , ,2 2 2

a b c

a = 2, b = 2, c = 2 8 = ± 32 × 6 = ± 24

60. Answer (B)

Solution : Let P = (, , ).Now PA PB ( – a) + (– b) + r = 0

a + b = 2 + 2 + 2

PBPC b + c = 2 + 2 + 2

1 1 1

a b c

2 2 2 2 2 2

,2 2

a b and

2 2 2

2

c

abc = ± 6 × 32

(2 + 2 + 2)3 = ± 192 × 8 × (x2 + y2 + z2) = ± 1536 xyz

Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions) All ... · 6/14 (B) H Cl 2 4 3 1 HO C H 25...

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Transcript of Mock Test - 1 (Paper - 2) (Code-C) (Hints & Solutions) All ... · 6/14 (B) H Cl 2 4 3 1 HO C H 25...