MOCK TEST - 3 (Paper-1) - Code-A...26. Answer (5) Hint : A and B = H 3 PO 4. Solution : C = H 4 P 2 O 7 PP OH OH HO OOH OO A = B = H 3 PO 4 P HO OH OH O 27. Answer (5) Hint : All carboxylic

Mock Test - 3 (Paper - 1) (Code-A) (Answers) All India Aakash Test Series for JEE (Advanced)-2019

1/13

1. (5)

2. (9)

3. (7)

4. (4)

5. (8)

6. (4)

7. (6)

8. (5)

9. (B, C)

10. (A, B, D)

11. (A, B)

12. (A, C)

13. (A, B)

14. (B, C)

15. (A, C)

16. (A, B, D)

17. (A, C)

18. (B, D)

19. A (T)

B (P, Q, R, S)

C (P, R, S, T)

D (Q)

20. A (Q)

B (P)

C (S, T)

D (R)

21. (2)

22. (7)

23. (6)

24. (2)

25. (7)

26. (5)

27. (5)

28. (7)

29. (A, B, C)

30. (B, C)

31. (A, D)

32. (B, C)

33. (A, B, C, D)

34. (B)

35. (A, D)

36. (A, B, C, D)

37. (A, C, D)

38. (B, C, D)

39. A (Q)

B (P, Q, R, S)

C (S, T)

D (S, T)

40. A (R, S, T)

B (P, Q, R, S)

C (Q, R, S)

D (P, Q, R, S)

41. (4)

42. (4)

43. (1)

44. (9)

45. (0)

46. (5)

47. (4)

48. (3)

49. (A, D)

50. (A, D)

51. (A, B, C)

52. (A, D)

53. (A, D)

54. (A, C, D)

55. (B, C, D)

56. (A, D)

57. (C, D)

58. (B, C, D)

59. A (Q, R, S)

B (P)

C (P, T)

D (S, T)

60. A (T)

B (T)

C (P)

D (Q, R, S)

ANSWERS

PHYSICS CHEMISTRY MATHEMATICS

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

Test Date: 28/04/2019

MOCK TEST - 3 (Paper-1) - Code-A

All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 1) (Code-A) (Hints & Solutions)

2/13

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (5)

Hint: Particle B will be on mid point of line joining A

and C

Solution : (vt – y) = y + t2

vy

t2

2y = (vt – t2)

0 2dy

v tdt

t = 5s

2. Answer (9)

Hint: B will be at rest when rod is horizontal.

Solution : B changes direction of motion at x = –15 m

3. Answer (7)

Hint: Acceleration of roller is always along incline.

Solution : Let acceleration be a

ma

2

cos2 3

l mlma

cmcos

2

la a

2

cm

3cos

4a a a

mg sin = m (acm

)

8

7

ga

4. Answer (4)

Hint: Apply gas equation

Solution : 0 0( )

2

lP P x l x

38l = (76 – x) (l – x) x = 19

44

19

x

5. Answer (8)

Hint: COM of half disc is at 4

3

R

distance from center.

Solution : d cos 45° = 4

3

R

d = 4 2

3

R

6. Answer (4)

Hint: Assign charges based on electrostatics of

conductors

Solution :

Q0

3

2

d

03

4

Q0

4

Q

2

d

2

0 0 0

0 0

3

2 4 2 4

Q Q QF

A

2

0

04

QF

A

n = 4

7. Answer (6)

Hint: Write down the expression for VI using V

o.

Solution : v = 40 cm

Mock Test - 3 (Paper - 1) (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019

3/13

m = 3

2

2

ˆ ˆ

I

V dx V dyV i j

dt U dtU

��

[∵ y = 0 ]

2

2

tan

V dy

U dx

V

U

1tan tan

3 (with (–ve) sign)

+ = 90°

tan tan 1

21tan 1

3

60

8. Answer (5)

Hint: Surface will take paraboloidal shape. Volume

under paraboloid will be equal to displaced liquid.

Solution : 2 2

2

xy

g

x

y

2

0

22

Rh

R xdxy

2 22

0

2

2 2

Rh x xdx

Rg

2 2 4

2 4

R h R

g

2

2 2 10 1 10

8

gh

R

= 5 rad/s

9. Answer (B, C)

Hint: Excess pressure will create stress on outer

boundary.

Solution : 0 0 0

5

4P V P V

0

4

5P P

0

0

4 2

5 2

P tP

r

03 2

10

P t

r

r t m2

04

And, 3 0

54

3 4

Vr

Vt

r m

0 015

4

PV

m

0 0 09

16

10. Answer (A, B, D)

Hint: Observe the motion from COM frame.

Solution : 2m

Tk

T = 2 sec

1 1

2 2

A V

A V

11. Answer (A, B)

For image to coincides with object in this case,

u = 10 cm or u = 15 cm

u = 10 cm

u = 15 cm


4/13

12. Answer (A, C)

Hint: Draw FBD of m in frame of M.

Solution : (T + Mg) sin = Ma

ma

mg

T

ma sin + T = mg

13. Answer (A, B)

Hint: Time constant for charging circuit is same as

discharging circuit.

Solution : Time constant, eff

0.5sR

L

01

t

i i e

41 A

4i

i6

1 3 1A

6 3 3

14. Answer (B, C)

Hint: dP P

ndV V

Solution : PVn = constant

dP Pn

dV V

3 1

4 2n

3

2n

15. Answer (A, C)

Hint: ( 1)

62

n n

(E1 – d

0) = 4(E

2 – d

0)

Solution : ( 1)

62

n n

n = 4

1

1 1513.6 1 13.6

16 16E

2

1 813.6 1 13.6

9 9E


Hint: Based on motion of charged particle in magnetic

field.

Solution : For V B� �

path is circular for ||V B� �

path

is straight line.

17. Answer (A, C)

Hint: I I I2

min 1 2

Solution : For, I = 3I0

,3 3

2

3

3

Dx

d

I = A2

I = 2AA

2 2

min 0I A A A

2

min

04

II

I

18. Answer (B, D)

Hint: Particle nearest to planet will fly when its

acceleration is 2(x – a)

Solution : 2

2 2( )

( )

GM Gmx a

x a a

mM

x

a

2

2

GMx

x

2 3 2( )

( )

GM GM Gmx a

x a x a

3 3

2 3 2

( )

( )

x x a GmGM

x a x a

2

5 2

3GM a x Gm

x a

1

33Mx a

m


5/13

19. Answer A (T); B (P, Q, R, S); C (P, R, S, T); D (Q)

Hint: 0 A

v v implies = 0

00

Av v

Solution : Angular momentum and linear momentum

of system remains conserved.

20. Answer A (Q); B (P); C (S, T); D (R)

Hint: Draw the circuit diagram.

Solution : Draw the circuit diagram

(A) All are in parallel

(B)1 2

4 3

(C)

(D)

PART - II (CHEMISTRY)

21. Answer (2)

Hint : Apply the law of equivalence.

Solution : 2 2 2 2

MnO 4HCl MnCl 2H O Cl

2 2 2 3 2 44Cl Na S O 5H O 8HCl 2NaHSO

8

MnO2

Na2S

2O

3

(mole × n-factor) = NV

n × 2 = (8 × 1) × 0.5

4.0n 2

2

22. Answer (7)

Hint : h = h0 + K.E. of the emitted electron.

Solution : Given, K.E. of the electron

= 1

4 of the irradiating photon

hc 1

h h4

hc 3h

4

8

7

16

4c 4 3 10 m/s 110 m

3 83 3.2 10

23. Answer (6)

Hint : Due to –NO2, it will show linkage isomerism.

Solution :

Pt Enantiomer+

ClCl

Cl

Py

H N3

O N2

Pt Enantiomer+

ClCl

Cl

Py

H N3

ONO

Pt Pt

Cl ClCl Cl

Py Py

Cl Cl

H N3

H N3

O N2

ONO

Pt Pt

Cl ClCl Cl

Py Py

Cl Cl

O N2

ONO

H N3

H N3

24. Answer (2)

Hint : Kp of the reaction remain same even in

simultaneous equilibria.

Solution : Kp(i)

= 2 × 2

Kp(ii)

= 4 × 4


6/13

in mixture of solids P and Q

(i)

1 1 2P P P

P(s) P (g) A(g)

��⇀↽��

(ii)

2 2 1P P P

Q(s) Q (g) A(g)

��⇀↽��

P1(P

1 + P

2) = 2 × 2

P2(P

2 + P

1) = 4 × 4

(P1 + P

2) (P

1 + P

2) = 20

2

1 2(P P )

210

25. Answer (7)

Hint : SiO2 is the impurity of Bauxite.

Solution : NaCN is used for leaching of Ag and Au.

SO2 is produced during auto reduction.

26. Answer (5)

Hint : A and B = H3PO

4.

Solution : C = H4P

2O

7

P P

OH OHHO O OH

O O

A = B = H3PO

4

P

OHHOOH

O

27. Answer (5)

Hint : All carboxylic acids and sulphonic acids will

liberate CO2 with NaHCO

3.

Solution : PhCOOH, PhSO3H, (NO

2)

3PhOH,

(CH3)2 NPhCOOH, CH

3COOH

28. Answer (7)

Hint :

C

OH

O bond is not cleaved by HIO

4.

Solution : 5 moles of HIO4 will be used and 2 moles of

CO2 will be produced along with other products.

29. Answer (A, B, C)

Hint : Since the rate determining step is the slow step.

Step-2 would be the RDS.

Rate [N2O

2] [H

2]

∵2 2

eq 2

[N O ]K (Fromstep-1)

[NO]

22 2 eqN O K [NO]

2

2Rate NO H

Solution : Order of reaction w.r.t. NO is two and

w.r.t. H2 is one, overall order of reaction is three.

30. Answer (B, C)

Hint : Total equivalents of acid = Total equivalents of

base.

Solution : Molarity w.r.t H2SO

4 = M

1

Molarity w.r.t H2C2O4 = M

2

When reacts with KOH solution

Total equivalents of acid = Total equivalents of base

15 × M1 × 2 + 15 × M

2 × 2 = 6 × 0.09 × 1

M1 + M

2 = 0.018

When treated with KMnO4 only H

2C2O4 will oxidise

into CO2.

Moles of H2C2O4 =

5 4 0.03

2 1000

2

100 5 4 0.03M ×

1000 2 1000

M2 = 0.003 and M

1 = 0.015

2 2 4H C O

W 0.27 g

2 4H SO

W 1.47 g

Wimpurity

= 4.26 g

31. Answer (A, D)

Hint : Group-II sulphides are insoluble in dil. HCl.

Solution : Nickel ions give red ppt with DMG.


7/13

32. Answer ( B, C)

Hint : Minimum boiling azeotrope has lower boiling point

than its both of the components and maximum boiling

azeotrope has greater boiling point than its both of the

components.

Solution : Since, Azeotropic composition of AZ-1 is

95% of ethanol, so its boiling point will be very close to

the boiling of ethanol. Actual boiling point of AZ-1 is

351.10 K

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

Hint : Conc. H2SO

4 acts as a dehydrating agent.

Solution : There are three possible isomers of given

compound

I : [Cr(H2O)

6]Cl

3

II : [Cr(H2O)

5Cl]Cl

2H2O

III : [Cr(H2O)

4Cl2]Cl2H

2O

As there is no effect of conc. H2SO

4 on A, this

implies all H2O molecules in A are present in

coordination sphere. Therefore isomer I is A

TB (B) > T

B (C)

No. of ions in B is more than that in C

So,

Isomer II is B

Isomer III is C

34. Answer (B)

Hint : 1 mole of QCl3 produce 2 mole and of RCl

2

produce 1 mole of cationic vacancy.

Solution :

2 × 10–6 × NA + 10–7 × N

A = cationic vacancies

QCl3

RCl2

= 2.1 × 10–6 NA

35. Answer (A, D)

Hint :

NaOH

CH – Cl3

Conjugate base (Planar)

O

Racemic mixture

OO

–

–

Solution :

(1) PCC oxidise 2°-alcohol into ketone

(2) Ring expansion take place

(3) Racemic mixture is produced

(4) Wolff Kishner reduction


Hint : Due to H-bonding one of the form may get

stabilised.

Solution :

H

O

HO

O

OH

CH3

37. Answer (A, C, D)

Hint : Sodium stearate is a soap solution with CMC

around 10–4 M.

Solution : Only soap is example of associate colloid.

38. Answer (B, C, D)

Hint : Those sugars which are hemiacetal can

reduce Tollens reagent.

Solution : Sugar (B), (C) and (D) have hemiacetal group.

39. Answer A(Q); B(P, Q, R, S); C(S, T); D(S, T)

Hint : Sn is more stable in Sn (IV).

Solution :

Readily

decompose

2 2NaNO HCl HNO NO

(A) H2 4 aq

HNO FeSO

1 1

2 5 4

(Brown ring)

Fe(H O) (NO) SO

(B) 2 2 2 2 4(White ppt)

2HgCl SnCl Hg Cl SnCl

(C) 2 2 4 2(Blackppt)

Hg Cl NH OH Hg Hg(NH )Cl


8/13

(D) 2

4 4

H S2

NH OH NH Cl (Black ppt)Ni NiS

40. Answer A(R, S, T); B(P, Q, R, S); C(Q, R, S); D(P, Q,

R, S)

Hint : Ph – CHO –ve Iodoform, –ve Fehling solution,

+ve Tollen’s test.

PART - III (MATHEMATICS)

41. Answer (4)

Hint : Convert it in definite integration

Solution:

n

n

n n n n n n

n n n n n

1

2 .(2 – 1).(2 – 2)...( 1). .( – 1)..3.2.1lim

( – 1)( – 2)....1. ( – 1)..3.2.1

�

n

n

n

n n n

n

n n n

1

1 2 – 12 2 – . 2 – .... 2 –

lim1 2 – 1

1 1– . 1– .... 1–

n

nr

r r

n n n

–1

0

1log lim log 2 – – log 1–

�

x x dx

1

0

log log(2 – ) – log(1– ) �

x 1 – x

x x dx

1

0

log log(1 ) – log �

x x x x x x1

0( 1)log( 1) – ( 1) – log

= 2 ln2 – 2 + 1 + 1 = 2 ln2

4 �

42. Answer (4)

Hint : ( ) '( ) ( )x xe f x f x dx e f x C

Solution :

x

xx x edx e dx

I

sin 22 sin

0(I) (II)0

cos

x x xxe e dx e dx

2 2sin sin sin2

00 0

–

e eI[ ] 4

2 2

43. Answer (1)

Hint : P1 + P

2 = 0

Solution : Equation of required plane is P1 + P

2 = 0

(x – y + z – 3) + (x – 2y – 3z) = 0

(1 + ) x + (–1 – 2 )y + (1 – 3) z – 3 = 0

is parallel to x + y + 9z + 3 = 0

1 –1– 2 1– 3

1 1 9

2–3

Equation of plane is x y z1 1

3 – 3 03 3

x + y + 9z – 9 = 0

a = 1, b = 9, c = – 9a + b + c = 1

44. Answer (9)

Hint : If abcdef is divisible by 7 then 1 × f + 3 × e + 2

× d + 6 × c + 4 × b + 5 × a is divisible by 7

Solution :

∵ (c + 3b + 2a + 6c + 4b + 5a) mod (7) = 0

7(a + b + c) mod (7) = 0 a {1, 2, ...9}

b, c {0, 1, 2,...9}

Number of required numbers = 9 × 10 × 10 = 900

Sum of digits = 9

Solution :

Ph CH PhNaCN

Benzoin

Condensation

CH

O

O +

Ph CH

OH

PhC

O

-hydroxy ketone also reduce Fehling solution and

Tollen’s reagent.


9/13

45. Answer (0)

Hint : Normal to the parabola at (1, –2) is also normal

to the circle at (1, –2)

Solution : y

x

y2

= c4

(1, –2) P

C r r ( , –3)r

y2 = 4x

2yy = 4

2

4 2N

y ym

mN = 1

Equation of normal to the parabola is

y = x – 3

Centre of circle is (r, r – 3)

∵ CP = r (r – 1)2 + (r – 1)2 = r2

2( 1)r r

2 2r ∵ r < 1 2 2r a – b = 0

46. Answer (5)

Hint : sin2x + sin2y = 1 cos2x = y2

cos –2

Solution : sin2x + sin2y = 1

sin2y = 1 – sin2x = cos2x y x2 2

cos – cos2

x–2 2

∵

y–2 2

∵

– x + y

– x – y

y n x–2

taking sign taking –

sign

y n x–2

y n x– –2

x y r r(2 1)2 2

x y n– –2

x y – ,2 2

x y– – ,2 2

y

xO

x

y = –

–

2

x

y =

+

2

x

y =

–

+

2 x

y =

–

2

A

21

42 2 2 2

A2

2 2

10 105

2

47. Answer (4)

Hint : Put x t4

Solution : dx

I

x

2

201– tan

4 2

2 2

0

4

1 tan 1 tan

dx dxI

x x

Let tanx = t

1

2 2 2 2

0 1(1 )(1 ) (1 )(1 )

dt dtI

t t t t

1 1

2 2 2 2

0 0 1 1

1

2 1 1 1 1

dt dt dt dtI

t t t t

t tI t

t t

1 1

0 0 1

1 1 1 1 1tan ln ln

2 2 1– 2 1

10

2 2 4I


10/13

48. Answer (3)

Hint : Find range

Solution :–1

0 cos3

x

∵

–1ln cos ln

3

x

–1lncos ln

3

x

x–1 –1– sin lncos2 3 2

Number of integers in range will be 3.

49. Answer (A, D)

Hint : z2 + 6 z – 6 + i (– 24 – 8z) = (z + 3 – 4i)2 + 1

Solution : z2 + 6z – 6 + i (–24 – 8z) = (z + z1)2 + z

2

z2 + 6z – 6 + i (–24 – 8z) = z2 + z1

2 + 2zz1 + z

2

6z – 8 i z = 2zz1z

1 = 3 – 4i

z2 + 6z – 6 + i (–24 – 8z) = (z + 3 – 4i)2 + z2

z2 + 9 – 16 – 24i + 6z – 8iz + z

2

z2 + 6z – 6 + i (– 24 – 8z) + (z

2 –1)

z2 = 1

put z + 3 – 4i = z

|z + i| = 1, ...(1)

|z2 + 1| = 1 ...(2)

|z – 3 + 4i| = 5 ...(3)

from (2), |(z + i) (z – i)| = 1 |z – i| = 1, from (1)

y

x

(3,– 4)

| – 3 + 4 | = 5z i| + | = 1z i

| – | = 1z i

O

(0,1)

(0,–1)

z = 0

z + 3 – 4i = 0

z = – 3 + 4i

50. Answer (A, D)

Hint : ab = 1 b = 0 or a = 1

Solution : (2 – cos2 x – sin x)1– sin2 x – cos x = 1

(sin2 x – sinx + 1)cos2x – cosx = 1

sin2 x – sinx + 1 = 1 or cos2x – cosx = 0

sin x = 0, sinx = 1 or cosx = 0, cosx = 1

x = n, or x = 2n n I,2

3 5, , 2 ,

2 2 2x


Hint : Break all the functions

Solution : If 1 < x < 2

x x

g x f t dt f t dt f t dt1

0 0 1( ) ( ) ( ) ( ) 1

0 1(2 – ) (1 2 – )

x

t dt t dt

xx

2

– 3 –12

If 2 < x < 3

x x

g x f t dt f t dt f t dt f t dt1 2

0 0 1 2( ) ( ) ( ) ( ) ( )

21 2

0 1 2(2 – ) (1 2 – ) (2 – 2) 1

2

x xt dt t dt t dt

If 3 < x < 4

x x

g x f t dt f t dt f t dt f t dt f t dt1 2 3

0 0 1 2 3( ) ( ) ( ) ( ) ( ) ( )

=x

x

2

– 22

(4, 10)

(2, 3)

31,

2

113,

2

1 2 3 4x

0

y


11/13

52. Answer (A, D)

Hint : f x x xx x

1( ) (sin cosec )

(sin cosec )

Solution :

x x x xf x

x x x x

4 2 2 2 2

2 2

sin 3sin 1 (1 sin ) sin( )

sin (sin 1) sin (1 sin )

x

x x

x

2

2

1 sin 1

sin 1 sin

sin

x x

x x

1(sin cosec )

(sin cosec )

Put sinx + cosecx = t

f x tt

1( ) ∵ A.M. G.M. sinx + cosecx = t 2

consider a function

2

1 1( ) , ( ) 1–f x x f x

x x

f(x) is increasing when

x 1∵ t = sinx + cosecx t is decreasing function.

∵ t 2 tt

1 is always increasing function

4 2

3

1 sin 3sin 1( ) is decreasing in 0,

2sin sin

x xf x t

t x x

f x min

1 3 1 5( )

1 1 2

53. Answer (A, D)

Hint : Length of tangent = dx

ydy

2

1

Solution : dx

y x ydy

2

2 21

dx x

dy y

2 2

2

dx dy

x y

ln x = lny + lnc

xy = c or x

cy


Hint : –1

–1( ( ))

( ( ))

dxd f x

f f x

Solution : Let f–1(x) = t x f t( ) ( )dx

f tdt

1

( )

dt

dx f t

–1

–1

1( )

( ( ))

df x

dx f f x

–1

–1( ( ))

( ( ))

dxd f x

f f x

33

3 3

–1

–1

– –

( )( ( ))

dxI f x

f f x

f f–1 3 –1 3( ) – (– ) 3 – 2

f(x) = 3 (x – 2)2 + cosx, when x f x0 , '( ) 02

,

when , ( ) 02

x f x

x f x x f x3 3

when , ( ) 0, when 2 , ( ) 02 2

,

when 2 , ( ) 0, when 0, ( ) 0f x x f x

( ) 0f x x R

f (x) = 6(x – 2) – sinx f (x) < 0, x < 2,

f (x) > 0 x > 2

x = 2 is only point of inflexion


Hint : Shortest distance between two skew lines

a b1 1

and 2 2

a b is a a b b

b b

1 2 1 2

1 2

–

Solution : ˆ ˆ ˆ ˆ: 2 2AB i j i k

CD i j k j kˆ ˆ ˆ ˆ ˆ:

i k i k j kl

i k j k

ˆ ˆ ˆ ˆ ˆ ˆ23

ˆ ˆ ˆ ˆ 6– 2


12/13

56. Answer (A, D)

Hint : i iP X , variance = i iP X2 2–

Solution : X can take value 0, 1 , 2

CP X

C

43

63

4 1( 0)

20 5

C CP X

C

4 22 1

63

. 12 3( 1)

20 5

C CP X

C

4 21 2

63

. 4 1( 2)

20 5

Mean, i iP X1 3 1

0 1 2 15 5 5

Variance, i iX P2 2 2

–

1 3 10 1 4 –1

5 5 5

2

5

57. Answer (C, D)

Hint : Point P divides C1C

2 in the ratio r

1 : r

2 internally

Solution :

2

BB

A

A

M

S1 = 0

C1 (0, 0) M'

1

C2(5, 0)

S2 = 0

P10

, 03

210 4

5 – –13 3

PA

3tan

4

C M2sin

1

C M23

5

1

6

5C M

9 165 –

5 5MM

'cos

2 1

AM A M

4 82

5 5AM

4

5A M

Area of quadrilateral

AB B Ais 1 8 16 16

2 5 5 5

192sq.units

25


Hint : n

n n

r

S s r2

1

Solution : ni j n

s i j

1

.

rt r r r n[( 1) ( 2) ... ]

n rr r n r

–. [2 2 – –1]

2

r n r n rr r n n r3 2( – ).( 1) 1

[– – ( 1) ]2 2

n

n n n n n n ns n n

21 ( 1) ( 1) ( 1)(2 1)

( 1). – –2 2 2 6

n nn n n n n2 2( 1)

6 6 – 3 – 3 – 4 – 224

n nn n2( 1)

(3 – – 2)24


13/13

n n n n( – 1) ( 1)(3 2)

24

n

n n n

r

n n nS s r s

2

1

( 1)(2 1)

6

n n n n

ni j i j

n ni j i j

2

1 1 1 1

( 1).

2

n

n nn

s n n

n n

( – 1)(3 2) 4 3 4 1lim lim

( 1) 24 24 2

n n

n n nS s

( 1)(2 1)–

6

n n

n n n ns

( – 1) ( 1)(3 2)–

24

59. Answer A(Q, R, S,); B(P); C(P,T); D(S,T)

Hint : Equation of normal to y2 = 4x is y = mx – 2m – m3

Solution :

(A) y = mx – 2m – m3 passes through C (3, 0)

0 = 3m – 2m – m3 m3 – m = 0

m = 0, 1

(B) Normal to the circles will pass through the centres

of the circles CCm m1 2

m3 –1

– 21– 2

(C) y mx m2

4 9 passes through 5,0

2

m m25

4 92

= 0 25 m2 = 16m2 + 36

m = ± 2

(D) xy = 4

xy+ y = 0

–

yy

x

N

xm

y

N

tm t

2/t

22

Equation of normal at tt

22 ,

y t x tt

2 3 2– 2

mN = t2 > 0

60. Answer A(T); B(T); C(P); D(Q,R,S)

Hint : System of linear equations in x2, y2, z2

Solution : = b(a + 1), 1 = b(b + 1),

2 = b (1 – ab),

3 = 0

• For infinitely many solutions, = 1 =

2 =

3 = 0

b = 0, a = {–2, –1, 0, 1, 2} or a = –1, b = –1

= 6

• For finitely many solutions, 0, b0, a –1

16

• For unique solution, 1 =

2 =

3 = 0, 0 there

is no such values of a and b = 0

• For no solution, = 0 and at least one of

1 ,

2

3 is non-zero

a = –1 and b = – 2, 1, 2 = 3

��

Mock Test - 3 (Paper - 1) (Code-A) (Answers) All India Aakash Test Series for JEE (Advanced)-2019

1/13

1. (5)

2. (6)

3. (4)

4. (8)

5. (4)

6. (7)

7. (9)

8. (5)

9. (B, D)

10. (A, C)

11. (A, B, D)

12. (A, C)

13. (B, C)

14. (A, B)

15. (A, C)

16. (A, B)

17. (A, B, D)

18. (B, C)

19. A (Q)

B (P)

C (S, T)

D (R)

20. A (T)

B (P, Q, R, S)

C (P, R, S, T)

D (Q)

21. (7)

22. (5)

23. (5)

24. (7)

25. (2)

26. (6)

27. (7)

28. (2)

29. (B, C, D)

30. (A, C, D)

31. (A, B, C, D)

32. (A, D)

33. (B)

34. (A, B, C, D)

35. (B, C)

36. (A, D)

37. (B, C)

38. (A, B, C)

39. A (R, S, T)

B (P, Q, R, S)

C (Q, R, S)

D (P, Q, R, S)

40. A (Q)

B (P, Q, R, S)

C (S, T)

D (S, T)

41. (3)

42. (4)

43. (5)

44. (0)

45. (9)

46. (1)

47. (4)

48. (4)

49. (B, C, D)

50. (C, D)

51. (A, D)

52. (B, C, D)

53. (A, C, D)

54. (A, D)

55. (A, D)

56. (A, B, C)

57. (A, D)

58. (A, D)

59. A (T)

B (T)

C (P)

D (Q, R, S)

60. A (Q, R, S)

B (P)

C (P, T)

D (S, T)

ANSWERS

PHYSICS CHEMISTRY MATHEMATICS

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

Test Date: 28/04/2019

MOCK TEST - 3 (Paper-1) - Code-B

All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 1) (Code-B) (Hints & Solutions)

2/13

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (5)

Hint: Surface will take paraboloidal shape. Volume

under paraboloid will be equal to displaced liquid.

Solution : 2 2

2

xy

g

x

y

2

0

22

Rh

R xdxy

2 22

0

2

2 2

Rh x xdx

Rg

2 2 4

2 4

R h R

g

2

2 2 10 1 10

8

gh

R

= 5 rad/s

2. Answer (6)

Hint: Write down the expression for VI using V

O.

Solution : v = 40 cm

m = 3

2

2

ˆ ˆ

I

V dx V dyV i j

dt U dtU

��

[∵ y = 0 ]

2

2

tan

V dy

U dx

V

U

1tan tan

3 (with (–ve) sign)

+ = 90°

tan tan 1

21tan 1

3

60

3. Answer (4)

Hint: Assign charges based on electrostatics of

conductors

Solution :

Q0

3

2

d

03

4

Q0

4

Q

2

d

2

0 0 0

0 0

3

2 4 2 4

Q Q QF

A

2

0

04

QF

A

n = 4

4. Answer (8)

Hint: COM of half disc is at 4

3

R

distance from center.

Solution : d cos 45° = 4

3

R

d = 4 2

3

R

5. Answer (4)

Hint: Apply gas equation

Solution : 0 0( )

2

lP P x l x

38l = (76 – x) (l – x) x = 19

44

19

x

Mock Test - 3 (Paper - 1) (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019

3/13

6. Answer (7)

Hint: Acceleration of roller is always along incline.

Solution : Let acceleration be a

ma

2

cos2 3

l mlma

cmcos

2

la a

2

cm

3cos

4a a a

mg sin = m (acm

)

8

7

ga

7. Answer (9)

Hint: B will be at rest when rod is horizontal.

Solution : B changes direction of motion at x = –15 m

8. Answer (5)

Hint: Particle B will be on mid point of line joining A

and C

Solution : (vt – y) = y + t2

vy

t2

2y = (vt – t2)

0 2dy

v tdt

t = 5s

9. Answer (B, D)

Hint: Particle nearest to planet will fly when its

acceleration is 2(x – a)

Solution : 2

2 2( )

( )

GM Gmx a

x a a

mM

x

a

2

2

GMx

x

2 3 2( )

( )

GM GM Gmx a

x a x a

3 3

2 3 2

( )

( )

x x a GmGM

x a x a

2

5 2

3GM a x Gm

x a

1

33Mx a

m

10. Answer (A, C)

Hint: I I I2

min 1 2

Solution : For, I = 3I0

,3 3

2

3

3

Dx

d

I = A2

I = 2AA

2 2

min 0I A A A

2

min

04

II

I


Hint: Based on motion of charged particle in magnetic

field.

Solution : For V B� �

path is circular for ||V B� �

path

is straight line.


4/13

12. Answer (A, C)

Hint: ( 1)

62

n n

(E1 – d

0) = 4(E

2 – d

0)

Solution : ( 1)

62

n n

n = 4

1

1 1513.6 1 13.6

16 16E

2

1 813.6 1 13.6

9 9E

13. Answer (B, C)

Hint: dP P

ndV V

Solution : PVn = constant

dP Pn

dV V

3 1

4 2n

3

2n

14. Answer (A, B)

Hint: Time constant for charging circuit is same as

discharging circuit.

Solution : Time constant, eff

0.5sR

L

01

t

i i e

41 A

4i

i6

1 3 1A

6 3 3

15. Answer (A, C)

Hint: Draw FBD of m in frame of M.

Solution : (T + Mg) sin = Ma

ma

mg

T

ma sin + T = mg

16. Answer (A, B)

For image to coincides with object in this case,

u = 10 cm or u = 15 cm

u = 10 cm

u = 15 cm


Hint: Observe the motion from COM frame.

Solution : 2m

Tk

T = 2 sec

1 1

2 2

A V

A V

18. Answer (B, C)

Hint: Excess pressure will create stress on outer

boundary.

Solution : 0 0 0

5

4P V P V

0

4

5P P

0

0

4 2

5 2

P tP

r

03 2

10

P t

r

r t m2

04

And, 3 0

54

3 4

Vr

Vt

r m

0 015

4

PV

m

0 0 09

16


5/13

PART - II (CHEMISTRY)

21. Answer (7)

Hint :

C

OH

O bond is not cleaved by HIO

4.

Solution : 5 moles of HIO4 will be used and 2 moles of

CO2 will be produced along with other products.

22. Answer (5)

Hint : All carboxylic acids and sulphonic acids will

liberate CO2 with NaHCO

3.

Solution : PhCOOH, PhSO3H, (NO

2)

3PhOH,

(CH3)2 NPhCOOH, CH

3COOH

23. Answer (5)

Hint : A and B = H3PO

4.

Solution : C = H4P

2O

7

P P

OH OHHO O OH

O O

A = B = H3PO

4

P

OHHOOH

O

24. Answer (7)

Hint : SiO2 is the impurity of Bauxite.

Solution : NaCN is used for leaching of Ag and Au.

SO2 is produced during auto reduction.

25. Answer (2)

Hint : Kp of the reaction remain same even in

simultaneous equilibria.

Solution : Kp(i)

= 2 × 2

Kp(ii)

= 4 × 4

in mixture of solids P and Q

(i)

1 1 2P P P

P(s) P (g) A(g)

��⇀↽��

(ii)

2 2 1P P P

Q(s) Q (g) A(g)

��⇀↽��

P1(P

1 + P

2) = 2 × 2

P2(P

2 + P

1) = 4 × 4

(P1 + P

2) (P

1 + P

2) = 20

2

1 2(P P )

210

26. Answer (6)

Hint : Due to –NO2, it will show linkage isomerism.

19. Answer A (Q); B (P); C (S, T); D (R)

Hint: Draw the circuit diagram.

Solution : Draw the circuit diagram

(A) All are in parallel

(B)1 2

4 3

(C)

(D)

20. Answer A (T); B (P, Q, R, S); C (P, R, S, T); D (Q)

Hint: 0 A

v v implies = 0

00

Av v

Solution : Angular momentum and linear momentum

of system remains conserved.


6/13

Solution :

Pt Enantiomer+

ClCl

Cl

Py

H N3

O N2

Pt Enantiomer+

ClCl

Cl

Py

H N3

ONO

Pt Pt

Cl ClCl Cl

Py Py

Cl Cl

H N3

H N3

O N2

ONO

Pt Pt

Cl ClCl Cl

Py Py

Cl Cl

O N2

ONO

H N3

H N3

27. Answer (7)

Hint : h = h0 + K.E. of the emitted electron.

Solution : Given, K.E. of the electron

= 1

4 of the irradiating photon

hc 1

h h4

hc 3h

4

8

7

16

4c 4 3 10 m/s 110 m

3 83 3.2 10

28. Answer (2)

Hint : Apply the law of equivalence.

Solution : 2 2 2 2

MnO 4HCl MnCl 2H O Cl

2 2 2 3 2 44Cl Na S O 5H O 8HCl 2NaHSO

8

MnO2

Na2S

2O

3

(mole × n-factor) = NV

n × 2 = (8 × 1) × 0.5

4.0n 2

2


Hint : Those sugars which are hemiacetal can

reduce Tollens reagent.

Solution : Sugar (B), (C) and (D) have hemiacetal group.


Hint : Sodium stearate is a soap solution with CMC

around 10–4 M.

Solution : Only soap is example of associate colloid.


Hint : Due to H-bonding one of the form may get

stabilised.

Solution :

H

O

HO

O

OH

CH3

32. Answer (A, D)

Hint :

NaOH

CH – Cl3

Conjugate base (Planar)

O

Racemic mixture

OO

–

–

Solution :

(1) PCC oxidise 2°-alcohol into ketone

(2) Ring expansion take place

(3) Racemic mixture is produced

(4) Wolff Kishner reduction

33. Answer (B)

Hint : 1 mole of QCl3 produce 2 mole and of RCl

2

produce 1 mole of cationic vacancy.

Solution :

2 × 10–6 × NA + 10–7 × N

A = cationic vacancies

QCl3

RCl2

= 2.1 × 10–6 NA


7/13


Hint : Conc. H2SO

4 acts as a dehydrating agent.

Solution : There are three possible isomers of given

compound

I : [Cr(H2O)

6]Cl

3

II : [Cr(H2O)

5Cl]Cl

2H2O

III : [Cr(H2O)

4Cl2]Cl2H

2O

As there is no effect of conc. H2SO

4 on A, this

implies all H2O molecules in A are present in

coordination sphere. Therefore isomer I is A

TB (B) > T

B (C)

No. of ions in B is more than that in C

So,

Isomer II is B

Isomer III is C

35. Answer ( B, C)

Hint : Minimum boiling azeotrope has lower boiling point

than its both of the components and maximum boiling

azeotrope has greater boiling point than its both of the

components.

Solution : Since, Azeotropic composition of AZ-1 is

95% of ethanol, so its boiling point will be very close to

the boiling of ethanol. Actual boiling point of AZ-1 is

351.10 K

36. Answer (A, D)

Hint : Group-II sulphides are insoluble in dil. HCl.

Solution : Nickel ions give red ppt with DMG.

37. Answer (B, C)

Hint : Total equivalents of acid = Total equivalents of

base.

Solution : Molarity w.r.t H2SO

4 = M

1

Molarity w.r.t H2C2O4 = M

2

When reacts with KOH solution

Total equivalents of acid = Total equivalents of base

15 × M1 × 2 + 15 × M

2 × 2 = 6 × 0.09 × 1

M1 + M

2 = 0.018

When treated with KMnO4 only H

2C2O4 will oxidise

into CO2.

Moles of H2C2O4 =

5 4 0.03

2 1000

2

100 5 4 0.03M ×

1000 2 1000

M2 = 0.003 and M

1 = 0.015

2 2 4H C O

W 0.27 g

2 4H SO

W 1.47 g

Wimpurity

= 4.26 g


Hint : Since the rate determining step is the slow step.

Step-2 would be the RDS.

Rate [N2O

2] [H

2]

∵2 2

eq 2

[N O ]K (Fromstep-1)

[NO]

22 2 eqN O K [NO]

2

2Rate NO H

Solution : Order of reaction w.r.t. NO is two and

w.r.t. H2 is one, overall order of reaction is three.

39. Answer A(R, S, T); B(P, Q, R, S); C(Q, R, S); D(P, Q,

R, S)

Hint : Ph – CHO –ve Iodoform, –ve Fehling solution,

+ve Tollen’s test.

Solution :

Ph CH PhNaCN

Benzoin

Condensation

CH

O

O +

Ph CH

OH

PhC

O

-hydroxy ketone also reduce Fehling solution and

Tollen’s reagent.

40. Answer A(Q); B(P, Q, R, S); C(S, T); D(S, T)

Hint : Sn is more stable in Sn (IV).

Solution :

Readily

decompose

2 2NaNO HCl HNO NO


8/13

PART - III (MATHEMATICS)

41. Answer (3)

Hint : Find range

Solution :–1

0 cos3

x

∵

–1ln cos ln

3

x

–1lncos ln

3

x

x–1 –1– sin lncos2 3 2

Number of integers in range will be 3.

42. Answer (4)

Hint : Put x t4

Solution : dx

I

x

2

201– tan

4 2

2 2

0

4

1 tan 1 tan

dx dxI

x x

Let tanx = t

1

2 2 2 2

0 1(1 )(1 ) (1 )(1 )

dt dtI

t t t t

1 1

2 2 2 2

0 0 1 1

1

2 1 1 1 1

dt dt dt dtI

t t t t

t tI t

t t

1 1

0 0 1

1 1 1 1 1tan ln ln

2 2 1– 2 1

10

2 2 4I

43. Answer (5)

Hint : sin2x + sin2y = 1 cos2x = y2

cos –2

Solution : sin2x + sin2y = 1

sin2y = 1 – sin2x = cos2x y x2 2

cos – cos2

x–2 2

∵

y–2 2

∵

– x + y

– x – y

y n x–2

taking sign taking –

sign

y n x–2

y n x– –2

x y r r(2 1)2 2

x y n– –2

x y – ,2 2

x y– – ,2 2

y

xO

x

y = –

–

2

x

y =

+

2

x

y =

–

+

2 x

y =

–

2

(A) H2 4 aq

HNO FeSO

1 1

2 5 4

(Brown ring)

Fe(H O) (NO) SO

(B) 2 2 2 2 4(White ppt)

2HgCl SnCl Hg Cl SnCl

(C) 2 2 4 2(Blackppt)

Hg Cl NH OH Hg Hg(NH )Cl

(D) 2

4 4

H S2

NH OH NH Cl (Black ppt)Ni NiS


9/13

A

21

42 2 2 2

A2

2 2

10 105

2

44. Answer (0)

Hint : Normal to the parabola at (1, –2) is also normal

to the circle at (1, –2)

Solution : y

x

y2

= c4

(1, –2) P

C r r ( , –3)r

y2 = 4x

2yy = 4

2

4 2N

y ym

mN = 1

Equation of normal to the parabola is

y = x – 3

Centre of circle is (r, r – 3)

∵ CP = r (r – 1)2 + (r – 1)2 = r2

2( 1)r r

2 2r ∵ r < 1 2 2r

a – b = 0

45. Answer (9)

Hint : If abcdef is divisible by 7 then 1 × f + 3 × e + 2

× d + 6 × c + 4 × b + 5 × a is divisible by 7

Solution :

∵ (c + 3b + 2a + 6c + 4b + 5a) mod (7) = 0

7(a + b + c) mod (7) = 0 a {1, 2, ...9}

b, c {0, 1, 2,...9}

Number of required numbers = 9 × 10 × 10 = 900

Sum of digits = 9

46. Answer (1)

Hint : P1 + P

2 = 0

Solution : Equation of required plane is P1 + P

2 = 0

(x – y + z – 3) + (x – 2y – 3z) = 0

(1 + ) x + (–1 – 2 )y + (1 – 3) z – 3 = 0

is parallel to x + y + 9z + 3 = 0

1 –1– 2 1– 3

1 1 9

2–3

Equation of plane is x y z1 1

3 – 3 03 3

x + y + 9z – 9 = 0

a = 1, b = 9, c = – 9a + b + c = 1

47. Answer (4)

Hint : ( ) '( ) ( )x xe f x f x dx e f x C

Solution :

x

xx x edx e dx

I

sin 22 sin

0(I) (II)0

cos

x x xxe e dx e dx

2 2sin sin sin2

00 0

–

e eI[ ] 4

2 2

48. Answer (4)

Hint : Convert it in definite integration

Solution:

n

n

n n n n n n

n n n n n

1

2 .(2 – 1).(2 – 2)...( 1). .( – 1)..3.2.1lim

( – 1)( – 2)....1. ( – 1)..3.2.1

�

n

n

n

n n n

n

n n n

1

1 2 – 12 2 – . 2 – .... 2 –

lim1 2 – 1

1 1– . 1– .... 1–


10/13

n

nr

r r

n n n

–1

0

1log lim log 2 – – log 1–

�

x x dx

1

0

log log(2 – ) – log(1– ) �

x 1 – x

x x dx

1

0

log log(1 ) – log �

x x x x x x1

0( 1)log( 1) – ( 1) – log

= 2 ln2 – 2 + 1 + 1 = 2 ln2

4 �


Hint : n

n n

r

S s r2

1

Solution : ni j n

s i j

1

.

rt r r r n[( 1) ( 2) ... ]

n rr r n r

–. [2 2 – –1]

2

r n r n rr r n n r3 2( – ).( 1) 1

[– – ( 1) ]2 2

n

n n n n n n ns n n

21 ( 1) ( 1) ( 1)(2 1)

( 1). – –2 2 2 6

n nn n n n n2 2( 1)

6 6 – 3 – 3 – 4 – 224

n nn n2( 1)

(3 – – 2)24

n n n n( – 1) ( 1)(3 2)

24

n

n n n

r

n n nS s r s

2

1

( 1)(2 1)

6

n n n n

ni j i j

n ni j i j

2

1 1 1 1

( 1).

2

n

n nn

s n n

n n

( – 1)(3 2) 4 3 4 1lim lim

( 1) 24 24 2

n n

n n nS s

( 1)(2 1)–

6

n n

n n n ns

( – 1) ( 1)(3 2)–

24

50. Answer (C, D)

Hint : Point P divides C1C

2 in the ratio r

1 : r

2 internally

Solution :

2

BB

A

A

M

S1 = 0

C1 (0, 0) M'

1

C2(5, 0)

S2 = 0

P10

, 03

210 4

5 – –13 3

PA

3tan

4

C M2sin

1

C M23

5

1

6

5C M

9 165 –

5 5MM

'cos

2 1

AM A M

4 82

5 5AM

4

5A M


11/13

Area of quadrilateral

AB B Ais 1 8 16 16

2 5 5 5

192sq.units

25

51. Answer (A, D)

Hint : i iP X , variance = i iP X2 2–

Solution : X can take value 0, 1 , 2

CP X

C

43

63

4 1( 0)

20 5

C CP X

C

4 22 1

63

. 12 3( 1)

20 5

C CP X

C

4 21 2

63

. 4 1( 2)

20 5

Mean, i iP X1 3 1

0 1 2 15 5 5

Variance, i iX P2 2 2

–

1 3 10 1 4 –1

5 5 5

2

5


Hint : Shortest distance between two skew lines

a b1 1

and 2 2

a b is a a b b

b b

1 2 1 2

1 2

–

Solution : ˆ ˆ ˆ ˆ: 2 2AB i j i k

CD i j k j kˆ ˆ ˆ ˆ ˆ:

i k i k j kl

i k j k

ˆ ˆ ˆ ˆ ˆ ˆ23

ˆ ˆ ˆ ˆ 6– 2


Hint : –1

–1( ( ))

( ( ))

dxd f x

f f x

Solution : Let f–1(x) = t x f t( ) ( )dx

f tdt

1

( )

dt

dx f t

–1

–1

1( )

( ( ))

df x

dx f f x

–1

–1( ( ))

( ( ))

dxd f x

f f x

33

3 3

–1

–1

– –

( )( ( ))

dxI f x

f f x

f f–1 3 –1 3( ) – (– ) 3 – 2

f(x) = 3 (x – 2)2 + cosx, when x f x0 , '( ) 02

,

when , ( ) 02

x f x

x f x x f x3 3

when , ( ) 0, when 2 , ( ) 02 2

,

when 2 , ( ) 0, when 0, ( ) 0f x x f x

( ) 0f x x R

f (x) = 6(x – 2) – sinx f (x) < 0, x < 2,

f (x) > 0 x > 2 x = 2 is only point of inflexion

54. Answer (A, D)

Hint : Length of tangent = dx

ydy

2

1

Solution : dx

y x ydy

2

2 21

dx x

dy y

2 2

2

dx dy

x y

ln x = lny + lnc

xy = c or x

cy


12/13

55. Answer (A, D)

Hint : f x x xx x

1( ) (sin cosec )

(sin cosec )

Solution :

x x x xf x

x x x x

4 2 2 2 2

2 2

sin 3sin 1 (1 sin ) sin( )

sin (sin 1) sin (1 sin )

x

x x

x

2

2

1 sin 1

sin 1 sin

sin

x x

x x

1(sin cosec )

(sin cosec )

Put sinx + cosecx = t

f x tt

1( ) ∵ A.M. G.M. sinx + cosecx = t 2

consider a function

2

1 1( ) , ( ) 1–f x x f x

x x

f(x) is increasing when

x 1∵ t = sinx + cosecx t is decreasing function.

∵ t 2 tt

1 is always increasing function

4 2

3

1 sin 3sin 1( ) is decreasing in 0,

2sin sin

x xf x t

t x x

f x min

1 3 1 5( )

1 1 2


Hint : Break all the functions

Solution : If 1 < x < 2

x x

g x f t dt f t dt f t dt1

0 0 1( ) ( ) ( ) ( ) 1

0 1(2 – ) (1 2 – )

x

t dt t dt

xx

2

– 3 –12

If 2 < x < 3

x x

g x f t dt f t dt f t dt f t dt1 2

0 0 1 2( ) ( ) ( ) ( ) ( )

21 2

0 1 2(2 – ) (1 2 – ) (2 – 2) 1

2

x xt dt t dt t dt

If 3 < x < 4

x x

g x f t dt f t dt f t dt f t dt f t dt1 2 3

0 0 1 2 3( ) ( ) ( ) ( ) ( ) ( )

=x

x

2

– 22

(4, 10)

(2, 3)

31,

2

113,

2

1 2 3 4x

0

y

57. Answer (A, D)

Hint : ab = 1 b = 0 or a = 1

Solution : (2 – cos2 x – sin x)1– sin2 x – cos x = 1

(sin2 x – sinx + 1)cos2x – cosx = 1

sin2 x – sinx + 1 = 1 or cos2x – cosx = 0

sin x = 0, sinx = 1 or cosx = 0, cosx = 1

x = n, or x = 2n n I,2

3 5, , 2 ,

2 2 2x

58. Answer (A, D)

Hint : z2 + 6 z – 6 + i (– 24 – 8z) = (z + 3 – 4i)2 + 1

Solution : z2 + 6z – 6 + i (–24 – 8z) = (z + z1)2 + z

2

z2 + 6z – 6 + i (–24 – 8z) = z2 + z12 + 2zz

1 + z

2

6z – 8 i z = 2zz1z

1 = 3 – 4i

z2 + 6z – 6 + i (–24 – 8z) = (z + 3 – 4i)2 + z2

z2 + 9 – 16 – 24i + 6z – 8iz + z

2

z2 + 6z – 6 + i (– 24 – 8z) + (z

2 –1)

z2 = 1

put z + 3 – 4i = z

|z + i| = 1, ...(1)


13/13

��

|z2 + 1| = 1 ...(2)

|z – 3 + 4i| = 5 ...(3)

from (2), |(z + i) (z – i)| = 1 |z – i| = 1, from (1)

y

x

(3,– 4)

| – 3 + 4 | = 5z i| + | = 1z i

| – | = 1z i

O

(0,1)

(0,–1)

z = 0

z + 3 – 4i = 0

z = – 3 + 4i

59. Answer A(T); B(T); C(P); D(Q,R,S)

Hint : System of linear equations in x2, y2, z2

Solution : = b(a + 1), 1 = b(b + 1),

2 = b (1 – ab),

3 = 0

• For infinitely many solutions, = 1 =

2 =

3 = 0

b = 0, a = {–2, –1, 0, 1, 2} or a = –1, b = –1

= 6

• For finitely many solutions, 0, b0, a –1

16

• For unique solution, 1 =

2 =

3 = 0, 0 there

is no such values of a and b = 0

• For no solution, = 0 and at least one of

1 ,

2

3 is non-zero

a = –1 and b = – 2, 1, 2 = 3

60. Answer A(Q, R, S,); B(P); C(P,T); D(S,T)

Hint : Equation of normal to y2 = 4x is y = mx – 2m – m3

Solution :

(A) y = mx – 2m – m3 passes through C (3, 0)

0 = 3m – 2m – m3 m3 – m = 0

m = 0, 1

(B) Normal to the circles will pass through the centres

of the circles CCm m1 2

m3 –1

– 21– 2

(C) y mx m2

4 9 passes through 5,0

2

m m25

4 92

= 0 25 m2 = 16m2 + 36

m = ± 2

(D) xy = 4

xy+ y = 0

–

yy

x

N

xm

y

N

tm t

2/t

22

Equation of normal at tt

22 ,

y t x tt

2 3 2– 2

mN = t2 > 0

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MOCK TEST - 3 (Paper-1) - Code-A...26. Answer (5) Hint : A and B = H 3 PO 4. Solution : C = H 4 P 2 O 7 PP OH OH HO OOH OO A = B = H 3 PO 4 P HO OH OH O 27. Answer (5) Hint : All carboxylic