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Test - 4A (Paper - I) (Code-A) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2018

1/8

1. (A)

2. (B)

3. (B)

4. (B, D)

5. (B, D)

6. (A, C)

7. (A, C)

8. (B, D)

9. (B, C, D)

10. (C)

11. (A)

12. (B)

13. (B)

14. (B)

15. (A)

16. A (Q, S)

B (P, S)

C (T)

D (T)

17. A (R)

B (P, S, T)

C (Q, S, T)

D (P)

18. (20)

19. (41)

20. (32)

21. (C)

22. (B)

23. (A)

24. (A, B, D)

25. (A, B, C, D)

26. (A, B, C)

27. (A, B, D)

28. (B, C, D)

29. (A, B)

30. (C)

31. (C)

32. (C)

33. (B)

34. (B)

35. (D)

36. A (R)

B (P, T)

C (Q, T)

D (S)

37. A (R, T)

B (S, T)

C (Q)

D (P, T)

38. (07)

39. (03)

40. (25)

41. (B)

42. (C)

43. (C)

44. (A, C)

45. (B, C)

46. (A, C, D)

47. (A, C, D)

48. (B)

49. (A, B, D)

50. (C)

51. (C)

52. (A)

53. (D)

54. (D)

55. (C)

56. A (R)

B (Q)

C (P)

D (S)

57. A (P, Q, R, T)

B (Q)

C (P, R, S, T)

D (S)

58. (09)

59. (02)

60. (05)

PHYSICS CHEMISTRY MATHEMATICS

Test Date : 04/02/2018

ANSWERS

TEST - 4A (Paper-1) - Code-A

All India Aakash Test Series for JEE (Advanced)-2018

All India Aakash Test Series for JEE (Advanced)-2018 Test - 4A (Paper - I) (Code-A) (Answers & Hints)

2/8

PART - I (PHYSICS)

ANSWERS & HINTS

1. Answer (A)

2. Answer (B)

3. Answer (B)

4. Answer (B, D)

5. Answer (B, D)

Maximum temperature of gas achieved at V = 0

3

2V

6. Answer (A, C)

On rotation liquid will take the shape as shown

Maximum pressure will be at A, C

Pressure will increase on moving away from central

axis

CA

7. Answer (A, C)

T2 R

3

3

3

3 3 3

84

A

B

T

T

Let 1 and 2 be angular velocities of A and B

(1t + 2t) = 2

1 = 3

27

GM

R

2 = 3

1

8

GM

R

8. Answer (B, D)

2 cos

Th

r g

9. Answer (B, C, D)

dTQ kA

dx �

2

1

0

0

TL

T

dxQ k A dT

x L

∫ ∫�

0 1 2ln2 ( )Q k A T T �

0 1 2( )

ln2

k A T TQ

�

10. Answer (C)

11. Answer (A)

Solution for Q. Nos. 10 and 11

Normal reaction will be uniformly distributed over the

periphery

N sin = Mg

dN cos = 2T2

d

dN 2

Nd

cos2

NdTd

dTT

dN cos

T = cos2 sin

Mg

T = cot2

Mg

l = 2

cot2

T l Mg r

AY AY

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

= cotMgr

AY

U = 1

2 × Stress × Strain × Total volume

=

21 (Stress)

(2 )2

A rY

=

2 2 2 2 2 2

2 2

1 cot cot(2 )

2 44

M g M g rA r

AYA Y

12. Answer (B)

13. Answer (B)


Process A B is isobaric process, since T = constant

T

V = constant

W = P(V)

= P(VB – VA)

= R(TB – TA) = 3RTA as TB = 4TA

U0 = 3

2A

RT

W = 2U0


3/8

Process B C is isothermal compression.

WB C = PdV∫

= lnC

B

B

VRT

V

⎛ ⎞⎜ ⎟⎝ ⎠

= –RTB ln 4

UB = 4U0 = +3

2B

RT

= 08

ln 43

UR

R

WB C = 0

16ln2

3U

14. Answer (B)

15. Answer (A)


Frequency observed by O1 is

=

5

4s

vf f

v v

wavelength observed = 4

5

v v

f

To calculate frequency observed by O2. We can divide

it into two parts.

First is calculate frequency detected by water surface

and then water surface will behave like a source.

5

4

5

vf f

vv

, frequency observed by O2 =

45 5

4 4

vv

fv

⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦

= 5 21

4 20f

= 21

16f

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

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

18. Answer (20)

Let acceleration of ball be a

Va0 = Vma

a = 1 m/s2

2

0

1( ) 8

2a a t

t = 2 seconds

x = 20

19. Answer (41)

Let M be mass of sphere

F1 = 2

4

GM

4

35

F3

F2 = F1 – 2F3 cos ...(i)

F3 = 2

64

5

MG⎛ ⎞⎜ ⎟⎝ ⎠

Put in equation (i), F2 = 2

123

125 4

GM

20. Answer (32)

Y = A sin (kx) cos (t)

= 2L

A = A sin (kx) = 2

A at x =

6

L

1

2 4 A

x = 0

x L =

24

2 A

= 16

A = 1

m32

x = 32


4/8

PART - II (CHEMISTRY)

21. Answer (C)

CH —CH==3

CH + BrCCl CH —CH—CHCCl2 3 3 3

Br CH3

R3N+—CH==CH2 + HCl H3N

+—CH2—CH2Cl

CH —CH==3

CH + X CH —CH—CH2 2 3 2

X X

CH —C==3

CH—CH + HX CH —C(X)—CH —CH3 3 2 3

CH3

CH3

22. Answer (B)

23. Answer (A)

24. Answer (A, B, D)

II is least stable due to incomplete octet

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

Reaction proceed through formation of carbocation in

which rearrangement may occur.

26. Answer (A, B, C)

N2H4 N = –2

NH2OH N = –1

N2O N = +1

N2O3 N = +3

Down the group boiling point increases but NH3 has

higher boiling point due to H-bonding.


(1) Down the group stability of hydride decreases.

(2) O2– has one unpaired electron

28. Answer (B, C, D)

Pb3O4 + 8HCl 3PbCl2 + Cl2 + 4H2O

Pb3O4 + 4HNO3 2Pb(NO3)2 + PbO2 + 2H2O

29. Answer (A, B)

30. Answer (C)

2 2

1

2

[AI] [4]Q 2

8[A ]

31. Answer (C)

At R equilibrium is reached

K = 12 12

246

G° = –RTlnK

Since K is +ve so G° is negative.

32. Answer (C)

33. Answer (B)


450 K

3 2 6

X

2BF 6NaH B H 6NaF

3B H + 6NH 3[BH (NH ) ] [BH ]2 6 3 2 3 2 4

+ –

2B N H + 12H3 3 6 2

(Z)

(Y)

34. Answer (B)

35. Answer (D)


P is CH —CH CH2 2

CH —CH CH2 2

CH —CHO + HCHO2

O3

Zn/H O2

CH —CH CH2 2

CH —CH—CH Br2 2

Br /Fe2

Br2

NBS

CH —CH CH2 2

CH—CH CH2

Br

(X)

Br

Br

(Y)

(Z)

CH —CH CH2 2

HClCH—CH—CH

3

+

H Hydrideshift

CH—CH CH2 3—

Cl–CH—CH CH

2 3

Cl

+

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

(A) meq of H2SO4 = 200 × 0.02

meq of KOH = 200 × 0.02

meq (Acid) = meq (Base) So pH = 7


5/8

(B) pH = pKa + 3

3

[CH COO ]log

[CH COOH]

= pKa +

2

1

10log

10

= 4.7 – 1 = 3.7

(C) pH = w a b

1[pK pK pK ]

2

= 1[14 4.7 5.7] 6.5

2

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

38. Answer (07)

39. Answer (03)

II, III, IV

40. Answer (25)

Cl CH—CH—CH CH2 2 3

CH3

CH —CH—CCl —CH3 2 3

CH3

CH —CH—CH —CHCl3 2 2

CH3

Cl—CH —CH—CH CH2 2 3

CH Cl2

Cl—CH —C(Cl)—CH CH2 2 3

CH3

Cl—CH —CH—CH(Cl)—CH2 3

CH3

Cl—CH —CH—CH CH Cl2 2 2

CH3

CH —C(Cl)—CHCl—CH3 3

CH3

CH —C(Cl)—CH CH3 2 2

Cl

CH3

CH —CH—CH(Cl)—CH Cl3 2

CH3

PART - III (MATHEMATICS)

41. Answer (B)

Let R be the circumradius

A

C

B

4

3

5

R = 5, R = 4, R = 3

12R⇒

62 12R R⇒ ⇒

Now, 21sin sin sin

2R

21 5 4 3sin sin sin

2R

R R R

⎧ ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

21 1 31

2 2 2R

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 3

9 3 1 3 2163 3abc

⇒

Also,

3

1 2 3 3

2 2 2 8 162 2 3p p p

a b c abc

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

and 2abc

Rr a b c

42. Answer (C)

A sin x + B cos(x + C) + B cos(x – C) = A sin x + 2B cos C cos x = The maximum value of A sin x + 2B cos C cos x is

2 2 24 cosA B C which increases as | cos C |

increases.

So the minimum value of | cos C | corresponds to a

situation when 2 2 24 cosA B C is just equal to

A2 + 4B2 cos2 C = 2

2 2

2

2cos

4

AC

B

43. Answer (C)

Locus of P is a branch of hyperbola

2

21

3

yx

For r = 2, the circle and branch of hyperbola intersect

in two points.

For r = 1, there is one point of intersection

If m is the slope of tangent then

m2 – 3 = m2r2 + m2


6/8

m2 =

2

2

3

1

r

r

Hence no common tangent for r > 1 and two common

tangent for r < 1

44. Answer (A, C)

As the tangents are perpendicular

(0, 1) lies on the director circle

2 21a b⇒

The equation of director circle of the ellipse

2 2

2 2 2 21

3 3

x y

a b b a

is 2 2 2 2

4 4x y a b

As (4, 3) lies outside 2 2

4x y

The angle is acute.

45. Answer (B, C)

S

R

P(1, 3)

Q

T

B

Equation of tangent at P is x + y = 4

Image of R(–5, 5) lies on PQ

Image of R in PT is (–1, 9)

Let PV cuts the axis in T

T = (–3, 7)

Equation of SP is x + 3y = 10

Let S = (10 – 3, )

Again ||TS PQ

7

13 3

⇒

9 33 4

1 1

⇒

S = (–2, 4)

46. Answer (A, C, D)

2

25 1710 24 20

2 2ae

⎛ ⎞ ⎜ ⎟⎝ ⎠

Also other focus is at (–5, 16)

As y-axis is a tangent 2

1 2p p b⇒

250b⇒

2 89

4a⇒

2 89a⇒


Any point on the line L through R(, 2) is

( + r cos , 2 + r sin )

L meets the given ellipse

2 2cos 2 sin

19 4

r r ⇒

2 2 24cos 9sin 4 2 cos 9sinr r⇒

2

4 0 2

2 2

4

4cos 9sinRA RD

⇒

Also the line meets the axes at B and C

2

sin cosRB RC

⇒

Now RA RD RB RC 2

2 2

4 2

sin cos4cos 9sin

⇒

2 sin2 5cos2 13⇒

2

131 1 | | 6

4 25⇒ ⇒

48. Answer (B)

As a = 18 cm, b = 24 cm, c = 30 cm

triangle ABC is right angled at C.

G = (6, 8)

P = (9, 12)

I = (6, 6)

(0, 24)A

24

C(0, 0) B(18, 0)18 cm


7/8

Area of triangle GIP = 3 cm2

Now 2 29 16 5, 3 6 3 5GP PI , GI = 2

7 3 52 5 2 3 5 7 3 5 cm

2s s

⇒

3 2 6 7 3 5 3 7 3 5cm

4 27 3 5r

s

Also, 30 5 5 5

cm4 4 3 2

abcR


2 3 1r r r r

s a s c a c

s s b b

⇒

2tan as ,

2 6 2

B a cB

b

⎛ ⎞⇒ ⎜ ⎟⎝ ⎠

1, 1

3

a c

b

⎛ ⎞ ⎜ ⎟⎝ ⎠

50. Answer (C)

ID BC

A

y

C(3, 0)DB( 3, 0)

y

I x y( , )

x = BD – BO = 5 – b b = 5 – x

c = 10 – b = 5 + x

8(8 6)(8 5 )(8 5 )x x

28 16(3 )(3 )y x x

2 24 9x y⇒

51. Answer (C)

Clearly H lies on the director circle of T.

The equation of director circle is 2 2 9

94

x y

52. Answer (A)

53. Answer (D)


Equation of any curve through P, Q, R, S is

3 3

1 1 1 2 2 24 2 4 2y t x t t y t x t t

28 0y x

As P, Q, R, S are concyclic

1 2 1 21 , 0t t t t

Also point of intersection of tangents at P and R is

1 2 1 2, ( )at t a t t , it lies on x-axis

Also, slope of PR =

2 1

2 2

2 1

4|| -axis

2

t tPR y

t t

⇒

54. Answer (D)

55. Answer (C)


y

F

D ( , )rr

C

E

P B(32, 0)A(0, 0)

(32, 1)x

(0, 40)

Let the equation of the circle be 22 2x r y r r

It passes through (32, 1)

266 1025 0r r 25, 41r⇒

Rejecting r = 41

r = 25

Now 1 1 125 5 15

tan 2 tan tan15 3 8

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

56. Answer A(R); B(Q); C(P); D(S)

H and E are confocal.

, 0 1, 0ae

12,

2a e⇒

Equation of ellipse is

2 2

12 1

x y

(A) Length of semilatus rectum =

2b

a=

1

2

(B)1

2e

(C) The points of intersection are 1

1,2

⎛ ⎞ ⎜ ⎟⎝ ⎠So area of the rectangle

= 2 2 2 2


8/8

(D) Also 2 2

4, 2h k

1 34 2

8 4⇒

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

( ) 2xf x

So, a = 8, b = 10, c = 12

1615 7,

4 7

abcR

7r

s

2 23 16 7cos , sec , tan

4 9 9A A A

58. Answer (09)

10 10

3

0 0

1cos cos 3cos

3 4 3r r

r rr

⎛ ⎞ ⎜ ⎟⎝ ⎠∑ ∑

I1 = 1

10

2

0

10 11cos sin

12 3 6cos

3 2sin

6

r

rI

⎛ ⎞ ⎜ ⎟⎝ ⎠ ∑

1 3 11

4 2 8I

⎛ ⎞ ⎜ ⎟⎝ ⎠

59. Answer (02)

Equation of director circle is

2 2 2 22x g y f g f c

As (0, 0) lies on it

2 2 2 22g f g f c

2 22g f c⇒

60. Answer (05)

Length AB is maximum when normal is parallel to

y-axis.

A

(2, 8)

B

So equation of normal is x = 2

28 2 2y y ⇒

2 2 2 2AB 4 2

4 1.414 5AB

� � �

Test - 4A (Paper - I) (Code-B) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2018

1/8

1. (B)

2. (B)

3. (A)

4. (B, C, D)

5. (B, D)

6. (A, C)

7. (A, C)

8. (B, D)

9. (B, D)

10. (C)

11. (A)

12. (B)

13. (B)

14. (B)

15. (A)

16. A (R)

B (P, S, T)

C (Q, S, T)

D (P)

17. A (Q, S)

B (P, S)

C (T)

D (T)

18. (32)

19. (41)

20. (20)

21. (A)

22. (B)

23. (C)

24. (A, B)

25. (B, C, D)

26. (A, B, D)

27. (A, B, C)

28. (A, B, C, D)

29. (A, B, D)

30. (C)

31. (C)

32. (C)

33. (B)

34. (B)

35. (D)

36. A (R, T)

B (S, T)

C (Q)

D (P, T)

37. A (R)

B (P, T)

C (Q, T)

D (S)

38. (25)

39. (03)

40. (07)

41. (C)

42. (C)

43. (B)

44. (A, B, D)

45. (B)

46. (A, C, D)

47. (A, C, D)

48. (B, C)

49. (A, C)

50. (C)

51. (C)

52. (A)

53. (D)

54. (D)

55. (C)

56. A (P, Q, R, T)

B (Q)

C (P, R, S, T)

D (S)

57. A (R)

B (Q)

C (P)

D (S)

58. (05)

59. (02)

60. (09)

PHYSICS CHEMISTRY MATHEMATICS

Test Date : 04/02/2018

ANSWERS

TEST - 4A (Paper-1) - Code-B

All India Aakash Test Series for JEE (Advanced)-2018

All India Aakash Test Series for JEE (Advanced)-2018 Test - 4A (Paper - I) (Code-B) (Answers & Hints)

2/8

PART - I (PHYSICS)

ANSWERS & HINTS

1. Answer (B)

2. Answer (B)

3. Answer (A)

4. Answer (B, C, D)

dTQ kA

dx �

2

1

0

0

TL

T

dxQ k A dT

x L

∫ ∫�

0 1 2ln2 ( )Q k A T T �

0 1 2( )

ln2

k A T TQ

�

5. Answer (B, D)

2 cos

Th

r g

6. Answer (A, C)

T2 R

3

3

3

3 3 3

84

A

B

T

T

Let 1 and 2 be angular velocities of A and B

(1t + 2t) = 2

1 = 3

27

GM

R

2 = 3

1

8

GM

R

7. Answer (A, C)

On rotation liquid will take the shape as shown

Maximum pressure will be at A, C

Pressure will increase on moving away from central

axis

CA

8. Answer (B, D)

Maximum temperature of gas achieved at V = 0

3

2V

9. Answer (B, D)

10. Answer (C)

11. Answer (A)


Normal reaction will be uniformly distributed over the

periphery

N sin = Mg

dN cos = 2T2

d

dN 2

Nd

cos2

NdTd

dTT

dN cos

T = cos2 sin

Mg

T = cot2

Mg

l = 2

cot2

T l Mg r

AY AY

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

= cotMgr

AY

U = 1

2 × Stress × Strain × Total volume

=

21 (Stress)

(2 )2

A rY

=

2 2 2 2 2 2

2 2

1 cot cot(2 )

2 44

M g M g rA r

AYA Y

12. Answer (B)

13. Answer (B)


Process A B is isobaric process, since T = constant

T

V = constant

W = P(V)

= P(VB – VA)

= R(TB – TA) = 3RTA as TB = 4TA

U0 = 3

2A

RT

W = 2U0


3/8

Process B C is isothermal compression.

WB C = PdV∫

= lnC

B

B

VRT

V

⎛ ⎞⎜ ⎟⎝ ⎠

= –RTB ln 4

UB = 4U0 = +3

2B

RT

= 08

ln 43

UR

R

WB C = 0

16ln2

3U

14. Answer (B)

15. Answer (A)


Frequency observed by O1 is

=

5

4s

vf f

v v

wavelength observed = 4

5

v v

f

To calculate frequency observed by O2. We can divide

it into two parts.

First is calculate frequency detected by water surface

and then water surface will behave like a source.

5

4

5

vf f

vv

, frequency observed by O2 =

45 5

4 4

vv

fv

⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦

= 5 21

4 20f

= 21

16f

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

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

18. Answer (32)

Y = A sin (kx) cos (t)

= 2L

A = A sin (kx) = 2

A at x =

6

L

1

2 4 A

x = 0

x L =

24

2 A

= 16

A = 1

m32

x = 32

19. Answer (41)

Let M be mass of sphere

F1 = 2

4

GM

4

35

F3

F2 = F1 – 2F3 cos ...(i)

F3 = 2

64

5

MG⎛ ⎞⎜ ⎟⎝ ⎠

Put in equation (i), F2 = 2

123

125 4

GM

20. Answer (20)

Let acceleration of ball be a

Va0 = Vma

a = 1 m/s2

2

0

1( ) 8

2a a t

t = 2 seconds

x = 20


4/8

PART - II (CHEMISTRY)

21. Answer (A)

22. Answer (B)

23. Answer (C)

CH —CH==3

CH + BrCCl CH —CH—CHCCl2 3 3 3

Br CH3

R3N+—CH==CH2 + HCl H3N

+—CH2—CH2Cl

CH —CH==3

CH + X CH —CH—CH2 2 3 2

X X

CH —C==3

CH—CH + HX CH —C(X)—CH —CH3 3 2 3

CH3

CH3

24. Answer (A, B)

25. Answer (B, C, D)

Pb3O4 + 8HCl 3PbCl2 + Cl2 + 4H2O

Pb3O4 + 4HNO3 2Pb(NO3)2 + PbO2 + 2H2O


(1) Down the group stability of hydride decreases.

(2) O2– has one unpaired electron

27. Answer (A, B, C)

N2H4 N = –2

NH2OH N = –1

N2O N = +1

N2O3 N = +3

Down the group boiling point increases but NH3 has

higher boiling point due to H-bonding.

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

Reaction proceed through formation of carbocation in

which rearrangement may occur.


II is least stable due to incomplete octet

30. Answer (C)

2 2

1

2

[AI] [4]Q 2

8[A ]

31. Answer (C)

At R equilibrium is reached

K = 12 12

246

G° = –RTlnK

Since K is +ve so G° is negative.

32. Answer (C)

33. Answer (B)


450 K

3 2 6

X

2BF 6NaH B H 6NaF

3B H + 6NH 3[BH (NH ) ] [BH ]2 6 3 2 3 2 4

+ –

2B N H + 12H3 3 6 2

(Z)

(Y)

34. Answer (B)

35. Answer (D)


P is CH —CH CH2 2

CH —CH CH2 2

CH —CHO + HCHO2

O3

Zn/H O2

CH —CH CH2 2

CH —CH—CH Br2 2

Br /Fe2

Br2

NBS

CH —CH CH2 2

CH—CH CH2

Br

(X)

Br

Br

(Y)

(Z)

CH —CH CH2 2

HClCH—CH—CH

3

+

H Hydrideshift

CH—CH CH2 3—

Cl–CH—CH CH

2 3

Cl

+

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

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

(A) meq of H2SO4 = 200 × 0.02

meq of KOH = 200 × 0.02


5/8

meq (Acid) = meq (Base) So pH = 7

(B) pH = pKa + 3

3

[CH COO ]log

[CH COOH]

= pKa +

2

1

10log

10

= 4.7 – 1 = 3.7

(C) pH = w a b

1[pK pK pK ]

2

= 1[14 4.7 5.7] 6.5

2

38. Answer (25)

Cl CH—CH—CH CH2 2 3

CH3

CH —CH—CCl —CH3 2 3

CH3

PART - III (MATHEMATICS)

41. Answer (C)

Locus of P is a branch of hyperbola

2

21

3

yx

For r = 2, the circle and branch of hyperbola intersect

in two points.

For r = 1, there is one point of intersection

If m is the slope of tangent then

m2 – 3 = m2r2 + m2

m2 =

2

2

3

1

r

r

Hence no common tangent for r > 1 and two common

tangent for r < 1

42. Answer (C)

A sin x + B cos(x + C) + B cos(x – C) = A sin x + 2B cos C cos x = The maximum value of A sin x + 2B cos C cos x is

2 2 24 cosA B C which increases as | cos C |

increases.

So the minimum value of | cos C | corresponds to a

situation when 2 2 24 cosA B C is just equal to

A2 + 4B2 cos2 C = 2

2 2

2

2cos

4

AC

B

43. Answer (B)

Let R be the circumradius

A

C

B

4

3

5

R = 5, R = 4, R = 3

12R⇒

62 12R R⇒ ⇒

Now, 21sin sin sin

2R

21 5 4 3sin sin sin

2R

R R R

⎧ ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

21 1 31

2 2 2R

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 3

9 3 1 3 2163 3abc

⇒

CH —CH—CH —CHCl3 2 2

CH3

Cl—CH —CH—CH CH2 2 3

CH Cl2

Cl—CH —C(Cl)—CH CH2 2 3

CH3

Cl—CH —CH—CH(Cl)—CH2 3

CH3

Cl—CH —CH—CH CH Cl2 2 2

CH3

CH —C(Cl)—CHCl—CH3 3

CH3

CH —C(Cl)—CH CH3 2 2

Cl

CH3

CH —CH—CH(Cl)—CH Cl3 2

CH3

39. Answer (03)

II, III, IV

40. Answer (07)


6/8

Also,

3

1 2 3 3

2 2 2 8 162 2 3p p p

a b c abc

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

and 2abc

Rr a b c


2 3 1r r r r

s a s c a c

s s b b

⇒

2tan as ,

2 6 2

B a cB

b

⎛ ⎞⇒ ⎜ ⎟⎝ ⎠

1, 1

3

a c

b

⎛ ⎞ ⎜ ⎟⎝ ⎠

45. Answer (B)

As a = 18 cm, b = 24 cm, c = 30 cm

triangle ABC is right angled at C.

G = (6, 8)

P = (9, 12)

I = (6, 6)

(0, 24)A

24

C(0, 0) B(18, 0)18 cm

Area of triangle GIP = 3 cm2

Now 2 29 16 5, 3 6 3 5GP PI , GI = 2

7 3 52 5 2 3 5 7 3 5 cm

2s s

⇒

3 2 6 7 3 5 3 7 3 5cm

4 27 3 5r

s

Also, 30 5 5 5

cm4 4 3 2

abcR


Any point on the line L through R(, 2) is

( + r cos , 2 + r sin )

L meets the given ellipse

2 2cos 2 sin

19 4

r r ⇒

2 2 24cos 9sin 4 2 cos 9sinr r⇒

2

4 0 2

2 2

4

4cos 9sinRA RD

⇒

Also the line meets the axes at B and C

2

sin cosRB RC

⇒

Now RA RD RB RC 2

2 2

4 2

sin cos4cos 9sin

⇒

2 sin2 5cos2 13⇒

2

131 1 | | 6

4 25⇒ ⇒


2

25 1710 24 20

2 2ae

⎛ ⎞ ⎜ ⎟⎝ ⎠

Also other focus is at (–5, 16)

As y-axis is a tangent 2

1 2p p b⇒

250b⇒

2 89

4a⇒

2 89a⇒

48. Answer (B, C)

S

R

P(1, 3)

Q

T

B

Equation of tangent at P is x + y = 4

Image of R(–5, 5) lies on PQ

Image of R in PT is (–1, 9)

Let PV cuts the axis in T

T = (–3, 7)

Equation of SP is x + 3y = 10

Let S = (10 – 3, )

Again ||TS PQ


7/8

7

13 3

⇒

9 33 4

1 1

⇒

S = (–2, 4)

49. Answer (A, C)

As the tangents are perpendicular

(0, 1) lies on the director circle

2 21a b⇒

The equation of director circle of the ellipse

2 2

2 2 2 21

3 3

x y

a b b a

is 2 2 2 2

4 4x y a b

As (4, 3) lies outside 2 2

4x y

The angle is acute.

50. Answer (C)

ID BC

A

y

C(3, 0)DB( 3, 0)

y

I x y( , )

x = BD – BO = 5 – b b = 5 – x

c = 10 – b = 5 + x

8(8 6)(8 5 )(8 5 )x x

28 16(3 )(3 )y x x

2 24 9x y⇒

51. Answer (C)

Clearly H lies on the director circle of T.

The equation of director circle is 2 2 9

94

x y

52. Answer (A)

53. Answer (D)


Equation of any curve through P, Q, R, S is

3 3

1 1 1 2 2 24 2 4 2y t x t t y t x t t

28 0y x

As P, Q, R, S are concyclic

1 2 1 21 , 0t t t t

Also point of intersection of tangents at P and R is

1 2 1 2, ( )at t a t t , it lies on x-axis

Also, slope of PR =

2 1

2 2

2 1

4|| -axis

2

t tPR y

t t

⇒

54. Answer (D)

55. Answer (C)


y

F

D ( , )rr

C

E

P B(32, 0)A(0, 0)

(32, 1)x

(0, 40)

Let the equation of the circle be 22 2x r y r r

It passes through (32, 1)

266 1025 0r r 25, 41r⇒

Rejecting r = 41

r = 25

Now 1 1 125 5 15

tan 2 tan tan15 3 8

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

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

( ) 2xf x

So, a = 8, b = 10, c = 12

1615 7,

4 7

abcR

7r

s

2 23 16 7cos , sec , tan

4 9 9A A A

57. Answer A(R); B(Q); C(P); D(S)

H and E are confocal.

, 0 1, 0ae

12,

2a e⇒

Equation of ellipse is

2 2

12 1

x y


8/8

� � �

(A) Length of semilatus rectum =

2b

a=

1

2

(B)1

2e

(C) The points of intersection are 1

1,2

⎛ ⎞ ⎜ ⎟⎝ ⎠So area of the rectangle

= 2 2 2 2

(D) Also 2 2

4, 2h k

1 34 2

8 4⇒

58. Answer (05)

Length AB is maximum when normal is parallel to

y-axis.

A

(2, 8)

B

So equation of normal is x = 2

28 2 2y y ⇒

2 2 2 2AB 4 2

4 1.414 5AB 59. Answer (02)

Equation of director circle is

2 2 2 22x g y f g f c

As (0, 0) lies on it

2 2 2 22g f g f c

2 22g f c⇒

60. Answer (09)

10 10

3

0 0

1cos cos 3cos

3 4 3r r

r rr

⎛ ⎞ ⎜ ⎟⎝ ⎠∑ ∑

I1 = 1

10

2

0

10 11cos sin

12 3 6cos

3 2sin

6

r

rI

⎛ ⎞ ⎜ ⎟⎝ ⎠ ∑

1 3 11

4 2 8I

⎛ ⎞ ⎜ ⎟⎝ ⎠

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