Test - 4 (Code-G) (Answers) All India Aakash Test Series for JEE (Main)-2018

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PHYSICS CHEMISTRY MATHEMATICS

31. (2)

32. (3)

33. (1)

34. (1)

35. (4)

36. (3)

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38. (4)

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84. (4)

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86. (3)

87. (2)

88. (1)

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Test Date : 14/01/2018

ANSWERS

TEST - 4 (Code-G)

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

All India Aakash Test Series for JEE (Main)-2018 Test - 4 (Code-G) (Hints)

2/7

1. Y = 4 sin2t sin2t

= 4(1 – cos2 ) sin2

2t t = 2 sin2t – sin4t

3. Vmax

= A = mg

K

= 10

40

g

= 1

4 m/s

4. Y = 2

8

( – 2 )x t

,

3

–16 2

( – 2 )

dy

dt x t

= 4 m/s

5.max

y

t

⎛ ⎞⎜ ⎟⎝ ⎠

= 2f

8. P0V

0 – V

0 = nRT

0

V0 =

0

0( – )

nRT

P , n = 1

w = P0V =

0 0

0( – )

P RT

P

9. U = 1

2 (stress × strain × volume)

10. Q = –dT

TAdx

0

lQ

dxA∫ =

2

1

–

T

T

T dT ∫

Q = 2 2

1 2( – )

2

AT T

l

12. = 2

13

= 120°

13. T = 20

21800

= 2s

3

t = 4

T =

1s

6

14.

45°

45°

A

A

A

Ar

= (2A cos45° + A) = ( 2 1)A15. Maximum possible will occur if it is around mean

position.

(Vavg

)max

= 2

4

A

T =

4 2A

T

PART - A (PHYSICS)

16.1

Ir

I A2

2

1

2

25

9

A

A

⎛ ⎞ ⎜ ⎟⎝ ⎠

1

2

5

3

A

A

17. T = 2m

K

T2 = 2 2

1 2T T 1 2

eq1 2

K KK

K K

⎛ ⎞⎜ ⎟⎝ ⎠18. (

1 –

2)T =

= 2 2 4

–5

TT T

⎡ ⎤⎢ ⎥⎣ ⎦

= 2

5

= 72°

19. dW∫ = P dv∫

= av2 dv = 3 3

2 1–

3

av v

W = 2 1( – )

3

RT T

21. WCD

= nRT2 ln

D

C

V

V

WAB

= nRT1 ln

B

A

V

V

VA = V

C, V

B = V

D

2WCD

= WAB

2T2 = T

1

1

2

T

T = 2

22.

4

A

B

T

T

⎛ ⎞⎜ ⎟⎝ ⎠

= 0.81

0.01 = 81

A

B

T

T = 3,

ATA =

BTB

23. 0 1 0 2–

mg mgP L P L

A A

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

1

101

99

L

L

24. T = Const.

4 4

2 1

1

16T T

T2 =

1

2

T

Test - 4 (Code-G) (Hints) All India Aakash Test Series for JEE (Main)-2018

3/7

25. TV r – 1 = Constant

–( –1)

dV V

dT T

–tan = 0–

( –1)V

T

( – 1) = 0

0tan

V

T

CV =

–1R

= 0

0

tanRT

V

26. f.L1 =

2

B

4.3

Lf =

1

B

1

4

3

L

L =

1

2

L1 =

1

2

4

3

L

27. 0.42

= 0.8 m

TV

= 8 m/s

f = 10 Hz

28. Q = U + W

W = 4

Q

3

4

Q = C

V dT

3

4

Q =

5

2R dT

C = 10

3R

29. On heating volume will increase. So, it will move

towards right.

30. TB

= B

A

A

VT

V = 600 K

WBC

= lnC

B

B

VnRT

V = 1200 R ln2

PART - B (CHEMISTRY)

35.

No rearrangement

36. CH CH + SeO2 CHO—CHO

3CH CH Red hot

iron tube

873 K

37. Heat of hydrogenation 1

stability of alkene

38. A = , B =

39.BH , THF

3

H

BH2

H

H

CH COOH3

(Alkane)

40.

OH

+ Zn (Dust)

(X)

CH3

+ CH Cl3

AlCl3

(Y)

CH3

alk. KMnO4

(Z)

COOH

NaOH

(U)

COONa

Sodalime

(X)

42. +I effect of D is more than of H.

All India Aakash Test Series for JEE (Main)-2018 Test - 4 (Code-G) (Hints)

4/7

45. 2R COOK + 2H2O R—R + 2CO

2 + H

2 + 2KOH

46. % of P = 2 2 7

wt. of Mg P O62100

222 wt. of compound

= 62 0.75

100222 0.6

= 35%

50.

X

Li/NH3

X

[with EDG]

Y

Li/NH3

Y

[with EWG]

due to stability of carbanion intermediates.

56.Br

2

Br

Br(A)

(i) alc. KOH

(ii) NaNH2

CH –C C–CH3 3

(B)

H /Pd

BaCO

2

3

C=CCH

3CH

3

(C) HH

59.

PART - C (MATHEMATICS)

61. If is angle between tangents, then

sin2

⎛ ⎞⎜ ⎟⎝ ⎠

= radius

distance between 3 3, 3 and (0, 0)

sin2

⎛ ⎞⎜ ⎟⎝ ⎠

= 3

6 =

1

2

2

=

6

=

3

So, cot = cot3

⎛ ⎞⎜ ⎟⎝ ⎠

= 1

3

62. Let P(h, k) be the point of intersection of E1 and E

2

2

2

21

hk

a h2 = a2(1 – k2) ...(i)

and

2 2

21

h k

a = 1 k2 = a2 (1 – h2) ...(ii)

Eliminating a from equation (i) and (ii), we get

2 2

2 21– 1–

h k

k h

h2(1 – h2) = k2 (1 – k2)

(h – k) (h + k) (h2 + k2 – 1) = 0

Hence, the locus is a set of curves consisting of the

straight lines y = x, y = –x and circle x2 + y2 = 1.

63. Equation of hyperbola is

2 2

– 1.9 16

x y

Equation of tangent is

29 – 16y mx m

29 – 16 2 5m m = ±2

a + b = sum of roots = 0

64. Make homogeneous 3x2 – y2 –2x + 4y = 0 with the

line y = mx + c

i.e., 2 2 – –

3 – – 2 4y mx y mx

x y x yc c

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 0

According to question coeff. of x2 + coeff. of y2 = 0

2 43 –1 0

m

c c

m + c + 2 = 0

which gives fixed point (1, –2) p + q = –1

65. The point 6

P⎛ ⎞

⎜ ⎟⎝ ⎠

is sec , tan6 6

a b ⎛ ⎞

⎜ ⎟⎝ ⎠

or

2,

3 3

a bP⎛ ⎞⎜ ⎟⎝ ⎠

.

Equation of tangent at P is – 13 3

2

x y

a b

Area of the triangle = 1 3

32 2

ab = 3a2

b

a = 4

e2 =

2

21 17

b

a

66. The two circles are x2 + y2 – 4x – 6y – 3 = 0 and

x2 + y2 + 2x + 2y + 1 = 0

Centres : C1(2, 3) and C

2(–1, –1)

radii : r1 = 4 and r

2 = 1.

We have, C1C

2 = 5 = r

1 + r

2, circles touch externally

therefore there are three common tangents to the

given circles.

Test - 4 (Code-G) (Hints) All India Aakash Test Series for JEE (Main)-2018

5/7

67. Centre and radius of circle x2 + y2 + 8x – 10y – 40 = 0

are C(–4, 5) and 9 respectively. Distance of the centre

C(–4, 5) from the point P(–2, 3) is CP = 4 4 =

2 2 .

a = CP + r = 2 2 + 9

b = |CP – r| = 9 – 2 2

a – b = 4 2

68. If S1 = 0 and S

2 = 0 are the equations, then

S1+ S

2= 0 is a second degree curve passing through

the points of intersection of S1 = 0 and S

2 = 0.

( + 4)x2 + 2( + 1)y2 – 2(3 + 10)x

– 12( + 1)y + (23 + 35) = 0 ...(i)

For it to be a circle, choose such that the coefficient

of x2 and y2 are equal

+ 4 = 2 + 2

= 2

This gives the equation of the circle as

6(x2 + y2) – 32x – 36y + 81 = 0 {(using (i))}

2 2 16 27– – 6 0

3 2x y x y

Its centre is 8, 3

3C⎛ ⎞⎜ ⎟⎝ ⎠

and radius is

r = 64 27

9 –9 2

= 1 47

3 2

69. The circle and the ellipse do not meet at any point.

Length of the common chord = 0.

70. Putting y = 2t2 in the equation of the given ellipse

2 2

1,9 4

x y we get

2 44

19 4

x t x2 = 9(1 – t4)

= 9(1 – t2) (1 + t2)

This will give real values of x if 1 – t2 0, i.e., |t| 1.

71. Clearly ax + by = 1

i.e.,– 1a

y xb b

is tangent to cx2 + dy2 = 1

2 2

11 1

x y

c d

2 2

1 1 – 1a

b c b d

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1 =

2 2a b

c d

72.

2 2

22

15 ( – 1) 2– 22

x y

aa a

2 2

22

13 ( – 1) 2( – 1)2

x y

aa

where, A2 = (a – 1)2 + 3

2, B2 = (a – 1)2 + 2

We know that,

A2 = B2 (1 – e2), (for vertical ellipse)

2 3( – 1)

2a = 2 1

( – 1) 2 1–4

a⎛ ⎞ ⎜ ⎟⎝ ⎠

4(a – 1)2 + 6 = 3(a – 1)2 + 6

(a – 1)2 = 0 a = 1

Equation of ellipse is

2 2

13 2

2

x y

So, length of L.R. =

22A

B =

32

2

2

=

9

2

73.

Pt

t(2 , 4 )

11

2

Q t t(2 , 4 )2 2

2

Let absciss of point P is 2

12 18t 2

19t

t1 = 3

We know that, If PQ is normal then 2 1

1

2– –t t

t

2

11–

3t

So, coordinates of P and Q are (18, 12) and

242 –44,

9 3

⎛ ⎞⎜ ⎟⎝ ⎠

respectively.

PQ =

2 2242 –44

– 18 – 129 3

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 80 10

9

74.

BR

P

X

AO

Y

Equation of tangent at P(a cos, b sin) is

cos sin 1x y

a b

All India Aakash Test Series for JEE (Main)-2018 Test - 4 (Code-G) (Hints)

6/7

It meets the major axis at A(a sec, 0) and the

minor axis at B(0, b cosec).

Equation of the OR, perpendicular to the tangent at

P is ax sin – by cos = 0.

Now PR = 2 2 2 2

( cos )sin – ( sin )cos

sin cos

a a b b

a b

=

2 2

2 2 2 2

( – ) sin cos

sin cos

a b

a b

Also, AB =

2 2 2 2sin cos

sin cos

a b

So, (AB) (PR) = a2 – b2

Here, a = 4, b = 3

(AB)(PR) = 16 – 9 = 7

75. The equation ax3 + 3bx2y + 3cxy2 + dy3 = 0

3 2

3 3 0y y y

d c b ax x x

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Let y

mx

dm3 + 3cm2 + 3bm + a = 0

If the roots are m1, m

2 and m

3, then m

1 + m

2 + m

3

= –3

,

c

d m

1m

2 + m

2m

3 + m

3m

1 =

3b

d and m

1m

2m3

= –a

d.

For three coincident lines, we have m1 = m

2 = m

3,

so, m = –c

d, m2 =

b

d and m3 =

–a

d

So,

3– –c a

d d

⎛ ⎞ ⎜ ⎟⎝ ⎠

c3 = ad2

76.

XO

Y

PQ

A (1, 3)

Let P(2cos, 2sin). The mid point Q(h, k) of AP

will be given by

h = 1 2cos

2

cos =

1–2

h

and k = 3 2sin

2

sin =

3–2

k

Eliminating , we get

2 21 3

– – 12 2

h k⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Hence, the locus is

2 21 3

– – 12 2

x y⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So, radius = 1.

77. Any point on the parabola is P(t2, 2t). The

circumcenter Q(h, k) of triangle PAB is midpoint of

P(t2, 2t) and the centre C(–3, 2) of the circle.

Hence,

2– 3

2

th and k = t + 1

Eliminating t, we get (k – 1)2 = 2h + 3

Locus of P(h, k) is (y – 1)2 = 2x + 3

Equation of tangent to parabola with slope 1 is

(y – 1) =

1

3 21

2 1x

⎛ ⎞ ⎜ ⎟⎝ ⎠

y = x + 3

78. The given parabolas 2y2 = 2x – 1 and 2x2 = 2y – 1

are symmetrical about the line y = x. Also, the

shortest distance occurs along the common normal

which is perpendicular to the line y = x. Then the

shortest distance is

1 1 1

16 16 2 2d

79. The locus of P is an ellipse, if K > AB

K > 50 K > 5 2

80. Ellipse 16x2 + 25y2 = 400

2 2

125 16

x y

Let the points on ellipse be P(5 cos1, 4 sin

1) and

Q(5cos2, 4sin

2).

Circle described on PQ as diameter touches x-axis

at (3, 0).

1 2cos cos

5 32

⎛ ⎞ ⎜ ⎟⎝ ⎠

and 1 2

sin sin4

2r

⎛ ⎞ ⎜ ⎟⎝ ⎠

1 24tan

5 2 3

r ⎛ ⎞ ⎜ ⎟⎝ ⎠

1 2 5

tan2 12

r ⎛ ⎞ ⎜ ⎟⎝ ⎠

slope of PQ = 1 2–4

cot5 2

⎛ ⎞⎜ ⎟⎝ ⎠

= –48

25r = –1

96

4825

2 25r

⎛ ⎞⎜ ⎟

⎜ ⎟⎝ ⎠∵

81. Equation of tangent on parabola y2 = 4x is

1y mx

m ...(i)

Test - 4 (Code-G) (Hints) All India Aakash Test Series for JEE (Main)-2018

7/7

It passes through point (–2, –1).

So, 1

–1 – 2mm

2m2 – m – 1 = 0

Roots of equation are m1, m

2.

Then, tan = 1 2

1 2

1 8

– 2

111–

2

m m

m m

tan = 3

82.

2 2( – 1) ( 3)1

4 16

x y

( 3) ,b

ye

a2 = b2 (1 – e2)

8

33

y

5

3y or

–11

3y

83. Foci of the parabolas are (0, a) and (2a, 0)

respectively, so equation of the circle described on

the line segment joining the foci of the parabolas as

diameter is

(x – 0) (x – 2a) + (y – a) (y – 0) = 0

x2 + y2 – 2ax – ay = 0

84. In slope form equation of tangent to 4x2 – 9y2 = 36 is

y = 2

9 – 4mx m ; m < 0

2

2

4 9 – 4

1

m m

m

= 4

–2

5

m

Required equation of tangent is

2 5 4 0x y 85. Equation of common chord on two circles is

S1 – S

2 = 0.

x2 + y2 – 6x – 4y + 9 – x2 – y2 + 8x + 6y – 23 = 0

2x + 2y – 14 = 0 x + y – 7 = 0

P

C2

(4, 3)

B

A

C1

(3, 2)

r1

C1P =

3 2 – 7

2

=

22

2

AB = 2AP = 2 2

1 12 – ( )r C P = 2 4 – 2 = 2 2

86. x2y2 – 9y2 – 6x2y + 54y = 0

y2 (x2 – 9) – 6y(x2 – 9) = 0

(x2 – 9) (y2 – 6y) = 0

(x + 3) (x – 3) (y) (y – 6) = 0

So, x = –3, x = 3, y = 0, y = 6

It will form a square, so

Area of quadrilateral = 36.

87. Let the direct common tangent intersect at P and

PN1 = l, PC

1 = .

b

C2

N2

a

C1

N1

Pl

PC1N

1 and PC

2N

2 are similar, so,

1 2

1 1 2 2

PC PC

C N C N

a

=

a b

b

=

2

–

a ab

b a

Now l2 = 2 – a2 =

22

2–

–

a aba

b a

⎛ ⎞⎜ ⎟⎝ ⎠

=

3

2

4

( – )

a b

a b

l =

3 1

2 22

( – )

a b

b a

89. Point of contact of y = mx + c is

2 2

2 2 2 2 2 2

,

a m b

a m b a m b

⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠

=

1 1–3 4,7 7

12 12

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

= –2 21 21

,21 14

⎛ ⎞⎜ ⎟⎝ ⎠

90. (x – 1)2 – (y – 2)2 – 1 + 4 – 7 = 0

2 2( – 1) ( – 2)– 1

4 4

x y

It is a rectangular hyperbola with eccentricity 2.

� � �

Test - 4 (Code-H) (Answers) All India Aakash Test Series for JEE (Main)-2018

1/7

1. (3)

2. (2)

3. (3)

4. (2)

5. (3)

6. (1)

7. (1)

8. (2)

9. (1)

10. (3)

11. (1)

12. (2)

13. (3)

14. (3)

15. (1)

16. (4)

17. (1)

18. (1)

19. (3)

20. (3)

21. (2)

22. (1)

23. (1)

24. (3)

25. (1)

26. (1)

27. (1)

28. (4)

29. (1)

30. (4)

PHYSICS CHEMISTRY MATHEMATICS

31. (1)

32. (4)

33. (3)

34. (4)

35. (2)

36. (4)

37. (2)

38. (2)

39. (2)

40. (3)

41. (3)

42. (4)

43. (2)

44. (3)

45. (4)

46. (2)

47. (1)

48. (4)

49. (1)

50. (3)

51. (2)

52. (3)

53. (4)

54. (3)

55. (3)

56. (4)

57. (1)

58. (1)

59. (3)

60. (2).

61. (2)

62. (2)

63. (1)

64. (2)

65. (3)

66. (4)

67. (4)

68. (2)

69. (1)

70. (4)

71. (1)

72. (2)

73. (2)

74. (3)

75. (1)

76. (4)

77. (3)

78. (1)

79. (3)

80. (3)

81. (2)

82. (4)

83. (4)

84. (3)

85. (2)

86. (1)

87. (4)

88. (4)

89. (1)

90. (3)

Test Date : 14/01/2018

ANSWERS

TEST - 4 (Code-H)

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

All India Aakash Test Series for JEE (Main)-2018 Test - 4 (Code-H) (Hints)

2/7

1. TB

= B

A

A

VT

V = 600 K

WBC

= lnC

B

B

VnRT

V = 1200 R ln2

2. On heating volume will increase. So, it will move

towards right.

3. Q = U + W

W = 4

Q

3

4

Q = C

V dT =

5

2R dT

C = 10

3R

4. 0.42

= 0.8 m

TV

= 8 m/s

f = 10 Hz

5. f.L1 =

2

B

4.3

Lf =

1

B

1

4

3

L

L =

1

2

L1 =

1

2

4

3

L

6. TV r – 1 = Constant

–( –1)

dV V

dT T

–tan = 0–

( –1)V

T

( – 1) = 0

0tan

V

T

CV =

–1R

= 0

0

tanRT

V

7. T = Const.

4 4

2 1

1

16T T

T2 =

1

2

T

PART - A (PHYSICS)

8. 0 1 0 2–

mg mgP L P L

A A

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

1

101

99

L

L

9.

4

A

B

T

T

⎛ ⎞⎜ ⎟⎝ ⎠

= 0.81

0.01 = 81

A

B

T

T = 3,

ATA =

BTB

10. WCD

= nRT2 ln

D

C

V

V

WAB

= nRT1 ln

B

A

V

V

VA = V

C, V

B = V

D

2WCD

= WAB

2T2 = T

1

1

2

T

T = 2

12. dW∫ = P dv∫

= av2 dv = 3 3

2 1–

3

av v

W = 2 1( – )

3

RT T

13. (1 –

2)T =

= 2 2 4

–5

TT T

⎡ ⎤⎢ ⎥⎣ ⎦

= 2

5

= 72°

14. T = 2m

K

T2 = 2 2

1 2T T 1 2

eq1 2

K KK

K K

⎛ ⎞⎜ ⎟⎝ ⎠

15.1

Ir

I A2

2

1

2

25

9

A

A

⎛ ⎞ ⎜ ⎟⎝ ⎠

1

2

5

3

A

A

16. Maximum possible will occur if it is around mean

position.

(Vavg

)max

= 2

4

A

T =

4 2A

T

Test - 4 (Code-H) (Hints) All India Aakash Test Series for JEE (Main)-2018

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PART - B (CHEMISTRY)

32.

35.Br

2

Br

Br(A)

(i) alc. KOH

(ii) NaNH2

CH –C C–CH3 3

(B)

H /Pd

BaCO

2

3

C=CCH

3CH

3

(C) HH

41.

X

Li/NH3

X

[with EDG]

Y

Li/NH3

Y

[with EWG]

due to stability of carbanion intermediates.

17.

45°

45°

A

A

A

Ar

= (2A cos45° + A) = ( 2 1)A

18. T = 20

21800

= 2s

3

t = 4

T =

1s

6

19. = 2

13

= 120°

21. Q = –dT

TAdx

0

lQ

dxA∫ =

2

1

–

T

T

T dT ∫

Q = 2 2

1 2( – )

2

AT T

l

22. U = 1

2 (stress × strain × volume)

23. P0V

0 – V

0 = nRT

0

V0 =

0

0( – )

nRT

P , n = 1

w = P0V =

0 0

0( – )

P RT

P

26.max

y

t

⎛ ⎞⎜ ⎟⎝ ⎠

= 2f

27. Y = 2

8

( – 2 )x t

,

3

–16 2

( – 2 )

dy

dt x t

= 4 m/s

28. Vmax

= A = mg

K

= 10

40

g

= 1

4 m/s

30. Y = 4 sin2t sin2t

= 4(1 – cos2 ) sin2

2t t = 2 sin2t – sin4t

45. % of P = 2 2 7

wt. of Mg P O62100

222 wt. of compound

= 62 0.75

100222 0.6

= 35%

46. 2R COOK + 2H2O R—R + 2CO

2 + H

2 + 2KOH

49. +I effect of D is more than of H.

51.

OH

+ Zn (Dust)

(X)

CH3

+ CH Cl3

AlCl3

(Y)CH

3

alk. KMnO4

(Z)

COOH

NaOH

(U)

COONa

Sodalime

(X)

All India Aakash Test Series for JEE (Main)-2018 Test - 4 (Code-H) (Hints)

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PART - C (MATHEMATICS)

61. (x – 1)2 – (y – 2)2 – 1 + 4 – 7 = 0

2 2( – 1) ( – 2)– 1

4 4

x y

It is a rectangular hyperbola with eccentricity 2.

62. Point of contact of y = mx + c is

2 2

2 2 2 2 2 2

,

a m b

a m b a m b

⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠

=

1 1–3 4,7 7

12 12

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

= –2 21 21

,21 14

⎛ ⎞⎜ ⎟⎝ ⎠

64. Let the direct common tangent intersect at P and

PN1 = l, PC

1 = .

b

C2

N2

a

C1

N1

Pl

PC1N

1 and PC

2N

2 are similar, so,

1 2

1 1 2 2

PC PC

C N C N

a

=

a b

b

=

2

–

a ab

b a

Now l2 = 2 – a2 =

22

2–

–

a aba

b a

⎛ ⎞⎜ ⎟⎝ ⎠

=

3

2

4

( – )

a b

a b

l =

3 1

2 22

( – )

a b

b a

52.BH , THF

3

H

BH2

H

H

CH COOH3

(Alkane)

53. A = , B =

65. x2y2 – 9y2 – 6x2y + 54y = 0

y2 (x2 – 9) – 6y(x2 – 9) = 0

(x2 – 9) (y2 – 6y) = 0

(x + 3) (x – 3) (y) (y – 6) = 0

So, x = –3, x = 3, y = 0, y = 6

It will form a square, so

Area of quadrilateral = 36.

66. Equation of common chord on two circles is

S1 – S

2 = 0.

x2 + y2 – 6x – 4y + 9 – x2 – y2 + 8x + 6y – 23 = 0

2x + 2y – 14 = 0 x + y – 7 = 0

P

C2

(4, 3)

B

A

C1

(3, 2)

r1

C1P =

3 2 – 7

2

=

22

2

AB = 2AP = 2 2

1 12 – ( )r C P = 2 4 – 2 = 2 2

67. In slope form equation of tangent to 4x2 – 9y2 = 36 is

y = 2

9 – 4mx m ; m < 0

2

2

4 9 – 4

1

m m

m

= 4

–2

5m

Required equation of tangent is

2 5 4 0x y 68. Foci of the parabolas are (0, a) and (2a, 0)

respectively, so equation of the circle described on

the line segment joining the foci of the parabolas as

diameter is

(x – 0) (x – 2a) + (y – a) (y – 0) = 0

x2 + y2 – 2ax – ay = 0

54. Heat of hydrogenation 1

stability of alkene

55. CH CH + SeO2 CHO—CHO

3CH CH Red hot

iron tube

873 K

56.

No rearrangement

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69.

2 2( – 1) ( 3)1

4 16

x y

( 3) ,b

ye

a2 = b2 (1 – e2)

8

33

y

5

3y or

–11

3y

70. Equation of tangent on parabola y2 = 4x is

1y mx

m ...(i)

It passes through point (–2, –1).

So, 1

–1 – 2mm

2m2 – m – 1 = 0

Roots of equation are m1, m

2.

Then, tan = 1 2

1 2

1 8

– 2

111–

2

m m

m m

tan = 3

71. Ellipse 16x2 + 25y2 = 400

2 2

125 16

x y

Let the points on ellipse be P(5 cos1, 4 sin

1) and

Q(5cos2, 4sin

2).

Circle described on PQ as diameter touches x-axis

at (3, 0).

1 2cos cos

5 32

⎛ ⎞ ⎜ ⎟⎝ ⎠

and 1 2

sin sin4

2r

⎛ ⎞ ⎜ ⎟⎝ ⎠

1 24tan

5 2 3

r ⎛ ⎞ ⎜ ⎟⎝ ⎠

1 2 5

tan2 12

r ⎛ ⎞ ⎜ ⎟⎝ ⎠

slope of PQ = 1 2–4

cot5 2

⎛ ⎞⎜ ⎟⎝ ⎠

= –48

25r = –1

96

4825

2 25r

⎛ ⎞⎜ ⎟

⎜ ⎟⎝ ⎠∵

72. The locus of P is an ellipse, if K > AB

K > 50 K > 5 2

73. The given parabolas 2y2 = 2x – 1 and 2x2 = 2y – 1

are symmetrical about the line y = x. Also, the

shortest distance occurs along the common normal

which is perpendicular to the line y = x. Then the

shortest distance is

1 1 1

16 16 2 2d

74. Any point on the parabola is P(t2, 2t). The

circumcenter Q(h, k) of triangle PAB is midpoint of

P(t2, 2t) and the centre C(–3, 2) of the circle.

Hence,

2– 3

2

th and k = t + 1

Eliminating t, we get (k – 1)2 = 2h + 3

Locus of P(h, k) is (y – 1)2 = 2x + 3

Equation of tangent to parabola with slope 1 is

(y – 1) =

1

3 21

2 1x

⎛ ⎞ ⎜ ⎟⎝ ⎠

y = x + 3

75.

XO

Y

PQ

A (1, 3)

Let P(2cos, 2sin). The mid point Q(h, k) of AP

will be given by

h = 1 2cos

2

cos =

1–2

h

and k = 3 2sin

2

sin =

3–2

k

Eliminating , we get

2 21 3

– – 12 2

h k⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Hence, the locus is

2 21 3

– – 12 2

x y⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So, radius = 1.

76. The equation ax3 + 3bx2y + 3cxy2 + dy3 = 0

3 2

3 3 0y y y

d c b ax x x

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Let y

mx

dm3 + 3cm2 + 3bm + a = 0

If the roots are m1, m

2 and m

3, then m

1 + m

2 + m

3

= –3

,

c

d m

1m

2 + m

2m

3 + m

3m

1 =

3b

d and m

1m

2m3

= –a

d.

For three coincident lines, we have m1 = m

2 = m

3,

so, m = –c

d, m2 =

b

d and m3 =

–a

d

So,

3– –c a

d d

⎛ ⎞ ⎜ ⎟⎝ ⎠

c3 = ad2

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77.

BR

P

X

AO

Y

Equation of tangent at P(a cos, b sin) is

cos sin 1x y

a b

It meets the major axis at A(a sec, 0) and the

minor axis at B(0, b cosec).

Equation of the OR, perpendicular to the tangent at

P is ax sin – by cos = 0.

Now PR = 2 2 2 2

( cos )sin – ( sin )cos

sin cos

a a b b

a b

=

2 2

2 2 2 2

( – ) sin cos

sin cos

a b

a b

Also, AB =

2 2 2 2sin cos

sin cos

a b

So, (AB) (PR) = a2 – b2

Here, a = 4, b = 3

(AB)(PR) = 16 – 9 = 7

78.

Pt

t(2 , 4 )

11

2

Q t t(2 , 4 )2 2

2

Let absciss of point P is 2

12 18t 2

19t

t1 = 3

We know that, If PQ is normal then 2 1

1

2– –t t

t

2

11–

3t

So, coordinates of P and Q are (18, 12) and

242 –44,

9 3

⎛ ⎞⎜ ⎟⎝ ⎠

respectively.

PQ =

2 2242 –44

– 18 – 129 3

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 80 10

9

79.

2 2

22

15 ( – 1) 2– 22

x y

aa a

2 2

22

13 ( – 1) 2( – 1)2

x y

aa

where, A2 = (a – 1)2 + 3

2, B2 = (a – 1)2 + 2

We know that,

A2 = B2 (1 – e2), (for vertical ellipse)

2 3( – 1)

2a = 2 1

( – 1) 2 1–4

a⎛ ⎞ ⎜ ⎟⎝ ⎠

4(a – 1)2 + 6 = 3(a – 1)2 + 6

(a – 1)2 = 0 a = 1

Equation of ellipse is

2 2

13 2

2

x y

So, length of L.R. =

22A

B =

32

2

2

=

9

2

80. Clearly ax + by = 1

i.e.,– 1a

y xb b

is tangent to cx2 + dy2 = 1

2 2

11 1

x y

c d

2 2

1 1 – 1a

b c b d

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1 =

2 2a b

c d

81. Putting y = 2t2 in the equation of the given ellipse

2 2

1,9 4

x y we get

2 44

19 4

x t

x2 = 9(1 – t4) = 9(1 – t2) (1 + t2)

This will give real values of x if 1 – t2 0, i.e., |t| 1.

82. The circle and the ellipse do not meet at any point.

Length of the common chord = 0.

83. If S1 = 0 and S

2 = 0 are the equations, then

S1+ S

2= 0 is a second degree curve passing through

the points of intersection of S1 = 0 and S

2 = 0.

( + 4)x2 + 2( + 1)y2 – 2(3 + 10)x

– 12( + 1)y + (23 + 35) = 0 ...(i)

For it to be a circle, choose such that the coefficient

of x2 and y2 are equal

+ 4 = 2 + 2

= 2

Test - 4 (Code-H) (Hints) All India Aakash Test Series for JEE (Main)-2018

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� � �

This gives the equation of the circle as

6(x2 + y2) – 32x – 36y + 81 = 0 {(using (i))}

2 2 16 27– – 6 0

3 2x y x y

Its centre is 8, 3

3C⎛ ⎞⎜ ⎟⎝ ⎠

and radius is

r = 64 27

9 –9 2

= 1 47

3 2

84. Centre and radius of circle x2 + y2 + 8x – 10y – 40 = 0

are C(–4, 5) and 9 respectively. Distance of the centre

C(–4, 5) from the point P(–2, 3) is CP = 4 4 =

2 2 .

a = CP + r = 2 2 + 9

b = |CP – r| = 9 – 2 2

a – b = 4 2

85. The two circles are x2 + y2 – 4x – 6y – 3 = 0 and

x2 + y2 + 2x + 2y + 1 = 0

Centres : C1(2, 3) and C

2(–1, –1)

radii : r1 = 4 and r

2 = 1.

We have, C1C

2 = 5 = r

1 + r

2, circles touch externally

therefore there are three common tangents to the

given circles.

86. The point 6

P⎛ ⎞

⎜ ⎟⎝ ⎠

is sec , tan6 6

a b ⎛ ⎞

⎜ ⎟⎝ ⎠

or

2,

3 3

a bP⎛ ⎞⎜ ⎟⎝ ⎠

.

Equation of tangent at P is – 13 3

2

x y

a b

Area of the triangle = 1 3

32 2

ab = 3a2

b

a = 4

e2 =

2

21 17

b

a

87. Make homogeneous 3x2 – y2 –2x + 4y = 0 with the

line y = mx + c

i.e., 2 2 – –

3 – – 2 4y mx y mx

x y x yc c

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 0

According to question coeff. of x2 + coeff. of y2 = 0

2 43 –1 0

m

c c

m + c + 2 = 0

which gives fixed point (1, –2) p + q = –1

88. Equation of hyperbola is

2 2

– 1.9 16

x y

Equation of tangent is

29 – 16y mx m

29 – 16 2 5m m = ±2

a + b = sum of roots = 0

89. Let P(h, k) be the point of intersection of E1 and E

2

2

2

21

hk

a h2 = a2(1 – k2) ...(i)

and

2 2

21

h k

a = 1 k2 = a2 (1 – h2) ...(ii)

Eliminating a from equation (i) and (ii), we get

2 2

2 21– 1–

h k

k h

h2(1 – h2) = k2 (1 – k2)

(h – k) (h + k) (h2 + k2 – 1) = 0

Hence, the locus is a set of curves consisting of the

straight lines y = x, y = –x and circle x2 + y2 = 1.

90. If is angle between tangents, then

sin2

⎛ ⎞⎜ ⎟⎝ ⎠

= radius

distance between 3 3, 3 and (0, 0)

sin2

⎛ ⎞⎜ ⎟⎝ ⎠

= 3

6 =

1

2

2

=

6

=

3

So, cot = cot3

⎛ ⎞⎜ ⎟⎝ ⎠

= 1

3