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Test - 7 (Code-G) (Answers) All India Aakash Test Series for JEE (Main)-2018

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PHYSICS CHEMISTRY MATHEMATICS

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Test Date : 25/02/2018

ANSWERS

TEST - 7 (Code-G)

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

All India Aakash Test Series for JEE (Main)-2018 Test - 7 (Code-G) (Hints)

2/8

1. ∵ Component of incident ray along normal is

ˆ ˆ( )i j .

to express reflected ray, ˆ ˆ( )i j reverses in

expression of incident ray.

ˆ ˆ ˆ– – –R i j k

�

= ˆ ˆ ˆ–( )i j k

ˆ ˆ ˆ–( )ˆ3

i j kR⎧ ⎫ ⎨ ⎬⎩ ⎭

2.

h = 2 m

i 90° O x

y

If is the angle of refraction for the horizontal rays.

(1 +

)

ay

1/2

1

ay

0

sin90

sin

sin = 1

1 ay, tan =

1

ay

1dx

dy ay

1 –12–2 2

0 0

x

dx a y dy∫ ∫

x =

2–1 1

2 20

2

⎡ ⎤⎢ ⎥⎣ ⎦a y =

1–1

–6 222 2 10 2 = 2000 m

4. The parallel rays will focus at focal point of concave

lens so after refraction from it, they become parallel.

5. When lenses are in contact

1 2

1 1 1

12f f ...(i)

When lenses are at separation

1 2 1 2

1 1 50– 0

f f f f ...(ii)

On solving equations (i) and (ii), we get

f1 = 20 cm, f

2 = 30 cm

PART - A (PHYSICS)

7. The focal length of the glass (lens) used 100

m,1.5

f

if y is the distance of near point, then

1 1 1.5––25 100y

{y = –40 cm}

The focal length of lens of microscope

fe =

1 100

20P = 5 cm

Magnifying power of microscope together with glass

M = 1

e

D

f =

251

5 = 6

In case of without glass, D = 40 cm = y

M = 1

e

D

f

= 40

15

= 9

8.

O

d

d

S

S

y

∵ Due to reflection. A phase difference of occurs.

Let the distance of first maxima from O is y

2

2 2

D

dy

⎛ ⎞⎜ ⎟ ⎝ ⎠

4

Dy

d

where, = fringe width.

9.

2r1

r2

e

i + e – A = 2, e =

1

sin2sinr

, 1

sinsin( – )r

on solving we get,

2–1

cos8

10. n11 = n

22

∵ n

11 = n

22

12 × 600 = n2 × 400

{n2 = 18}

Test - 7 (Code-G) (Hints) All India Aakash Test Series for JEE (Main)-2018

3/8

11.

r1

r2

i2

i1

sin

sin

i

r

⎡ ⎤ ⎢ ⎥⎣ ⎦

= (i1 – r

1) + (i

2 – r

2)

For minimum deviation, i1 = i

2, r1 = r

2

sin60

sin30

= , = 3

12.

x

ir

y

∵ i + r = 90° and i = r

i = r = 45°, i + = 90°

= 90° – i = 90° – 45° = 45°

∵ Given equation of surface, y2 = 2x

2y

dy

dx = 2

1dy

dx y = tan45°

y = 1

x =

21

2 2

y

Coordinates are 1,1

2

⎛ ⎞⎜ ⎟⎝ ⎠

13. ∵ A1 = 4°,

1 = 1.54,

2 = 1.72

Condition for no deviation

A2 =

1

1

2

( – 1)–( – 1)

A

1.54 – 1

– 41.72 – 1

⎛ ⎞ ⎜ ⎟⎝ ⎠

A2 = –3° |A

2| = 3°

15.

O

36 c

m

24 c

m

x

C

For lens:

∵ v = –24 cm, u = –x

1 1 1–24 x f

, 1 1 1–24 40x

x = 15 cm

17. 3 cos = 2

cos = 2

3

3

tanX

D

X = 5

2D

18. According to diagram given in question

OP = d sec, CO = OP cos2 = d sec cos2 Path difference, x = CO + OP

x = d sec + d sec cos2Effective path difference

xeff

= d(sec + seccos2) + 2

For constructive interference, xeff

=

d sec (1 + cos2) + 2

=

or cos4d

19. For A:

2.

2 4

⎡ ⎤ ⎢ ⎥ ⎣ ⎦IA = I + I + 2I cos = 0

For B:

–4

x PB QB⎛ ⎞ ⎜ ⎟

⎝ ⎠ = (PB – QB) +

4

= 04 4

2

IB = I + I + 2I cos

2

= 2I

For C:

2– 0

2 4

⎛ ⎞ ⎜ ⎟ ⎝ ⎠IC = I + I + 2I cos0° = 4I

20. ∵ ,

h

mv so with the increase in velocity of

electron, wavelength decreases and so fringe width

decreases.

22. I1 = 4I

0, I2 = I

0, Imax

= 9I0, Imin

= I0

23. The deviation produced by the prism

= ( – 1)A = (1.5 – 1) × 4° = 2°

Thus the angle through which mirror should be rotated

is 1°.


4/8

PART - B (CHEMISTRY)

32. LiAlH4 reduce ester group and H

2/Ni reduce C C

also.

33.

+ CH —CH=CH3 2

H+

CH

CH3

CH3

(A)

O2

CH —C—O—OH3

CH3

(B)

H+

/H O2

CH —C—CH3 3

+

O

OH

(C)

24. If P is equivalent power of the combination, then

2 1

15

P ,

1–

m

Pf

By using mirror equation

1 1 1

–20 –7.5v

– 12 cmv

25. For first lens,

1

1 1 1––30 10v

v1 = 15 cm at I

1

For second lens, u2 = 15 – 5 = 10 cm

5 cm 10 cm 30 cm

I1

I

O

The focal length of second lens is also 10 cm, thus

it forms the image at infinity.

Finally the third lens forms the image at its focal point

i.e., 30 cm from it.

26. Image formed by lens should be on the pole of mirror.

28. ∵ Fringe width

D

d

and new fringe width

= (1.1 )

(1.1 )

D D

d d

= 0.2 mm

29.

d = 10 cm

I1 I

3

I2

For refraction through concave lens:

1 1 1–

v u f

, 1 1 –1–– 10v

v = –10 cm

I2 is 20 cm behind the plane mirror.

1 1 –1––(30) 10v

1 –1 1 –4–

10 30 30v

= –7.5 cm

I3 is formed at 2.5 cm in front of mirror.

30. I =2 20

cos cos (90 – )2

I⎡ ⎤ ⎢ ⎥⎣ ⎦

I = 2 20

cos sin2

I =

20sin 2

8

I

34.CH MgBr

3

OMgBr+

CH3

H O2

OHO

41. CH —CH—CH—CH3 3

CH3

OH

H+

CH —CH—CH—CH3 3

CH3

(+)

CH —C—CH CH3 2 3

CH3

+

Rearrange

CH —C—CHCH3 3

CH3

–H O2

42. CH3—CH

2Br

LiAlH4 CH3—CH

3 + LiAlH

3Br

47. The structure that can form the more stable

carbocation will undergoes SN1 faster.


5/8

PART - C (MATHEMATICS)

61. I = cos8 cos7

1 2cos5

x x

x

∫ dx

= 2

152sin sin

2 2

53 4sin

2

x x

x

∫ dx

=

3

2

5 52 3sin 4sin sin

2 2 2

53 4sin

2

x x x

x

⎛ ⎞ ⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

∫ dx

= 5

2sin sin2 2

x xdx ∫

= cos3 cos2x x dx∫

= 1 1sin3 sin2

3 2x x c

62. Let 1

sec tx

63. I = 3 5

4 5

x x

x x

e edx

e e

∫ =

2

2

3 5

4 5

x

x

edx

e

∫

=

2

2

3 154 5 5

4 4

4 5

x

x

e

dxe

∫

49.

Cl

, Cl,

Cl

,

Cl

,

Cl

,

Cl

51. X

Br /h2

Racemic mixture Presence of

alkyl part⇒

Br /Fe2 Product Presence of benzene ring⇒

Br , CCl2 4 No reaction No C=C is present in

side chain⇒

CH CH2 3

X =

53.

O

CH CH2 3

HBr

OH

+ C H Br2 5

54.

OH

SOCl2

S iN

Cl

(Retention ofconfiguration)

58. NH2

OH HNO2

N2

OH

+

–N2

OH

+

ring

expansion

:OH

+

–H+

O

= 23 35 1

ln 4 54 4 10

xx e c

= 23 7 7

ln(4 5)4 8 4

xx e x c

= –x + 7

8 ln| 4e

2x – 5 | + c

64. Given equation is

2

1 3dy

dx

⎛ ⎞⎜ ⎟⎝ ⎠ =

33

3

34

d y

dx

⎛ ⎞ ⎜ ⎟⎝ ⎠

Degree = 3

65. Required area =

12 2

0

1212

xx dx

⎛ ⎞⎜ ⎟⎝ ⎠∫

= 48 sq. units

66. The tangent at p(x, y) is : Y – y = (X – x)dy

dx

It meet x-axis at A = , 0

dxx y

dy

⎛ ⎞⎜ ⎟⎝ ⎠

∵ OP2 = PA

2 x2 + y2 = y2

2dx

dy

⎛ ⎞⎜ ⎟⎝ ⎠

+ y2

dy

xdx

= ±y

xy = c2 or y = cx, where c is constant


6/8

67.

( 5, 0)

2(1,0)1

( 5 )

0,

Required area

=

2

2

1

1 1(5 ) 2 2 1 1

2 2x dx

⎧ ⎫⎪ ⎪ ⎨ ⎬⎪ ⎪⎩ ⎭∫ sq. units

=

2

2 1

1

5 55 sin

2 2 25

x xx

⎡ ⎤ ⎢ ⎥⎣ ⎦

= 5 5 5 2

24 2 4

⎛ ⎞ ⎜ ⎟⎝ ⎠ sq. units

68. I =

ln2 1

y

x

dx

e ∫

Let ex –1 = t2

exdx = 2tdt

I =

1

2

1

2

( 1)

ye

tdt

t t

∫

= 11

12 tan

ye

t

= 1

2 tan 14

ye

⎛ ⎞ ⎜ ⎟⎝ ⎠

12tan 1

ye

= 6 2

1tan 1

ye

= 3

ey = 3 + 1

y = ln4

69. I = 21

dx

x x x ∫

Let x = sin2 dx = 2sin cos d

I = 21 sin

d ∫

= 22(sec sec tan )d ∫

= 2(tan – sec) + c

= 2 1

1

xc

x

70. Let ln x = t dx = etdt

I =

2

2

1

1

tte dt

t

⎛ ⎞ ⎜ ⎟⎝ ⎠∫ = 2 2 2

1 2

1 (1 )

tdt

t t

⎛ ⎞⎜ ⎟⎝ ⎠ ∫

= 2 21 1 (ln )

te x

c ct x

71. I = 3

tan 2 sec2x xdx∫Let sec2x = t

tan 2x sec2xdx = 1

2dt

I = 21

tan 22

xdt∫

= 211

2t dt∫

= 31 1

sec 2 sec26 2

x x c

72. Let t = tanx

I = 2

2 2 2

2 4 2 4

0

( )

(1 )( )

∫a t b dt

t a t b

= 2 2

2

a b

2 2

2 4 2 4

0

1

1

a bdt

t a t b

⎛ ⎞⎜ ⎟

⎝ ⎠∫

=

2

1 1

2 2 20

2tan tan

a tt

a b b

⎛ ⎞

⎜ ⎟⎝ ⎠= 2 2

2

a b

73. 2 2 2 2cos

x xx y d x y d

y y

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

2 2sin

xx y C

y

⎛ ⎞ ⎜ ⎟⎝ ⎠

74. I =

2017

1

sin( ln )e

xdx

x

∫

Let lnx = t dx

x

= dt

I =

2017

0

sin 1 cos(2017 )tdt

∫

= 2 = even integer

75. |cos x| is periodic function of period and

0

| cos | 2x

∫


7/8

76. ∵ I1 =

1 1

0

tan xdx

x

∫

Let x = tan dx = sec2d

I1

= 24 4

0 0

sec 2

tan sin2

d d

∫ ∫

Now let 2 = t d = 1

2dt

I1 =

2

0

1

2 sin

tdt

t

∫

I1 = 2

1

2I

1

2

1

2

I

I

77. 4(x2 – 2x) + y2 + 4y – 4 = 0

2( 1)

3

x +

2( 2)

12

y = 1

This is an ellipse and its area = 3 12 = 6

78. ∵

4

( ) sin cos 24

f x dx

∫

On differentiating both sides w.r.t we get

f() = sin + cos – 4

sin + 2

1 1 1

24 4 42 2 2

f ⎛ ⎞ ⎜ ⎟⎝ ⎠

= 1

22

= 3

2=3 2

2

80. The equation of circle is

x2 + y2 –2hx = 0 ..... (i)

where h R– is a parameter

On differentiating : x + y dy

dx= h ..... (ii)

On eliminating h from equation (i) and (ii) we get

2dy

x x ydx

⎛ ⎞⎜ ⎟⎝ ⎠ = x2 + y2

x2 – y2 = – 2xy

dy

dx

81. f (x) = 6(x – 1) f (x) = 3(x – 1)2 + a

The slope of tangent at (2, 1) is 3

f (2) = 3 a = 0

f (x) = 3(x – 1)2

f(x) = (x – 1)3 + b

But f(2) = 1 b = 0

f(x) = (x – 1)3

82.2 2

lnxdy ydx xc

x x

2

lny xdxd

x x

⎛ ⎞ ⎜ ⎟⎝ ⎠

1 lny x

x x x

f (x) = 2

1

x

83. A.T.Q: dx

y x ady

dy

y=

dx

a x lny = lnc – ln|(x – a)|

y = ,| |

c

x a where x = 2a, y = a c = a2

y =

2

| |a

x a

84. ∵ y = cx2

dy

dx=

2y

x

Now replace dy

dx by

dx

dy we get

2ydy + xdx = 0

y2 +

2

2

x

= d1, i.e., x2 + 2y

2 = d

86. ∵

1

1lim

⎛ ⎞⎜ ⎟⎝ ⎠∑

n

nr

rf

n n =

1

0

( )f x dx∫

=

1

0

| sin2 |x dx∫

=

1

0

cos2

2

x⎡ ⎤⎢ ⎥⎣ ⎦

= 1 cos2

2


8/8

� � �

87. f(x) = ex

0

( )

x

te f t dt ∫ (x

2 – x + 1)ex

Again on differentiating w.r.t. x

f (x) = ex(2x + 1)

f(x) = (2x + 1) ex – 2

xe dx∫

f(x) = (2x – 1)ex

1

2f⎛ ⎞⎜ ⎟⎝ ⎠ = 0

88. ∵ f(x) = 3

4

cos( )

x

t dt∫

f (x) = cos(x3)

f (4) = cos 64 = 1

f(4) =

4

3

4

cos( ) 0t dt ∫

Equation of tangents : y – 0 = 1 (x – 4)

x – y = 4

89. I = 8 7( 2)

dx

x x ∫

= 7 8

7

21

dx

x xx

⎛ ⎞ ⎜ ⎟⎝ ⎠

∫

Let 7

21

x

= t

8

dx

x =

14

dt

I = 1

28

11 dt

t

⎛ ⎞⎜ ⎟⎝ ⎠∫

= 1

287 7

2 2ln 1 1

x x

⎛ ⎞ ⎜ ⎟⎝ ⎠ + c

= 1

287 7

2 2ln 1

x x

⎛ ⎞ ⎜ ⎟⎝ ⎠ + c

90. I =

534 ( 1) 2

dx

x x ∫

= 3

42 1( 2)

2

dx

xx

x

⎛ ⎞ ⎜ ⎟⎝ ⎠

∫

Let 1

2

x

x

= t4

2( 2)

dx

x =

4

3t3dt

I =

3

3

4 4

4

3( )

t dt

t

∫ =4

3t + c

Test - 7 (Code-H) (Answers) All India Aakash Test Series for JEE (Main)-2018

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27. (2)

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29. (2)

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PHYSICS CHEMISTRY MATHEMATICS

31. (3)

32. (1)

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61. (4)

62. (1)

63. (3)

64. (2)

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66. (4)

67. (4)

68. (1)

69. (1)

70. (2)

71. (2)

72. (2)

73. (3)

74. (3)

75. (3)

76. (3)

77. (3)

78. (2)

79. (4)

80. (1)

81. (1)

82. (4)

83. (2)

84. (3)

85. (4)

86. (3)

87. (4)

88. (4)

89. (3)

90. (1)

Test Date : 25/02/2018

ANSWERS

TEST - 7 (Code-H)

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

All India Aakash Test Series for JEE (Main)-2018 Test - 7 (Code-H) (Hints)

2/8

1. I =2 20

cos cos (90 – )2

I⎡ ⎤ ⎢ ⎥⎣ ⎦

I = 2 20

cos sin2

I =

20sin 2

8

I

2.

d = 10 cm

I1 I

3

I2

For refraction through concave lens:

1 1 1–

v u f

, 1 1 –1–– 10v

v = –10 cm

I2 is 20 cm behind the plane mirror.

1 1 –1––(30) 10v

1 –1 1 –4–

10 30 30v

= –7.5 cm

I3 is formed at 2.5 cm in front of mirror.

3. ∵ Fringe width

D

d

and new fringe width

= (1.1 )

(1.1 )

D D

d d

= 0.2 mm

5. Image formed by lens should be on the pole of mirror.

6. For first lens,

1

1 1 1––30 10v

v1 = 15 cm at I

1

For second lens, u2 = 15 – 5 = 10 cm

5 cm 10 cm 30 cm

I1

I

O

PART - A (PHYSICS)

The focal length of second lens is also 10 cm, thus

it forms the image at infinity.

Finally the third lens forms the image at its focal point

i.e., 30 cm from it.

7. If P is equivalent power of the combination, then

2 1

15

P ,

1–

m

Pf

By using mirror equation

1 1 1

–20 –7.5v

– 12 cmv

8. The deviation produced by the prism

= ( – 1)A = (1.5 – 1) × 4° = 2°

Thus the angle through which mirror should be rotated

is 1°.

9. I1 = 4I

0, I2 = I

0, Imax

= 9I0, Imin

= I0

11. ∵ ,

h

mv so with the increase in velocity of

electron, wavelength decreases and so fringe width

decreases.

12. For A:

2.

2 4

⎡ ⎤ ⎢ ⎥ ⎣ ⎦IA = I + I + 2I cos = 0

For B:

–4

x PB QB⎛ ⎞ ⎜ ⎟

⎝ ⎠ = (PB – QB) +

4

= 04 4

2

IB = I + I + 2I cos

2

= 2I

For C:

2– 0

2 4

⎛ ⎞ ⎜ ⎟ ⎝ ⎠IC = I + I + 2I cos0° = 4I

13. According to diagram given in question

OP = d sec, CO = OP cos2 = d sec cos2 Path difference, x = CO + OP

x = d sec + d sec cos2Effective path difference

xeff

= d(sec + seccos2) + 2

For constructive interference, xeff

=

d sec (1 + cos2) + 2

=

or cos4d

Test - 7 (Code-H) (Hints) All India Aakash Test Series for JEE (Main)-2018

3/8

14. 3 cos = 2

cos = 2

3

3

tanX

D

X = 5

2D

16.

O36 c

m

24 c

m

x

C

For lens:

∵ v = –24 cm, u = –x

1 1 1–24 x f

, 1 1 1–24 40x

x = 15 cm

18. ∵ A1 = 4°,

1 = 1.54,

2 = 1.72

Condition for no deviation

A2 =

1

1

2

( – 1)–( – 1)

A

1.54 – 1

– 41.72 – 1

⎛ ⎞ ⎜ ⎟⎝ ⎠

A2 = –3° |A

2| = 3°

19.

x

ir

y

∵ i + r = 90° and i = r

i = r = 45°, i + = 90°

= 90° – i = 90° – 45° = 45°

∵ Given equation of surface, y2 = 2x

2ydy

dx = 2

1dy

dx y = tan45°

y = 1

x =

21

2 2

y

Coordinates are 1,1

2

⎛ ⎞⎜ ⎟⎝ ⎠

20.

r1

r2

i2

i1

sin

sin

i

r

⎡ ⎤ ⎢ ⎥⎣ ⎦

= (i1 – r

1) + (i

2 – r

2)

For minimum deviation, i1 = i

2, r1 = r

2

sin60

sin30

= , = 3

21. n11 = n

22

∵ n

11 = n

22

12 × 600 = n2 × 400

{n2 = 18}

22.

2r1

r2

e

i + e – A = 2, e =

1

sin2sinr

, 1

sinsin( – )r

on solving we get,

2–1

cos8

23.

O

d

d

S

S

y

∵ Due to reflection. A phase difference of occurs.

Let the distance of first maxima from O is y

2

2 2

D

dy

⎛ ⎞⎜ ⎟ ⎝ ⎠

4

Dy

d

where, = fringe width.


4/8

PART - B (CHEMISTRY)

33. NH2

OH HNO2

N2

OH

+

–N2

OH

+

ring

expansion

:OH

+

–H+

O

24. The focal length of the glass (lens) used 100

m,1.5

f

if y is the distance of near point, then

1 1 1.5––25 100y

{y = –40 cm}

The focal length of lens of microscope

fe =

1 100

20P = 5 cm

Magnifying power of microscope together with glass

M = 1

e

D

f =

251

5 = 6

In case of without glass, D = 40 cm = y

M = 1

e

D

f

= 40

15

= 9

26. When lenses are in contact

1 2

1 1 1

12f f ...(i)

When lenses are at separation

1 2 1 2

1 1 50– 0

f f f f ...(ii)

On solving equations (i) and (ii), we get

f1 = 20 cm, f

2 = 30 cm

27. The parallel rays will focus at focal point of concave

lens so after refraction from it, they become parallel.

29.

h = 2 m

i 90° O x

y

If is the angle of refraction for the horizontal rays.

(1 +

)

ay

1/2

1

ay

0

sin90

sin

sin = 1

1 ay, tan =

1

ay

1dx

dy ay

1 –12–2 2

0 0

x

dx a y dy∫ ∫

x =

2–1 1

2 20

2

⎡ ⎤⎢ ⎥⎣ ⎦a y =

1–1

–6 222 2 10 2 = 2000 m

30. ∵ Component of incident ray along normal is

ˆ ˆ( )i j .

to express reflected ray, ˆ ˆ( )i j reverses in

expression of incident ray.

ˆ ˆ ˆ– – –R i j k

�

= ˆ ˆ ˆ–( )i j k

ˆ ˆ ˆ–( )ˆ3

i j kR⎧ ⎫ ⎨ ⎬⎩ ⎭

37.

OH

SOCl2

S iN

Cl

(Retention ofconfiguration)

38.

O

CH CH2 3

HBr

OH

+ C H Br2 5

40. X

Br /h2

Racemic mixture Presence of

alkyl part⇒

Br /Fe2 Product Presence of benzene ring⇒

Br , CCl2 4 No reaction No C=C is present in

side chain⇒

CH CH2 3

X =


5/8

PART - C (MATHEMATICS)

61. I =

534 ( 1) 2

dx

x x ∫

= 3

42 1( 2)

2

dx

xx

x

⎛ ⎞ ⎜ ⎟⎝ ⎠

∫

Let 1

2

x

x

= t4

2( 2)

dx

x =

4

3t3dt

I =

3

3

4 4

4

3( )

t dt

t

∫ =4

3t + c

62. I = 8 7( 2)

dx

x x ∫

= 7 8

7

21

dx

x xx

⎛ ⎞ ⎜ ⎟⎝ ⎠

∫

42.

Cl

, Cl,

Cl

,

Cl

,

Cl

,

Cl

44. The structure that can form the more stable

carbocation will undergoes SN1 faster.

49. CH3—CH

2Br

LiAlH4 CH3—CH

3 + LiAlH

3Br

50. CH —CH—CH—CH3 3

CH3

OH

H+

CH —CH—CH—CH3 3

CH3

(+)

CH —C—CH CH3 2 3

CH3

+

Rearrange

CH —C—CHCH3 3

CH3

–H O2

57.CH MgBr

3

OMgBr+

CH3

H O2

OHO

58.

+ CH —CH=CH3 2

H+

CH

CH3

CH3

(A)

O2

CH —C—O—OH3

CH3

(B)

H+

/H O2

CH —C—CH3 3

+

O

OH

(C)

59. LiAlH4 reduce ester group and H

2/Ni reduce C C

also.

Let 7

21

x

= t

8

dx

x =

14

dt

I = 1

28

11 dt

t

⎛ ⎞⎜ ⎟⎝ ⎠∫

= 1

287 7

2 2ln 1 1

x x

⎛ ⎞ ⎜ ⎟⎝ ⎠

+ c

= 1

287 7

2 2ln 1

x x

⎛ ⎞ ⎜ ⎟⎝ ⎠

+ c

63. ∵ f(x) = 3

4

cos( )

x

t dt∫

f (x) = cos(x3)

f (4) = cos 64 = 1

f(4) =

4

3

4

cos( ) 0t dt ∫

Equation of tangents : y – 0 = 1 (x – 4)

x – y = 4


6/8

64. f(x) = ex

0

( )

x

te f t dt ∫ (x2 – x + 1)ex

Again on differentiating w.r.t. x

f (x) = ex(2x + 1)

f(x) = (2x + 1) ex – 2x

e dx∫f(x) = (2x – 1)ex

1

2f⎛ ⎞⎜ ⎟⎝ ⎠ = 0

65. ∵

1

1lim

⎛ ⎞⎜ ⎟⎝ ⎠∑

n

nr

rf

n n =

1

0

( )f x dx∫

=

1

0

| sin2 |x dx∫ =

1

0

cos2

2

x⎡ ⎤⎢ ⎥⎣ ⎦ =

1 cos2

2

67. ∵ y = cx2 dy

dx=

2y

x

Now replace dy

dx by

dx

dy we get

2ydy + xdx = 0

y2 +

2

2

x

= d1, i.e., x2 + 2y2 = d

68. A.T.Q: dx

y x ady

dy

y=

dx

a x

lny = lnc – ln|(x – a)|

y = ,| |

c

x a where x = 2a, y = a c = a2

y =

2

| |a

x a

69.2 2

lnxdy ydx xc

x x

2

lny xdxd

x x

⎛ ⎞ ⎜ ⎟⎝ ⎠

1 lny x

x x x

f (x) = 2

1

x

70. f (x) = 6(x – 1) f (x) = 3(x – 1)2 + a

The slope of tangent at (2, 1) is 3

f (2) = 3 a = 0

f (x) = 3(x – 1)2

f(x) = (x – 1)3 + b

But f(2) = 1 b = 0

f(x) = (x – 1)3

71. The equation of circle is

x2 + y2 –2hx = 0 ..... (i)

where h R– is a parameter

On differentiating : x + y dy

dx= h ..... (ii)

On eliminating h from equation (i) and (ii) we get

2dy

x x ydx

⎛ ⎞⎜ ⎟⎝ ⎠ = x2 + y2

x2 – y2 = – 2xydy

dx

73. ∵

4

( ) sin cos 24

f x dx

∫

On differentiating both sides w.r.t we get

f() = sin + cos – 4

sin + 2

1 1 1

24 4 42 2 2

f ⎛ ⎞ ⎜ ⎟⎝ ⎠

= 1

22

= 3

2=3 2

2

74. 4(x2 – 2x) + y2 + 4y – 4 = 0

2( 1)

3

x +

2( 2)

12

y = 1

This is an ellipse and its area = 3 12 = 6

75. ∵ I1 =

1 1

0

tan xdx

x

∫

Let x = tan dx = sec2d

I1

= 24 4

0 0

sec 2

tan sin2

d d

∫ ∫

Now let 2 = t d = 1

2dt


7/8

I1 =

2

0

1

2 sin

tdt

t

∫

I1 = 2

1

2I

1

2

1

2

I

I

76. |cos x| is periodic function of period and

0

| cos | 2x

∫

77. I =

2017

1

sin( ln )e

xdx

x

∫

Let lnx = t dx

x

= dt

I =

2017

0

sin 1 cos(2017 )tdt

∫

= 2 = even integer

78. 2 2 2 2cos

x xx y d x y d

y y

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

2 2sin

xx y C

y

⎛ ⎞ ⎜ ⎟⎝ ⎠

79. Let t = tanx

I = 2

2 2 2

2 4 2 4

0

( )

(1 )( )

∫a t b dt

t a t b

= 2 2

2

a b

2 2

2 4 2 4

0

1

1

a bdt

t a t b

⎛ ⎞⎜ ⎟

⎝ ⎠∫

=

2

1 1

2 2 20

2tan tan

a tt

a b b

⎛ ⎞

⎜ ⎟⎝ ⎠= 2 2

2

a b

80. I = 3

tan 2 sec2x xdx∫Let sec2x = t

tan 2x sec2xdx = 1

2dt

I = 21

tan 22

xdt∫

= 211

2t dt∫

= 31 1

sec 2 sec26 2

x x c

81. Let ln x = t dx = etdt

I =

2

2

1

1

tte dt

t

⎛ ⎞ ⎜ ⎟⎝ ⎠∫ = 2 2 2

1 2

1 (1 )

tdt

t t

⎛ ⎞⎜ ⎟⎝ ⎠ ∫

= 2 21 1 (ln )

te x

c ct x

82. I = 21

dx

x x x ∫

Let x = sin2 dx = 2sin cos d

I = 21 sin

d ∫

= 22(sec sec tan )d ∫

= 2(tan – sec) + c

= 2 1

1

xc

x

83. I =

ln2 1

y

x

dx

e ∫

Let ex –1 = t2

exdx = 2tdt

I =

1

2

1

2

( 1)

ye

tdt

t t

∫

= 11

12 tan

ye

t

= 1

2 tan 14

ye

⎛ ⎞ ⎜ ⎟⎝ ⎠

12tan 1

ye

= 6 2

1tan 1

ye

= 3

ey = 3 + 1

y = ln4

84.

( 5, 0)

2(1,0)1

( 5 )

0,

Required area

=

2

2

1

1 1(5 ) 2 2 1 1

2 2x dx

⎧ ⎫⎪ ⎪ ⎨ ⎬⎪ ⎪⎩ ⎭∫ sq. units


8/8

� � �

=

2

2 1

1

5 55 sin

2 2 25

x xx

⎡ ⎤ ⎢ ⎥⎣ ⎦

= 5 5 5 2

24 2 4

⎛ ⎞ ⎜ ⎟⎝ ⎠ sq. units

85. The tangent at p(x, y) is : Y – y = (X – x)dy

dx

It meet x-axis at A = , 0

dxx y

dy

⎛ ⎞⎜ ⎟⎝ ⎠

∵ OP2 = PA2 x2 + y2 = y2

2dx

dy

⎛ ⎞⎜ ⎟⎝ ⎠

+ y2

dy

xdx

= ±y

xy = c2 or y = cx, where c is constant

86. Required area =

12 2

0

1212

xx dx

⎛ ⎞⎜ ⎟⎝ ⎠∫

= 48 sq. units

87. Given equation is

2

1 3dy

dx

⎛ ⎞⎜ ⎟⎝ ⎠ =

33

3

34

d y

dx

⎛ ⎞ ⎜ ⎟⎝ ⎠

Degree = 3

88. I = 3 5

4 5

x x

x x

e edx

e e

∫

=

2

2

3 5

4 5

x

x

edx

e

∫

=

2

2

3 154 5 5

4 4

4 5

x

x

e

dxe

∫

= 23 35 1

ln 4 54 4 10

xx e c

= 23 7 7

ln(4 5)4 8 4

xx e x c

= –x + 7

8 ln| 4e2x – 5 | + c

89. Let 1

sec tx

90. I = cos8 cos7

1 2cos5

x x

x

∫ dx

= 2

152sin sin

2 2

53 4sin

2

x x

x

∫ dx

=

3

2

5 52 3sin 4sin sin

2 2 2

53 4sin

2

x x x

x

⎛ ⎞ ⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

∫ dx

= 5

2sin sin2 2

x xdx ∫

= cos3 cos2x x dx∫

= 1 1sin3 sin2

3 2x x c

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