KERALA ENGINEERING / MEDICAL ENTRANCE EXAMINATIONS

KEAM 2014

Solutions

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SOLUTIONS & ANSWERS FOR KERALA ENGINEERING ENTRANCE EXAMINATION - 2014 – PAPER I I

VERSION – B1

[MATHEMATICS] 1. Ans: 185 Sol: 2 ⊕ 3 = 4 + 9 = 13 (2 ⊕ 3) ⊕ 4 = 13 ⊕ 4 = 132 + 42 = 169 + 16 = 185. 2. Ans: 520 Sol:

30 = ( )30x10010 +

⇒ x = 270 and

30 = ( )y3010012 +

⇒ y = 220 x + y + 30 = 270 + 220 + 30 = 520. 3. Ans: f is not a 1 − 1function Sol: f(x) = |x − 2| f(1) = |1 − 2| = 1 f(3) = |3 − 2| = 1 Not one-one. 4. Ans: 1680

Sol: 8P4 = !4!8

5. Ans: No answer Sol: No correct option is given. Options (A) and

(D) are same. If one of them were [1, ∞), that would have been the correct option.

6. Ans:

Sol: (fοg) (x) = 3x2 −

If the question were domain of (fοg) (x) is, then (E) would be the correct option.

.

7. Ans: 2

Sol: ( ) ( )

( )223

i68

4i3i3

+++

( )( )

36i966416i249i33i933

−+++−−−+=

= ( )( )

22 9628

i9628192i56i9628

192i56

+−−=

+−

= 9216784

i1843253765376i1568+

+−+

= i210000

i20000 =

∴ |z| = 222 = . 8. Ans: Re(z) = 0 Sol: Let ω = u + iv

∴ z = 1ivu1ivu

11

++−+=

+ω−ω

= ( )( )

( )[ ] ( )[ ]( )[ ] ( )[ ]iv1uiv1u

iv1uiv1uiv1uiv1u

−+++−++−=

+++−

= ( ) ( )

22

222

v1u2u

v1uiv1uiv1u

++++++−−−

= 1u2vu

iviv1vu22

22

+++++−+

u22iv20

++=

=+ 1vu 22Q

= u1

iv+

.

9. Ans: 3i

e3π

−

Sol: 3i2

eπ

= cos ω=π+π3

2sini

32

∴ 1 + z + 3z2 + 2z3 + 2z4 + 3z5 = 1 + ω + 3ω2 + 2ω3 + 2ω4 + 3ω5 = 1 + ω + 3ω2 + 2 + 2ω + 3ω2 = 3 + 3ω + 6ω2 = 3(−1 + ω + ω2) + 3ω2

= 3ω2 = 3 3i4

eπ

C

y 30 x

M

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=

π+π34

sini34

cos3

=

π+π+

π+π3

sini3

cos3

= 3i

e33

sini3

cos3π

−=

π−π− .

10. Ans: −128i

Sol: z1 = 22 (1 + i), z2 = 1 + 3 i ∴z1

2 = 4 × 2 (1 + i)2 = 8(1 + 2i − 1) = 16i

z23 = ( )3i31+

= ( )3i33333i31 +×−+

= 1 + i 33i933 −− = −8 ∴ z1

2 z23 = 16i × −8 = −128 i.

11. Ans: 0 Sol: (z3 − z1) = (z2 − z1) [cos90 + isin 90] (z3 − z1) = i(z2 − z1) (z3 − z1)

2 = −(z2 − z1)2

⇒ (z3 − z1)2 + (z2 − z1)

2 = 0. 12. Ans: a2 − 4b = 4 Sol: Let α, α + 2 be two consecutive odd root ∴ α + α + 2 = a and α(α + 2) = b ∴ 2α + 2 = a

∴ α = 2

2a − and α2 + 2zα = b

since α2 + 2α = b

⇒ ( )

b2

2a2

42a 2

=

−+−

⇒ b2a4

4a4a2=−++−

⇒ a2 − 4a + 4 + 4a − 8 = 4b

⇒ a2 − 4b = 4. 13. Ans: a2 − 2b2 Sol: α + β = a, αβ = b2 α2 + β2 = (α + β)2 − 2αβ = a2 − 2b2.

14. Ans: 43

Sol: α + β = −3 αβ = −4

43

4311 =

−−=

αββ+α=

β+

α.

15. Ans: 2

Sol: ( ) 081323 2xx2 =+− +

⇒ 0813.3.23 2xx2 =+−

⇒ 0813183 xx2 =+−

⇒ ( ) 0813183 x2x =+−

⇒ (3x − 9)2 = 0 ⇒ 3x − 9 = 0 ⇒ 3x = 32

⇒ x = 2.

16. Ans: ( )

4

2β−α

Sol: α + β = −2b ⇒ b = 2−β+α

αβ = c

∴ b2 − c = αβ−

−β+α 2

2

= ( ) ( )

444 22 β−α=αβ−β+α

17. Ans: 4x2− 5x + 1 = 0 Sol: 2x2 + 3x + 1 = 0

α + β = 23−

, αβ = 21

∴ α2 + β2 = ( ) αβ−β+α 22

= 21

223

2

×−

−

= 45

149 =−

∴ Required quadratic equation is

041

x45

x2 =+− .

18. Ans: 201

Sol: ( )( ) ( )( )∑∑== ++

−=++

17

8n

17

8n3n2n

233n2n

1

= ( ) ( )( )( )∑

= +++−+17

8n3n2n2n3n

= ∑=

+−

+

17

8n3n

12n

1

= ....121

111

111

101 +

−+

−

…..

−+201

191

= 201

2012

201

101 =−=−

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19. Ans: 3 : 1

Sol: a : b = 3 + 2 2 : 3 − 2 2

326

2ba ==+

11ab ==

∴ 13

M.GM.A = ⇒ A.M : G.M = 3 : 1.

20. Ans: 1020 Sol: x1 + x4 + x9 + x11 + x20 + x22 + x27 + x30 = 272 ⇒ a + a + 3d + a + 8d + a + 10d + a + 19d + a + 21d + a + 26d + a + 29d = 272 ⇒ 8a + 116d = 272 ⇒ 4(2a + 29d) = 272

x1 + x2 + ……. + x30 = S30 = [ ]d29a22

30 +

= 15 × 4

272

= 1020.

21. Ans: 2

189

Sol: ar = 24, ar4 = 3

243

arar4

= ⇒ r3 = 81

⇒ r = 21

∴ a + ar + ar2 + ar3 + ar4 + ar5 = a (1 + r + r2 + r3 + r4 + r5)

= 48

++++52 2

1......

2

121

1

= 48 2

189

21

2

1148

21

1

21

16

6

=

−=

−

−

.

22. Ans: 35

Sol: [ ]d74a22

75 + = 2625

75[a + 37d] = 2625

a + 37d = 3575

2625 = .

23. Ans: −3 Sol: 2k = −6 k = −3. 24. Ans: 9 Sol: Tn = nC3 Tn + 1 − Tn = (n + 1)C3 − nC3 = nC2 = 36

⇒ ( )

362.1

1nn =−

⇒ n(n − 1) = 72

⇒ n2 − n − 72 = 0 ⇒ (n − 9) (n + 8) = 0 ⇒ n = 9. 25. Ans: 10C5

Sol: Tr + 1 = 10Cr rr10

10x

x10

−

T5 + 1 = 5510

510

10x

x10

C

−

= 55

510

10x

x10

C

= 10C5. 26. Ans: −1275 Sol: x49(−1 − 2 −3 − ……. − 50) = −x49(1 + 2 + 3 + …… + 50)

= −x49

×2

5150

= −1275 x49. 27. Ans: 729 Sol: Put x = 1

Then ( ) 666

321x2x1 =+=

+

= 729.

28. Ans: 2

2nn2 −+

Sol: 2P1 + 3P1 + …… + nP1 = 2 + 3 + …… + n

= ( )

12

1nn −+

= 2

2nn2 −+.

29. Ans: 400 Sol: a can take the values 1, 3, 5, 7, 9 b can take 0, 3, 6, 9 c can take 0, 2, 4, 6, 8 d can take 2, 3, 5, 7 ∴ Total number of 4 digit numbers = 5C1 × 4C1 × 5C1 × 4C1 = 5 × 4 × 5 × 4 = 400. 30. Ans: 0

Sol:

0aaaa

0aaaa

1aa

1aa

1aa

1aa

3423

2312

21

43

32

21

−−−−=

R2 → R2 − R1

R3 → R3 − R2

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= 0

0dd

0dd

1aa 21

= .

31. Ans: 81

Sol: Area of the triangle

1cy

cx

1by

bx

1by

ax

21

33

22

11

=

cyx

byx

ayx

abc21

33

22

11

c2yx

b2yx

a2yx

abc1

41

33

22

11

= C3 → 2C3

= 81

2abc

abc1

41 =×× .

32. Ans: a ≠ 2 Sol: 3x + y − z = 2 x + 0 y − z = 1 2x + 2y + a = 5

0

a22

101

113

≠−−

⇒ 3(0 + 2) −1(a + 2) − 1(2 − 0) ≠ 0 ⇒ 6 − a − 2 − 2 ≠ 0 ⇒ 2 − a ≠ 0 ⇒ a ≠ 2. 33. Ans: 4

Sol: 0k31

2k2=

−−

(since matrix is singular)

⇒ (2 − k) (3 − k) − 2 = 0 ⇒ 6 − 5k + k2 − 2 = 0 ⇒ k2 − 5k + 4 = 0 ⇒ −k2 + 5k − 4 = 0 ⇒ 5k − k2 = 4. 34. Ans: zero

Sol:

1logclogc1

log

c1

log1logb1

log

clogblog1log

caa

aba

aaa

=

1logclogclog

clog1logblog

clogblog1log

−−−

= ,0

0clogclog

clog0blog

clogblog0

=−

−− because

the determinant of a skew symmetric matrix of odd order is zero.

35. Ans: 1 Sol: 2x + y − 4 = 0 3x + 2y − 2 = 0 x + y + 2 = 0

( ) ( ) ( )23426242

211

223

412

−−+−+=−−

= 12 − 8 − 4 = 0 ∴ The system of equation has a unique

solution. 36. Ans: 0 Sol: 37. Ans: 14x + 5y ≥ 70, y ≤ 14 and x − y ≥ 5 38. Ans: ~[p ∨ (~q)] ≡ (~p) ∧ q 39. Ans: F, F, T 40. Ans: p ∨ (~q) Sol: ~ [(~p) ∧ q] = ~(~p) ∨ ~q = p ∨ ~q.

41. Ans: sinθ + cosθ ≤ 2

Sol: a sinθ + b cosθ ≤ 22 ba +

⇒ sinθ + cosθ ≤ 22 11 +

≤ 2

42. Ans: 2π

Sol:

2 2 3

1 2 2

3 1

−2 2 0

f(x) = −2x + 2 f(x) = 2x

4

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Given sum = ( )

+ −−

22

1tan22tan 11

= ( ) ( )2

22cot22tan 11 π=+ −− .

43. Ans: ab1ba

−+

Sol: Given sum = 2 tan−1a + 2 tan−1b

= 2 xtan2ab1ba

tan 11 −− =

−+

⇒ x = ab1ba

−+

44. Ans: °2sin

2

Sol: Given sum = tan 1° + cot 1°

= °°°+°

1cos.1sin1cos1sin 22

= °° 1cos.1sin2

2

= °2sin

2.

45. Ans: 0

Sol: S5 = 02

cos105

cos =π=π.

46. Ans: 5

3π

Sol:

π−π=

π −−5

32coscos

57

coscos 11

= 5

35

3coscos 1 π=π− .

47. Ans: 15 Sol: Given sum = 1 + tan2 (tan−1 3) +

1 + cot2 ( )2cot 1−

= 1 + 32 + 1 + 22 = 15. 48. Ans: 2 Sol: sinθ + cosec θ = 2 ⇒ sinθ = 1 ∴ sin6θ + cosec6 θ = 16 + 16 = 2. 49. Ans: 2 Sol: ( )[ ]xfLt

0x +→

++

+++=

+→x

x3sin12x

x5sin6x

x7sinx

x2sin12x

x4sin18x

x6sin7x

x8sin

Lt0x

= 2 Only option (D) matches. 50. Ans: –4 Sol:

M is

=

++3,

27

215

,2

52

Y = 2x + K passes through

3,

27

⇒ K = –4 51. Ans: (5, 2) Sol: Circum centre is the mid point of

Hypotenuse

⇒

+−+2

62,

282

⇒ (5, 2) 52. Ans: 1 :1 Sol: The required ratio is

( ) ( )( ) ( )

−−+−+−−=

++++−

7556277542

cbyaxcbyax

22

11

= 1 :1 53. Ans: 4 Sol: a2 − b2 = 512 ⇒ (a + b) (a − b) = 29 ⇒ (a + b, a − b) = (28, 2), (27, 22), (26, 23), (25, 24) (Q a > b, a + b > a − b), the other

combinations like (24, 25) etc cannot be accepted) 29, 1 also cannot be accepted since a and b are positive integers.

M

D

A (2, 5)

B

C (5, 1)

R (8, 6)

Q (8, –2) P (2, –2)

C

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54. Ans: 51

Sol: The line is ( ) ( )41y

31x + = 1

⇒ 3x + 4y − 1 = 0

P = ( ) ( )

22 43

10403

+

−+ =

51

.

55. Ans: 27−

Sol: Centre : (4, −1) Given parabola is y = (x − 2)2 + 6

⇒ (x − 2)2 = 4 . ( )6y41 −

∴ Vertex is (2, 6)

∴ slope = 27−

.

56. Ans: 21

Sol: The required line is x − 2y + k = 0 passes through (1, 1) ⇒ 1 − 2 + k = 0 ⇒ k = 1 ∴ x − 2y + 1 = 0

y intercept = 21

21

bc =

−−=−

.

57. Ans: a2

Sol: p = θ+θ

−22 eccossec

a

θθθ+θ

=

22

22

cossin

cossin

a

= asinθ cosθ = θ2sin2a

q = θ=θ+θ

θ−2cosa

sincos

2cosa22

∴ 4p2 + q2 = a2(sin2 2θ + cos22θ) = a2 . 58. Ans: 2 Sol: C1 : (1, −2), C2 (−2, 2) r1 = 1 r2 = 2

C1 C2 = 21 rr5169 +>=+

∴ Two circles do not intersect ∴ dmin = C1 C2 − (r1 + r2) = 5 − (1 + 2) = 2. 59. Ans: (9, 8)

Sol:

( )5,52

2k,

21h =

++

⇒ h= 9 and k = 8. 60. Ans: 2 Sol: The two circles touch internally ⇒ C1C2 = |r1 − r2|

⇒ 22 kh + = 22 kh4 +−

⇒ 4kh2 22 =+ ⇒ 2kh 22 =+

∴ r = 2. 61. Ans: (x – 2)2 + (y – 2)2 = 8 Sol: Let the required circle be x2 + y2 + 2gx + 2fy + c = 0 ––––(1)

8cfg 22 =−+ ⇒ g2 + f2 = 8 –––(2)

16 + 8g = 0 ⇒ g = –2 ⇒ f2 = 4 ⇒ f = –2

∴ Req uired circle is x2 + y2 – 4x – 4y = 0 or (x – 2)2 + (y – 2)2 = 8 62. Ans: x = cosθ – 1, y = 2sinθ + 1

Sol: ( ) ( )

12

1y11x

2

22

=−++

⇒ x + 1 = 1 cosθ and y – 1 = 2sinθ ⇒ x = cosθ – 1 and y = 2sinθ + 1 63. Ans: 10

Sol: 53

ePMPF ==

⇒ PM = 10635

PF35 =×=×

64. Ans: 16y

3x 22

=−

(0, 1) (h, K)

x2 + y2 = 16

(0, 0)

(h, K) (1, 2) 5 5

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Sol: a32b34ab2 2

2=⇒=

2a = 3a32 =⇒

⇒ b2 = 6332 =×

⇒ 16y

3x 22

=−

65. Ans: 524

Sol: Focus : (ae, 0) =

0,

4a5

2x + 3y − 6 = 0 passes through

0,

4a5

⇒ ( ) 06034a5

2 =−+

⇒ 62a5 = ⇒ a =

512

∴ 2a = 524

.

66. Ans: Sol: a = cosα

Given ellipse is 19y

16x 22

=+

Foci ⇒ ( ) ( )0,70,916 ±=−± ⇒ a2 + b2 = 7 ⇒ b2 = 7 – cos2α

∴ 1cos7

y

cos

x2

2

2

2=

α−−

α

67. Ans: 4

15

Sol: A = θ=× sinb.a21

ba21

= 4

1521

544121 =××++

68. Ans: a3brr

Sol: cos6θ = ( )

aba

abarrr

rrr

•+

•+

= a.ba

a

aba

b.aa22

rrr

r

rrr

rrr

+=

•+

+

= ba

arr

r

+, where

222babarrrr

+=+

∴ a3b

ba

a

21

22

rr

rr

r

==+

=

69. Ans: Sol:

jkCA −=r

ikAB −=

cosθ = 21

2.2

0001 =+−−

⇒ θ = 60°

θ =3π

70. Ans: 7

Sol: 222

bb.a2abarrrrrr

+−=−

⇒ 7 = 14 –2 22

bbrr

+

⇒ 2

br

= 7

⇒ 7b =r

71. Ans: 222

cbarrr

++

Sol: +++=++2222

cbacbarrrrrr

[ ]c.bc.ab.a2rrrrr

+++

( ) 0cb.a.e.icba =++⊥rrrrrr

–––(1)

Similarly ( ) 0ac.b =+rrr

––––(2)

& ( ) 0ba.c =+rrr

–––––(3) (1) + (2) + (3)

( ) 0c.bc.ab.a2 =++rrrrrr

∴ 2222

cbacbarrrrrr

++=++

72. Ans: –25

Sol: 2222

wvuwvurrrrrr

+++++

[ ]w.uw.vv.u2rrrrrr

+++ = 0

∴ 9 + 16 +25 = –2 [ ]w.uw.vv.urrrrrr

++

[ ]w.uw.vv.urrrrrr

++ = 25250 −=−

73. Ans: 71±

Sol: ( ) 1k6j2i3 =−+λ

A B

C ( )ki +

( )kj + ( )ji +

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= ( ) 1k6j2i3 =−+λ

⇒ 13649 =++λ

|λ| 49 =1

⇒ 71

49

1 ±==λ

74. Ans: k2ji2 +− Sol: The required vector

= ( )knjmi3 ++l

where 194

9a

94 2

=++

⇒ a2 = 1 ⇒ a = 1

∴ vector is

+− k32

j31

i32

3

= k2ji2 +−

75. Ans: ( )k11j3i5.r −+r

= 2

135

Sol: Mid point of PQ is

−29

,27

,23

DR of the normal is 5, 3, –11

∴ Plane is

029

z1127

y323

x5 =

+−

−+

−

⇒ 5x + 3y – 11z = 2

135

⇒ ( )k11j3i5.r −+r

= 2

135

76. Ans: 2π

Sol: Ιst line is 2

5z

23

23

y

11x +=

+=−

2nd line is 0

2z21y

32x −=

−+=−

cosθ = 0049

449

1

033 =++++

−−

θ = 90°

θ =2π

77. Ans: 3

1z43y

52x −=

−−=

−−

Sol: Let DR of the line of intersection of the

planes be a, b, c

⇒ a – 2b – c = 0 ––––(1) a + b + 3c = 0 ––––(2)

3c

4b

5a ==

− ⇒ a = –5k, b = –4k, c= 3k

⇒ 3

1z43y

52x −=

−−=

−−

78. Ans: ( )k4ji326

1 −−

Sol: DR of the line : 3, –1, –4

DC’ S : 26

4,

26

1,

26

3 −−

79. Ans: cos–1

32

Sol: DR’s of the normal : 2, −1, 2 DR’s of Z axis : 0, 0, 1

cosθ = ( )( ) ( ) ( )( )( )

32

100414

120102 =++++

+−+

⇒ θ = cos−1

32 .

80. Ans: (2, –3, 4)

Sol: c

zzb

yya

xx 111 −=−=−

( )

222111

cba

czbyax

++++−=

∴ 12929

4z

3y

2x ===

−=

⇒ x = 2, y= –3, z = 4 81. Ans: 13

Sol: 1325144 =+ 82. Ans: 2 Sol: ∑ ∑ =⇒=− 54x945x ii

∑ =×+×− 459255410x 2i

⇒ ∑ = 360x 2i

⇒ σ = 29

549

3602

=

−

83. Ans: 8

13

Sol: Events: (1, 5), (1, 6), (2, 4), (2, 5), (2, 6) (3, 3), (3, 4), (3, 5), (3, 6), (4, 2) (4, 3), (4, 4), (4, 5), (4, 6), (5, 1) (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

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Probability = 8

1366

26 =×

84. Ans: [1 – P(A)] . P(B’) Sol: P (A ∪ B)’ = P(A’ ∩ B’) = P(A’) P(B’) = [1 – P(A)] . P(B’) 85. Ans: 14

Sol: σ = 12

1nd

2 −

= 12

1497

−

= 7 × 2 = 14

86. Ans: 2165

Sol: 2165

38

357

4=

××

87. Ans: 5

Sol: ( ) ( )

( ) ( )1xx1x

1x1xlim

22

35

1x +−−+−

→

= 1x1x

lim55

1x −−

→ = 5 × 14 = 5

88. Ans: 53

Sol:

( )53

x5x5

1ex3

x3x31log

limx5

2

2

0x=

−

×+

→

89. Ans: x

Sol: 1

1x36x3

21x3

2x

1x32x

f−

−+

+−

+

=

−+

= x

90. Ans: –1 and 2 Sol:

2xlim→

ax + 3 = 2x

lim→

ax2 – 1

⇒ 2a + 3 = 2a2 – 1 ⇒ 2a2 = 2a – 4 = 0 ⇒ a = –1, 2 91. Ans: f’(x) = 0 Sol: |f(x) – f(y)|2 ≤ |x – y|3 –––(1) |f(y) – f(x)|2 ≤ |y – x|3 –––(2) If x > y, 2nd inequality will be wrong

If x < y. 1st inequality will be wrong So only possiblity is f(x) = k(a constant

function) f’(x) = 0 92. Ans: 1

Sol: f(x) = ( )dttcos121

x

1∫ −

= [ ]x1tsint21 −

= [ ]1sin1xsinx21 +−−

f’(x) = [ ]xcos121 −

f’(π) = [ ] 11121 =+

93. Ans: 10

Sol: ( ) x22x'fdxdy 2 +=

1xdxdy

= = 5 x 2 × 1

= 10 94. Ans: –4 Sol: f’(x) = 2x + b 10 + b = 2 × ( 7 + b) = 14 + 2b b = –4

95. Ans: tcos

13

Sol: tcos

1tcostsec

dxdy

3

2==

96. Ans: θ2sec

Sol: θθ−

θθ=sincosa3

cossina3dxdy

2

2

= tanθ

θ=

+ 22

secdxdy

1

97. Ans: 2x1

2

−

Sol: x = sinθ ⇒ y = 2sin–1x

⇒ 2x1

2dxdy

−=

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98. Ans: (–1, 1)

Sol: 2y 4dxdy = ⇒

y2

dxdy =

1y2y2 =⇒=

⇒ 1 = 4x + 5 x = –1 (–1, 1) 99. Ans: x∈ (2, 3) Sol: f’(x) = 6x2 – 30x + 36 = 6(x2 – 5x +6) = 6(x – 2) (x – 3) < 0 x∈ (2, 3) 100. Ans: 15

Sol: y2 exy dxdy

y2.eydxdy

x xy+

+ = 9e–3 . 2x

9. e–3 33 e18dxdy

6.e3dxdy −− −=+

+−

–9y1 + 27 + 6y1 = –18 –3y1 + 27 = –18 3y1 = 45 y1 = 15 101. Ans: –6 Sol: V = πr2h

0dtdr

r2.hdtdh

rdtdv 2 =

+π=

⇒ ( ) 05.30dtdh

.25 =+

625150

dtdh −=−=

102. Ans: f(x) is not differentiable at x = 4 Sol: f(x) is not differentiable at x = 4

103. Ans: 41

Sol: y = x2 − 2x

1

3x

2x2

dxdy += = −2 + −2 = −4

41

dxdy

1−−=

−=

41

.

104. Ans: 2−

Sol: 211 22 −=+−

105. Ans: ( )

Cxx1 4

14

++−

Sol: ∫

+4

3

45

x

11x

dx

1 + ∫−⇒=4

34t

dt41

tx

1

= C4

1t

41 4

1

+−

−−

= Cx

11

41

4+

+−

( )

Cxx1 4

14

++−

106. Ans: – cot (x ex) + C Sol: x ex = t ⇒

∫ dtectcos 2 = – cot t + C

= – cot (x ex) + C

107. Ans: C1x

ex+

+

Sol: ( )∫

+−

+dx

x1

1x1

1e

2x

= C1x

ex+

+

108. Ans: ex sin2 x + C

Sol: ( )∫ + dxxcosxsin2xsine 2x

= ex sin2x + C

109. Ans: 2 C2x

sin2 +

Sol: ∫ dx2xcos2 2

= ∫ dx2x

cos2

= C2

12

xsin2 +

= C2x

sin22 +

110. Ans: Cxsec1x 12 +−− −

Sol: x = sec θ ⇒ ∫ θθθθθ

secdtansectan

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= ( )∫ θ−θ d1sec2

= tan θ - θ + C

= Cxsec1x 12 +−− −

111. Ans: Cx

51

151 2

3

2+

+−

Sol: x = 5 tan θ ⇒ ∫+

dxx

x54

2

= ∫ θθθθ

4

2

tan25

dsec5sec5

= ∫ θθθ

dsin

cos51

4

= ∫ 4t

dt51

= Ct

1151

3+

cosec x = 2

2

x

5x +

= Cx

51

151 2

3

2+

+−

112. Ans:

+2

e1log

e1

Sol: ( )∫ ∫−

− +=

+

1

0

e

1xx

x

x eee

dxe

e1

dx

ex = t ex dx = dt

( )∫ ∫

+−=

+=

e

1

e

1et

1t1

e1

ettdt

= e

1ett

loge1

+

=

+−

e11

loge2

elog

e1

=( )

+ e11

21

loge1

=

+2

e1log

e1

113. Ans: e + 2e1 −

Sol: ex = ex ⇒ x = 0

A = ( )∫−−

1

0

xx dxee

= (ex – e–x) dx

= e + 2e1 − .

114. Ans: 21

Sol: ∫+

e

1

dxx3

xlog1

= ∫2

1

dtt31

=

22

2t

31

= ( )21

1461 =−

115. Ans: 16π

Sol: ∫ +⇒=

1

08

34

x1

dxxtx

= ∫ +

1

02t1

dt41

= ( )101 ttan41 −

= 1644

1 π=π×

116. Ans: ( )22log23

Sol: ∫ ∫=4

2

4log

2log

dxxt

dttlog

=

4log2

2log

2

2x

= ( ) ( )[ ]22 2log2log421 −

= ( )22log321 ×

= ( )22log23

117. Ans: x3 + y3 = 3 k xy + C Sol: k [x dy + y dx] = x2 dx + y2 dy

⇒ k [d (x y) = ( )33 yxd31 +

⇒ k x y = k3

yx 33++

⇒ x3 + y3 = 3 k xy + C = x3 + y3 = 3 k xy + C

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118. Ans: y = x + ex + C Sol: dy = ex + 1) dx y = ex + x + C y = x + ex + C 119. Ans: 2, 2

Sol: 2

223

2

dx

ydy

dxyd −=

2

2

23

dx

ydy

dxdy

−=

= 2, 2 120. Ans: sin2x

Sol: ( ) ecxcosyxcot2dxdy =+

I.F = ∫dxxcot2

e

= xsine 2)xlog(sin2 =

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