Kerala-engg-entrance-2016-paper2-solution.pdf

SOLUTIONS & ANSWERS FOR KERALA ENGINEERING ENTRANCE EXAMINATION-2016 – PAPER II VERSION – B1 [MATHEMATICS] 1.

2.

Ans: Sol:

(2 ∗ 3) ∗ 2 = (23) ∗ 2 = 8 ∗ 2 = 82 = 64.

Ans:

(−∞, ∞)

Sol:

f(x) =

(x + 3)(x − 3) , if x (x − 3)

Ans:

0, ±

−1

= 2tan−1 (2 − 3 ) = 2 × 15° = 30°.

≠3

x + 3, if x ≠ 3 x ∈ (−∞, ∞). 3.

π     cos 3  = 2 tan π   1 + sin  3 

64

8.

Ans:

1

Sol:

(1 + i)(2 + 3i)(3 − 4i) (2 − 3i)(1 − i)(3 + 4i)

3

2 . 13 . 25

Sol:

3

gf(a) = g(a ) = 3

a3

( )

a f(g(a)) = f(3 ) = 3a 3 3a = a a=0 2 ⇒a =3

a=± 4.

3

Ans:

2

Sol:

 x +1  f  = 2x  2x − 1 

1 = a2 + b2 2 2 ⇒ a + b = 1.

= 33 a

9.

Ans: Sol:

 1+ 1  f  = 2 ×1  2 × 1− 1 ⇒ f(2) = 2.

5.

6.

7.

Ans:

{e, f, c, d}

Sol:

A/B = {a, b}, B/A = {c, d}, A ∩ B = {e, f} A − B = {a, b} B − A = {c, d} A ∩ B = {e, f} B = {c, d, e, f}.

Ans:

A⊂B

Sol:

A ⊂ B (Not onto)

Ans: Sol:

a2 + b2

=

13 . 2 . 25

3.

π 6

argz = arg(Nr) − arg(Dr) π   π     cos 3   −1 cos 3  −1 = tan   − tan    1 + sin π   1 + sin π    3   3 

= a + ib

10. Ans: Sol:

π 4

z + iω = 0 ⇒ z = −iω ⇒ arg z = arg(−i) + arg (ω) π ⇒ arg z = − + arg(ω) ----- (1) 2 arg(zω) = π ⇒ arg(z) + arg(ω) = π ---- (2) π Solving (1) and (2), arg(z) = . 4 1 z=

2−i i

(2 − i)2

4 − 4i − 1 = 4i − 3 −1 i 2 2 Re(z ) = −3 Im(z ) = 4 Re(z2) + Im(z2) = −3 + 4 = 1.

z2 =

11. Ans: Sol:

2

=

In the region x < 0 |z + 1| < |z − 1| |x + iy + 1| < |x + iy − 1| |(x + 1) + iy| < |(x − 1) + iy|

(x + 1)2 + y2

<

(x − 1)2 + y2

(x + 1) + y < (x − 1)2 + y2 2x < −2x ⇒ 4x < 0 ⇒ x < 0. 2

2

12. Ans:

Sol:

3

Sol:

3 3  |z| =  z −  + z 2  ≤ z−

|z| ≤ 2 + |z| −

3 3 + z |z|

|z|2 − 3 ≤ 2 |z| |z|2 − 2|z| − 3 ≤ 0 (|z| − 3) (|z| + 1) ≤ 0 −1 < |z| < 3 13. Ans:

Sol:

14. Ans:

a + 3  4= −   2  a + 3 = −8 a = − 11.

3 |z|

3 ≤2 |z|

18. Ans:

Sol:

2048 a=2 a a a a a × × × × × a × ar × r5 r 4 r3 r2 r ar 2 × ar 3 × ar 4 × ar 5 a11 = 211 = 2048.

3 2m + n + k n k = 2+ + m m m = 2 + tan33 + tan12 + tan33 × tan12 =2+1 = 3. tan(45) = tan(33 + 12) tan 33 + tan 12 1= 1 − tan 33 tan 12 1 − tan33 tan12 = tan33 + tan12 1 = tan33 + tan12 + tan33 tan12

19. Ans:

Sol:

3 2 243 3 =   32 2 3 a= . 2

1 − 4α 1 4 1 . β= − 4α

αβ = −

a 4 − a1 a3 − a 2 a2 − a1 a2 − a1

=

= 15. Ans:

Sol:

8 x2 − 6x + c = 0 α + α2 = 6 α(α + 1) = 6 ⇒2×3=6 α=2 α3 = c c = 8.

21. Ans:

Sol:

22. Ans:

Sol: 16. Ans:

Sol:

37 6

α+β= 17. Ans:

−11

b 1 37 = 6+ = . a 6 6

23. Ans:

Sol:

a3 − a 2

+

a3 − a 2

+

a 4 − a3 a 4 − a3

a2 − a1 + a3 − a2 + a 4 − a3 a2 − a1 a 4 − a1 a2 − a1

=

a 4 − a1 a3 − a 2

.

450 a1 + a20 = 45 a1 + a2 + a3 + ….. a20 = 10 × 45 = 450. 3025 1 × 1 + 2 × 22 + 3 × 32 + 4 × 42 + ..… + 10 × 102 = 13 + 23 + 33 + ….. 103  10 × 11 =    2  = (55)2 = 3025.

α=6 αβ = 1 1 β= 6

5

a5 =

20. Ans:

Sol: Sol:

2x2 + (a + 3)x + 8 = 0 α2 = 4 α=2 a + 3 2α = −    2 

12 5d = 60 d = 12.

2

24. Ans:

Sol:

25. Ans:

Sol:

2

31. Ans:

9 C3 36 k3 = 9C4 35 k4 k = 2.

Sol:

4320 X

X

X

X

X

X

X

6 ways 6 ways 5

4

3

2

1

Total ways = 6 × 6! = 6 × 720 = 4320. 26. Ans:

8 n

P4

n

P3

Sol:

Sol: =5

Sol:

28. Ans:

Sol:

29. Ans:

Sol:

( )

336

34. Ans:

Sol:

2 C2 + nC3 + 6C3 n=5 5 C x = 5C 3 x = 2.

[nCr + nCr − 1 = n + 1Cr]

35. Ans:

Sol:

9 0+

1 18

 18 1 +  

C1

+

2 18

C2

+ ..... +

8 19 98 A = 2 1 2 = 38 − 98 = −60 1 0 0

−3 7x + x(x − 1) = −9 x2 + 6x + 9 = 0 (x + 3)2 = 0 x = −3. −10 Expand along R1 we get coefficient of x2 = −10. 299A  1 ω2  A=   ω 1   1 ω2   1 ω2  A2 =     ω 1  ω 1 

18 18

C18

1 1  + + .... + 18 18 C1 C8 

 2 ω2 + ω2  =   2  ω + ω

9 18

C9

 + C8 

1 18

 2 2ω2  =   2ω 2   1 ω2  =2   = 2A ω 1  A3 = A2 A = 2A × A = 2A2 = 4A An = 2n − 1A ⇒ A100 = 299A.

a 18

C9

36. Ans:

1 − a2 a

Sol:

a2 + ab ab − ab  a b  a b      =  2 ab − ab a + ab  a − a a − a

1 0 =  0 1

2

a + ab = 1 ⇒ ab = 1− a2 b=

3600

1 1  + ..... + 18 18 C1 C18 

 1 = 2a 1 + 18 + ..... +  C1  ∴ 2a = 18 ⇒ a = 9.

Sol:

Sol:

22016 = 26 = (64 )336 = (63 + 1)336 Remainder = 1.

 1 + = a  18  C 0 

30. Ans:

33. Ans:

1

n

 1 [1 5 + 5x 2 + x] 1  x  1 + 5 + 5x + 2x + x2 = 0 x2 + 7x + 6 = 0 (x + 6) (x + 1) = 0 x = −6, −1.

A2 = (−60)2 = 3600.

(n − 3)! = 5 n! ⇒ × (n − 4)! n! n−3=5 n = 8. 27. Ans:

32. Ans:

−1, −6

1 − a2 . a

37. Ans:

Sol:

3 396 − 19 + x < 376 − 19 + 9x 396 + x < 376 + 9x 20 < 8x [−2, 6] x < 1 ⇒ −2x + 4 ≤ 8 ⇒ x ≥ −2 1≤x<3 ⇒ 2 ≤ 8 inequality satisfied x ≥ 3 ⇒ 2x − 4 ≤ 8 x≤6

x ∈ [−2, 1) ∪ [1, 3) ∪ [3, 6] x ∈ [−2, 6]. 38. Ans:

Sol:

39. Ans:

Sol:

~p ∨ (q ∧ r) p−F q−T r−T check options ~p ∨ (q ∧ r)

Sol:

~(p ∨ q) ∨ ~(p ∨ r) = (~p ∧ ~q) ∨ (~p ∧ ~r) = (~p) ∧ (~q ∨ ~r). ~p ∨ ~q

Sol:

~p ∨ ~q

Sol:

45. Ans:

(~p) ∧ (~q ∨ ~r)

40. Ans:

41. Ans:

Sol:

The given expression π 3π 3π   = sin2 + sin2 + sin2  π − + 8 8 8  

46. Ans:

Sol:

tanA − tanB = x 1 1 − =y tan B tan A tan A − tan B ⇒ =y tan A tan B x ⇒ tanA tanB = y 1 tan(A − B ) 1 + tan A tan B = tan A − tan B

4 2 3 1 − = sin 15 cos 15

1 2

sinx + cosx = 2 ⇒ (sinx + cosx)2 = 2 ⇒ 1 + 2sinx cosx = 2 ⇒ 2sinx cosx = 1 1 ⇒ sinx cosx = 2 3

x 1 +    y  = (x + y ) = x xy

3 cos 15 − sin 15 sin 15 cos 15

3 1 cos 15 − sin 15 sin(60 − 15 ) 2 2 = = 1 1 (sin 30) sin 15 cos 15 2 4 4 sin 45 = =4 2. sin 30

44. Ans:

1 1 + x y

cot (A − B) =

= 2.

Sol:

7 2 tan−1x = tan−1  − tan−1  4 3

2

π π  = 2sin2 + cos2  8 8 

43. Ans:

1 2

 7 2  −   4 3  = tan 7 2   1+ .  4 3  1 ⇒x= 2

π 3π   = 2 sin2 + sin2  8  8 

Sol:

2.

−1

π  sin2  π −  8 

42. Ans:

1 4 2 = = tan(2θ) = 1 − tan2 θ 1 − 1 3 4 4 1 + tan(2θ + φ) = 3 3 = 3. 4 1 1− . 3 3 2 tan θ

=

47. Ans:

Sol:

1 1 + . x y

π 3

cot −1 x + cot −1 y

π  π  =  − tan−1 x  +  − tan−1 y  2  2  2π π = π− = . 3 3

48. Ans:

Sol:

15 units. 2

a=

75

h=

49. Ans:

Sol:

54. Ans:

3 3 15 × 75 = a= . 2 2 2

(4, −1)

Sol:

•A(3, −2)

x + y − 8x − 14y + 52 = 0 2

2

It is circle (x − x1) (x − x2) + (y − y1) (y − y2) = 0 ⇒ (x − 2) (x − 6) + (y − 4) (y − 10) = 0 ⇒ x2 + y2 − 8x − 14y + 52 = 0.

50. Ans:

• (h, k)

(3, −4)

Sol:

(h, k) is given by solving x − x1 y − y1 −(ax1 + by1 + c ) = = a b a2 + b2 x − 3 y + 2 −(3 − 2 − 3 ) ⇒ = = 1 1 2 ⇒ (x, y) = (4, −1)

C(−1, 0)

D

B(−6, 5)

A(−2, 1)

D = A + C − B = (−2 + −1 − −6, 1 + 0 − 5) = (3, −4)

55. Ans:

(8, 6) •

Sol: 51. Ans:

4x + 3y − 25 = 0

⇒ 4 = 3(−a) ⇒ a = − 53. Ans:

Sol:

•

56. Ans: 4 . 3

4x + 3y = 3 The equation is (y − 1) = m(x − 0) ⇒ mx − y + 1 = 0 1 3 25 = ⇒ m2 + 1 = 2 5 9 m +1 16 9 4 ⇒m=± 3 4 ⇒ − x − y +1= 0 3 ⇒ 4x + 3y − 3 = 0.

⇒ m2 =

(h, k)

(h, k) is given by solving x − x1 y − y1 −2(ax1 + by1 + c ) = = a b a2 + b2 x − 0 y − 0 −2(−25 ) ⇒ = = 4 3 25 ⇒ x = 8, y = 6

4 − 3

4x − ay − 8 = 0 c −c − = 3× b a 1 3 ⇒ = ⇒ a = 3b b a

Sol:

(0, 0)

2 5

 1   −1  Sol:   x +   y − 3 = 0  10   4  −1 2 −a m= = 10 = . −1 b 5 4

52. Ans:

x+y−3=0

Sol:

−16 x2 + y2 + 2x − 4y + r= =

g2 + f 2 − c

1+ 4 −

λ 4

λ  πr2 = π 5 −  = 9π 4  λ ⇒5− =9 4 λ ⇒ = −4 4 λ = −16

λ =0 4

57. Ans:

5 2

8 4 = PM 5

⇒ PM =

Sol:

(2, 7) •

61. Ans:

(–6, 0)

• Centre

h+o 0+0 ,   = (− 3, 0 ) 2   2

Sol: (2,• 3)

(h, o)

=5

(–3, 0)

(0, 0)

5 . 2

r=

58. Ans:

•(5, 3)

(5 − 2)2 + (7 − 3)2

d=

40 = 10 . 4

⇒ h = –6 ∴ F(–6, 0)

3

Sol:    ± 5 , 0   6  

62. Ans: r2 r1

•(2, 1)

• (1, 3)

x2 y2 + =1  1  1     4 9

Sol:

d=

(1 − 2)2 + (3 − 1)2

r1 =

g2 + f 2 − c = 1 + 9 − 6 = 2

r2 =

r12 + d2 = 5 + 4 = 9 = 3.

59. Ans:

Sol:

2 k passes through (1, 1) 2

x2 + y2 =

2

⇒x +y =2=

∴r=

24

Sol:

k ⇒k=4 2 2

1 1  − ,0 4 9 

  5 = ± , 0   6  

63. Ans:

⇒2=

60. Ans:

   F :  ± a2 − b2 , 0  =  +   

= 1+ 4 = 5

F

( 2)

P

2

(4, 3)

2.

10 x+ 8 = 0

Sol: PF = 2a = 12 ⇒ 4a = 24

P •

M

• Directrix

PF =e PM

F

64. Ans:

Sol:

2 x2 y2 + =1 4 a e = 1−

b2 a

2

= 1−

a 1 = 4 2

⇒ 1−

b2 = 2 b= 65. Ans:

⇒ − 2kˆ = λkˆ ⇒ λ = –2

2

71. Ans:

3 b

2 Sol:

( 6)

2

Sol:

(kˆ )+ (− kˆ )+ −2kˆ = λ(kˆ )

a 1 a 1 = ⇒ = 4 2 4 2 ⇒a=2

Substituting ,

a

6 a2

−

9 b2

−

2

3

2

b

2

b

=1

60°

= 1 ––––––– (1)

a b

2b2 18 9a = ⇒ b2 = ––––– (2) a 5 5 a Solving, a2 = 1, b2 = 5 ∴ 2a = 2

66. Ans:

Sol:

cos60 =

⇒

θ=

2

⇒ 5π 6

72. Ans:

Sol:

68. Ans:

Sol:

73. Ans:

Sol:

–7 2 −1 −m = = 2 4 −2     7 7     7 −m ⇒ ⇒ m = –7 = 2 2

Sol:

(

) (

)

r r r r 3a − 5b • 4a × 5b r r r r = 20 3a − 5b • a × b r r r r r r = 20 3a • a × b − 5b • a × b

( ) ( ) [ ( ) ( )]

=0 70. Ans:

Sol:

= 4b

= 3b

2

2

⇒ a = 3 b

29

(a + b )• (a − b ) = 0 = 29

x=z=0 Represents y axis ∴ DC: 0, 1, 0

0

2

2

= 4b

⇒ a = b = 9 + 16 + 4

0, 1, 0

69. Ans:

a

2

2

⇒ a2 − b2 = 0

r −9 5π = a × cos 6 3 r ⇒ a =6

Sol:

a+b

b 1 = 2 a+b

⇒

a+b

a +b

6

r r a .b r −9 r = a cos θ = 5 b

67. Ans:

a+b

74. Ans:

Sol:

1 5   , , 0 2 4  x −1 y −1 z −1 = = =λ 2 −1 4 ⇒ (x, y, z) = (2λ +1, – λ + 1, 4λ + 1) ⇒ 4λ + 1 = 0 −1 ⇒λ= 4 1 5  ∴ (x, y, z) =  , , 0  2 4 

(

)

r • 2ˆi − 5ˆj + 7kˆ + 76 = 0

The required plane: 2x – 5y + 7z + k = 0 and passes through (–1, 5, –7) ⇒ 2x – 5y + 7z + 76 = 0

75. Ans: 90°

Sol:

P(2, 4, 5)

–2 r r r a + 2b − c = 0 r r r Let a = ˆi, b = ˆj ⇒ c = ˆi + 2ˆj Substituting,

b A

θ

a Q(3, 3, 1)

(1, 2, 3)

a.b

cosθ =

=

a b =

76. Ans:

Sol:

77. Ans:

Sol:

(2ˆi + ˆj − 2kˆ )• (ˆi + 2ˆj + 2kˆ ) 4 + 1+ 4

=

1+ 4 + 4

5 5 ⇒ (a – 6)2 + (b – 6)2 = 13 –––––– (2) ⇒ a = 3, b = 4 ∴ ab = 12

2+2−4 =0 9

–5

83. Ans: 34

DR1: 2, a, 4 DR2: 1, 2, 2 (2) (1) + (a)(2) + (4)(2) = 0 ⇒ 10 + 2a = 0 ⇒ a = –5

Sol:

14

The point in the line is (2λ – 1, 3λ – 1, 4λ – 1) which lies on x + 2y + 3z = 14 ⇒ λ = 1 ∴ point: (1, 2, 3) ∴d=

1 + 4 + 2 = 14

79. Ans:

Sol:

47 66

Sol:

P=

x−3 y+4 z−5 = = =λ −1 −2 2 3−x 4+y z−5 , = =µ −1 2 7 Solving (3, –4, 5)

(

) (

r = 2ˆi + 0ˆj + kˆ + λ ˆi − 3ˆj − 2kˆ x− 2 y − 0 z − 1 = = 1 3 −2 r = 2ˆi + 0ˆj + kˆ + λ ˆi − 3ˆj − 2kˆ

) (

85. Ans:

Sol:

81. Ans:

Sol:

)

86. Ans:

Sol:

2 + 1− 6 − 5 4 + 1+ 9

=

8 14

7 12

C2 + 4 C2 +5 C2 12

)

C2

(+ , + , –) or (–, –, –) 5

C2.

6

C1 + 6 C 3

11

C3 16 33

3 4

 sin 3 x  lim  

x →0  tan 4 x 

87. Ans: f(x) is continuous at x= 0

sum = 4, 6, 8, 9, 10, 12 ⇒ (1, 3), (3, 1), (2, 2), (3, 3), (2, 4), (4, 2), (5, 1), (1, 5), (2, 6), (6, 2), (5, 3), (3, 5), (4, 4), (3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (6, 6)

Sol:

a + b + 23 =6 5 ⇒ a + b = 7 –––––– (1)

∑ (x i − x )

2

n

lim f ( x ) = lim cosh = 1

x →0 +

h →0

lim f ( x ) = lim − cos(− h) = lim − (cosh) = −1

x →0 −

82. Ans: 12

σ2 =

3

  sin 3 x      x  3  = lim = x →0   tan 4 x   4     x  

Point (1, 1, 2) and plane 2x + y – 3z–5 = 0

88. Ans:

Sol:

(

=

8

d=

C2 −

16 33

⇒ P=

)

14

Sol:

12

47 66

(

80. Ans:

Total = (10 × 100) + (2 × 0) + 72n 1000 + 72n = 76 Class average = n + 12 ⇒ 1000 + 72n = 76n + 912 4n = 88 ⇒ n = 22 ∴ Total = 22 + 12 = 34

84. Ans:

78. Ans: (3, –4, 5)

Sol:

(a − 6)2 + (b − 6)2 + 4 + 1 + 16 = 34

Sol:

h →0

−5 6

 n C 3 − n P3   lim  n→ ∞   n3 

 n C 3 − n C 3 3!   lim  n→ ∞   n3 

h →0

 n(n − 1) (n − 2)  = −5 × lim   n→∞  1. 2 . 3. n 3     1  2  −5 lim 1. 1 −  1 −  = 6 n→ ∞   n  n  −5 = 6

'

Sol:

f f ' g − f . g' f ' ( x ). g( x ) − f ( x ) g' ( x )   = = g g2 [g( x )]2   '

f   (4 ) = g

(5) (6) −  1 . (12) 3 36

95. Ans: 2 89. Ans: 3x4 – 6x2 + 5

Sol:

90. Ans:

Sol:

(

Sol:

)

f o g x2 − 1 = f[g(x2 – 1)] = f[(x2 – 1)2 – 1] = f[x4 – 2x2] = 3(x4 – 2x2 )+ 5 = 3x4 – 6x2 + 5

96. Ans:

π 4

Sol: π π = a 4

Period =

y’ = a. e3x + (ax – 5) 3e3x =a(1) + (–5) (3) (1) = a – 15 = –13 ⇒ a = 2 1 2

y = tan–1[secx + tanx]   π x  = tan −1 tan +    4 2  =

y’ =

91. Ans: –1

Sol:

2x log2 + 2y log2 y’ = 2(x + y) (1 + y’) log2 x y x y ⇒ 2 + 2 y’ = 2 . 2 (1 + y’) ⇒ 2 + 2y’ = 4(1 + y’) = 4 + 4y’ ⇒ 2y’ = –2 ⇒ y’ = –1

Sol:

S = cos–1(2x2 – 1) = 2cos–1x t = 1− x2 −2 ds = dt

−1

y 1 − x = sin x y (−2x ) 1 + 1 − x 2 y' = ⇒ 2 1− x 2 1− x2 2 ⇒ –xy + (1 – x )y’ = 1 2

98. Ans: 8

Sol:

f(x) = f ‘(x) =

1 210 1

x 10

10

2 1

10 x 9

=

1

210 =1 13 94. Ans: 18

[

10

y = 2x3 – 9x2 + 12x + 4 dy = 6 x 2 − 18 x + 12 = 0 dx ⇒ x2 – 3x + 2 = 0 ⇒ x = 2 or 1 d2 y

= 12x − 18 dx 2 ∴ x = 2 is point of minima ymin = 16 – 36 + 24 + 4 = 8

8

10. 9. x etc 210 ∴ The given sum 1  10 10.9 10.9.8 = + + 1 + 1! 2! 3! 210 

f”(x) =

1 − x2 = 4 . −x 1 − x2

93. Ans: 1

Sol:

1 2

97. Ans: 4

92. Ans: 1

Sol:

π x + 4 2

10 !  + .........  10 !  C 0 + 10 C1 + 10 C 2 + .... + 10 C10

99. Ans: x = 0

Sol:

]

y = excosx dy S= = e x (− sin x + cos x ) dx ds = e x [− cos x − sin x + − sin x + cos x ] dx =ex[–2sinx] = 0 ⇒ x = 0 d2 s

= −2e x [sin x + cos x ] < 0 dx 2 ∴x=0

 x 106. Ans: − cos + 10  + C 7  

100. Ans: 3

Sol:

f ‘(x) = =

f (b) − f (a) b−a

16 − − 2 18 = =3 6−0 6

1 x  sin + 10 dx 7 7  x  − cos + 10  1 x  7  = = − cos + 10  + C 1 7 7   7

∫

Sol:

101. Ans: 6x – y – 11 = 0 dy = 3x 2 − 6 = 6 dx ∴ Tangent is (y – 1) = 6(x – 2) ⇒ 6x – 12 – y + 1 = 0 ⇒ 6x – y – 11 = 0

Sol:

107. Ans: = a log

∫

Sol:

102. Ans: (2, –9)

=

dy = 6 x 2 − 10 x − 4 = 0 dx ⇒ 3x2 – 5x – 2 = 0

Sol:

5 ± 25 + 24 5 ± 7 = 6 6 ⇒ x=2 ⇒ y = 16 – 20 – 8 + 3 = –9

(x

2

−a a

x+a +C x

)

− a2 − x2 1 dx = −a 2 x (x + a ) x (x + a ) 2

∫

1

1 

x

∫  x − x + a dx = −a log x + a + C = a log

⇒x=

103. Ans: 0.8 m2/sec

x+a +C x

5 1 sine (x )  + C    5

108. Ans:

∫x

Sol:

4

5 5 e (x ). cos e (x )  dx  

5 1 sine (x )  + C    5 (by inspection)

Sol:

5 θ

109. Ans: xtanx + C 8

1 × 8 × 5 × sin θ = 20 sin θ 2 dA dθ π = 20 cos θ = 20 × cos × 0.08 dt dt 3 = 0.8 m2/sec

A=

104. Ans:

Sol: `

(2, 3)

2x

∫ 2 cos 2 x + ∫ tan x dx

Sol:

∫

∫

∫

= x tan x − tan x + tan xdx = x tan x + C

110. Ans: log |tanx|+ C

dy = 24 x 2 − 120 x + 144 dx = 24(x2 – 5x + 6) < 0 ⇒ (x – 2) (x – 3) < 0 ⇒ x∈(2, 3)

=

Sol:

sec m+ 2 x +C m+2

The given integral

∫ (sec x ) sec x tan x = ∫ sec m+ 2 x tan x dx m

=

=

2

m+ 2

sec x +C m+2

1

∫ sin x cos x dx = log tan x + C (by inspection)

Sol: Sol:

∫

= x sec 2 x + tan x dx

111. Ans: 105. Ans:

(splitting)

dx

=

1 tan −1 (3 tan x ) + C 3

=

sec 2 x

dx

∫ 8 sin 2 x + 1∫ 8 tan 2 x + sec 2 x dx

sec 2 xdx

dt

∫ 1 + (3 tan x )2 = ∫ 1 + (3t )2 =

=

1 tan −1 (3t ) + C 3

1 tan −1 (3 tan x ) + C 3

112. Ans: 0

4

π 2

π 2

∫

∫

0

(y − 2)dy 2

2

Ι = log(cot x )dx = log(tan x )dx

Sol:

∫

A=

=

0

2Ι = 0 ⇒Ι=0

4   1  y 2  4   − [y ]2  = 1 2  2   2   y

117. Ans: x + y + 4 = Ce 113. Ans: 17

Sol:

2

∫

Sol:

0

4 x 2 x dx =

−1

∫ 4x

2

dx = x+y+3 dy Let x + y + 3 = t dx dt +1= ⇒ dy dy

(− x )dx

−1 2

∫

+ 4 x 2 . x dx

⇒

dt dt −1= t ⇒ = 1+ t dy dy

⇒

∫ 1 + t = ∫ dy

0

0

2

x4  x4  = −4   + 4   = 17  4  −1  4  0

log(1 + t) = y + C ⇒ 1 + x + y + 3 = Cey y ⇒ x + y + 4 = Ce

114. Ans: 4π

Sol:

118. Ans: order 2, degree 1 circle

Sol:

4

A=

1 2 1 πr = π × 4 2 = 4π 4 4

115. Ans: 0

∫ (x − 3) (x

2

− 6x + 8

)

∫ (x − 3) [(x − 3) 4

=

2

]

− 1dx

2

∫ [(x − 3) 4

2

]

− (x − 3 ) dx

2

4

 (x − 3 )4 (x − 3 )2  = −  2   4 2

=0 116. Ans: 1

Sol:

Sol:

y = 2x + 2 (0, 2)

x c−x

ydx – xdy = –y2dx dy ⇒y−x = −y 2 dx dy    y1− x  dx   ⇒ = −1 y2 ⇒

d x   = −1 dx  y 

⇒

x = −x + C y

y=

4

(–1, 0)

y By inspection, sin −1   − 5 x 2 = C x

120. Ans: y =

2

a2 ⇒ yy"+ (y')2 = 0 2

y 119. Ans: sin −1   − 5 x 2 = C x Sol:

4

Sol:

y2 = a2x + ab 2yy’ = a2 ⇒ yy’ =

4

=

dt

x c−x