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ENGINEERING MATHS SOLUTIONS

ex - e-x1. Given f(x) = = tanh x

ex + e-x

1 1 + x tanh-1 x = log

e , x < 12 1 - x

Ans: (1)2. The domain of the function f(x) = x - 1 + 2 3 - x

x - 1 0 and 3 - x 0

x 1 and x 3

x [1, 3]Ans: (2)3. 12 + 32 + 52 + ......... + (2n - 1)2

= Sum of squares of n odd natural numbers

= [{12 + 22 + 32 + 42 + 52 + ....... + (2n - 1)2 + (2n)2} - {22 + 42 + 62 + ........+ (2n)2}]

2n(2n + 1)[2(2n) + 1] n(n + 1)(2n + 1)= - 22 [ ]6 6

n(2n + 1)(4n + 1) 2[n(n + 1)(2n + 1)]= -

3 3

n(2n + 1)= [4n + 1 - (2n + 2)]

3

n(2n + 1)(2n - 1)=

3

n(4n2 - 1)=

3

n(4n2 - 1)or ( (2n)2 - 22 (n)2 = )3

Ans: (3)

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4. (1 + x)21 + (1 + x)22 + (1 + x)23 + ....... + (1 + x)30

= (1 + x)21 [1 + (1 + x) + (1 + x)2 + .... + (1 + x)9][(1 + x)10 - 1]

= (1 + x)21 (1 + x) - 11

= [(1 + x)31 - (1 + x)21]x

Coefficient of x5 is coefficient of x6 in (1 + x)31 - (1 + x)21 i.e. 31C6 - 21C6Ans: (4)

1 a65. The coefficient of x7 in (ax2 + )11 is 11C5 (1)bx b51 a5The coefficient of x-7 in (ax - )11 is 11C6 (2)bx2 b6

From (1) and (2)a6 a511C5 =

11C6 b5 b6

ab = 1

Ans: (1)6. Here det A = 10

adj AA-1 =

det A

and is the co-factor of 1 of matrix A i.e. 3rd row 2nd element

= 5

Ans: (3)7. adj (adj A) = An -1 A Here n = 3

A3-1 A = A2 AAns: (4)8. C1 C1 + C2 + C3 + +

+ + is the coefficient of x i.e. 0Ans: (1)

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9. We have Zk = 1, k = 1, 2, 3, ......., n1

Zk2 = 1 Zk Zk = 1 Zk = ; k = 1, 2, ......., nZk1 1 1Given + + ........ + = 1Z1 Z2 Zn

Z1 + Z2 + ......... + Zn = 1Ans: (2)

n n10. (1 + i)n = 2n/2 (cos + i sin )4 4

n n(1 - i)n = 2n/2 (cos - i sin )4 4n n(1 + i)n + (1 - i)n = 2n/2 (2 cos ) = 2 n + 22 cos 4 4

n x = 4

Ans: (3)11. Here = 1 + i 3, = 1 - i 3

1 3 nn = (1 + i 3)n = 2n ( + i )n = 2n (cos + i sin )n2 2 3 3

n n= 2n (cos + i sin )3 3

n nSimilarly n = 2n (cos - i sin )3 3

n n n - n = 2n 2i sin = 2n + 1 i sin

3 3

Ans: (4)1 - i (1 - i)2 -2i 3

12. Z = = = = -i = i3/2 = (Cis )3/2 = Cis 1 + i 2 2 2 43

Arg Z = 4

Ans: (1)

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13. Here = qr, = rp, = pq = pqr

Ans: (2)14. ( + )( +) = 2 + ( + ) +

and 2 + p - q = 0, + = p, = q 2 + p = q2 + p + = q + q = 2q

Ans: (3)x2 + 14x + 9

15. Let = y R then we get y2 + y - 20 0x2 + 2x + 3

y [-5, 4]p = -5, q = 4 (p, q) = (-5, 4)

Ans: (4)16. Here f (x ) = 0 (x)3 + 3 (x )2 - 7x + 6 = 0

(x - 7) x = -3(x + 2)Squaring on both sides

x3 - 23x2 + 13x - 36 = 0

Ans: (1)17. Here x4 + x2 + 1 = x4 + 2x2 + 1 - x2

(x2 + 1)2 - x2 = (x2 + 1 + x)(x2 + 1 - x) k = -1

Ans: (2)18. Total number of triplets without restriction = n3

n!Total number of triplets with all different co-ordinates = nP3 = (n - 3)!

= n(n - 1)(n - 2)The required number of triplets = n3 - n(n - 1)(n - 2)

= n[n2 - (n - 1)(n - 2)]= n(n2 - n2 + 3n - 2)

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= n(3n - 2)= 3n2 - 2n

Ans: (3)19. The number of diagonals in n sided polygon is

n(n - 1) n2 - n - 2nnC2 - n = - n = 2 1 2

n2 - 3n n(n - 3)= =

2 2

n(n - 3) = 275

2

n(n - 3) = 550 = 25 22 n = 25

Ans: (1)20. Here sin ( - ) = a sin ( - ) - b sin ( - )

sin2 ( - ) = a2 sin2 ( - ) + b2 sin2 ( - ) - 2ab sin ( - ) sin ( - )cos ( - ) = cos ( - ) cos ( - ) + sin ( - ) sin ( - )

= ab + sin ( - ) sin ( - )2ab cos ( - ) = 2a2b2 + 2ab sin ( - ) sin ( - )sin2 ( - ) + 2ab cos ( - ) = a2(1 - b2) + b2(1 - a2) + 2a2b2

= a2 + b2

Here ( - ) = ( - ) - ( - )Ans: (2)

3 + 3 3 + 1 6 + 4321. tan A tan B tan C = = = 23 + 33 - 1 3 + 1 2= 3 (2 + 3)

Ans: (3)22. 5 cos + 3 cos ( + ) + 3 3

= 5 cos + 3 [cos cos - sin sin ] + 33 3

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1 3= 5 cos + 3 cos - 3 sin + 3

2 2

3 33= 5 cos + cos - sin + 32 2

10 + 3 33= () cos - sin + 32 2

169 27 196Maximum value = 3 + + = 3 + 4 4 414

= 3 + 2

= 10Ans: (4)

23. Here (sin-1 x) = 6

2(sin-1 x)2 = 361 1 1 2(sin-1 x)2 [ cos-1 ()]2 = ()2 = ]9 2 9 3 81

Ans: (1)24. 2 (2 cos x . cos 3x) cos 2x = 1

2 (cos 4x + cos 2x) cos 2x = 12 (2 cos2 2x - 1 + cos 2x) cos 2x = 14 cos3 2x + 2 cos2 2x - 2 cos 2x - 1 = 0

(2 cos2 2x - 1)(2 cos 2x + 1) = 0cos 4x (2 cos 2x + 1) = 0

1cos 4x = 0, cos 2x = -

2 24x = (2n + 1) 2x = 2n 2 3

x = (2n + 1) x = n 8 3 3 5 7 2

x = , , , x = , 8 8 8 8 3 3 Total six.

Ans: (2)

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3 coth x + coth3x 525. coth 3x = Given coth x =

1 + 3 coth2x 3

5 1253 + 3 27

=

251 + 3 9

135 +125=

3(9 + 75)260

= 252

65= 63

Ans: (3)26. (a + b + c)(b + c - a) = b2 + c2 + 2bc - a2 = bc

b2 + c2 - a2 = ( - 2)bcb2 + c2 - a2 - 2

cos A = = ()2bc 2 - 2

-1 < cos A < 1 -1 < < 12

-2 < - 2 < 2 0 < < 4

Ans: (4)A B C

27. r1 = 2r2 = 3r3 s tan = 2s tan = 3s tan 2 2 2A C

tan tan 2 tan B 2

= = = k6 3 2A B C

since + + = 90, We have2 2 2

A B B C C Atan tan + tan tan + tan tan = 12 2 2 2 2 2

(6k)(3k) + (3k)(2k) + (2k)(6k) = 1 36k2 = 1

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1 k = 6

A2 tan

2 12 k 5sin A = = = 1 =

A 1 + 36 k2 51 + tan2 ()2B

2 tan 2 6 k 4

and sin B = = = B 1 + 9 k2 51 + tan2 ()2

a sin A 5now = = a : b = 5 : 4b sin B 4

C2 tan

2 2 2 k 4 k 3sin C = = = =

C 1 + 4 k2 1 + 4 k2 51 + tan2 ()24b 5 4Now =

c 3 35

b : c = 4 : 3

Ans: (1)A - B + C

28. 2ac sin ()2A + C B

= 2ac sin ( - )2 2

= 2ac sin ( - B)2A + B + C A + B + C A + C - B [ = - B = - B = - B]2 2 2 2 2 2

(a2 + c2 - b2)= 2ac cos B = 2ac

2ac

= a2 + c2 - b2

Ans: (2)

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29.

AD = 2 BC = 2 bCD = CA + AD = AD -AC = 2 b -( a + b) = b - aCE = CD + DE = b - a - a = b - 2 a

Ans: (3)30. 3 A = 2 B

(3x + 12y) a + (6x + 3y + 3)b = (2y - 4x + 4) a + (4x - 6y - 2)b7x + 10y = 4

2x + 9y = -5

then x = 2, y = -1

Ans: (4)x 2x -3x

31. [r1 r2 r3] = 2x + 1 2x + 3 x + 13x + 5 x + 5 x + 2= x(15x2 + 31x + 37) and 15x2 + 31x + 37 0for all x real x 0

15x2 + 31x + 37 0 for x R

Since = 961 - 2220 = -1259 < 0( = b2 - 4ac)

Ans: (3)

A B

C

DE

F

a

b a

a

b

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[AC b d]32. Shortest distance =

b d

-10 -2 -3

[AC b d] = 1 -2 23 -2 -2= -10 [4 + 4] +2 [-2 - 6] - 3 [-2 + 6]= -80 - 16 - 12

= - 108i j k

b d = 1 -2 2 3 -2 -2

= i [4 + 4] - j [-2 - 6] + k [-2 + 6]

= 8i + 8j + 4k[AC b d]

S.D. = 8i + 8j + 4k108 108

= = = 9144 12Ans: (2)

1 1 1

33. [a b c ] = 2 -1 31 -1 0

= 1(0 + 3) - 1(0 - 3) +1(-2 + 1) = 5d. a = 6 + 2 + 3 = 11, d. b = 12 - 2 + 9 = 19, d. c = 6 - 2 = 4d = x (b c ) + y (c a ) + z (a b)

d. a d. b d. cx = , y = , z = [a b c ] [a b c ] [a b c ]

11 19 4x = , y = , z = 5 5 5

Ans: (3)

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34. Here a - b = 30 a - b 2 = 900 a2 + b2 - 2 a. b = 900

a . b = -125

a . b - 125 -125

cos = cos (a, b) = = = a b 11 23 253-125 125

= cos-1 ( ) = - cos-1 ( )253 253Ans: (4)

1 n

135. Variance = 2 = xi2 - ( xi)2n

i = 1 n

1 n(n + 1)(2n + 1) 1 n2 (n + 1)2= -

n 6 n2 4

(n + 1)(2n + 1) (n + 1)2= -

6 4

(n + 1) 2n + 1 (n + 1)= [ - ]2 3 2

(n + 1) 4n + 2 - 3n - 3= []2 6

(n + 1)(n - 1)=

12

n2 - 1=

12

Ans: (2)(ax1 + b) + (ax2 + b) + ....... + (axn + b)36. Mean =

n

a(x1 + x2 + ......... + xn) nb= +

n n

= a x + b

Ans: (3)

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37. Here np = 10 and npq = 5

1 q =

2

1p = , n = 20

2

Ans: (1)38. Here P(X = 2) = 3P(X = 3)

e- 2 e- 3

= 3 2! 3!

e- 2 = e- 3

= 1Ans: (4)

7 3 139. Here P(A) = , P(A B) = and P (A B) = 10 10 51P(AB ) = 5

1 4 P(A B) = 1 - P(AB ) = 1 - = 5 5

3

B P(B A) 10 3 P() = = = A P(A) 7 7

10

P(A B) = P(A) + P(B) - P(A B)4 7 3 = + P(B) -5 10 10

2 P(B) = 5

3

A P(A B) 10 3 10 3P() = = = = B P(B) 2 10 4 4

5 1

A P( A B) 1 - P(A B) 5 1P() = = = = B P( B ) 1 - P(B) 3 35

Ans: (1)

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40. The total number of five digit numbers formed by 1, 2, 3, 4 and 5 (withoutrepetition) = 5 = 120A number is divisible by 4 if the last two digit number (i.e. tens and unit place)is divisible by 4.

The last two digit number must be: 12, 24, 32 and 52.

With last two digits 12, the other three places can be arranged in 3! (= 6) ways. The number of favourable cases = 3! 4 = 24

24 1 Probability = = .

120 5

Ans: (2)41. E1 : Event if spade card missing

E2: Event if non spade card missing

13 1 39 3 P(E1) = = ; P(E2) = = 52 4 52 4 E = Event if drawn card is spade

E 12C2 22P() = = E1 51C2 25 17

E 13C2 2 13P( ) = = E2 51C2 25 17

EP(E1) P()E1 E1

P() = E E EP(E1) P() + P(E2) P()E1 E2

1 22 4 25 17

=

1 22 3 13 2 + 4 25 17 4 25 17

22 11= =

100 50

Ans: (3)

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x y 2x 142. Here + = = +

a b a x y 1 2y 1 - = = - a b b x 1 1 y 1 1

= [ + ], = [ - ]a 2 b 2 x2 y2 1 1 1

- = {[2 + + 2] - [2 + - 2]}a2 b2 4 2 2x2 y2

- = 1 (Hyperbola).a2 b2

Ans: (4)T1 T2 T3

-i43. A(4, 1) (1, 4) (4, 4) (4 + 4i) e 6

3 1 (4 + 4i) (cos - i sin ) = (4 + 4i) ( - i )6 6 2 24 3 4 4 3 4

= [ + , ( - ) i] = (23 + 2, 23 - 2)2 2 2 2Ans: (1)

5 344. Eliminate c to get a(x - ) + b(y - ) = 02 25 3

The point is ( , )2 2Ans: (2)45. Let B (, ) be the image of A(1, 2) with respect to the line 2x + 3y + 4 = 0

- 1 - 2 (2 1 + 3 2 + 4) = = -2

2 3 22 + 32 - 1 - 2 -24 = =

2 3 1348 72 = 1 - , = 2 - 13 13

-35 -46 = , = 13 13

-81 + =

13Ans: (3)

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-346. Here m1 = -1, m2 = , m3 = + 3, m4 = + 12

m2 < m1 < m4 < m3Ans: (4)47. x2 sin2 + y2 sin2 = x2 cos2 + y2 sin2 - 2xy sin cos

x2(cos2 - sin2 ) - 2xy sin cos = 0x2 cos 2 - xy sin 2 = 0

2 h2 - abtan = a + b

-sin 22 ()22 sin 2

tan = = = tan 2cos 2 cos 2

= 2Ans: (1)48. Slopes m1, 2m1

3m1 = -2h, 2m12

= 8 m1 = 2

-2h asince m1 + m2 = , m1m2 = b b

-2h = 6 -h = 3, h = 3, -3

Ans: (2)49. Let centre of the circle C (2, 1)

r = 22 + 12 + 20 = 5P(10, 7), C(2, 1); CP = 82 + 62 = 10

S.D. P(10, 7)

C(2, 1) Q CP - r = QP 10 - 5 = QP = 5

Ans: (3)50. Equation of the circle is

(x - 2)(x + 6) + (y - 4)(y - 8) = 0x2 + 4x - 12 + y2 - 12y + 32 = 0

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x2 + y2 + 4x - 12y + 20 = 0

Intercept made by the circle on Y - axis is

2 2 - c = 2 (-6)2 - 20 = 2 4 = 8Ans: (4)51. The inverse point of (-2, 3) w.r.t. the circle

x2 + y2 - 4x - 6y + 9 = 0 is the point of intersection of polar and the linejoining centre (2, 3) and (-2, 3).Polar of (-2, 3) w.r.t. the circle x2 + y2 - 4x - 6y + 9 = 0 is-2x + 3y - 2(x - 2) - 3(y + 3) + 9 = 0- 2x + 3y - 2x + 4 - 3y - 9 + 9 = 0

- 4x + 4 = 0

- 4x = = 1

- 4

3 - 3Equation of the line y - 3 = (x - 2) y = 3-2 - 2

(1, 3)Ans: (1)52. Length of the common chord 2 r2 - P.D.2

Circle S' = x2 + y2 - 2ax - 2by + a2 + b2 - c2 = 0

S'' = x2 + y2 - 2bx - 2ay + a2 + b2 - c2 = 0

Common chord is S' - S'' = 2(b - a)x + 2(a - b)y = 0= 2(b - a) x - 2 (b - a)y = 0

S' - S'' = x - y = 0

P.D. from C1(a, b) to x - y = 0 isa - b

P.D. = 2r = a2 + b2 - a2 - b2 + c2 = cLength of the common chord = 2 r2 - P.D.2

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(a - b)2= 2 c2 - 2= 4c2 - 2(a - b)2

Ans: (2)x y

53. + = 1 touches the circle x2 + y2 = a2 Then x + y - = 0 and P.D. from (0, 0) to x + y - = 0 is

- = a(radius)

2 + 22 2

= a22 + 22 + 2 1 1 1 1

= + = 22 a2 2 2 a2

1 1 1 ()2 + ()2 = a2

1 1 1Locus of ( , ) is x2 + y2 = is a circle. a2Ans: (3)54. The ends of any double ordinate are A(at2, 2at), B(at2, -2at)

-2atCoordinates of point of trisection 1 : 2 is (at2, )3Locus is 9y2 = 4ax

Ans: (4)1

55. Equation of the normal at t is y + xt = 2at + at3 (a = )41 1

y + xt = () t + () t32 4 This passes through (c, 0)

1 1 ct = t + t32 4

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1 1 t = 0 or t2 = c - 4 2

1t = 2 c - 2

1 t is real if c > 2

Ans: (1)x2 y2

56. y = mx + 49 m2 + 4 is a tangent to + = 149 4This is tangent to x2 + y2 = 16

49 m2 + 4 = 16 (1 + m2)12 233 m2 = 12 m2 = , m = 33 11

4 24049 m2 + 4 = 49() + 4 = 11 11415

c = 49m2 + 4 = 112x 415Tangent is y = + or11 11-2x 415y = - 11 11

Ans: (2)57. SO + S'O = 2a

72 = 2a Also SS' = 2ae 49 + 1 = 2ae 52 = 2ae

2ae 52 5 = =

2a 72 7Ans: (3)

x2 y258. Hyperbola is - = 1

144 81() ()25 25

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81 225e2 = 1 + = 144 144

15 e = 12

12 15Focus = ( , 0) = (3, 0)5 12Focus of ellipse is (16 - b2, 0) 16 - b2 = 3 16 - b2 = 9

b2 = 7

Ans: (4)59. A = (2, 3, 5), B = (h, k, l), P = (-1, 4, 3)

m : n = 2 : 3

2h + 6 -11 = -1 h = 5 2

2k + 9 +11 = 4 or k = 5 2

2l + 15 = 3 or l = 05

-11 11 B = ( , , 0)2 2

Ans: (1)2 2 1

60. Answer is (2) i.e. ( , , )9 9 961. Equation of the plane through O, A, B is 5x - y - 3z = 0

Equation of the plane through O, A, C is x - y + z = 0

5 1 + (-1) (-1) + (-3) 1cos =

52 + (-1)2 + (-3)2 12 + (-1)2 + 123 3

= =

35 3 35

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3 = cos-1 35

Ans: (3)62. lim (x2 + 2x - 1 - x)

x

1Put x = . Then y 0 as x y

1 2 1lim [ + - 1 - ]y 0 y2 y y1 + 2y - y2 - 1

= lim [ ]y 0 y1 (1 + 2y - y2 - 1)

= lim [ ]y 0

y (1 + 2y - y2 + 1)y (2 - y) 2

= lim [ ] = = 1y 0 y 1 + 2y - y2 + 1 2

Ans: (4)-1

63. lim f(x) = lim f(x) = f(0) = = Px 0- x 0+ 2

Ans: (1)f(x) - f(a)

lim x a x - a f ' (a)64. = g(x) - g(a) g' (a)lim x a x - a

d3 [ (Tan-1 x)]dx x = a

=

d2 [ (Tan-1 x)]dx

x = a

3= 2

Ans: (2)

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dy x x2 xn-1 xn xn65. = 1 + + + ........ + + - dx 1! 2! (n - 1)! n! n!dy xn = y -dx n!

Ans: (3)x16 - 166. (x + 1)(x2 + 1)(x4 + 1)(x8 + 1) =

x - 1

d (x16 - 1) (x - 1) 16x15 - (x16 - 1) 1 = dx (x - 1) (x - 1)2

16x16 - 16x15 - x16 + 1=

(x - 1)2

15x16 - 16x15 + 1=

(x - 1)2Ans: (1)67. f(x) = log

e x

f(3) = loge

3, f(1) = loge

1

f(3) - f(1) loge

3 - loge

1 loge

3f'(c) = = =

3 - 1 2 2

1f'(x) = x log e

1 1f'(c) = = loge

3c 2

c = 2 log3 e

Ans: (3)68. Let f(x) = (sin x)sin x

and s = log f(x) = sin x . log sin xds cos x = sin x . + (log sin x) cos xdx sin x

= cos x (1 + log sin x)d2s 1 = cos x ( . cos x) - sin x (1 + log sin x)dx2 sin x

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cos2 x= - sin x(1 + log sin x)

sin x

dsNow = 0 cos x = 0 or 1 + log sin x = 0dx

x = or sinx = e-1

2

d2s( ) = 0 - 1 < 0 f(x) has maximum valuedx2 x = 211 -

d2s e2 1( ) = - (1 + log e-1)dx2 sin x = e-1 1 e

e

1= e(1 - ) > 0e2

1 f(x) has minimum when sin x =

e

1Minimum value = [f(x)]

sin x = 1e= () 1ee

= e

1-e

Ans: (4)69. Here V = 288

4 r3 = 288 r = 63

4 dV drV = r3 = 4r2 3 dt dt

dr dV 1 when = 4 is dt dt 36

Ans: (3)dy 2

70. y2 = 4(x + 1) = (first curve)dx ydy -18

y2 = 36(9 - x) = (second curve)dx y

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1Slope of the tangent to the first curve y2 = 4(x + 1) at P(8, 6) is m1 = 3Slope of the tangent to the second curve y2 = 36(9 - x) at P(8, 6) is m2 = -3

1 -8then m1 + m2 = - 3 = 3 3

Ans: (2)1x2 + (x2 - 1) x2

71. dx = A tan-1 ( ) + c (x2 + 1) 1 + x4 21

x2 (1 - )x2 - 1 x2

I = dx = dx (x2 + 1) x4 + 1 1 1 x2 (x + ) x2 + x x2

1Put x + = t, we havex

dtI =

t t2 - 2Again putting t2 - 2 = s2

2t dt = 2s ds

s ds ds 1 yI = = = tan-1 () + c(s2 + 2)s s2 + 2 2 2

1 t2 - 2= tan-1 () + c2 2

1x2 + 1 x2= tan-1 () + c2 2

1 A = 2

Ans: (3)

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(1 + log x)572. dx = t5 dtx

1Put (1 + log x) = t, dx = dtx

t6 (1 + log x)6= + c = + c6 6

Ans: (4)

73. (1 + tan2 x) + (tan4 x + tan6 x) dx= [(1 + tan2 x) + tan4 x (1 + tan2 x)] dx= (1 + tan2 x)(1 + tan4 x) dx = sec2 x (1 + tan4 x) dx

tan5 x= (sec2 x + sec2 x tan4 x) dx = tan x + + c5

tan x= (5 + tan4 x) + c5

Ans: (4)74. y = 2 - x2

y = x2 = 2 - x2

2x2 = 2, x2 = 1

x = 1, y = 1

Area bounded by two parabolas be

x3 x3= 2 [ 1

0(2 - x2) dx - 1

0x2 dx] = 2 {[2x - ]10 - ( )10 }3 3

1 1 2 4 8 = 2 [(2 - ) - ] = 2[2 - ] = 2 = sq. units.3 3 3 3 3

Ans: (2)75. Here I1 + I3 - 2 I2

ex[(1 + x2) - 2x]dx 1 2x

= = ex ( - )dx(1 + x2)2 1 + x2 (1 + x2)2

B (-1, 0)(0, 2)

X

Y

A (1, 0)

y = 2 - x2

y = x2

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We know that ex [f(x) + f'(x)]dx = ex f(x) + ce x

= + c1 + x2

Ans: (1) cos x dx sin x dx

76. Here I = /20

= //20

1 + cos x + sin x 1 + sin x + cos x

sin x + cos x 2I = //2

0 dx1 + cos x + sin x

dx 1 + sin x + cos x 2I + //2

0 = //2

0 dx = /2

01 dx

1 + cos x + sin x 1 + sin x + cos x

= 2

dx /2

0 = - 2I 1 + cos x + sin x 2

Ans: (3)n

77. /20

(r + sin )2 cos d r = 1

Let sin = t cos d = dtWhen = 0 sin 0 = 0 t = 0

= sin = 1 t = 12 2n

10

(r + t)2 dtr = 1

= 10

(1 + t)2 dt + 10

(2 + t)2 dt + ....... + 10

(n + t)2 dt

(1 + t)3 (2 + t)3 (n + t)3= [ ]10 + [ ]10 + ..... + [ ]103 3 3

23 1 33 23 n3 (n - 1)3 (n + 1)3 n3= - + - + ...... + - + - 3 3 3 3 3 3 3 3

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(n + 1)3 1 n3 + 3n2 + 3n + 1 - 1= - =

3 3 3

n(n2 + 3n + 3)=

3

Ans: (2)cos x dx ey78. On separation of the variables we get + dy = log csin x ey+ 1

log sin x + log (ey + 1) = log c(ey + 1) sin x = c; P ( , 0)6

(1 + 1) sin = c6

1 2 = c c = 12

(ey + 1) sin x = 1Ans: (3)

1 dy79. y2 + (x - ) = 0y dx

1x - 1 dy dx y - x 1 (x - ) = - y2 = = + y dx dy - y2 y2 y3

dx 1 1 + x = dy y2 y3

I.F. = e 1y2 dy

= e-

1y

e-

1y

x. e-

1y

= dyy31 1

Let - = z dy = dzy y2

x. e-

1y

= - z ez dz = - (zez - ez) + c = - ez (z - 1) + c

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1x. e

-

1y

= - e-

1y (- - 1) + cy

1x e

-

1y

= e-

1y ( + 1) + cy

x (1 + y) = + c

e

1y ye

1y

xy = (1 + y) + c ye1y

Ans: (3)

d2y dy80. [ + ()3] 65 = 6ydx2 dx

d2y dy + ()3 = (6y) 56dx2 dx Order = 2, degree = 1

Ans: (4)

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ENGINEERING PHYSICS SOLUTIONS

mass m81. d = = volume r2l

d m r l 100 = 100 + 2 100 + 100d m r l

0.003 0.005 0.06= 100 + 2 100 + 100

0.3 0.5 6

= 1% + 2% + 1% = 4% Ans: (4)82. Horizontal component of velocity = u cos = 62 + 82 = 10

Vertical component of velocity = u sin = 20u2 sin 2 2(u sin )(u cos ) 2 20 10

R = = = = 40 m Ans: (3)g g 1083. Due to rotation of vector, components will change, but magnitude remains

constant.

F = (1)2 + (P + 1)2 = P2 + 421 + (P2 + 2P + 1) = P2 + 16

or 2P = 14, P = 7 Ans: (4)84. Total work done W = W1 + W2 + W3

1W1 = Area of triangle = (5)(10) = 25 N2W2 = Area of rectangle = 10 5 = 50 N

1W3 = Area of trapezium = h (a + b)2

1= (10)(10 + 20) = 150

2

W = 25 + 50 + 150 = 225 J Ans: (1)85. Apparent weight will be less when lift moves down with acceleration or lift

moves up with retardation. Ans: (3)86.

20 N

10 N

5 m 20 m10 m0

2 kg 14 Nfk fk5 kg

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muApplied force is more than s (mu + mB)g.mBSo, the 2 blocks move separately with different accelerations.

fk = kmug = (0.1)(2)(10) = 2 Nfk acts as reaction force on bottom block and moves.

fk 2aB = = = 0.4 ms

-2 Ans: (2)mB 5

87. For ring on rolling axis, I1 = Mr12

3For disc about a tangent normal to plane, I2 = M r22

2

Here I1 = I23Mr1

2= Mr2

22

r1 3 r1 3()2 = , = Ans: (3)r2 2 r2 22R R R388. T = = 2R = 2 v0 GM GM

R3 3= 2 = 2 4 G4G R33

So, T is independent of radius. Ans: (3)89. The rod is under the action of gravitational force which acts vertically

downwards. So, centre of mass moves vertically downwards. Ans: (2)90. Tension produced in the rope is F = 200 N

Fl 200 2Y = = = 2 108 Pa Ans: (1)

Ae 2 10-4 1 10-2

91. Liquid level rises until, volume falling in to the tank = Volume leaving the tank.

V = av = a 2ghV2 = a2 (2gh)

V2 70 70 5h = = = cm Ans: (4)

a2 (2g) 1 1 2 980 2

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92. The water film formed between the glass plates is such that,pressure inside is less than outside. Then atmospheric air

pushes the glass plates towards each other. Hence force is to be applied equivalent to force exerted by atmosphere. Ans: (3)

T A93. At time , particle will have displacement of . Then at that displacement 12 2

1 m2 (A2 - x2)KE 2

= PE 1 m2x22

A2A2 -

4 3= = Ans: (2)

A2 1

4

294. For each small sphere, I = mr25

2For big sphere, I ' = MR25

Here M = 27 m

R = n13 r = (27 13 ) r = 3 r

2 2I' = (27 m)(9 r2) = ( mr2) 2435 5

= 243 I Ans: (4)95.

R R = r

F = ( i + H j) mg (- j) = mg (- k)2 21 u2 sin 2 1 2 (u sin )(u cos )

= mg (-k) = mg (-k)2 g 2 g= (u sin )(u cos ) m (-k)

r F

R2

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= (10)(10)(1)(-k) = -100 k Ans: (2)d 1 + 296. = K ( - 0)dt 212 = K (66 - 0) (1)108 = K (56 - 0) (2)10(1) 12 10 66 - 0 gives = (2) 10 8 56 - 0

3 66 - 0 =

2 56 - 0132 - 20 = 168 - 30

0 = 36C Ans: (4)d 1 cm97. R = = ()(t) 5 10-6 100

1 10-2 m=

5 10-4

= 20 m Ans: (1)98. WAB = Work during isochoric = 0

WBC = Work done by gas during isobaric

= nR (dT) = (1)(R)(2400 - 800) = 1600 RWCD = Work during isochoric = 0

WDA = Work done by during isobaric

= nR dT = (1)(R)(400 - 1200) = - 800 RTotal work = 1600 R - 800 R = 800 R Ans: (3)

99. In a bottle with opened mouth, pressure and volume of gas inside bottle remainsconstant.

mUsing PV = RT,M

Since P, V, M, R are rate constants,

A

BC

D

T

P

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m1T1 = m2T2

m1 m 5T2 = T1 = 320 = 320 = 400 K = 127 C Ans: (4)m2 4 4 m5

v v v101. (i) - = = 22l 4l 4lv v v v 7v 7 v 7 (ii) - = - = = ( ) = (2) = 7 Ans: (3)2 l 4(2l) l 8l 8l 2 4l 22

TA

vAnA 102. = = vB nB TB

A TA 4 g 1or = = = B TB 16 g 2 B = 2 A = 0.02 Ans: (2)

v nv103. n1 = n ( ) v - vs = (1)v - vs n1v nv

n2 = n ( ) v + vs = (2)v + vs n21 1 n1 + n2(1) + (2) gives 2v = nv ( + ) = nv ( )n1 n2 n1 n2

2n1n2or n = Ans: (3)

n1 + n21104. 1 sin i = 2 sin r sin r = (sin i) 2

(sin i) is same in all cases.1

sin r 21 1 3(a) = = = 0.752 4 4

3

A

B

4 kg

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41 3 8(b) = = = 0.92 3 9

231 2 9(c) = = = 1.1 Ans: (1)2 4 8

3

fliq g - 1105. = f

air g

- 1l

-0.5 0.5 -1 1 = or = +0.2 1.5 0.2 1.5

- 1

- 1l l1

-5 = 1.5

- 1l

-7.5 7.5 7.5 15 + 5 = 1 or = 4 or l = = Ans: (2)l l 4 8

106. (1) L = fo

+ fe

= 100 + 5 = 105 cm

fo 100(2) m = = = 20fe

5

(3) Final image is astronomical telescope is inverted. Ans: (4)107. y = n11 = n22

n1 1D n2 2 Dor = d d

or n1 1 = n2 2n2 2 (4)(700)

n1 = = = 7 Ans: (2)1 (400)

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108.

I'I1 = 2

I'I2 = I1 cos2 = cos2 2

10012.5 = cos2 212.5 2 1

cos2 = = 100 4

1cos = , = 60 Ans: (2)2

Q inside closed surface109. E = 0

Since charge inside is not changed, E remains same. Ans: (1)

110.

V1 C2 2 F 2 = = =

V2 C1 1 F 1

Hence potentials developed are in the ratio V1 : V2 = 2 : 1

i) If V1 = 6 kV, V2 = 3 kV, V = V1 + V2 = 9 kVIn this case, no capacitor will be damaged.

ii) If V2 = 4 kV, V1 = 8 kV, V = V1 + V2 = 12 kV, but first capacitor will be damaged. Ans: (4)

I' I1 I2P A

C1 C2

V1 V2

V

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111. Equivalent circuit

E1r2 - E2r1Effective emf = r1 + r2

rE (r) - E ()4=

r5 4

3 Er4 3E

= = = 0.6 E Ans: (2)r 5

5 4

112. At balancing position,

x 40 =

6 60

x = 4 Resistance of meterbridge wire

= 0.1 100 = 10 Total resistance of circuit = 10 , 10 in parallel = 5

5 Vi = = 1 amp. Ans: (3)5

BV113. (i) tan = BH

BV(ii) tan ' = BH cos tan ' 1

= tan cos

tan tan 45 2tan ' = = =

cos cos 30 32' = tan-1 () Ans: (1)3

rE 4

E r

G

40

5 V

60

x 6

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0 i 0 fe114. Due to rotating charge, magnetic induction at centre is B = = 2 r 2 r4 10-7 6.25 1015 1.6 10-19

=

2 0.5 10-10

4 10-10=

10-10

= 4 tesla. Ans: (1)115. Due to current ip in outer loop, at centre of inner loop

0 ipB = 2R

0 ip = ()r22R0 r

2= () iP2R

But magnetic flux through secondary is

= M iP0 r

2

M = 2R

r2 M Ans: (4)R

116. VLC = VL VC = 50 V 50 V = 0 Ans: (3)117. Using E = W + KMax

hc hc 2hc hc = + k or = + k (1) 0 0

2

hc hc 3hc hc = + 2k or = + 2k (2) 0 0

3

4hc 2hc(1) (2) gives, = + 2k (3) 0

R

r

ip

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hc hc(3) - (2) gives, = 0or 0 = Ans: (1)

118. Energy absorbed by nucleus to remove neutron

= Decrease of B.E.

= Total initial B.E. - Total final B.E.

= (7 MeV) 4 - (2.5 MeV) 3= 28 MeV - 7.5 MeV

= 20.5 MeV Ans: (2)119. Maximum distance between transmitting and receiving antennas,

D = 2RehT + 2RehRD' = 2Re2hT + 2Re2hR

= 2 [2RehT + 2RehR ]= 2 D Ans: (2)

IC 10 mA - 5 mA 5 mA120. = = = IB 150 A - 100 A 50 A5 10-3 5000

= = = 100 Ans: (1)50 10-6 50

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ENGINEERING CHEMISTRY SOLUTIONS

i - 1 2.6 - 1121. % = 100 = 100 = 80% Ans:( 2)

n - 1 3 - 1

122. Ptotal = PA . XA + PB . XB Ans:( 2)2 3560 mm = PA . + PB. 2 + 3 2 + 3

2 PA + 3 PB = 5 560 = 2800 mm (1)After adding 2 moles of B,

2 5600 mm = PA . + PB . 2 + 5 2 + 5 2 PA + 5 PB = 7 600 = 4200 mm (2)

(2) - (1) gives 2 PB = 1400 mm PB = 700 mm

By substituting PB in (2)2 PA + 5 700 = 4200 mm2 PA = 4200 - 3500 = 700 mm

PA = 350 mmn molar mass

123. Density = Ans:( 1)Volume Avogadro number

n 752 =

(5 10-8)3 6 1023n = 2 So it crystallises in B.C.C.

3 3Radius of atom = a = 5 = 2.165 A4 40.0591

124. Ecell = log10 Kc Ans:( 1)n

2 1.10log10 Kc = = 37.230.0591 Kc = 10

-37

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125. For 1st rate = K[A]x [B]y Ans:( 2)21 = K[A]x [2B]y

21 = 2y

y = 1

For 2nd rate = K[A]x [B]y

8 = K[2A]x [2B]y

23 = 2x . 2y = 2x + y

x + y = 3

x = 3 - y = 3 - 1 = 2

rate = K[A]2 [B]1126. 100 ml standard gold sol 10-4 g or 10-1 mg gelatin Ans:( 3 )

10 ml standard gold sol ?

10 10-1= = 10-2

100

127. Everson Ans:(2 )128. Helium has unusual property of diffusing through glass or rubber or plastic.

Ans:(1 )129. Cr+2 = [Ar]3d4 = 4 unpaired e- Ans:(4)

Fe+2 = [Ar]3d6 = 4 unpaired e-

Co+2 = [Ar]3d7 = 3 unpaired e-

V+2 = [Ar]3d3 = 3 unpaired e-

Cu+2 = [Ar]3d9 = 1 unpaired e-

Ti+3 = [Ar]3d1 = 1 unpaired e-

130. If 3 donor atoms of same ligands occupy adjacent positions at the corners of theoctahedral face, the isomer is called fac or facial. Ans:(2 )

131. SnO2 & PbO2 Amphoteric Ans:(3)GeO2 Acidic

132. All are carcinogenic. Ans:(4 )133.

.

CH3 + .

CH3 C2H6 Ans:(1 )

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Anhyd. AlCl3134. C6H6 + Cl2 C6H5Cl + HCl Ans:(3 )Anhyd. AlCl3C6H6 + 6 Cl2 C6Cl6 + 6 HCldark, cold

UVC6H6 + 3 Cl2 C6H6Cl6500 K

28 Vol. of N2 at STP 100135. % of N = Ans:(2 )22400 wt. of organic compound28 32 100

= = 16%22400 0.25

136. 2 MnO4- + 6 H+ + 5 H2O2 2 Mn

+2 + 8 H2O + 5 O2 Ans:(3 )5 moles of H2O2 reduces 2 moles of MnO4

-

1 mole of H2O2 reduces - ?1 2

= = 0.4 moles.5137. Initially alkali dissolved liquid NH3 is paramagnetic and blue coloured. But

concentrated solution on warming, blue colour changes to bronze colour andbecomes diamagnetic. Ans:(1 )

138. I.E. of B = 801, Tl = 589 Ans:(2 )Ga = 579, Al = 577, In = 558 KJ mol-1

139. AgBr Ag+ + Br- Ans:(3 )s s

Ksp = S2

Precipitate starts when ionic product > Solubility productKsp 5 10-13

[Br-] = = = 10-11[Ag+] 0.05 Mass of KBr = 120 10-11 = 1.2 10-9 grams.

140. Knew = (Kold)change Ans:(4 )KP2 = (KP1)2 KP1 = KP2

141. K = 226 pm; Na = 186 pm Ans:(3 )Li = 152 pm; Br = 114 pm

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142. G = H - T S Ans:(2 )0 = 170 103 J - T 340 JK-1

170 1000 T = = 500 K

340

T = 500 - 273 = 227 C143. Bond order of Li2 = 1; Li2+ = 0.5 Ans:(4 )

Li2- = 0.5B.O. StabilityThough Li2+ & Li2- have same bond order,Li2- is least stable as it has maximum number of electrons (3) in antibondingorbital & having large ionic size.

Equivalent mass of Oxygen Mass of metal144. Equivalent mass of metal =

Mass of Oxygen

8 60= = 1240

Atomic mass of metal = Valency Equivalent mass

= 4 12 = 48 Ans:(1 )C 1 145. E = h; = ; = Ans:(3 )

E = hc

1 1 1 1= hc . RH [ -]= hc . RH [ -]n12 n22 12 22

3= hcRH4

146. At low pressure 'b' is neglected Ans:(1 )a a(P + )V = RT; PV + = RTV2 V

aPV = RT - if divided by RT,VPV RT a = - RT RT RTV

a Z = 1 - RTV

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aAt high pressure is neglected

V2

P(V - b) = RT; PV = RT + PbPV RT PbIf divided by RT, = + RT RT RT

Pb Z = 1 + RT

147. At constant temperature, K.E. of 1 mole of any gas is same. Ans:(3 )148. Energies of the orbitals in the same sub shell decreases with increase in the

atomic number. Ans:(2 )PV149. No. of moles of phosphorous (n) = Ans:(2 )RT

1 34.05 10-3= = 5 10-4 [1 bar = 0.987 atom]0.0831 819.15

Massn =

Gram molar mass

Mass 0.062gram molar mass = = = 124 g

n 5 10-4

124No. of atoms = = 431

Molecular formula = P4150. SSS bond angle in S8 Crown is 107 Ans:(1 )

SSS bond angle in S6 Chair is 102.2

151. Aquaregia dissolves gold and platinum Ans:(3 )I2O5 is good oxidising agent

ClO2 bleaching agent

O2F2 is oxidising agent & removes plutonium.

....

.

.

152. F S F F Cl F Ans:(4 )

F F F

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153. Average oxidation states of S in S4O6-2 are Ans:(1)4x + 6(-2) = -24x = -2 + 12 = 10

10 x = = 2.54Average oxidation states of 4s atoms = 2.5, 2.5, 2.5, 2.5

154. n H2N - (CH2)6 NH2 + HOOC - (CH2)4 - COOH Ans:(4)

[ HN - (CH2)6 - NHCO - (CH2)4 - CO ] + n H2OAmide n

Small molecules like H2O are removed, so it is condensation polymer. It isformed from 2 different monomers, so it is copolymer.

155. Tyrosine is obtained from cheese. Ans:(2 )156. Penicillin & Ofloxacin kill bacteria. So they are bactericidal antibiotics.

Ans:(1)

H H H H Slow

157. H3C C C CH3 H3C C C CH3 Hydride + ShiftBr CH3 CH3 H OH H OH- +

H3C C C CH3 H3C C C CH3 Ans:(3) hydrolysis H CH3 H CH3

OH OH OH OH

158. < < < Ans:(2)

CH3 NO2 NO2

CH3 is electron donating group. NO2 is electron withdrawing group (more tency found at para position).

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O HOH-

OH H

159. H C + H C CHO H C C CHO Ans:(1)H H H H

-H2O

OHHCN

H C = C C C N H C C = CHO

H H H O H2 H H HO H

H3O+ OH

H C = C C COOH + NH3H H H

H2 O O H2

160. CH3 CN + CH3MgBr H3C C = N Mg Br Ans:(4)CH3

NH4Br + MgO + H3C C = O H3O+

CH3

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