mm 3rd mid online

MM


JNTU ONLINE EXAMINATIONS
[Mid 3 - MM]

1. Solving the differential equation 1 + y
, y(0) = 0 by Taylor's series method, y(0.2)
2
=
1. 0.10266
2. 0.00266
3. 0.30266
4. 0.20266
2. Solving the differential equation x + y, y(1) = 0 h = 0.1, by Taylor's series method,
y(1.1) =
1. 0.110338
2. 1.110338
3. 1.210338
4. 1.310338
3. Solving y
+ x, y(0) = 1 by Taylor's series method y(0.1) =
2
1. 0.1164
2. 1.1164
3. 2.1164
4. 3.1164
4. Solving , y(4) = 4 by Taylor's series method y(4.2) =
1. 3.0098
2. 2.0098
3. 1.0098
4. 4.0098
5. Solving the differential equating x
+ y
, y(0) = 1 h = 0.1 by Taylor's series
2
2
method. y(0.1) =
1. 0.11145
2. 2.11145
3. 1.11145
4. 3.11145
6. To find the solution of the x
y - 1, y(0) = 1 by Taylor series y (0) =
2
1. -6
2. 6
3. 8
4. -8
7. Solving x - y
, y(0) = 1 by Taylors series method y(0.1) =
2
1. 0.7138
2. 0.8138
3. 0.9138
4. 0.6138
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8. For the differentia equation y(x0), y(x0)= y0 the solution by Taylor's series
is given by yn+1 =
1.
2.
3.
4.
9. To solve the differential equation y(0) = 1 by Taylor's series method y
(0) =
1. 3
2. 4
3. 2
4. 1
10. To find the solution of the , y(1) = 1 by Taylor's series method y (o) =
1.
2.
3.
4. 1
11. Solving the differential equation 1 + xy, y(0) = 1 by Picard's method y(0.1) =
1. 0.1053
2. 0.2053
3. 1.1053
4. 1.2053
12. To solve the differential equation , y(0) = 0 by Picard's method, the
approximation =y
1.
2.
3.
4.
13. The solution of the differential equation , y(0) = 0 by Picard method is
1.
2.
3.
4.
14. The formula to find yn by Picard's method is given by y
=
n
1.
2.
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3.
4.
15. If x + y, y(0) = 1, the first approximation by Picard's method in given by y1 =
1.
2.
3.
4.
16. Using Picards method, the solution of the equation y, y(0) = 1 is given by y4 =
1.
2.
3.
4.
17. The solution of xy + 1, y(0) = 1 by Picard's method is given by y
=
2
1.
2.
3.
4.
18. Using Picard's method, the solution of the equation y
+x
. y(0) = 1 is given by y
2
2
1
=
1.
2.
3.
4.
19. To solve the differential equation , y(0) = 1 by Picard method, y
=
1
1. x + x log (1 + x)
2. 1 - x + 2 log (1 + x)
3. 1 - x log (1 + x)
4. 1 + x + 2 log (1 + x)
20. To solve the differential equation f(x, y), y(x
) = y
the formula by Euler's method
0
0
is given by y =
1. y
- hf(x
, y
)
n
n
n
2. y
= hf(x
, y
)
n
n
n
3. y
= hf (x , y )
n
4. y
+ hf(x , y )
n
21. Using modified Euler's method, the second approximation of the differential equation
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1 - y , y (0) =0, h = 0.1, y_1
=
2
1. 1.09525
2. 2.09525
3. 3.09525
4. 0.09525
22. Using Euler's method, the solution of the differential equation with y(0) = 1
when x = 0.1 is given by y(0.1) =
1. 0.0200
2. 1.0200
3. 2.0200
4. 3.0200
23. Using modified Euler's method, the solution of the differential equation 2xy, y(0)
=1, h = 0.25 is given by y(0.25) =
1. 1.0645
2. 0.0645
3. 2.0645
4. 1.2645
24. Solving the differential equation 1-2xy, y(0) = 0, h =0.2 by Euler's method y(0.6)
=
1. 1.4748
2. 0.4748
3. 2.4748
4. 3.4748
25. Solving the differential equation by modified Euler's method y(2)
=
1. 4.051
2. 5.051
3. 6.051
4. 3.051
26. Solving the differential equation x+ y + xy, y(0) = 1, h = 0.025 by Euler's method,
y(0.1) =
1. 0.1448
2. 2.1448
3. 1.1448
4. 1.2448
27. Solving the differential equation -xy
, y(0) = 2, h = 0.1 by modified Euler's method
2
y(0.2) =
1. 1.9227
2. 0.9227
3. 2.9227
4. 3.9227
28. Solving the differential equation y + e
, y(0) = 0 by modified Eule's method y(0.2)
x
=
1. 0.2468
2. 0.4268
3. 1.2468
4. 1.4268
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29. To solve the differential equation = f(x, y) y(x
) = y
, the formula by modified
0
0
Euler's method
1.
2.
3.
4.
30. To solve the differential equation f(x, y), y(x
) = y
for the formula by Runge Kutta
0
0
method is gives by K =
1.
2.
3.
4.
31. Give x
+ y
, y(1) = 0, he solution of y when x = 1.1 by Runge-Kutta method is
2
2
given by
1. 0.11072
2. 1.11072
3. 0.21072
4. 1.21072
32. The solution of differential equation , y(0)=0 by Runge Kuta method, y(0.2) =
1. 0.2027
2. 1.2027
3. 1.0227
4. 0.0227
33. The solution of differential equation - y, y(0) = 1 by Runge Kutta method, y(0.1) =
1. 1.9048
2. 0.9048
3. 0.0948
4. 1.0948
34. The solution of differential equation = x + y, y(0) = 1, h = 0.1 by Runge Kutta
method y(0.1) =
1. 0.11034
2. 2.11034
3. 1.11034
4. 3.11034
35. Using Runge-Kutta method, to solve the differential equation x + y, y(0) = 1, h =
0.1, K
=
4
1. 1.12105
2. 0.12105
3. 1.22105
4. 0.22105
36. To solve the differential equation + y + xy
= 0, y(0) = 1, h = 0.1 by Runge Kutta
2
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method, K
=
2
1. 0.0995
2. 1.0995
3. - 1.0995
4. - 0.0995
37. To solve the differential equation , y(0) = 1 by Runge Kutta method K
=
2
1. 1.19672
2. 0.19672
3. 0.16972
4. 1.16972.
38. Which is the best for solving initial value problems
1. Runge Kutta method
2. Euler's method
3. Modified Euler's method
4. Taylor's method.
39. Milne's Preductor formula
1.
2.
3.
4.
40. Milne's corrector formula
1.
2.
3.
4.
41. Solving differential equation , y(0) = 1, y(0.1)=1.01, y(0.2) = 1.022, y(0.3) =
1.023 y_0
= 0 y_1
= 0.0505, y_2
= 0.1022, y_3
= 0.1535 by Adams - Moulton
1
1
1
1
formula, y_4
= y(0.4) =
c
1. 1.0410
2. 1.0410
3. 2.0408
4. 1.2408
42. Solving differential equation 1 + y
, y(0) = 0, y(0.2) = 0.2027, y(0.4) = 0.4228, y
2
(0.6) = 0.6841, y_0
= 1.000, y_1
= 1.0411, y_2
= 1.1787, y_3
= 1.4681 by Milne's
1
1
1
1
formula, y_4
= y(0.4) =
c
1. 0.0294
2. 1.0294
3. 0.1294
4. 1.1294
43. If y(0.1) = 1.06, y(0.2) = 1.12, y(0.3)=1.21, y_1
= 0.567,
1
y_2
= 0.6522, y_3
= 0.7980 by Milne's formula, = y(0.4) =
1
1
1. 1.2797
2. 0.2797
3. 1.7297
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4. 0.7297
44. Given x - y
, y(0) = 0, y(0.2) = 0.200, y(0.4) = 0.0795, y(0.6) = 0.1762 y_1
=
2
1
0.1996, y_2
= 0.3937, y_3
= 0.5989 by Milne's formula, = y(0.8) =
1
1
1. 0.3046
2. 0.4046
3. 1.3046
4. 1.4046
45. Solving differential equation , y(0) = 1, y(0.1)=1.01, y(0.2) = 1.022, y(0.3) =
1.023 y_0
= 0 y_1
= 0.0505, y_2
= 0.1022, y_3
= 0.1535 by Adams - Moulton
1
1
1
1
formula, y_4
= y(0.4) =
p
1. 1.0408
2. 0.0408
3. 2.0408
4. 1.2408
46. Solving differential equation 1 + y2, y(0) = 0, y(0.2) = 0.2027, y(0.4) = 0.4228, y
(0.6) = 0.6841, y_0
= 1.000, y_1
= 1.0411, y_2
= 1.1787, y_3
= 1.4681 by Milne's
1
1
1
1
formula, y_4
= y(0.4) =
P
1. 1.0239
2. 0.0239
3. 1.1239
4. 0.1239
47. If y(0.1) = 1.06, y(0.2), y(0.3)=121, y_1
= 0.567, y_2
=
1
1
0.6522, y_3
= 0.7980 by Milne's formula, = y(0.4) =
1
1. 0.2772
2. 1.2772
3. 0.7225
4. 1.7225
48. Given x - y
, y(0) = 0, y(0.2) = 0.200, y(0.4) = 0.0795, y(0.6) = 0.1762 y_1
=
2
1
0.1996, y_2
= 0.3937, y_3
= 0.5689 by Milne's formula, = y(0.8) =
1
1
1. 0.0346
2. 1.0346
3. 1.3046
4. 0.3046
49. Adams Moluton predictor formula
1.
2.
3.
4.
50. Adams-Moulton corrector formula
1.
2.
3.
4.
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51. Given x
+ y
, y(1) = 0, y(1.1) = 1.1107, y(1.2) = 0.24631, y(1.3) = 0.41357, y_1
2
2
1
= 0.11072, y_2
= 1.50067, y_3
= 1.86104 by Adams-Moulton method, y_4
= y(1.4)
1
1
1
=
1. 1.62208
2. 1.26208
3. 0.62208
4. 0.25208
52. Given x
+ y
, y(1) = 0, y(1) = 0, y(1.1) = 1.11073, y(1.2) = 0.24631, y(1.3) =
2
2
0.4157, y_1
= 0.11072, y_2
= 1.50067, y_3
= 1.86104 by Adams-Moulton method,
1
1
1
= y(1.4) =
1. 0.62274
2. 1.62274
3. 0.26274
4. 1.26274
53. Which is a step by step method
1. Taylor's Series
2. Adams Bashfortion
3. McGauran's Series
4. Modified Euler's method
54. Given 1 + y
, y(0) = 0, y(0.2) = 0.2027, y(0.4) = 0.4228, y(0.6) = 0.6841, y_0
=
2
1
1.0, y_1
= 1.041099 y_2
= 1.17876, y_3
= 1.46799 by Adams-Moulton method, y_4
1
1
1
c
= y(0.8) =
1. 0.0296
2. 2.0296
3. 1.0296
4. 3.0296
55. Given y(0.2) = 2027, y(0.4) = 0.4228, y(0.6) = 0.6841, ,
y_1
= 1.041099, y_2
= 1.17876, y_3
= 1.46799 by Adam=Moulton method, y_4
= y
1
1
1
P
(0.8) =
1. 0.02337
2. 1.02337
3. 0.3237
4. 1.3237
56. The period of sin2x is
1. 2
p
2.
p
3.
p
/2
4.
p
/4
57. In the Fourier expansion of f(x) defined in the internal ( + 2
p
), a
=
n
1.
2.
3.
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4.
58. If a function f(x) in defined in the internal ( + 2
p
) the Fourier expansion is
1.
2.
3.
4.
59. If x = a is a point of discontinuity of the function f(x), then the Fourier series of f(x) is
convergent to
1. point itself
2. f(a)
3.
4.
60. If x = a is a point of continuity of the function f(x) then the Fourier series of f(x) is
convergent to
1. f(a)
2. point itself
3.
4.
61. In the Fourier expansion of f(x) = 2x - x
in (0, 3) a
=
2
n
1.
2.
3.
4.
62. In the Fourier expansion of f(x) = in (-
p
,
p
), b
=
n
1.
2.
p
3.
4. 0
63. In the Fourier expansion of = in -
p
x
p
, a
=
0
1.
p
e
p
2. 1
3.
4. 0
64. In the Fourier expansion of f(x) = cos
a
x in (0. 2
p
)
a
being non-integral a
=
0
1.
2.
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3.
4.
65. The Fourier series of f(x) = 0, -
p
x 0
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = sinx, 0 x
p
at x =
p
/2 is convergent to
1. 0
2.
p
3. 1
4. -
p
66. If
x then the series at x = 0 becomes
1.
2.
3.
4.
67. In the Fourier expansion of f(x) = e defined in 0 x 2
p
, a
=
n
1. )
2. )
3. )
4.
68. The Fourier series of f(x) = -
p
, -
p
x 0
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = 0, 0 x
p
at x = 0 is convergent to
1. -
p
/2
2.
p
/2
3. 0
4.
p
69. In the Fourier expansion of f(x) in (c, c + 2l) b
=
n
1.
2.
3.
4.
70. In the Fourier expansion of f(x) = l - x, 0 x
=
l
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = 0, l x
=
2l
a
=
0
1. /2
2.
3. -
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4. - /2
71. The Fourier expansion f(x) = -k, -
p
x 0
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = k, 0 x x
is given by
1.
2.
3.
4.
72. In the Fourier expansion of =
1. 0
2.
3.
4.
p
73. In the Fourier expansion of =
1.
2.
3.
p
4. 0
74. The function
1. odd
2. even
3. constant function
4. neither even nor odd
75. The function is
1. odd
2. even
3. constant function
4. neither even nor odd
76. Expanding the function f(x) = x sin x as a fourier series in the interval
a
=
1
1.
2.
3.
4.
77. The Fourier expansion of an even function f(x), - x contains
1. Sine terms only
2. Both sine and cosine terms
3. Cosine terms only
4. constant terms
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78. The Fourier expansion of an odd function f(x), in - x contains
1. cosine terms only
2. both sine and cosine terms only
3. sine terms only
4. constant terms
79. The even function f(x) in the interval - x is symmetrical about
1. y - axis
2. x - axis
3. y = x
4. both the axis
80. The odd function f(x) in the interval - x is symmetrical about
1. y - axis
2. y = x
3. x - axis
4. in opposite quadrants
81. For half range cosine expansion of f(x) in (- , ) an =
1.
2.
3.
4.
82. The half-range cosine expansion of x in 0 x 2 is given by
1.
2.
3.
4.
83. The half range sine series for e
in 0 x 1 in given by
x
1.
2.
3.
4.
84. In the half range cosine expansion of f(x) = x
in 0, 1) a
=
3
n
1.
2.
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3.
4.
85. If f(x) = (x - 1)
in 0 x 1, the half range cosine series is given by
2
1.
2.
3.
4.
86. Fourier integral of f(x) is
1.
2.
3.
4.
87. Fourier sine integral is
1.
2.
3.
4.
88. Fourier cosine integral is
1.
2.
3.
4.
89. For half range sine expansion of f(x) in (- , ), b
=
n
1.
2.
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3.
4.
90. Fourier Transform of f(x) is
1.
2.
3.
4.
91. Fourier sine transform of f(x) is
1.
2.
3.
4.
92. If F
(s) and F
(s) are the Fourier sine and cosine transform of f(x) respectively then F
[f
s
c
s
(x)]=
1
1. sF
(s)
c
2. -sF
(s)
c
3. -sF
(s)
s
4. sF
(s)
s
93. Fourier cosine transform of f(x) is
1.
2.
3.
4.
94. Inverse Fourier sine transform of F
(s) is
s
1.
2.
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3.
4.
95. Inverse Fourier cosine transform of F
(s) is
c
1.
2.
3.
4.
96. In F
(s) is the Fourier sine transform of f(x),
s
1.
2.
3.
4.
97. If F
(s) and F
(s) are the Fourier sine and cosine transform of f(x) respectively then Fc
s
c
[x f(x)] =
1.
2.
3.
4.
98. If F
(s) and F
(s) are the Fourier sine and cosine transform of f(x) respectively then Fs
s
c
[x f(x)]
1.
2.
3.
4.
99. If Fc (s) is the Fourier Cosine transform of f(x) F
[f(x) sin ax].
c
1.
2.
3.
4.
100. If F
(s) and F
(s) are the Fourier sine and Cosine transform of f(x) respectively, then F
s
c
s
[f(x) sin ax]
1.
2.
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3.
4.
101.
1.
2.
3.
4.
p
102. F
(e ) =
c
1.
2.
3.
4.
103. The Fourier transform of is
1.
2.
3.
4.
104. Fourier transform of is
1.
2.
3.
4.
105. Fourier cosine transform of f(x) where is
1.
2.
3.
4.
106. Fourier cosine transform of is
1.
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2.
3.
4.
107. Fourier transform of is
1.
2.
3.
4.
108. If F(s) is the Fourier transform of f(x), then F[f (ax)]
1.
2.
3.
4.
109. If F
(s) is the Fourier cosine transform of f(x) then
c
1. F
(a s)
c
2. - F (as)
3.
4.
110. =
1.
2.
3.
4.
111.
1.
2.
3.
4.
112. Fourier sine transform of is
1.
2.
3.
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4.
113. F
(e ) =
s
1.
2.
3.
4.
114. F
[f(x)] where f(x) = 2e + 5e is
s
1.
2.
3.
4.
115. F
[f(x)] where f(x) = 2e + 5e is
c
1.
2.
3.
4.
116. If then =
1.
2.
3.
4.
117. If
1.
2.
3.
4.
118. F
[f(x)] where is
s
1.
2.
3.
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4.
119. If , then f(x) =
1. e
x
2. e
3. e
4. e
120. If inverse Fourier sine transform of as then
1.
2.
3. 1
4. 0
121. If , then F[f(x)] =
1.
2.
3.
4.
122. If
then F[f(x)]
1.
2.
3.
4.
123. Finite Fourier sine transform of f(x) = x
, 0 x 4 is
2
1.
2.
3.
4.
124. Finite Fourier sine transform of f(x) = x, 0 x 4 is
1.
2.
3.
4.
125. Finite Fourier cosine transform of f(x) x, 0 x 4 is
1.
2.
3.
4.
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126. Finite Fourier cosine transform of in (0,
p
) is
1.
2.
3.
4.
127. Finite Fourier sine transform of , 0 x
p
is
1.
2.
3.
4.
128. Finite Fourier sine transform of f(x) in 0 x l is
1.
2.
3.
4.
129. Finite Fourier cosine transform of f(x) in 0 x l is
1.
2.
3.
4.
130. Finite cosine transform of f(x) = x
, 0 x 4
2
1.
2.
3.
4.
131. Finite sine transform of , 0 x
p
is
1.
2.
3.
4.
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132. The partial differential equation formed by eliminating the arbitrary constants a and b
from z = (x
+ a) (y
+ b)
2
2
1. xyz = pg
2. yz = p
3. 4xyz = pq
4. xyz = q
133. The partial differential equation formed by eliminating the arbitrary constants a and b
from (x - a)
+ (y - b)
+ z
= r
2
2
2
2
1. z
(p
- q
+ 1) = r
2
2
2
2
2. z
(p
+ q
+1) = r
2
2
2
2
3. z
(p
+ q
- 1 ) = r
2
2
2
2
4. z
(p
q
1) = r
2
2
2
2
134. Partial differential equation of all planes passing though the origin is
1. z = py + qx
2. z = px - qy
3. z = px + qy
4. z = py - qx
135. Partial differential equation of all spheres whose centers lie on z-axis given radius r is
1. xp - yq
2. xq - yp = 0
3. xp + yq = 0
4. xq + yp = 0
136. Partial differential equation formed from z = f(x
+ y
+ z
) is
2
2
2
1. py = qx
2. px = qy
3. xy = pq
4. xyp = q
137. The partial differential equation formed by eliminating the arbitrary constant a and b
from z = ax + by + ab is
1. z = p + q + pq
2. Z = px + qy + pq
3. z = py + qx
4. z = py + qx + pq
138. Partial differential equation formed from z = f(x
+ y
) is
2
2
1. py - xq = 0
2. py - qx = 0
3. py + xq = 0
4. px + qy = 0
139. Partial differential equation formed from z = is
1. px - qy = 0
2. px - qy = 0
3. py + qx = 0
4. px + qy = 0
140. Partial differential equation formed from z = f
(x) f
(y) is
1
2
1. pq = zp
2. pq = zq
3. pq = zs
4. pq = zr
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141. Partial differential equation forme from z = f(x + it) + g( x- it) is
1.
2.
3.
4.
142. Partial differential equation formed from Z = is
1. px + qy = 0
2. py - qx = 0
3. px - qy = 0
4. py + qx = 0
143. The partial differential equation formed from xyz = Ø (x + y + z ) is
1. x(y + z) + y(z + x) = z(x + y)
2. x(y - z) + y(z - x) = z(x - y)
3. y(y z) + x (z- x) = (x y)
4. z(y z) + z (z- x) = (x y)
144. The solution of px + qy = z is
1.
2.
3.
4.
145. One of the solutions of the differentia equation
x(y
- z
)p + y (z
- x
)q = z (x
- y
) taking x, y, z as multipliers
2
2
2
2
2
2
1. x
+y
-z
= a
2
2
2
2. x
+y
+z
=a
2
2
2
3. x
-y
-z
= a
2
2
2
4. x
-y
+z
= a
2
2
2
146. The solution of (y
+ z
)p - xyq = -zx is
2
2
1. Ø(xd+y+z, xyz) = 0
2. Ø(x+y+z, ) = 0
3. Ø(x
+y
+z
, ) = 0
2
2
2
4. Ø(x
+y
+z
, ) = 0
2
2
2
147. One of the solutions of the differential equation
x(y - x)p + y(z - x)q = z(x- y) taking 1, 1, 1 as multiplexes is
1. x+y+z =a
2. x-y-z = a
3. x + y - z = a
4. x - y + z = a
148. One of the solution of the differential equation
x(y - z)p + y(z - x) q = z (x - y) taking as multipliers is
1. xy = za
2. xz = ay
3. yz = ax
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4. xyz = a
149. One of the solutions of the differential equation
x(y
- z
)p + y (z
- x
)q = z (x
- y
) taking as multipliers is
2
2
2
2
2
2
1. xy = za
2. xz = ay
3. xyz = a
4. yz = ax
150. The solution is
1.
2.
3.
4.
151. The solution px + qy = nz is
1.
2.
3.
4.
152. The solution of xzp + y = q = xy
1.
2.
3.
4.
153. The solution of z = pq is
1. (x + ay + c)
= 4 az
2
2. (x + ay + c)
= 4 ax
2
3. (x + ay + c)
= 4 ay
2
4. (x + ay + c)
= 4 axy
2
154. The solution of pq = y is
1.
2.
3.
4.
155. The solution of p + q = 1 is
1. Ø (x - z, y - z) = 0
2. Ø (x+z, y-z) = 0
3. Ø (x + z, y + z) = 0
4. Ø (x - z, y + z) = 0
156. Complete integral of p
+ q
= 1 is
2
2
1.
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2.
3.
4.
157. The singular integral of pq = k is
1.
2.
3. z = ax + y
4. no singular integral
158. The complete solution of is
1.
2.
3.
4.
159. The complete solution of p = q
is
2
1. z = bzx + by + c
2. z = -bzx + by + c
3. z = -bzx - by+c
4. z = bzx - by + c
160. The solution of pe
= qe
is
y
x
1. z = ae
+ c
x
2. z = ae
+ c
y
3. z = ae + c
4. z = a(e
+ e
) + c
x
y
161. The solution of x
p
+ y
q
= z
2
2
2
2
2
1. y = a logx
2.
3. x
z
+ y
b
= z
2
2
2
2
2
4. y = a logx + logy + c
162. Solution of p
+ q
= x + y is
2
2
1.
2.
3.
4. x + y = a
+ b
2
2
163. The solution of px = qy is
1. z = logx + logy + c
2. z = logx
+ logy
+ c
a
a
3. z = xyz b + c
4. z = xy + c
164. The solution of z = px + qy - 2 is
1.
2.
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3.
4.
165. The solution of z
= pqxy is
2
1.
2.
3. axy
4. abxy
166. The solution is
1.
2.
3.
4.
167. Solution by p
- q
= x - y is
2
2
1.
2.
3.
4.
168. The general solution of dx = dy = dz is
1. f(x - y, y - z) = 0
2. f(x + y, y +z) = 0
3. f(x + y z, x+ y) = 0
4. f(x + y +2z, x - y ) = 0
169. The general solution of p
= qz is
3
1. 4z=a(x+ay-b)
2
2. 4z=a(x-ay-b)
2
3. 4z=a(ay-x-b)
2
4. 4z=a(b-ay-x)
2
170. If the solution of the equation by the method of separation of variables
is z = X(x) Y(y) then X(x)
1.
2.
3.
4.
171. z(n
) =
2
1.
2.
3.
4.
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172. z(cosn
) =
1.
2.
3.
4.
173. z(sin n
) =
1.
2.
3.
4.
174. The general solution of p cot x + q cot y = cot z is
1.
2.
3.
4.
175. If the solution of the equation = 0 by the method of separation of variables
is z = X(x) Y(y) the Y(y) =
1.
2.
3.
4.
176. If the solution of the equation by the method of separation of variables
is u = X(x) Y(y) then X(x)
1.
2.
3.
4.
177. If z(4
) = there z(a u
) =
n
n
1.
2.
3.
4.
178. If z(4
) = then z(a
u
) =
n
n
n
1.
2.
3.
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4.
179. z(-1)
=
n
1.
2.
3.
4.
180. z[4z
+ 5(-1)
]
n
n
1.
2.
3.
4.
181. z(coshn
) =
1.
2.
3.
4.
182. z(n
) =
4
1.
2.
3.
4.
183. Z(sinh n
) =
1.
2.
3.
4.
184.
1.
2.
3.
4.
185.
1.
2.
3.
4.
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186.
1. (n - 1)a
n
2. (n + 1)a
n
3. na
n
4. a
n
187.
1.
2.
3.
4.
188.
1. 1
2. 4
3. 2
4. 0
189. 1 * 1 * 1 =
1. 3
2. 2
3. 1
4. 0
190. for 2 > > 3 is given by
1. and
2. and
3. and
4. and
191. for > 3 is given by
1.
2.
3. and 0 , n < 0 and 0, n< 0 and 0, n < 0
4. 0, n 1
192.
is given by
1.
2.
3.
4.
193. If
1.
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2.
3.
4.
194. Solution of difference equation = 1, y0 = 0 is
1.
2. - 2(-1)
3.
4. 2(-1)
195. Solution of difference equation y_{n + 2} - 4y_{n - 1} + 3y_n = 5
is
n
1.
2.
3.
4.
196. Solution
by z-transform is
1.
2.
3.
4.
197. The solution of by z-transform is
1.
2.
3.
4.
198. where
1.
2.
3.
4.
199. =
1. 3
2. 4
3. 2
4. 1
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