MODEL ANSWER
Subject: ENGINEERING MATHEMATICS
Que.
Sub.
No.
Que.
1.
Subject Code: 17216
Model answers
Marks
Attempt any TEN of the following:
Total
Marks
09
(a)
Ans.
x 1 i
y 2 i
6 0
x xi 2 y iy 6 0
x 2y 6
x
y i
0
½
By equality
x 2y 6
0
and
x
y
By solving above equations.
x
2
y
2
0
½
1
1
03
(b)
Ans.
x 1 , y
r
x2
3
y2
tan
1
= tan
1
= tan
1
3
12
2
3
2
y
x
1
1
03
3
1
3
or 600
1
c)
Ans.
Let f x
f 2
f 3
x3 6 x 3
23 6 2
33 6 3
3
1 0
2
3 12 0
root lies between 2 & 3
1
03
No.
1.
Sub.
Model answers
Que.
Marks
Total
Marks
(d)
Ans.
1
18 2 y 3z
5
1
y
22 x 3z
7
1
z
22 2 x y
6
Initial iterations: x0
1
x
03
y0
z0
0
First iteration:
x1
y1
z1
1
18 2 y0 3z0 3.6
5
1
22 x1 3z0
3.6571
7
1
22 2 x1 y1 1.8572
6
2
Que.
Sub.
No.
Que.
2.
Model answers
Marks
Attempt any FOUR of the following:
Total
Marks
16
(a)
Ans.
z z
3 4i
1 i
1
3 4i
=
½
1 i
3 4i
2
1 i
2
9 24i 16i 2 1 2i i 2
=
3 4i - 3i - 4i 2
9 24i 16 1 2i 1
=
3 i 4
7 22i
=
7 i
=
=
=
=
=
½
1 i 3 4i
7 22i 7 i
7 i
7 i
49 7i 154i 22i 2
72 i 2
49 161i 22
49 1
27 161i
50
27 161
i
50 50
z
z
3
1
3
=
1
3
=
4i
i
4i 1 i
i 1 i
3i 4i 4i 2
12 i 2
3 7i 4
1 7i
=
=
1 1
2
1 i
3 4i
1
1 i 3 4i
3 4i 3 4i
3 4i 3i 4i 2
=
=
32
4i
2
i2
1
1
½
½
i2
1
½
½
OR
½
1
½
04
Que.
Sub.
No.
Que.
Model answers
Marks
Total
Marks
2.
3 7i 4
9 16
1 7i
=
25
1 7i
1 7i
z1
2
25
25+175i 2 14i
=
50
27+161i
=
50
27 161
=
i
50 50
=
z
1
½
½
(b)
Ans.
Let z=1+i
12 12
r
2
x +ve and y +ve
1
= tan
=tan
1
y
x
1
1
=tan 1 1
1
4
Polar form of a complex no.is,
z
r cos
1 i
i sin
2 cos
4
i sin
4
Similarly,
1 i
2 cos
4
i sin
4
1
04
Que.
Sub.
No.
Que.
2.
Model answers
1 i
8
Marks
8
i sin
4
8
2 cos
4
8
8
=
2
Marks
8
1 i
2 cos
Total
cos
i sin
4
4
i sin
4
8
8
2
4
cos
i sin
4
½
4
By De-Moivre's theorem
=16 cos
8
4
8
4
i sin
=16 cos 2
16 cos
i sin 2
8
4
16 cos 2
i sin
8
4
1
i sin 2
=32cos2
½
= 32
04
(c)
Ans.
e 2i
cos 2
e
2i
=
=
=
=
=
ei
sin 2
e
2
2
i
ei
e
2i
i
2
e 2i
2ei e
4
i
e
2i
e 2i
2ei e i
4i 2
e
2i
e 2i
2ei e
4
i
e
2i
e 2i
2ei e
4
e
2i
e 2i
2ei e
i
e
2i
e 2i
4
2e 2i
2e
cos 2
2ei e
i
i
e
1
i2
1
1
2i
½
2i
½
4
2i
e
e 2i
=
2
From (1) and (2)
cos 2
½
...(1)
2
Consider cos 2
...(2)
½
04
sin 2
Que.
Sub.
No.
Que.
2.
(d)
Model answers
Marks
Total
Marks
½
Ans.
cos 3
i sin 3
cos 4
i sin 4
cos 4
i sin 4
cos 5
i sin 5
cos 3
i sin 3
cos 4
i sin 4
cos
i sin
cos
i sin
= cos
i sin
= cos
i sin
= cos 40
(a)
3.
Ans.
4
5
3
4
4
3
12
12
cos
i sin
cos
i sin
cos
i sin
i sin
cos 4
i sin 4
cos 5
i sin 5
i sin
cos
i sin
20
12
20
cos
cos
12
½
½
½
½
5
4
20
20
04
1
20 20
40
1
i sin 40
Using Bisection method, find the approximate root of
(three iterations only.)
xex 1
Let f x
f 0
1 0
f 1
1.71 0
1
the root lies in (0,1)
x1
a b
2
0 1
2
1
0.5
04
0.17 <0
f 0.5
the root lies in (0.5,1)
x1 b
2
x2
f x2
0.5 1
2
1
0.75
0.5 0
the root lies in (0.5,0.75)
x3
x2 b
2
0.5 0.75
2
0.625
1
(b)
Ans.
Using Regula – Falsi method, find the root of
(three iterations only).
x2 2 x 1
Let f x
f 2
1 0
f 3
2
0
1
the root lies in (2,3)
af b
f b
x1
bf a
f a
2 2
2
3
1
7
3
1
2.333
1
0.223 <0
f 2.333
04
the root lies in (2.333,3)
x1 f b
f b
x2
f x2
bf x1
f x1
1
2.333 2 3 0.223
2
0.223
2.40
0.04 0
the root lies in (2.4,3)
x2 f b
f b
x3
(c)
bf x2
f x2
2.4 2
2
3 0.04
0.04
2.412
1
Using Newton –Raphson method, find approximate root of
(three iterations only).
x3 2x 5 0
Let f x
f 2
1
f 3
16
f 2
1
f 3
16 f (2) is nearer to zero
f' x
3x2 2
f' 2
x1
x2
x0
1/2
2
½
10
x0
f x0
=2
f ' x0
x1
f x1
f ' x1
1
10
2.1
2.1
3
1
2 2.1
2.1
3 2.1
5
2.095
2
1
2
04
x3
(d)
x2
f x2
f ' x2
2.095
3
2 2.095
2.095
3 2.095
2
5
2.095
2
Using Newton –Raphson method, find the approximate value of
(three iterations only).
1
Ans.
3
Let, x
100
x3 100
x3 100 0
x3 100
f x
f 4
36
f 5
25
1
f (5) is nearer to zero
x1
f' x
3x 2
f' 5
75
x0
f x0
25
=5
'
75
f x0
x0
5
4.667
1
f 4.667
x2
x1
f 4.642
x3
x2
and
1.651
f x1
f ' x1
4.667
4.642
0.027
64.644
65.343
4.642
and
0.027
f x2
f ' x2
1.651
65.343
f ' 4.667
f ' 4.642
64.644
1
04
4.642
1

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