vidyalankar applied mathematics - iii prelim question...
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1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 1
Vidyalankar S.E. Sem. III [BIOM]
Applied Mathematics - III Prelim Question Paper Solution
f(t) =
1
2
1L
s 2s 2 =
12
1L
s 1 1 =
t 1
2
1e L
s 1
= et sin t
s1
2
eL
s 2s 2 = e(t) sin (t ) H(t )
ux = ex cos y + 3x2 3y2
uxx = ex cos y + 6x uy = ex sin y 6xy
uyy = ex cos y 6x uxx + uyy = ex cos y + 6x ex cos y 6x
= 0 u satisfies Laplace equation u is Harmonic.
If n is an even integer then since, n m 1 = (m + n)! , we have
Jn(x) =
m 2m n
m 0
( 1) (x / 2)m!(m n)!
Jn (x) =
m 2m n
2m nm 0
( 1) ( x)
m!(m n)!2
If n is even, (2m + n) is even & (x)2m+n = x2m+n
Jn (x) =
m 2m n
m 0
( 1) (x / 2)m!(m n)!
= Jn(x)
Hence, Jn(x) is even. Again, if n is odd then (2m + n) is odd & hence,
(x)2m+n = 2m n(x )
Jn(x) =
m 2m n
m 0
( 1) ( 1) (x / 2)m!(m n)!⋅ ⋅
=
m 2m n
m 0
( 1) (x / 2)m!(m n)!
= Jn(x)
Hence, Jn(x) is odd
1. (a)
1. (b)
1. (c)
Vidyala
nkar
Vidyalankar : S.E. – Maths III
1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 2
We have,
u = ax4 + bx2 y2 + cy4 + dx2 2y2
ux
= 4ax3 +2bxy2 + 2dx
uy
= 2bx2y + 4cy3 4y
v = 4x3y exy3 + 4xy vx
= 12xy2 ey3 + 4y
vy
= 4x3 3exy2 + 4x
f(z) is analytic. It satisfies Cauchy Riemann equations.
ux
= vy
4ax3 + 2bxy2 + 2dx = 4x3 3exy2 + 4x Comparing both sides, we get
4a = 4 a = 1 2b = 3e …(1) 2d = 4 d = 2
Also uy
=
vx
2bx2y + 4cy3 4y = 12x2y + ey3 4y
Comparing both sides, we get 2b = 12 b = 6 4c = e …(2)
From (1), we get 2b = 3e
12 = 3e e = 4
From (2), we have 4c = e 4c = 4 c = 1
a = 1, b = 6, c = 1, d = 2, e = 4
a0 =
0
2f(x) dx =
0
2x dx =
2
0
2 x2
=
an =
0
2 n xf(x) cos dx =
0
2 n xx cos dx
1. (d)
2. (a)
Vidyala
nkar
Prelim Question Paper Solution
1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 3
=
2 2
20
n x n xsin cos
2 xn n
=
n
2 22 2
22
1( 1) 02 0n nn
=
n 2 2
2 2 2 2
2 ( 1)
n n =
n2
2 2
2 1 1
n
=
n
2 2
( 1) 12
n
=
2 2
4 , if n is oddn
0, if n is even
The Half Range cosine series is given by
f(x) =
0n
n 1
a n xa cos
2
=
n
2 2n 1
( 1) 1 n x2 cos
2 n
The curve C is the boundary of the triangle in
xoy plane as z = 0.
F = 2 2ˆ ˆ ˆy i x j (x z) k where z = 0
= 2 2ˆ ˆ ˆy i x j x k
F =
2 2
ˆ ˆ ˆi j k
x y z
y x x
= ˆ ˆ ˆ0 i j ( 1) 2(x y) k
= ˆ ˆj 2(x y) k
(F ) . n̂ =
ˆ ˆj 2(x y) k . k̂ as n̂ = k̂ .
= 2(x y) We have by Stoke’s theorem.
C
F. dr =
S
ˆ( F) . n ds
= S
2(x y) ds = R
2(x y) dx dy
= 1 x
0 0
(x y) dy dx =
x1 2
0 0
y2 xy dx
2
2. (b)
x
y
z
A B
C
O (0, 0, 0)
(1, 0, 0)
(1, 1, 0)
O x
y
B
A
(1, 0) (0, 0)
C y=x
(1, 1)
Vidyala
nkar
Vidyalankar : S.E. – Maths III
1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 4
=
1 2
2
0
x2 x dx
2 =
1 2
0
x2 dx
2
=
13
0
x3
= 13
(i)
3L H t
2 =
3
s21
es
as1
1 eH(t a)s
L sin t H t2
=
s2e L sin t
2 L f(t)H(t a) = ase L f(t a)
=
s2e L cos t
=
s2
2
se
s 1
3
L sinH t H t2 2
=
3
L sin t H t L H t2 2
=
s 3s
2 22
s 1e e
ss 1
(ii) Since, n
nd
x J xdx
= xn Jn + 1 (x)
nn
dx J x dx
dx =
nn 1x J x dx + C
xn nJn (x) = n
n 1x J x dx + C … (1)
Put n = 2 in equation (1), 2
3x J x dx = x2J2 (x) + C … (2)
Putting n = 1 in equation (1), 1
2x J x dx = x 1J1 (x) + C … (3)
Now, 3J x dx = x2 (x2) J2 (x) + 122 x J x dx + C
3J x dx = J2 (x)
12J x
x + C
Find the Fourier series for f(x) = | x | in (, ).
Hence deduce that 2
8 =
2 2 2 2
1 1 1 1...
1 3 5 7
a0 =
1
f(x) dx =
0
0
1x dx x dx
3. (a)
2. (C)
Vidyala
nkar
Prelim Question Paper Solution
1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 5
=
02 2
0
1 x x2 2
=
2 210
2 2
= 2
1 =
an =
1
f(x) cos nx dx =
1
| x | cos nx dx =
0
2x cos nx dx
=
20
2 sinnx cos nxx (1)
n n =
2 2
2 cos n 1
n n
=
n
2 2
2 ( 1) 1
n n =
n
2
2 ( 1) 1
n
f(x) is even function bn = 0.
f(x) =
0
n 1
aan cos nx
2
| x | =
n
2n 1
2 ( 1) 1cos nx
2 n =
2 2 2
2 2 2 2cos x cos 3x cos 5x ...
2 1 3 5
= 2 2 2
4 cos x cos 3x cos 5x...
2 1 3 5
Putting x = 0, we get
0 = 2 2 2
4 1 1 1...
2 1 3 5
2
8 =
2 2 2
1 1 1...
1 3 5
t tr e cos t i e sin t j i.e. x = et cos t ; y = et sin t
dx = et (cos t – sin t) dt, dy = et (sin t + cos t) dt t tdr [e (cos t sin t) i e sin t cos t) j ] dt x2 + y2 = e2 t (cos2 t + sin2 t) = e2 t
(x2 + y2)3/2 = (e2 t)3/2 = e3 t
F .dr =
t
3 t
e (cos t i sin t j)
e. et [ (cos t sin t) i + (sin t + cos t) j ] dt
= 2t
3t
e{cos t (cos t sin t) sin t (sin t cos t)} dt
e
= et dt
0t 0-t t
11
e 1F .dr e dt e 1
1 e
1 e
F .dre
3. (b)
Vidyala
nkar
Vidyalankar : S.E. – Maths III
1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 6
f (x)
0
n nn 1
aa cosnx b sinnx
2 … (1)
an =
1
f(x)dx =
0 π
π 0
1 π dx x dxπ
=
20
0
1 xπxπ 2
=
221 π
π 2
=
21π 2
= 2
an = π
-
1f(x) cosnx dx
π =
0
0
1 cos nx dx x cos nx dx
π
=
0
20
1 sinnx sinnx cosnxx 1
n n n
= 2 2
1 cosn 10 0 (0) 0
n n
=
n
2 2
11 1
n n =
n
2
11 1
n
bn = π
-
1f(x) sin nx dx
π =
0
0
1 sin nx dx x sin nx dx
π
=
0
20
1 cosnx cosnx sinnxx 1
n n n
=
1 1 cosn cosnx0
n n
=
11 cosn cosn
n n =
11 2cosn
n
= n1
1 2 1n
Putting these values in (1), we get
f (x) =
n n
2n 1 n 1
1 1 1 2 11cosnx sinnx
4 nn
= 2 2 2
2 cos x cos3x cos5x.....
4 1 3 5
sinx 13 sin2x sin3x .....
1 2
… (2)
3. (c)
Vidyala
nkar
Prelim Question Paper Solution
1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 7
Now f(x) is discontinuous at x=0. At a point of discontinuity x = c
f (x) =
x c x c
1lim f(x) lim f(x)
2
f (0) = 1
02
= 2
Hence substituting x=0 in equation (2), we get
2
= 2 2 2
2 1 1 1.........
4 1 3 5
2
8 =
2 2 2
1 1 1........
1 3 5
=
2n 1
1
2n 1
L{sin 3u} = 2
3
s 9
L{u sin 3u} = 1
2
d 31
ds s 9 nL t f(t) = n d
1 L f(t)ds
=
22
3(2s)
s 9
=
22
6s
s 9
t
0
L u sin u du =
22
1 6ss s 9
=
22
6
s 9
t
0
(s)L f(u)du
s
t
4t
0
L e u sin 3u du =
22
6
s 4 9 =
22
6
s 8s 16 9
=
22
6
s 8s 25
(ii) L{1cos t} = L{1} L{cos t}
= 2
1 ss s 1
1 cos tL
t =
2s
1 sds
s s 1
s
f(t)L (s) ds
t
=
2
s
1log s log s 12
= 2
s
12 logs log s 12
4. (a)
Vidyala
nkar
Vidyalankar : S.E. – Maths III
1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 8
=
2 2s
1log s log s 1
2 =
2
2s
1 slog
2 s 1
=
2
2
1 slog
2 s 1 =
2
2
1 slog
2 s 1
Let u = 3x2y + 2x2 y3 2y2
ux
= 6xy + 4x
uy
= 3x2 3y2 4y
We know that
dv =
v v
dx dyx y
=
u u
dx dyy x
= 2 23x 3y 4y dx 6xy 4x dy
= 2 23x 3y 4y dx 6xy 4x dy Integrating, we get
v = x3 + 3xy2 + 4xy
orthogonal trajectory of family of curves is given by x3 + 3xy2 + 4xy = C1 x3 3xy2 4xy = C when C = C1
(i) We know that x +iy = z …(1)
x iy z …(2) From (1) and (2) we get
1
x z z2
, i
y (z z)2
x 1z 2
y iz 2
x 1
2z
y i
2z
We know that
z
=
x yx z y z
=
1 ix 2 y 2
=
1i
2 x y … (3)
z
=
x yx yz z
=
1 ix 2 y 2
=
1i
2 x y … (4)
4. (b)
4. (c)
Vidyala
nkar
Prelim Question Paper Solution
1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 9
From (3) and (4) we get
2
z z =
z z
=
1i i
4 x y x y
=
2 2 2 2
2 2
1i i
4 x y x yx y =
2 2
2 2
14 x y
2
4z z
=
2 2
2 2x y
(ii) Note that, log f ' (z) = 21log f ' (z)
2 =
1log [f '(z). f '(z)
2
= 1 1
log f '(z) log [f '(z)]2 2
L.H.S. =
2 2
2 2log| f '(z)|
x y
= 4
2 1 1log f ' (z) log f '(z)
2 2z z
= 2
log f '(z) 2 logf '(z)z z zz
= 2
(0) 2 (0) 0
z z (z, z are independent)
By Green’s Thm.
C
pdx Q dy =
R
Q Pdx dy
x y
Now C
f d r = 2 2
C
x y i xy j dx idy
= 2 2
C
x y dx xy dy
By comparison P = x2y, Q = xy2
Qx
= y2, Py
= x2
C
f d r = 2 2
R
y x dx dy
put x = r cos, y = r sin, dx dy = r dr d
=
r a(1 cos )
2
0 r 0
2 r r dr d
5. (a)
y
x
0
r = a(1+cos)
Vidyala
nkar
Vidyalankar : S.E. – Maths III
1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 10
=
a (1 cos )4
00
r2 d4
=
4 4
0
1a (1 cos ) d
2
=
442
0
a2 cos d
2 2 =
4 8
0
8a cos d2
Put 2
= t, d = 2dt & using reduction formula,
C
f d r =
/2
4 8
0
8a 2cos t dt
=
4 7 5 3 116a
8 6 4 2 2
= 435 a
16
Since w =
5 4z4z 3
We have 4wz 3w = 5 4z z(4w + 4) = 5 + 3w
z =
5 3w4(w 1)
when |z| = 1,
5 3w4(w 1)
= 1
5 3(u iv)(u iv 1)
= 4
5 3u i3vu 1 iv
= 4
2 25 3u 9v = 2 216 u 1 v
25 + 30u + 9u2 + 9v2 = 16(u2 + 2u + 1 + v2) 16u2 9u2 + 32u 30u + 16 25 + 16v2 9v2 = 0 7u2 + 2u 9 + 7v2 = 0 7u2 + 2u + 7v2 = + 9
2 22u u v
7 =
97
2 22 1u u v
7 49 =
9 17 49
221
u v7
= 6449
centre =
1,0
7, radius =
87
5. (b)
Vidyala
nkar
Prelim Question Paper Solution
1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 11
i is root of J0(x) = 0 Let the Bessel fourier series be,
x2 = f(x) = 0Ci J ( ix)
Multiplying by xJ0 (nx) on both sides and integrating from 0 to 1.
1
20
0
x x J ( nx) = 1
0 n 0 ii 1 0
Ci x J ( x) J ( x) dx
=
12
n 0 n0
C x J ( x)
= 21 n
CnJ ( )
2
Cn = 1
30 n2
1 n 0
2x J ( x) dx
J ( )
nn
dx J (x)
dx = xn Jn1 (x)
xnJn (x) = nn 1x J (x) dx
1
30 n
0
x J ( x) dx =
131 n
n 0
x J ( x)
= 1 n
n
J ( )
Cn = 21 n
2
J ( ) 1 n
n
J ( )
= n 1 n
2J ( )
x2 = 2i
0 i3i 1 i 1 i
42 J ( x)
J ( )
Consider f (t) = cos 4t cos 3t
L
f(t)t
= 2 2 2 2s
s sds
s 4 s 3
=
2 2 2 2
s
1log (s 4 ) log (s 3 )
2 =
2 2
2 2s
1 s 3log
2 s 4
=
2 2
2 2
1 s 3log
2 s 4 …(1)
By definition of Laplace Transform The equation (1) means
st
0
cos4t cos3te dt
t =
2 2
2 2
1 s 3log
2 s 4
Putting s = 0
0
cos4t cos3tdt
t =
2
2
1 3log
2 4 =
3log
4
6. (a)
5. (c)
Vidyala
nkar
Vidyalankar : S.E. – Maths III
1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 12
= 3z + 4 + 2i Let = u + iv and z = x + iy u + iv = 3 (x + iy) + 4 + 2i
= 3x + 3iy + 4 + 2i u + iv = (3x + 4) + i (3y + 2) u = 3x + 4 and v = 3y + 2
Now | z | = k i.e. 2 2x y k
x2 + y2 = k2 This is a circle with centre (0, 0) and radius k.
x = 2 2k y
x = k if y = 0 (k, 0) x = 0 if y = k (0, k) u = 3k + 4 and v = 3k + 2 and u = 4 and v = 2 u = 3k + 4 and v = 2 and u = 4 and v = 3k + 2 are the images of (k, 0) and (0, k) for | z | = k.
a0 =
2
0
1f(x) dx =
22
0
1 xdx
2 =
23
0
1 ( x)4x 3
=
3 31
12 =
3212
= 2
6
an =
2
0
1f(x) cos nx dx =
22
0
1 xcos nx dx
4
=
22
2 30
1 sinnx cos nx sinnxx 2 2( 1)( 1)x 1
4 n n n
= 2 2
1 cos 2 n 20 2 0 0 0
4 n n
= 2 2
1 2 24 n n
= 2
1
n [ cos 2n = 1]
bn =
2
0
1f(x) sinnx dx =
2 2
0
1 ( x)sinnx dx
4
=
22
2 30
1 cos nx sinnx cos nxx 2 x 1 2( 1)( 1)
4 n n n
=
2 2
3 3
1 cos 2 n 2 cos 2n 20 0
4 n nn n
6. (b)
6. (c)
Vidyala
nkar
Prelim Question Paper Solution
1113/Engg/SE/Pre Pap/2013/BIOM/Maths III_Soln 13
bn =
2 2
3 3
1 2 24 n nn n
= 0
Now, f(x) =
0
n 1
aan cos nx bn sinnx
z
2x
2 =
2
2n 1
cos nx12 n
=
2
2 2 2 2
cos x cos 2x cos 3x...
1 1 2 3
Put x = 0, we get
2
4 =
2
2 2 2 2 2
1 1 1 1...
1 1 2 3 4
2
6=
2 2 2 2
1 1 1 1...
1 2 3 4
Vidyala
nkar