fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
MATHEMATICAL PHYSICS SOLUTIONS
GATE-2010
Q1.
Consider an anti-symmetric tensor Pij with indices i and j running from 1 to 5. The
number of independent components of the tensor is
(a) 3
Ans:
(b) 10
(c) 9
(d) 6
(b)
Solution: The number of independent components of the tensor
1 2
N  N   1 25  5  10
 N  5 
2
2
e z sin  z 
The value of the integral 
dz , where the contour C is the unit circle: z  2  1 ,
2
z
C
=
Q2.
is
(a) 2 i
Ans:
(c)  i
(b) 4 i
(d) 0
(d)
Solution:  z  2  1  1  z  3 i.e. the pole z  0 does not lie inside the contour.

Q3.
e z sin z
C z 2 dz  2 i  0  0 .
2

The eigenvalues of the matrix  3
0

(a) 5, 2, -2
Ans:
3
2
0
(b) -5, -1, -1
0

0  are
1 
(c) 5, 1, -1
(d) -5, 1, 1
(c)
Solution: The characteristic equation of the matrix A , A  I  0
2
3
0
3
2
0
0
0
1 
 A  I 
2
 0  1     2     9   0    1, 2    3


   5,1,  1
Q4.
0
If f  x   
x  3
(a) s 2 e3s
for x  3,
then the Laplace transform of f(x) is
for x  3
(b) s 2 e3s
(c) s 2
(d) s 2 e 3s
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Ans:
(d)

Solution: L f  x    e
0
 sx
3
f  x  dx   e
0
e  sx
L  f  x    x  3
s
Q5.
s x

3
f  x  dx   e
3
 sx
f  x  dx    x  3 e  sx dx
3


 e sx 
1  sx
1  e sx 
  1 



 s 2 e3 s
dx
0
e
dx




s3
s  s 3
 s 
3
The solution of the differential equation for y t  :
(a)


initial conditions y 0   0 and
d2y
 y  2 cosh(t ) , subject to the
dt 2
dy
 0 , is
dt t  0
1
cosh t   t sinh t 
2
(b)  sinh t   t cosh t 
(c) t cosh t 
Ans:

(d) t sinh t 
(d)
Solution: For C.F D 2  1y  0  m  1  C.F .  C1e t  C 2 e t
 e t  e t
1
1
2 cosh t   2 2
P.I .  2
D 1
D 1  2
 y  C1e t  C 2 e t 

t
t
1
1
  2
et  2
e t  e t   e t
2
2
D 1
 D 1
 
 


t t t t
e  e
2
2
As, y  0   0  C1  C2  0............ 1
1
1
dy
t
t
 C1e t  C 2 e t  e t  e t  e t  e t
2
2
2
2
dt
Also,
dy
1
1
 0  C1  C2  0   0   0  C1  C2  0...........  2 
dt t 0
2
2
From equation (1) and (2),
C1  0, C2  0 .
t
t
Thus y  et  e  t  y  t sinh t
2
2
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
GATE-2011
Q6.
Two matrices A and B are said to be similar if B = P-1AP for some invertible matrix P.
Which of the following statements is NOT TRUE?
Ans:
(a) Det A = Det B
(b) Trace of A = Trace of B
(c) A and B have the same eigenvectors
(d) A and B have the same eigenvalues
(c)
Solution: If A and B be square matrices of the same type and if P be invertible matrix, then
matrices A and B = P-1AP have the same characteristic roots.
Then, B  I  P 1 AP  P 1IP  P 1  A  I P where I is identity matrix.
B  I  P 1  A  I P  P 1 A  I P  A  I P 1 P  A  I PP 1  A  I
Thus, the matrices A and B (= P-1AP) have the same characteristic equation and hence
same characteristic roots or eigen values. Since, the sum of the eigen values of a matrix
and product of eigen values of a matrix is equal to the determinant of matrix, hence third
alternative is incorrect.
Q7.
If a force F is derivable from a potential function V(r), where r is the distance from the
origin of the coordinate system, it follows that
(a)   F  0
Ans:
(b)   F  0
(c)  V  0
(d)  2 V  0
(a)


Solution: Since, F is derivative of potential V(r) and F  V r 
 
   F    V  0 .
Q8.
A 33 matrix has elements such that its trace is 11 and its determinant is 36. The
eigenvalues of the matrix are all known to be positive integers. The largest eigenvalues of
the matrix is
(a) 18
Ans:
(b) 12
(c) 9
(d) 6
(d)
Solution: We know that for any matrix
1. The product of eigenvalues is equals to the determinant of that matrix.
2. 1   2  3  .......  Trace of matrix
1   2  3  11 and 12 3  36 . Hence, the largest eigen value of the matrix is 6.
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Q9.
The unit vector normal to the surface x2 + y2 – z = 1 at the point P(1, 1, 1) is
iˆ  ˆj  kˆ
(a)
Ans:
(b)
3
2iˆ  ˆj  kˆ
6
(c)
iˆ  2 ˆj  kˆ
(d)
6
2iˆ  2 ˆj  kˆ
3
(d)
Solution: The equation of the system is f x, y, z   x 2  y 2  z  1  0


 ˆ  ˆ 2
f   iˆ 
j  k x  y 2  z  1  2 xiˆ  2 yjˆ  kˆ

x

y
z 


f
2iˆ  2 ˆj  kˆ
.
Hence, unit normal vector at (1, 1, 1)   
3
f
Q10.
Consider a cylinder of height h and radius a, closed at both ends, centered at the origin.
Let r  iˆx  ˆjy  kˆz be the position vector and n̂ be a unit vector normal to the surface.
The surface integral  r  nˆ ds over the closed surface of the cylinder is
S
z
O
y
x
2
(a) 2πa (a + h)
Ans:
(b)
2
(b) 3πa h
(c) 2 πa2h
(d) zero
 
Solution:  r.nˆ ds   .r d  3 d  3a 2 h
S
Q11.
V
V
The solutions to the differential equation
dy
x
are a family of

dx
y 1
(a) circles with different radii
(b) circles with different centres
(c) straight lines with different slopes
(d) straight lines with different intercepts on the y-axis
Ans:
(a)
Solution:
x2 y2
dy
x


 y  C1  x 2  y 2  2 y  2C1
 xdx  ydy  dy  0 
dx
y 1
2
2
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Phone: 011-26865455/+91-9871145498
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
  x  0    y  1  2C1  1  C
2
2
which is a family of circles with different radii.
Q12.
Which of the following statements is TRUE for the function f  z  
z sin z
 z   2
?
(a) f  z  is analytic everywhere in the complex plane
(b) f  z  has a zero at z  
(c) f  z  has a pole of order 2 at z  
(d) f  z  has a simple pole at z  
Ans:
(c)
Solution: f  z  
z sin z
z  
2
has a pole of order 2 at z  
Q13. Consider a counterclockwise circular contour z  1 about the origin. Let f  z  
then the integral
(a) –iπ
Ans:
z sin z
 z   2
,
 f z dz over this contour is
(c) iπ
(b) zero
(d) 2iπ
(b)
Solution: Since, pole z   does not lie inside the contour, hence
 f  z  dz  0
GATE-2012
Q14.
Ans:


Identify the correct statement for the following vectors a  3iˆ  2 ˆj and b  iˆ  2 ˆj


(a) The vectors a and b are linearly independent


(b) The vectors a and b are linearly dependent


(c) The vectors a and b are orthogonal


(d) The vectors a and b are normalized
(a)


Solution: If a  3iˆ  2 ˆj, b  iˆ  2 ˆj are linearly dependent, then
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Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics


a  mb  0, for some values of m but here,
3 + m = 0 and 2 + 2m = 0 , do not have any solution. So, they are linearly independent.
 
 
a  b  0 (Not orthogonal); a  b  0 (Not normalized)
Q15.
The number of independent components of the symmetric tensor Aij with indices
i, j  1, 2,3 is
(a) 1
Ans:
(b) 3
(c) 6
(d) 9
(c)
 A11
Solution: For symmetric tensor, Aij   A21

 A31
A12
A22
A32
A13 
A23 
A33 
 A12  A21 , A23  A32 , A13  A31 , hence there are six independent components.
Q16.
0
The eigenvalues of the matrix 1
0

0
1  are
0 
1
0
1
(b) 0, 2 , 2
(a) 0, 1, 1
(c)
Ans:
1 1
,
,0
2 2
(b)

Solution: A  I  0  1
0
Q17.
Ans:
2 , 2 ,0
(d)
1

1
0
1  0   2  1    0    0,  2 ,  2



GATE-2013

  

If A and B are constant vectors, then  A  B  r is
 
 

(a) A  B
(b) A  B
(c) r
 

(d) zero
(d)

Solution: Let A  A0  xˆ  yˆ  zˆ  , B  B0  xˆ  yˆ  zˆ  and r  xxˆ  yyˆ  zzˆ .
  
 
B  r  xˆ  z  y B0  yˆ  z  x B0  zˆ  y  x B0   A  B  r  0 .
 

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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Q18.
For the function f  z  
16 z
z  3z  12
, the residue at the pole z  1 is (your answer
should be an integer) ____________.
3
Ans:
2
1 d 21   z  1 16 z 
Solution: At z  1 , pole is of order 2. So, residue is

 =3.
2  1 dz 21   z  3 z  12 

 z 1
 4  1  1
Q19. The degenerate eigenvalue of the matrix  1 4  1 is (your answer should be an
 1  1 4 
integer) ____________
Ans:
2,5,5
4  
 1

 1
Q20.
1
4
1
1 
1 
 1 1


1   0  (2   )  0 5  
0  = (2   )(5   )2  0    2,5,5 .
 0
4   
0
5   
The number of distinct ways of placing four indistinguishable balls into five
distinguishable boxes is ___________.
Ans:
120
Solution: 4  C 45 =120 ways
GATE-2014
Q21.
The unit vector perpendicular to the surface x 2  y 2  z 2  3 at the point (1, 1, 1) is
(a)
Ans:
xˆ  yˆ  zˆ
3
(b)
xˆ  yˆ  zˆ
3
(c)
xˆ  yˆ  zˆ
3
(d)
xˆ  yˆ  zˆ
3
(d)

Solution: Let, f  x 2  y 2  z 2  3  0   f  2 xxˆ  2 yyˆ  2 zzˆ

2 xˆ  2 yˆ  2 zˆ xˆ  yˆ  zˆ
f
 nˆ   at 1,1,1 

12
3
f
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Q22.
The matrix
A
1  1 1  i

 is
3 1  i  1 
(a) orthogonal
(b) symmetric
(c) anti-symmetric
(d) Unitary
(c) 4 3i
(d) 4 2 i
Ans. : (d)
Solution: Unitary A† A  I
Q23.
The value of the integral
z2
 z dz
C e 1
where C is the circle z  4 , is
(a) 2 i
(b) 2 2 i
Ans. : (c)
Solution: Pole e z  1  e z  ei 2 m 1 where m  0,1, 2,3.....
  z
2
For z  i , Res  lim
  i   2
z i    z 
e
Similarly, for z  i , Res   2
I  2 i  2   2   4 3i
Q24.
The solution of the differential equation
d2y
 y  0 , subject to the boundary conditions
d t2
y 0   1 and y    0 is
(a) cos t  sin t
(b) cosh t  sinh t
(c) cos t  sin t
(d) cosh t  sinh t
Ans:
(d)
Soluiton:
D 2  1  0  D  1  y  t   c1et  c2 e  t
Applying boundary condition,
y 0   1  1  c1  c2 and y     0  0  c1e  c2 e   c1  0, c2  1
 y  t   e t  y  t   cosh t  sinh t
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Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
GATE-2015
Q25.
Consider a complex function f  z  
1
. Which one of the following
1

z  z   cos  z 
2

statements is correct?
(a) f  z  has simple poles at z  0 and z  
1
2
(b) f  z  has second order pole at z  
1
2
(c) f  z  has infinite number of second order poles
(d) f  z  has all simple poles
Ans.: (a)
Solution: f  z  
1
1

z  z   cos  z 
2

For nth order pole, Res.  lim  z  a  f  z   finite
n
z a
At z  0 , lim zf  z   finite  z  0 is a simple pole.
z 0
2
1
1


z 
z 
1
1
2
2

 lim
 lim 
At z   , lim
1
1
1
2 z  1 
z  1.cos z  z.   sin z 
z  z cos z
2
2 z z 
2
 cos z
2

1
1
2
 lim1

   finite


z  cos z  z sin z

2
2
1
 f  z  has second order pole at z  
2
3
Q26.
The value of  t 2 3t  6 dt is_______________ (upto one decimal place)
0
Ans.: 1.33
3
3
3
1
4
Solution:  t   3t  6  dt   t  3  t  2   dt   t 2  t  2  dt 
30
3
0
0
2
2
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Q27.
If f  x   e  x and g  x   x e  x , then
2
2
(a) f and g are differentiable everywhere
(b) f is differentiable everywhere but g is not
(c) g is differentiable everywhere but f is not
(d) g is discontinuous at x  0
Ans.
(b)
Solution: f ( x)  e  x is differentiable but g ( x)  x e  x is not differentiable.
2
2
 xe  x ; x  0
g ( x)  
 x2
 xe ; x  0
2
Left hand Limit lim g  x  h     x  h  e
 x h
2
h 0
Right hand Limit lim g  x  h    x  h  e
 xh
2
h 0
 lim g  x  h   lim g  x  h 
h 0
Q28.
h 0
Consider w  f  z   u  x, y   iv x, y  to be an analytic function in a domain D . Which
one of the following options is NOT correct?
(a) u  x, y  satisfies Laplace equation in D
(b) v x, y  satisfies Laplace equation in D
z2
(c)
 f z dz is dependent on the choice of the contour between z and z
1
2
in D
z1
(d) f  z  can be Taylor expended in D
Ans.: (c)
Solution: w  f ( z )  u  x, y   iv  x, y  to be an analytic function in a domain D ,
z2
 f  z  dz
is
z1
independent of the choice of the contour between z1 and z2 in D .
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Q29.
1, for t  0
The Heaviside function is defined as H  t   
and its Fourier transform
1, for t  0
is given by  2i /  . The Fourier transform of
 
sin  
2
(a)
 
cos 
2
(b)

 
(c) sin  
2

2
1
H t  1 / 2  H t  1 / 2 is
2
(d) 0
2
Ans.: (a)
Solution: H  f  

 H t  e
 i 2 ft
dt , for a function H  t  , H  f   

For H  t  t0  , Fourier Transform is e
 i 2 ft0
2i

Hf
Shifting Theorem



 i  2i
 i  2i
1  i 2
1  1
 1  1  i 2
2
For  H  t    H  t     e  e 
 e  e 2 
i
2   2

2i 
 2  2 

 
 
sin  
1
2 .
The Fourier transform of  H  t  1/ 2   H  t  1/ 2   

2
2
Q30.
A function y  z  satisfies the ordinary differential equation y 
1
m2
y  2 y  0, where
z
z
m  0, 1, 2, 3, ..... Consider the four statements P, Q, R, S as given below.
P: z m and z  m are linearly independent solutions for all values of m
Q: z m and z  m are linearly independent solutions for all values of m  0
R: ln z and 1 are linearly independent solutions for m  0
S: z m and ln z are linearly independent solutions for all values of m
The correct option for the combination of valid statements is
(a) P, R and S only
(b) P and R only
(c) Q and R only
(d) R and S only
Ans.: (c)
Solution: y  
1
m2
y   2 y  0  z 2 y  zy  m 2 y  0 , m  0,1, 2,3,...,
z
z
z  ex , D 
d
dx
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If m  0 ; z 2 y  zy  0 ,  D  D  1  D  y  0   D 2  D  D  y  0
D 2 y  0  y  c1  c2 x  y  c1  c2 ln z
( R is correct)


And if m  0, m  0 , then m  0 , then D 2  m 2 y  0  D   m
y  c1e mx  c2 e  mx  c1e m log z  c2 e  m log z  c1 z m  c2 z  m
or if m  0, m  0 , then
y  c1 cosh  m log  z    ic2 sinh  m log  x   , m  0
GATE-2016
Q31.
Consider the linear differential equation
dy
 xy . If y  2 at x  0 , then the value of y at
dx
x  2 is given by
(a) e 2
(b) 2e 2
(c) e 2
(d) 2e 2
Ans.: (d)
2
dy
1
x2
Solution:
 xy  dy  xdx  ln y   ln c  y  ce x / 2
2
y
dx
If y  2 at x  0  c  2  y  2e x
2
/2
.
The value of y at x  2 is given by y  2e 2
Q32.
Which of the following is an analytic function of z everywhere in the complex plane?
 
(a) z 2
(b) z *
2
(c) z
2
(d)
z
Ans.: (a)
Solution: z 2   x  iy   x 2  y 2  i  2 xy   u  x 2  y 2 and v  2 xy
2
Cauchy Riemann equations
Q33.
u v
v
u

 2 x,

 2 y satisfies.
x y
x
y

1
1
The direction of f for a scalar field f  x, y, z   x 2  xy  z 2 at the point P1,1,2  is
2
2
(a)
 ˆj  2kˆ
5
(b)
 ˆj  2kˆ
5
(c)
 ˆj  2kˆ
5
(d)
 ˆj  2kˆ
5
Ans.: (b)
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

Solution: f   x  y  iˆ  xjˆ  zkˆ  nˆ  


Q34.

f

f

 ˆj  2kˆ
 

5
1,1,2
A periodic function f  x  of period 2 is defined in the interval    x   
 1,    x  0
f x   
0 x 
 1,
The appropriate Fourier series expansion for f  x  is
4
(a) f  x    sin x  sin 3x  / 3  sin 5 x  / 5  ...
 
4
(b) f  x    sin x  sin 3x  / 3  sin 5 x  / 5  ..
 
4
(c) f  x    cos x  cos 3x  / 3  cos 5 x  / 5  ...
 
4
(d) f  x    cos x  cos 3x  / 3  cos 5 x  / 5  ...
 
Ans.: (a)
 1,    x  0
Solution: f  x   
0 x 
 1,

Let f  x   a0    an cos nx  bn sin nx 
n 1

a0 
 a0 
1
2
1
2




  f  x dx

f  x dx 

1  0
1 

0


 x    x 0   0
1
dx
1 dx  












0
 2
2 
This can also be seen without integration, since the area under the curve of f  x  between
 to  is zero.
1

f  x  cos nxdx
 

an 
 an 
0


1   sin nx 
1 0
 sin nx  




1
cos
1
cos
nxdx
nxdx


 
 


 0
0  
    n   n 0 
  

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bn 

 bn 
1


  f  x  sin nxdx


1 0
1 sin nxdx   1 sin nxdx 


0

  
n
n
0

1   cos nx 
 cos nx   1  1  1  1 1  1

 
 bn   
 
   
   n   n 0    n
n
n
n  
 2 2  1n 
 

n 
 n
0; n  even

 bn   4
 n ; n  odd
Thus, Fourier series is f  x  
4
1
1

sin x  sin 3 x  sin 5 x  ...

3
5


GATE-2017
Q35.
The contour integral
dz
 1  z
2
evaluated along a contour going from  to  along the
real axis and closed in the lower half-plane circle is equal to………….. (up to two
decimal places).
Ans. : 

Solution:
1
1
1
C 1  z 2 dz   1  x 2 dx  C 1  z 2 dz
Poles, 1  z 2  0  z  i , z  i is inside C
 Res  z  i   lim  z  i 
z  i
1
 z  i  z  i 

1
1

i  i 2i

1
1
dx    2 i  
2
1 x
2i

(Since, here we use lower half plane i.e., we traversed in clockwise direction, hence we

have to take 2 i )
Q36.
The coefficient of eikx in the Fourier expansion of u  x   A sin 2  x  for k  2 is
(a)
A
4
(b)
A
4
(c)
A
2
(d)
A
2
Ans.: (b)
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Solution: Since, sin  x  
ei x  e  i x
ei 2 x  2  e 2i x
 sin 2  x  
2i
 4 
Since, 2   k , hence sin 2  x  
Hence, ck 
A
2

2
 sin  x  dx  

e  ikx  2  eikx
 4 




A   ikx ikx
 ikx
 ikx ikx
2


e
e
dx
e
dx
e
e
dx




8  







A  2ikx
 ikx

  e dx  2  e dx   dx 
8  



The first two integrals are zero and the third integral has the value 2 .
Thus,
ck  
Q37.
A
A
 2   
8
4
The imaginary part of an analytic complex function is v  x, y   2 xy  3 y . The real part of
the function is zero at the origin. The value of the real part of the function at 1  i
is ……………... (up to two decimal places)
Ans. : 3
Solution: The imaginary part of the given analytic function is v  x , y   2 xy  3 y . From the
Cauchy – Riemann condition
v u

 2x  3
y x
Integrating partially gives
u  x , y   x 2  3x  g  y 
From the second Cauchy – Riemann condition
u
u
v
 2 y,   x, y    y 2  g  x 

, we obtain
y
y
x
dg  y 
 2 y  g  y    y 2  c
dy
Hence, u  x , y   x 2  3 x  y 2  c
Since, the real part of the analytic function is zero at the origin.
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Hence, 0  0  0  0  c  c  0
Thus, u  x , y   x 2  3 x  y 2
 f  z    x 2  3 x  y 2   i  2 xy  3 y 
Thus, the value of real part when
z  1  i ,i.e. x  1 and y  1 is u  x, y   1  3 1  1  3 .
2
Q38.
Let X be a column vector of dimension n  1 with at least one non-zero entry. The
number of non-zero eigenvalues of the matrix M  XX T is
(b) n
(a) 0
(c) 1
(d) n  1
Ans. : (c)
0
0
 
a 
Let X    , then X T   0 0 a... 0
0
0
 
 0 
Solution:
Here, X is an n  1 column vector with the entry in the i th row equal to a. X T is a row
vector having entry in the i th column equal to a. Then, XX T is an n  1 matrix having
the entry in the i th row and i th column equal to a 2 .
Hence,
0 0 0...0...0 0 
0 0 0...0...0 0 
 ith row
T
XX  0 0 0...0...0 0 
.....................
..................... 
0 0 0...0...0 0 

ith row
Since this matrix is diagonal, its eigenvalues are a 2 , 0, 0.....0 . Hence, the number of non
zero eigenvalues of the matrix XX T is 1 .
Q39.
Consider
the
differential
equation
dy
 y tan  x   cos  x  .
dx
If
 
y  0   0, y  
3
is …………... (up to two decimal places)
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Ans.: 0.52
Solution: The given differential equation is a linear differential equation of the form
dy
 p  x  y  cos x
dx
Integrating factor  e 
p  x  dx
Thus integrating factor  e 
tan x dx
 I  F  eln sec x  sec x
Thus the general solution of the given differential equation is
y  sec x   sec x  cos xdx  c
 y sec x  x  c
-(i)
It is given that y  0   0  0  sec 0  0  c  c  0
Thus the solution satisfying the given condition is
x
y sec x  x  y 
sec x
 
Thus the value of y   is
3
y
 /3
 /3 

  0  52
sec  / 3
2
6
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Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
GATE-2018
Q40.
The eigenvalues of a Hermitian matrix are all
(a) real
(b) imaginary
(c) of modulus one
(d) real and positive
Ans. : (a)
Solution: Eigenvalue of Hermitian matrix must be real.
Q41.
In spherical polar coordinates  r ,  ,   , the unit vector ˆ at 10,  / 4,  / 2  is
(a) k̂
(b)

1 ˆ ˆ
jk
2

Ans. : (d)

1
 ˆj  kˆ
2
(c)
(d)

1 ˆ ˆ
jk
2

z
Solution: ˆ  cos 45 ˆj  sin 45 kˆ
0
r̂
0

1 ˆ ˆ
 ˆ 
jk
2
Q42.

 /4

ˆ
y
The scale factors corresponding to the covariant metric tensor gi j in spherical polar
coordinates are
(a) 1, r 2 , r 2 sin 2 
(b) 1, r 2 ,sin 2 
(d) 1, r , r sin 
(c) 1,1,1
Ans. : (d)
Q43.
  



Given V1  iˆ  ˆj and V2  2iˆ  3 ˆj  2kˆ , which one of the following V3 makes V1 ,V2 , V3


a complete set for a three dimensional real linear vector space?


(a) V3  iˆ  ˆj  4kˆ
(b) V3  2iˆ  ˆj  2kˆ


(d) V3  2iˆ  ˆj  4kˆ
(c) V3  iˆ  2 ˆj  6kˆ
Ans. : (d)
 

Solution: Let A be the matrix formed by taking V1 , V2 and V3 as column matrix i.e.,
 1 2 2 
A  V1 V2 V3    1 3 1   A  2 . Here V3  2iˆ  ˆj  4kˆ
 0 2 4 
 

Since, A  0 , hence, V1 , V2 and V3 form a three dimensional real vector space.


Hence, option (d) is correct.
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Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Q44.
Given
d 2 f  x
dx 2
2
df  x 
dx
 f  x  0 ,
and boundary conditions f  0   1 and f 1  0 , the value of f  0.5  is __________ (up
to two decimal places) .
Ans. : 0.81
Solution:
d 2 f  x
dx
2
2
df  x 
dx
 f  x  0
Auxiliary equation is,
m
2

 2m  1  0   m  1  0  m  1,1
2
Hence, the solution is
f  x    c1  c2 x  e x
using boundary condition,
f  0   c1e0  c1  1
(i)
f 1   c1  c2  e  0
(ii)
From (i) and (ii), c2  1
Hence, f  x   1  x  e x  f  0.5   1  0.5  e0.5  0.81
Q45.
The absolute value of the integral
5 z 3  3z 2
 z 2  4 dz ,
over the circle z  1.5  1 in complex plane, is __________ (up to two decimal places).
Ans. : 81.64
Solution: f  z  
5 z 3  3z 2
 z  2  z  2 
Pole, z  2, 2
z  2 is outside the center
2  1.5  1 So, will not be considered
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Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
 5z  3z   52
Now, Re s  2   lim  z  2 
 z  2  z  2 
3
2
3
z 2
 322 40  12

 13
4
4
I  2 i  residue  2 i  13  26  3.14  I  81.64
GATE-2019
Q46.
For the differential equation
d2y
y
 n  n  1 2  0 , where n is a constant, the product of
2
dx
x
its two independent solutions is
(a)
1
x
(c) x n
(b) x
(d)
1
x n 1
Ans. : (b)
Q47.
During a rotation, vectors along the axis of rotation remain unchanged. For the rotation
0 1 0


matrix  0 0 1 , the vector along the axis of rotation is
 1 0 0 




(b)
1 ˆ ˆ ˆ
i  jk
3

(d)
1 ˆ
2i  2 ˆj  kˆ
3
1 ˆ ˆ
2i  j  2kˆ
3
(c)
1 ˆ ˆ ˆ
i  jk
3



(a)


Ans. : (b)
Q48.
The pole of the function f  z   cot z at z  0 is
(a) a removable pole
(b) an essential singularity
(c) a simple pole
(d) a second order pole
Ans. : (c)
Solution: f  z   cot z at z  0
f  z 
1
tan z
z  0 is a simple pole f  z  

Q49.
The value of the integral


(a)

a
e  ka
(b)
cos  kx 
x2  a2
2  ka
e
a
1 1 2

1  z  ....

z 3

dx , where k  0 and a  0 , is
(c)

2a
e  ka
(d)
3  ka
e
2a
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Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Ans. : (a)

Solution:
cos kx
dx
2
 a2
x

f  z 
eikx
eikz

z 2  a 2  z  ia  z  ia 
I  Re.2 i 
Q50.
eik  ia   e  ka

2ia
a
Let  be a variable in the range      . Now consider a function



 
1 for
    
2
2
0 otherwise
if its Fourier-series is written as      m  Cm e  im , then the value of C3

2
(rounded off to three decimal places) is__________
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