CSIR DECEMBER 2015 QUESTION AND SOLUTION
8.
PART ‘A’
1.
A circle drawn in the x-y coordinate plane passes
through the origin and has chords of lengths 8 units
and 7 units on the x and y axes, respectively. The
coordinates of its centre are
(a)
(8, 7)
(b)
(-8, 7)
(c)
(-4, 3.5)
(d)
(4, 3.5)
9.
A wheel barrow with unit spacing between its wheels
is pushed along a semi-circular path of mean radius
10. The difference between distances covered by
the inner and outer wheels is
(a)
0
(b)
10
(c)
(d)

2
The missing number is
5
8
The probability that a ticketless traveler is caught
during a trip is 0.1. If the traveler makes 4 trips, the
probability that he/she will be caught during at least
one of the trips is:
(a)
(c)
4
1   0.9 
(b)
4
1 10.9 
(d)
10.9 4
 0.9
The statement: “The father of my son is the only
child of your parents”
(a)
can never be true
(b)
is true in only one type of relation
(c)
can be true for more than one type of relations
(d)
can be true only in a polygamous family
4.
The base diameter of a glass is 20% smaller than the
diameter at the rim.The glass is filled to half th height.
The ratio of empty to filled volume of the glass is
(c)
10  9
9 8
102 92
98
(b)
109
9 8
(d)
103 93
93 83
5.
The number of diagonals of a convex deodecagon
(12-gon) is
(a)
66
(b)
54
(c)
55
(d)
60
6.
One is required to tile a plane with congruent regular polygons. With which of the following polygons
is this possible?
(a)
6-gon
(b)
8-gon
(c)
10-gon
(d)
12-gon
7.
5
10.
2
9
9
-5
9
(a)
-19
(c)
9
Decode
?
(b)
(d)
-5
-9
4
3.
(a)
7
Suppose three meetings of a group of professors
where arranged in Mumbai, Delhi and Chennai. Each
professor of the group attended exactly two meetings. 21 professors attended Mumbai meeting, 27
attended Delhi meeting and 30 attended Chennai
meeting. How Many of them attended both the
Chennai and Delhi meeting
(a)
18
(b)
24
(c)
26
(d)
Cannot be found from the above information
G
I
L
L
E
T
11.
E
S
I
M
E
N
(A)
N
T
S
T
U
S
O
L
V
D
I
S
P
A
E
H
T
R
B
N
L
B
O
L
T
I
Y
B
E
S
GENT STUDENT CAUSE LITTLE
HEART BURNS
(B)
STUDENTS ARE INTELLIGENT BUT
PROBLEM IS NOT SOLVABLE
(C)
THIS PROBLEM IS UNSOLVABLE BY
ANY STUDENT
(D)
THIS PROBLEM IS SOLVABL BY
INTELLIGENT STUDENTS
Three circles of equal diameters are placed such
that their centres make an equilateral triangle as in
the figure
Within each circle, 50 points are randomly scattered.
The frequency distribution of distances between
all possible pairs of points will look as
1.
Frequency
2.
Distance
Frequency
2.
16.
3.
Frequency
Distance
(b)
cos r < cos g , cos d
(c)
cos r < cos d < cos g
(d)
cos g < cos d < cos r
A vendor sells articles having a cost price of Rs. 100
each .He sells there articles at a premium price during first eight months, and at a sale price which is
half of the premimum price during next four months
He makes a net profit of 20% at the end of the year.
Asuming that equal nmbers of articles are sold each
month what is the premimum price of the article?
(a)
122
(b)
144
(c)
150
(d)
160
17.









4.
Frequency
Distance
Distance
12.
13.
14.
18.
How many digits are there in 316 when it is expressed
in the decimal form?
(a)
Three
(b)
Six
(c)
Seven
(d)
Eight
Most Indian tropical fruit trees produce fruits in
April-May. The best possible explanation for this
is
(a)
optimum water availability for fruit production
(b)
The heat allows quicker ripening of fruit.
(c)
animals have no other source of food in
summer.
(d)
the impending monsoon provides optimum
conditions for propagation.
There is an inner circle and an outer circle around a
square. What is the ratio of the area of the outer
circle to that of the inner circle?
19.
20.
(C)
22.
15.
2
(b)
3
If A is a 5  5 real matrix with trace 15 and if 2 and 3
are eigenvalues of A, each with algebraic multiplicity 2, then the determinant of A is equal to
(a)
0
(b)
24
(c)
120
(d)
180
Let f :    be a twice continuously differentiable
function, with f (0)  f (1)  f '(0)  0. Then
(a)
f " is the zero function.
2
(c)
(d)
2 2
3/ 2
Write d = 1 degree r = 1 radian and g =1 grad. Then
which of the following is true? (100 grad = a right angle)
(a)
cos d < cos r < cos g
(D)
5
PART ‘B’
21.
(a)
The minimum number of straight lines required to
connect the nine points above without lifting the
pen or retracing is
(a)
3
(b)
4
(c)
5
(d)
6
“The clue is hidden in this statement” real the note
handed to Sherlock by Moriarty, who hid the stolen
treasure in one of the ten pillars. Which pillar is it?
(a)
X
(b)
II
(c)
III
(d)
IX
Three boxes are coloured red, blue and green and
so are three balls. In how many ways can one put
the balls one in each box such that no ball goes into
the box of its own colur?
(a)
1
(b)
2
(c)
3
(d)
4
Let A, B be the ends of the longest diagonal of the
unit cube. The length of the shortest path from A to
B along the surface is
(a)
(b)
3
1 2
23.
(b)
f " (0) is zero
(c)
f " (x)= 0 for some x  (0,1) .
(d)
f " never vanishes
1
k
Let Sn   nk 1 . Which of the following is true?
24.

S
(b)
Sn is bounded sequence.
(c)
|S n S
(d)
Sn
 1 as n  .
n
2n
2
real matrix P.
n
for every n  1.
2
(a)
2n 1
(c)
4 4 real matrix P..
(d)
| 0 as n  .
29.
T is a linear transformation and dim
range(T) = 5
(b)
T is a linear transformation and dim
range(T) = 3.
(c)
T is a linear transformation and dim
range(T) = 2.
(d)
T is not a linear transformation.
Let A be a real 3 4 matrix of rank 2. Then the rank of
30.
27.
31.
32.
S = {31, 59}
S is infinite
Let f :[0, )  [0,  ) be a continuous function.
Which of the following is correct?
x2  y2 . Then the
resulting
lim
n 
1
1
1  1


...

 is
4 6
2n  2n  2 
n  2 4
(a)
2
(b)
(c)
2 1
(d)
1
2
1
2 1
Let A be a closed subset of  , A   , A   .
Then A is
(a)
The closure of the interior of A
(b)
a countable set.
(c)
a compact set
(d)
not open
A group G is generated by the elements x, y with the
(a)
There is x0 [0, ) such that f ( x0 )  x0 .
(b)
If f ( x)  M for all x[0, ) for some M > 0,
2
relations x3  y 2   xy   1. The order of G is
then there exists x0 [0, ) such that
(a)
(c)
f ( x0 )  x0
If f has a fixed point then it must be
unique
(d)
f does not have a fixed point unless it is
differentiable on (0,  )
Consider the quadratic form Q(v)  vt Av, where
(c)
28.
Trace (A) = Rank (A).
(c)
Rank (A) + Rank ( In  A)  n.
(d)
The eigenvalues of A are each equal to 1.
For (x, y)  2 with (x, y)  (0, 0), let  (x, y) be
the unique real number such that     and (x, y)
function  : 2 \{(0, 0)}   is
(a)
differentiable
(b)
continuous, but not differentiable
(c)
bounded but not continuous
(d)
neither bounded, nor continuous.
 91 31 0 
 29 31 0 
 has an
the property that the matrix 
 79 23 59 


inverse in the field  p . Then
(a)
S = {31}
(b)
(c)
S = {7, 13, 59}
(d)
( I n  A)2  In  A
= (r cos , r sin ) where r 
At A , where At denotes the transpose A, is
26.
4 4 real matrix P..
Let A  I n be an n  n matrix such that A2 = A,
(a)
(b)
(a)
(A)
exactly 2
(b)
exactly 3
(c)
exactly 4
(d)
at most 2 but not necessarily 2
Let S denote the set of all the prime numbers p with
x2  y2  zw  Q(Pv) for some invertible
where I n is the identity matrix of order n. Which of
the following statements is false?
For a positive integer n, let Pn denote the vector
space of polynomials in one variable x with real coefficients and with degree  n. Consider the map
T : P2  P4 defined by T ( p( x))  p( x2 ). . Then
25.
xy  y2  z2  Q(Pv) for some invertible
1

0
0

 0
(a)
(b)
0
1
0
0
0
0
0
1
0

0
1  , v  ( x, y , z, w) . Then

0 
Q has rank 3
xy  z2  Q(Pv) for some invertible 4 4
33.
34.
4
8
(b)
(d)
6
12
Let a, b , c, d   be such that ad  bc  0. Consider
the Mobius transformation
Ta,b, c, d ( z ) 
az  b
. Define H   {z   : Im( z )  0} ,
cz  d
H   {z   : Im( z)  0} ,
R  {z   : Re( z )  0}, R  {z   : Re( z )  0} .
Then Ta,b,c, d maps
(a)
H  to H 
(b)
H  to H 
(c)
R to R
35.
R to R
(d)
What is the total number of positive integer solutions to the equation
( x1  x2  x3 ) ( y1  y2  y3  y4 )  15?
36.
(a)
1
(b)
2
(c)
3
(d)
4
Which of the following is an irreducible factor of
x12 1 over  ?
(a)
x8  x4  |1 .
(b)
x 4 1 .
(c)
x4  x2 1 .
(d)
x5  x4  x3  x 2  x 1 .
38.
and z n  1 for any 0 < n < 98}?
(a)
0
(b)
12
(c)
42
(d)
49
For a subset A of the topological space X, let Â
denote the union of the set A and all those connected components of X \ A which are relatively
compact inX (i.e the closure is compact). Then for
every A  X .
40.
A is compact
ˆ
Aˆ  Aˆ ,
(b)
(c)
A is connected (d)
Aˆ  X .
Let R be a Euclidean domain such that R is not a
field. Then the polynomial ring R[X] is always
(a)
a Euclidean domain
(b)
a principal ideal domain but not a Euclidean domain
(c)
a unique factorization domain, but not a
principal ideal domain
(d)
not a unique factorization domain
Consider the following power series in the complex
variable z.
en n
z . If r, R are the
n
(a)
r  0, R  1 .
(b)
r  1, R  0
(c)
r  1, R  
(d)
r  , R  1.
45.
1
y  C1e2 x  C2 e2 x  cos x
2
The resolvent kernel R(x, t,  ) for the Volterra inte-
iˆ 16 ˆj  9kˆ
(b)
iˆ 16 ˆj  9kˆ
(c)
iˆ 16 ˆj  9kˆ
The functional
(c)
iˆ 16 ˆj  9kˆ
I ( y( x)  ab ( y 2  y ,2  2 y sin x)dx ,
(a)
y
 C1e2 x
 C2
1
 sin x
2
u
u
 ( y  x  u)  u,
x
y
(a)
u2 (x  y  u)  ( y  x  u)  0 .
(b)
u2 ( x  y  u)  ( y  x  u)  0 .
(c)
u2 ( x  y  u)  ( y  x  u)  0 .
(d)
u2 ( x  y  u)  ( y  x  u)  0 .
Let f ( x)  ax 100 for a   . then the iteration
46.
(a)
a  5
(b)
a  1
(c)
a  0.1
(d)
a  10
The PDE
(a)
(b)
(c)
(d)
47.
48.
 2u
 2u  2u
2

 x. has
2
xy y 2
x
only one particular integral
a particular integral which is linear in x
and y
a particular integral which is a quadratic
polynomial is x and y.
more than one particular integral.
Consider the ODE on  y '( x )  f ( y ( x )). If f is an
even function and y is an odd function, then
(A)
 y( x) is also a solution
(b)
y (-x) is also a solution
(c)
-y(x) is also solution
(d)
y(x)y (-x) is also a solution
Consider the system of ODE in
2 ,
dY
 AY , Y (0) 
dt
 , t  0
 1 1 
 and Y (t ) 
 0 1
where A  
(a)
(b)
e 2 x
e ( x t )
(c)
(d)
e( x  t )
ext
The solution of the initial value problem
has the following extremal with c1 and c2 as arbitrary
constatnts.
(b)
e ( x  t )
xn 1  f ( xn ) for n  0 and x0 0 converges for
A force 5iˆ  2 ˆj  3kˆ acts on a particle with position
vector 2iˆ  ˆj  2kˆ . The torque of the force about
the origin is
42.
(d)
u( x,0)  1, satisfies
radii of convergence of f and g respectively then
(a)
1
y  C1e x  C2e x  sin x
2
(x  y)
2
f ( z )   n 1 n log n z n , g ( x)   n 1
41.
(c)
gral equation ( x)  x  ax (s)ds , is
44.
What is the cardinality of the set {z   | z98  1
39.
1
y  C1e x  C2e x  sin x
2
(a)
37.
(a)
43.
(b)
0
1
 . Then
y1 (t )
y2 ( t )
y1 (x) and y2 (t ) are monotonically
increasing for t > 0
y1 (t ) and y2 (t ) are monotonically
increasing for t > 1.
prior distribution of  is normal with mean 0 and
y1 (t ) and y2 (t ) are monotonically
decreasing for t > 0
(d)
y1 (t ) and y2 (t ) are monotonically
decreasing for t > 1
From the six letters A, B, C, D, E and F, three letters
are chosen at random with replacement. What is
the probability that either the word BAD or the word
CAD can be formed from the chosen letters?
(c)
49.
(a)
51.
(b)
6
216
2
(b)
Posterior mode  given X1, X 2 , X 3 and
(d)
  4i 1 X i 

 4 
54.
covariance matrix V ( I n , the nth order identity matrix). Also let A be a symmetric matrix of order n.
Which of the following statements is true?
(a)
x ' Ax follows a central chi-square distribution if an only if ( AV )2  AV
(b)
55.
5x1  3x2  15
 x1  x2 1
2 x1  5x2 10
x1, x2 0
where 1, 2 and 3 are unknown parameters.
56.
estimator of 2 is
(b)
(c)
1 2 2
(e  e ).
4 1 2
(d)
e12  e22 .
Let X1, X 2 , X 3 and X 4 independent and identically distributed random variables with common distribution normal with mean  and variance 2. If the
bution if and only if A2  A
(c)
The mean of x ' Ax is  ' A  tr ( AV ) where
tr(.) denotes the trace of a square matrix.
(d)
x ' Ax always has a central chi-square
distributiopn with n degree of freedom
Consider the following Linear Programming Prob2
E(Y3 )  1 2  E(Y4 )
(a)
x ' Ax follows a central chi-square distri-
5
lem. Max x1  x2 subject to
E(Y1)  1 2 3  E(Y2 ),
1 2 2
(e  e ).
2 1 2
Let the n1 vector x follow an n-variate normal
distribution with mean vector   0  and variance
with common unknown variance 2 and expectations given by
1 2 2
(e  e ).
2 1 2
Posterior variance of  given X1, X 2 , X 3
and X 4 is 
Let Y1, Y2 , Y3 and Y4 be uncorrelated observations
1
1
Define e1  2 Y1 Y2  and e2  2 Y3 Y4  An unbiased
Posterior median of  given X1 , X 2 , X 3
4
vations resulted in observations x1 , x2 ,..., xk once
each so that the modified (new) sample of size N+k
has observation xi with frequency f i 1
(a)
The new mean is necesarily less than or
equal to the original mean.
(b)
The new median is necessarily more than
or equal to the original median.
(c)
The new variance is necessarily less than
or equal to the origion variance
(d)
The new mode will be same as the original
mode.
 4i 1 X i
8
4 X
and X 4 is i 1 i
(d)
f1 , f 2 ,... f k so that ik1 fi  N Another k obser--
53.
The prior distribution is not a conjugate
prior
(c)
ues x1 , x2 ,..., xk with respective frequencies
52.
(a)
X 4 is
3
216
12
216
Let X be a radom variable which is symmetric about
0. Let F be the cumulative distribution function of
X. Which of the following statements is always true?
(a)
F(x) + F(-x) = 1 for all x  .
(b)
F(x) - F(-x) = 0 for all x  .
(c)
F(x) + F(-x) = 1 + P(X=x) for all x  .
(d)
F(x) + F(-x) = 1 -P(X=-x) for all x  .
A set of N observations resulted in k distinct val(c)
50.
1
216
1
variance , then which of the following is true?
The problem
(a)
Has no feasible solution
(b)
has infinitely many optimal solutions
(c)
has a unique optimal solution
(d)
has an unbounded solution
Let Xi ' ’s be independent random variables such
that Xi ' s are summetric about 0 and Var ( X i )  2i 1,
P( X 1  X 2  ...  X n  n long )
for i 1. Then, nlim

57.
(a)
does not exist
(b)
equals 1/ 2
(c)
equals 1.
(d)
equals 0.
Consider a randomized block design involving 3
treatments and 3 replicates and let ti denote the
c4  c5 where c4 and c5 are such that the
effect of the i th treatment (i =, 2, 3). If 2 denotes
the variance of an observation which of the following statements is true?
(a)
The variance of the best linear unbiases
estimators (BLUE) of (t1  t2 ) / 2 and
test is of size  .
PART ‘C’
61.
(t1  2t2  t3 ) / 6 are equal .
(b)
The varince of the BLUE of
ti t j,(i  j, i, j 1, 2, 3, ) is 2 \ 3
(d)
(a)
limn  pn ( x) exists for all x  .
(b)
limn  pn '( x) exists for all x  .
(c)
limn  pn
(d)
limn  pn '
The variance of the BLUE of (t1  2t2  t3 ) is
2 / 6
58.
Let Y1, Y2 be two independent random variables
taking value -1 and +1 with probability
1
each.
2
62.
Define
(b)
n  3. Then
59.
P( X 8  1, X 9  1, X10  1) 
(c)
1
P( X 8  1, X 9  1, X10  1) 
8
(d)
P( X 8  1, X 9  1, X10  1) 
(c)
1
4
(d)
2
ogy for some fixed positive real number a.
Let f :    be a differentiable function such
64.
that sup x | f '( x) |  . Then
(a)
f maps a bounded sequence to a bounded
sequence.
(b)
f maps a Cauchy sequence to a Cauchy
sequence.
(c)
f maps a convergent sequence to
a convergent sequence
(d)
f is uniformly continuous.
For (x, y)   2 consider the series
n
X  n   max{X1, X 2 ,..., X n} . Maximum likelihood esti-
mator of  is
(c)
60.
X 1
5
X 1
(b)
(d)
X n
X n 
lim  n, k  0
n 
5
k 2 x k y
Then the series converges for
!
(x, y) in
Consider the problem of testing H0 : X ~ Normal
1
with mean 0 and variance against H1 : X ~ Cauchy
2
(0, 1) . Then for testing H 0 against H1 , the most
powerful size  test
(a)
does not exist
(b)
rejects H0 if and only if |x| > c2 where c2 is
such that the test is of size 
(c)
rejects H0 if an only if |x| < c3 where c3 is
such that the test is of size 
(d)
rejects H0 if and only if |x| < c4 or |x| > c5 ,
{z   : | Re z | a} in the Euclidean topol-
63.
, 5 ,   0. Define X   min {X , X ,..., X } and
(a)
n 1 An with product topology, where
2,3,...
1
8
1
{( z1, z2 , z3 )  3 : z12  z22  z32  1} in the
An  {0, 1} has discrete topology for n = 1,
Let X1, X 2 ,..., X n be a random sample from uniform
1
{( x, y, z)  3 : x 2  y 2  z 2  1} in the
Euclidean topology.
1
4
(b)
( 0 1)
does not exist.
2
Euclidean topology.
X1  Y1, X 2  Y2 , X 3  X 2 X 1,..., X n  X n 1 X n 2 for
P( X 8  1, X 9  1, X10  1) 
( 0 1)
does not exist.
2
Which of the following sets are compact?
(a)
(a)
be a sequence of quadratic
polynomials where an , bn  for all n  1. Let 0 , 1
be distinct nonzero real numbers such that
limn  pn (0 ) and limn  pn (1) exist Then
The covariance between the BLUE of
t1  t3 and the BLUE of t1  2t2  t3 is 22 / 3
(c)
Let
pn ( x)  an x 2  bn x
65.
(a)
(1, 1)  (0, )
(b)
  (1, 1)
(c)
(1, 1)  (1, 1)
(d)

Let f : (0, 1)   be continuous . Suppose that
| f ( x)  f ( y) |  | cos x  cos y | of all x, y  (0,1) . Then.
(a)
(b)
(c)
f is discountinuous at least at one point
in (0, 1).
f is continuous everywhere on (0, 1) but
not unidormly continuous on (0, 1)
f is uniformly continuous on (0, 1)
66.
Let
(d)
lim x  0 f ( x) exists.
(d)
f : 2   2
be the function
f (r, )  (r cos , r sin ).
Then for which of the open subsets U of 2 given
below f restricted to U admits an inverse?
(a)
U  2
(b)
U  {( x, y)   2 : x  0, y  0}
(c)
U  {( x, y)   2 : x2  y 2  1}
(d)
67.
U  {( x,
y)   2
limn  .0 f n ( x)dx  0 f ( x)dx.
71.
Then for any compactly supported continuous function f on n which of the following are correct?
S is discrete in 2
The set of limit points of S is the set
72.
{(m, n) : m, n   }.
(c)
S c is connected but not path connected
(d)
S c is path connected.
Let A  {( x, y)  2 : x  y  1} Define
(b)
(c)
(d)
69.
the deternimant of the jacobian of f does
not vanish on A.
f is infinitely differentiable on A.
f is one to one
f(A) = 2
73.
f ( x, y)  (3x  2 y  y 2  | xy |, 2 x  3 y  x2  | xy |).
Then
(b)
(c)
f is discontinuous at (0, 0)
74.
f is continuous at (0, 0) but not differentiable at (0, 0)
f is differentiable at (0, 0)
f is differentiable at (0, 0) and the derivative Df(0, 0) is invertiable.
Consider all sequences { f n} of real valued continuous function on[0,  ). Identify which of the following statements are correct.
(a)
If { f n} converges to f pointwise on [0,  ),
75.
 n f ( x  y)dx   n f ( x)dx, for some y 2
(d)
n
 n f (tx)dx  n f ( x)t dx.
If { f n } converges to f unifromly on [0,  ),
If { f n} converges to f uniformly on [0 ,  )
then f is continuou on [0 ,  )
Let G1 and G2 be two subsets of 2 and f : 2  2
be a function . Then
(a)
f 1(G1  G2 )  f 1 (G1)  f 1(G2 )
(b)
f 1(G1c )  ( f 1(G1 ))c
(c)
f (G1  G2 )  f (G1)  f (G2 )
If G1 is open and G2 is closed then
open nor closed.
Let A and B be n  n matrices over  . Then
(a)
AB and BA always have the same set of
eigenvalues.
(b)
If AB and BA have the same set of eigen
values then AB=BA.
(c)
If A1 exists then AB and BA are similar..
(d)
The rank of AB is always the same as the
rank of BA.
Let V be a finite dimensional vector space over  .
Let T : V  V be a linear transformation such that
(a)
Kernal (T 2 )  kernal (T)
(b)
(c)
Range (T 2 )  = Range (T)
Kernal (T)  Range (T) = {0}.
(d)
Kernal (T 2 )  Range (T 2 ) {0} .
Let V be the vector space of polynomials over  of
degree less than or equal to n. For
mation T :V V b (Tp)( x)  an  an 1x  ...  a0 xn . Then
(a)
T is one to one (b)
T is onto.
(c)
T is invertible
(d)
det T  1.
then limn  0 f n ( x)dx 0 f ( x)dx
(c)
(c)
t a
p(x)  a0  a1 x  ...  an1 xn in V, define a linear transfor--
then limn  0 f n ( x)dx 0 f ( x)dx
(b)
Ba f (tx)dx  B n f ( x)t dx
rank (T 2 )  rank (T ) . Then,
(d)
70.
(b)
G1  G2  {x  y : x  G1, y  G2} is neither
Let f : 2  2 be given by the formula
(a)
n
Ba f (tx)dx  Bta f ( x)t dx
(d)
y
x
f : A  2 by f (x, y)  (1 x y , 1 x y ) . Then
(a)
(a)
: x  1, y  1}
1
1 

S  {  m
,n 
: m, n, p, q, }
4|
p
|
4|
q| 

68.
Let t and a be positive real numbers Define
Ba  {x  ( x1, x2 ,..., xn )   n | x12  x22  ...  xn2  a 2}.
Let S   2 be defined by
Then
(a)
(b)
There exists a sequence of continuous
functions { f n} on [0 ,  ) such that { f n}
converges to f uniforlmy on [0,  ) but
76.
2 2 1 
Consider the matrices A   0 2 1 and
0 0 3 


82.
2 1 0
B   0 2 0 
. Then
 0 0 3


(a)
77.
78.
79.
A and B are similar over the field of rational numbers  .
(b)
A is diagonalizable over the field of rational numbers  .
(c)
B is the jordan canonical form of A.
(d)
The minimal polynomial and the characteristic polynomial of A are the same.
Let A be an m  n real matrix and b   m with b  0.
(a)
The set of all real solutions of Ax=b is a
vector sapce.
(b)
If u and v are two solutions of Ax = b, then
u  (1  )v is also a solution of Ax = b.
for any    .
(c)
For any two solutions u and v of Ax=b, the
linear combination u  (1 )v is also a
solution of Ax = b only when 0    1.
(d)
If rank of A is n, then Ax  b has at most
one solution
Let A be an n  n matrix over  such that every
nonzero vector of n is an eigenvector of A. Then
(a)
All eigenvalues of A are equal
(b)
All eigenvalues of A are distinct.
(c)
A  I for some    where I is then n  n
identity matrix.
(d)
If  A and mA denote the characteristic
polynomial and the minimal ploynomial respectively then  A  mA
Which of the following statements is /are true?
(a)
There exists a continuous map f :   
(d)
and g respectively in A. Then f(0). g(0) = 0.
83.
2
2
Let   cos  i sin . Let K  (2 ) and let
10
L
 (2 ).
(a)
10
Then
[ L :] 10
(b)
[L :K ]  2
(d)
L= K
[ K : ]  4
Let G be an simple group of order 60. Then
(a)
G has six Sylow -5 subgroups
(b)
G has four Sylow-3 subgroups
(c)
G has a cyclic subgroup of order 6.
(d)
G has a unique element of order 2.
Which of the following quotient ring are fields?
(c)
84.
85.
(a)
F3[ X ]/( X 3  X  1), where F3 is the finite
field with 3 element
(a)
[ X ]/( X  3)
(c)
[ X ]/( X 2  X 1)
F2 [ X ] /( X 2  X 1) F2 is the finite field
with 2 elements.
Let f be an anlytic function in  Then f is constant
if the zero set of f contains the sequence
an  1/ n
(a)
(d)
86.
n 1
1
n
(b)
(b)
such that f ( )   .
There exists a continuous map f :   
such that f (  )  
an  (1)
(c)
an 
(c)
There exists a continuous map f :   2
(d)
an  n if 4 does not divide n and an 
(d)
such that f ()  {( x, y )   2 : x2  y 2  1}.
There exists a continuous map
4 divides n
For n  1 let ( / n)* be the group of units of
87.
Which of the following intervals contains an integer satisfying the following three congruences:
(a)
(c)
[401, 600]
[801, 10000]
(b)
(d)
1
2n
1
if
n
( / n) . Which of the following groups are cyclic
x  2(mod 5), x  3(mod 7) and x  4(mod 11).
81.
Let f , g be in [ X ] such that f  g  0 in
A. Here f and g denote the image of f
f : 0,1   2,3  {0,1}.
80.
Let A denote the quotient ring [ X ]/( X 3). Then
(a)
There are exactly three distinct proper ideals in A
(b)
There is only one prime ideal in A.
(c)
A is an integral domain
[601, 800]
[1001, 12000]
88.
a5  50
(b)
a4 14
(c)
a5  40
(d)
a4 11
( /10)*
(b)
( / 23 )*
(c)
( /100)*
(d)
( /163)*
1
Let f ( z )  n 1 for all z  such that e z  1. Then
e
(a)
(b)
(c)
Let an denote the number of those permutation 
on {1, 2,...,n} such that  is a product of exactly
two disjoint cycles Then.
(a)
(a)
(d)
89.
f is meromorphic
the only singularities of f are poles
f has infinitely many poles on the imaginary exis.
Each pole of f is simple.
1
Consider the function f ( z)  on the annulus
z
A  {z   :
1
| z | 2} . Which of the following is/
2
93.
are true?
(a)
(b)
There is a sequence {pn (z)} of polynomialsthat approximate f(z) uniformly on compact subset of A.
There is a sequence {rn ( z)} of rational
function whose poles are contained in
 / A and which approximates f(z) uni-
formly on compact subsets of A.
99.
(c)
No sequence { pn ( z )} of polynomials approximate f(z) uniformly on compact subsets A.
(d)
No sequence {rn ( z)} of rational funciton
94.
(a)

K
k

K
(d)

K
(c)
 0
u(t)  
4
(t 1)
(d)
 0
u(t)  
4
(t 10)
for 0  t 10
for
t 10.
m 2 2 2 m
(r  r  ) 
. Then
2
r
(b)
are pr  mr and p0  mr 2
The hamiltonian of the system is
(c)
(d)
2
1  2 p   1 m
 pr  2  
r  2 r
2m 
The hamiltonian of the system is
H
2
1  2 p   m
 pr  2  
r  r
2m 
The generalized momenta of the system
are pr  mr and p  mr 2 
95.
Let u ( x, y) be the solution of the equation
 2u  2u

 0, which tends to zero as y   and
x 2 y 2
is a norm on C() for every compact
K   with non-empty interior..
has the value sin x when y = 0. Then

(a)
K
is a norm on H(  ) for every compact
Let y :[0, ) [0, ) be a continuously differentiable
y 2 (0)  0t
(b)
(a)
(b)
y2 (t )  y 2 (0)  20t y 2 (s)ds.
(c)
y (s)ds.
(c)
y2 (t) 
y 2 (0)  20t y(s)ds.
(d)
y2 (t ) 
y 2 (0)  (0t y(s)ds).
2
u   n 1 an sin(nx bn )e  n y , where a1  1 and
an (n  1), bn are non-zero constants.
2
y2 (t ) 
u   n 1 an sin(nx bn )e  ny , where an are
arbitrary and bn are non-zero constant
function satisfying y(t )  y(0)  0t y(s)ds for t  0.
Then
92.
t 1.
The generalized momenta of the system
K  with non-empty interior
91.
for
(a)
is a norm on C () for every compact
is an norm on H () for every com-
for 0  t 1
Consider a mass m moving in an inverse square
central force with characteristic coefficient  and
described by the Lagragian :
sup
| f ( z) |. Then
zK
pact K  .
(c)
u(t)  t 4
H
K  .
(b)
(b)
L(r , r , , ) 
whose poles are contained in  /A, approximate f(z) uniformly on compact subsets of
A.
Let C() denote the vector sapce of continuous
complex valued functions on  and H(  ) denote
the vector sapce of entire functions. For any function f in C(  ) or H(  ), and for nay compact subset K of  , define. f
(d)
Both H and L are quadratic in v.
Let u(t) be a continuously differentiable function
taking nonnegative values for t > 0 and satisfying
u '(t )  4u3 / 4 (t); u(0)  0. Then
(a)
u(t) = 0
u   n 1 an sin(nx bn )e  ny where
a1  1, an  0 for n > 1 and bn  0 for n 1.
(d)
u   n 1 an sin(nx  bn )e n
2y
where
bn  0 for n  0 and an are all nonzero.
+ 2 y(0)0t y  s ds.
Consider the hamiltonian (H) and the lagrangian (L)
for a free particle of mass m and velocity v. Then
(a)
H and L are independent of each other
(b)
H and L are related but have different dependence on v.
(c)
H and L are equal.
96.
Let f ( x) 
x 3 for x  3 . Consider the iteration
xn 1  f ( xn ), x0  0 ; n  0 . The possible limits
of the iteration are
(a)
-1
(b)
(c)
0
(d)
3
3 3 3...
97.
Consider the boundary value problem
102.
u "( x)  2u ( x) ; x  (0,1) u(0)  u (1)  0.
If u and u’ are continuous on [0, 1], then
(a)
u '2 ( x) 2u 2 (x)  u '2 (0)
98.
(b)
1 2
2 1 2
0 u ' ( x)dx   0 u ( x)dx0
(c)
u '2 ( x) 2u2 (x)  0
(d)
2 1 2
2
1 2
0 u ' ( x)dx  0 u ( x)dx u ' (0)
Let 1,  2 be the characteristic numbers and f1, f 2
be the corresponding eigen functions for the homogeneous integral equation
1
( x)   (2 xt  4 x 2 )(t )dt 0. Then
0
(a)
1  2
(b)
1  2
To shown the existence of a minimizer for the func-
(c)
0 f1( x) f 2 ( x)dx0
tional J[ y]  ab f ( x, y, y ')dx, for which there is a
(d)
0 f1 ( x) f 2 ( x)dx1
minimizing sequence (n ) , it is enough to have
(a)
(n ) is convergent and J is continuous
(b)
(n ) is convergent and J is differentiable
(c)
(n ) has a convergent subsequence and J
103.
S : {1, 2,..., 23} with transition probability given by
pi ,i  1  p i , i 1 
p1,2  p1,2,3 
Let u( x, t) satisfy the wave equation
(a)
u( x, t )  eixeit
(b)
u( x, t )  eixeit
(c)
 eix eit 
u(x, t)  eix 



2


(d)
u( x, t )  t 
101.
104.
x2
2
A solution of the PDE
2
(b)
(Xn)n0
(c)
( X n  1) 
(d)
(Xn)n0
is irreducible
1
23
is recurrent.
A fair coin tossed repeatedly. Let X be the number
of fails before the first Head occurs. Let Y denote
the number of Tails observed between the occurrence of the first and the second Head Let X+Y =N.
Then which of the following statements are true?
(a)
X and Y are independent random variables
with
2
u
u  u   u 
 y        u  0 represetns
x
y  x   y 
2(k 1)
P( X  k )  P(Y  k )  
 0
(a)
an ellipse in the x -y plane
(b)
an ellipsoid in the xyz space
(c)
a parabola in the u-x palne
(d)
a hyperbola in the u-y plane
The iteration
2
1
xn 1   xn 
xn
2
1
2
Then, which of the following statements are true?
(Xn)n0 has a unique stationary distribution.
(a)
u(x, 0) = eix for some   . Then
x
1
2
p2,3,1  p23,22 
 2u  2u

; x  (0, 2), t  0
t 2 x 2
100.
1
 2  i  22
2
(n ) has a convergent subseqeunce and J
is differentiable.
99.
1
Let ( X n )n0 be a Markov chain on the state space
is continuous.
(d)
1
(b)
N has probability mass function given by
(k 1)2 k
P( N  k )  
 0

, n  0

(c)
for a given x0  0 is an instance of
(d)
(a)
fixed point iteration for f (x)  x2  2
(b)
Newton’s method for f (x)  x2  2
(c)
x2 2
fixed point iteration for f ( x) 
(d)
Newton’s method for f (x) 
2x
x2  2
for k  2,3,4,...
otherwise
Given N = n, the conditional distribution
of X and Y are independent
Given N = n,
 1

P{X  k}   n 1
 0

105.
for k 0,1,2,...
otherwise
for k  0,1,2,...,n
otherwise.
Suppose that (X, Y) has a joint distribution with the
marginal distribution of X being N (0, 1) and
E(Y | X  x)  x3 for all x  . Then which of the
106.
107.
following statements are true?
(a)
Corr(X,Y) = 0.
(b)
Corr(X, Y) > 0.
(c)
Corr(X, Y) < 0.
(d)
X and Y are independent.
An urn has 3 red and 6 black balls. Balls are drawn
at random one by one without replacement. The
probability that second red ball appers at the, fifth
drawn is
(a)
1
9!
(c)
 6!4! 
4

 9! 
E[eitX isY ]  E eitX  E[eisY ] for all real s and t.
 
(d)
110.
A 24 experiment involving factors F1, F2, F3 and F4 ,
each at two levels coded 0 and 1 is conducted in
blocks of size 4 each. The block contents are as
below:
Block I
Block II
(b)
4!
9!
F1 F2 F3 F4

0 0 0 0
0 1 1 0
F1 F2 F3 F4

0 0 0 1
0 1 1 1
(d)
6!4!
9!
1 0 1 1
1 1 0 1
1 0 1 0
1 1 0 0
Let X1, X 2..., X n denote a random sample from a
distribution having a probability density function
f ( x ; )   x1,) 0  x  1 zero elsewhere ;   0 .
The set {( x1, x2 ,..., xn ) : 1n log( xi )c} where c is asuitably chosen real number, is a uniformly most pow-
108.
H 0 :   1 against H1 :   1.
(b)
H 0 :   1 against H1 :   4.
(c)
H0 :   4 against H1 :   1.
(d)
H0 :   4 against H1 :   1.
111.
109.
(c)
sn
 0 as n with probability 1
log n
Suppose
sider the test
1 if
  
0
  is a random vector such that the marX
Y
P(X  0, Y  0)  14
in1 X i  cn
otherwise
then which of the following statements are true?

n

P  Sn  occurs for infinitely many n   1.
3


ginal distribution of X and the marginal distribution of Y are the same and each is mormally distributed with mean 0 and variance 1. Then which of the
following conditions imply independence of X and
Y?
(a)
Cov (X, Y) = 0
(b)
aX+bY is normally distributed with mean 0
and variance a 2  b2 for all real a and b.
(c)
1
3
To test H 0 : p vs H A : p , with size   0.01 con2
4

P Sn  n  occurs for infinitely many n   1


The desing is connected .
The desing is disconnected.
X1, X 2 ,..., X n are independently and identically dis-
tributed random variables, which follow Bin(1, p) .
sn
 0 as n   with probability 1.
n log n
(b)
(d)
The confounded effects are
(c)
(d)
(0, 1). Let
for n  1. Then, which of the
following statements are true?
2n 
1 0 0 0
1 1 1 0
F1F2 F3 , F2 F3 F4 , F1F4.
Sn in1 X i
 
1 0 0 1
1 1 1 1
F1F2 F3 , F1F2 F4 , F3 F4.
(b)
Let X1, X 2 ,... be independent and identically distributed, each having a uniform distribution on
(a)
Block IV
F1 F2 F3 F4

0 0 1 1
0 1 0 1
Then which of the following statements are true?
1.
The confounded effects are
erful region for testing H 0 against H1 when
(a)
Block III
F1 F2 F3 F4

0 0 1 0
0 1 0 0
112.
(a)
1
As n   power of the test converges to .
(b)
As n   power of the test converges to
1
2
(c)
As n   power of the test converges to
3
4
(d)
As n   power of the test converges to 1.
4
A system consists of 3 components arranged as in
the figure below:
c1
c3
c2
Each of the components C1, C2 , C3 has independent
and identically distributed lifetimes whose distribution is exponential with mean 1. Then the survival
function, S(t), of the system is given by
(a)
S (t )  e3t , for t > 0.
113.
(b)
S (t )  (1 e1)2 et , for t > 0
(c)
S (t )  (1 e2t )et , for t > 0
(d)
S(t) = (1-(1- et )2 )et , for t > 0
116.
Let Y1, Y2 ,..., Yn be random variable with common
unknown mean  . The variance covariance matrix
V of the vector (Y1, Y2 ,.., Yn ) is such that the inverse
of V has all its diagonal elements equal to c and all
its off-diagonal elements equal to d. Let T1 be the
best linear unbaised estimator of  and T2 be the
ordinary least squares estimator of  . Which of the
following statement are true?
Let X1,..., X n be independent and identically distributed random variables with N (,1) distribution.
Assume that  [0, ). Let ̂ be the MLE of  Then
which of the following statements are true?
114.
(a)
ˆ  max( X n ,0).
(b)
̂ is unbaised for  .
(c)
X n is sufficient for  .
(d)
̂ is a consistent estimator fo  .
1
T1   in1Yi T2
n
2.
T2  nY and T1   in1Yi Y where Y is the
mean of the Yi is
3.
There are exactly (n  1) linearly indepen-
4.
dent linear functions of Y1 , Y 2 ,...,Y n each
with zero expectation
There are exactly (n-2) linearly independent linear functions of Yi , Y2 ,...,Yn each
with zero expectation
Let X1, X 2 ,..., X n be a random sample from
U (, 1). If X 1  X  2   ...  X  n  denote the
117.
ordered value of X1, X 2 ,..., X n , then which of the
following statements are true?
115.
1.
(a)
 X  , X  1 is joinly sufficient statistic for 
(b)
X (n)  1
(c)
 X   , X    is a joinly sufficinet statistic for 
(d)
X 1 is a sufficient statistic for 
1
1
n
is a sufficinet statistic for 
Consider an M/MI queue with arrivals as a Poisson
process at a rate of 8 per hour and a service time
which is exponentially distributed at a rate of 6 minutes per customer. The waiting time of a customer in
the queue
1.
has a gamma distribution with p.d.f
( 10 )8 x7 e10 x
f ( x )  
for x  0.
7!

n
2.
1(0.8)e2x for x0
F(x)  
0
otherwie.

A finite papulation has N units, labelled U1,U 2 ,...,U n
and the value of a study variable on unit Ui is
Yi (i  1, 2, ..., N ). Let Y
 iN1Yi
and Y 
p1, p2 ,..., pN ;0  pi  1, i  1,2,..., N and iN1 pi 1 . De-
1
fine T  is Yi / pi where the sum extends over the
n
units in the sample. Then, which of the following
statements are true?
(a)
T is an unbaised estimator of Y ?
(b)
T is an unbaised estimator of Y
(c)
The variance of T is zero if Yi is propor-
(d)
tional to pi for all i, i  1, 2,..., N .
An unbaised estimator of the variance of
T is
 Yi

1
is  T 
p
n(n1)
 i

2
3.
4.
1 N
 i 1Yi , A
N
sample of size n 1 is drwan fron the papulation
with probability proportational to size with replacement with selection prababilities
has distribution function given by
118.
has mean 4 minutes.
has mean 24 minutes .
Let X be a 4 1 random vector with Multivariate
normal distribution with mean  and dispersion
matrix  . Suppose, the eigenvalue of  are
1  6,  2 ,  3, 3  2., 4  1. Let Y1, Y2 , Y3Y,4 be the
four principal components.Which of the following
statements are correct?.
1.
The percentage of variation explained by
the first two components is  95%
2.
The percetange of variation expalined by
the first three components is  95%
3.
Y1, Y2 , Y3, Y4 are independent
4.
Y1, Y2 , Y3, Y4 have identical distribution.
119.
Consider a region R, which is a triangle with vertices (0,0) , (0,  ), (0,  ). where  > 0. A sample of
size n is selected at random from this region R. Denote the sample as {( X i , Yi ): i 1, 2,..., n} Then denoting
120.
Let X  ( X1, X 2 , X 3 , X 4 )' be 4 1 random vector such
that
X  n  max( X1, X 2 ,..., X n ) and Y n   max(Y1, Y2 ,..., Yn )
which of the following statemetns are true?
1.
X n  and Y n  and independent
2.
MLE of  is
3.
MLE of  max1i  n  X i Yi 
4.
MLE of  is max {X  n  , Y n }
X ~ N4 (O,  )
1


 
`


wher
  

1  
 1 

  1 
is positive define. Then which of the folowing
statemetns are true?
1.
X X , X X
1 2
2 3
and have identical distri-
bution
X  n  Y n 
2
2
2.
 X1  X 2  ~ F
1, 1
2
 X1  X 3 
3.
{( X1  X 3 )2  ( X 2  X 4 )2}.
4.
 X1  X 2  ~ F
1,1
 X 3  X 4 2
1
~ 22 .
2(1)
2
****************************************** Pi
AIM*******************************************
CSIR DECEMBER 2015 SOLUTION
PART ‘A’
1.(4)
diagonals =
Y
12(12 3)
= 54
2
6.(1)
A plane can be filled with the help of tiles which are
in regular hexagonal shape.
7.(1)
( 4, 3 .5)
Total number of professors =
X
21 27 30
78

 39
2
2
Number of professors who attended both th chennai
and delhi meetings = 39 21  18
As perpendicular bisector of the chord of a circl
passes
through the centre of the circle, hence centre will be
8.(3)
Distance covered by inner and outer wheel is .1
(Path is semicircular and there is unit spacing)
9.(1)
8 7
 ,  i.e, (4, 3.5)
2 2
Each number is arithmetic mean of the numbers
in next level between which it lies
2.(1)
Probability of being caught in atleast one trip =1 
probability being canght in no trip
4
 11 0.1  1 0.9 
So,
9 z
  5  z  19 .
2
10.(4)
4
Letter wise we are capable of finding implicit statement as “THIS PROBLEM IS SOLVABLE BY INTEL
LIGENT STUDENTS”
3.(2)
B (Father)
C (Mother) Speaker)
11.(3)
Graph in option (3) correctly represents distance
versus frequency relation
A (Son)
Statement is true in only one type of relation.
12.(4)
316  43046721 which has 8 digits
4.(4)
13.(4)
Most Indian tropical fruit trees produce fruits in
April -May because the impending monsoon provides optimum conditions for propagation.
14.(2)
The ratio of empty to filled volume of the glass
=
volume of empty frustum
volume of filled trustum
1 2
r h 103 93
3
1 2 3 3
r h 9 8
3


 10 9
 = 9 8
3
3
diameter of outer circle = diagonal of square. Also,
side of square =diameter of inner circle So, if r is the
radius of inner circle r 2 will be radius of outercircle.
So, Ratio of area of outer and inner circle will be
3
3

r 2
where r and h are radius and height of the cone
from which glass been extracted.
5.(2)
Number of diagonals of a convex polygon of n sides
n(n3)
=
so for deodecagon (12 gon), number of
2
(r )2

2
 2
15.(3)
As cosx is monotonically decreasing in
cos r < cos d < cos g.
0,  2  so.
16.(2)
23.(1)
If premium price isx, then
 2   12(10020)
8x4 x
n
Sn  
1
k 1 k
1 1
1
 1  ...
2 3
n
 10 x  1440  x  144
17.(3)
Minimum numbers of lines required for the required
purpose is 5.
1
1
1
1 1
1
1
... n

...
n 1
3
4
5
8
2
2

1
2




1
1
4
2n 1

 2.
.
 n
2
4
8
2
1
 S
18.(4)
One of the ten pillars mean IX numbered pillar.
2n
on R.H.S. there are n blocks each of which sums
19.(2)
n
2
to a value greater than1/2. So, S2n  for every n  1.
R
B
B G

G R
Boxes Balls
G R B 
24.(2)
T ( p( x))  p( x 2 ) , p( x)  1  p( x 2 )  1
There are only two possibilities
so, T (1)  1
20.(3)
p( x)  x  p( x 2 )  x 2
so, T ( x)  x 2
B
p(x) = x2 => p(x2)  ( x 2 ) 2
so, T ( x 2 )  x 4
T (a bx  cx 2 )  a  bx 2  cx 4
T (a bx  cx 2 )  0  a  bx 2  cx 4  0
 a  b  c  0
1 - x
C
So, Nullity (T) = 0
x
& Rank (T) = dim  p2 (R)   0  3  0  3
A
so, dim Range (T) = 3
Total distance covered from A to B along AC and
25.(1)
then CB will be AC+CB = 1 x 2  1 (1 x) 2 which
Rank  A3 4   2, so A has 2 linearly independent
will be minimum when x  1x  x  12 So, minimum
rows (columns).Hence At A will be reduced to
1
2

1
 I2

0
2
distance  1   1 1 
 2
 2

5 5


2 2
5
0
 by suitable elementary transformations
0
.Hence Rank of At A  2
26.(4)
PART ‘B’
 91 31 0 
A   29 31 0 
 79 23 59 
21.(4)
As sum of the eigen values of A is equal to trace of
A, so, 5th eigen value of A is 15 (22  23)  5
Now as determinant of A is product of eigen values
 determinant of A , det A or
of A, so det( A)  22335  180
|A| = 5931(91 29)  593162  2(31) 2 59
By Rolle’s theorem there exist C(0, 1) such that
 det (A)  0 if P is any prime number other than
2, 31 or 59 . so, S is infinite set.
22.(3)
f '(c)  0
Now for Rolle’s theorem on f ’(x) there exist a point
d (0, c) such that f "(d )  0 d(0, c)  d(0, 1) .
So, f "( x)  0 for some x(0, 1)
27.(2)
According to the fixed point theorem for functions
continuous in [a, b] it can be applied to function f
from [0, ] to [0, ] if f ( x) is bounded because in
this case if M is an upper bound then f(x) is continuous
in [0, M ] and codomain is also [0, M]
28.(4)
1
0
A  
0

 0
0 0 0
1 0 0
0 0 1

0 1 0

By congruent operations
Now.
0
1
(cx  d )2  (cy )2
(ad bc)
.y
(cx  d )2  (cy )2
V 
1
1
R3  R3 & C3  c3 we get A as congruent to
2
2
1
0

B  PT AP  0


0
(ax b)(cx  d ) acy 2 i  ad bc  y
y > 0  V > 0 & y <0  V > 0
So, T maps H to H and H to H 
35.(4)
0
0
0 
0 
0 0  1  whichhas
2

0 1
0 
2

(x1  x2  x3)( y1  y2  y3  y4)  15  35
 x1  x2  x3  1 & y1  y2  y3  y4  5
So, any one yi is equal to 2 and remaining yi equals
to 1. either y1  2 or y2  2 or y3  2 or y4  2
so there are 4 different solutions.
correspondingquadratic form as Q(PU) x2  y2  ZW
36.(3)
29.(4)
idempotent So,  I n  A    I n  A  . Trace (A) = K
 

  x 1 x  x 1 x  x 1
 A has K eigenvalues 1, so n-K eigenvalues are
 ( x 1)( x 1)( x 2 1)( x 4  x 2 1)( x 4  x 2 1) which are
A2
x12 1  x 4 1 x8  x 4 1
 A  A is idempotent  I n  A is also
2
0. As idempotent matrices are diagonalisable so.
G.M of 0 = n -K =n - Rank (A) . Rank (A) = K = Trace
(A) Also, Rank (A) + Rank
 I n  A)  n . As, A  I
4
37.
4
2
irreducible factors.
3
98  27 2
30.(3)




Z 98  1  Z  (1)
such that ( x, y )  
1
1
1
1 

lim

...

n n  2  4
4 6
2n  2n  2 
1 1
 
n  2  


4 2 
 
6  4  ...
As, Z n  1 only for those multiples of 2 which
are multiples of 2 or 7 from 0, 1, 2, ..., 97
No.
of
M(2) = 49
M(7) = 14
"
"
M(14) = 7
"
"
No. of M ( 2 or 7) = 49 + 14 - 7 = 56.So, |S| = 98 - 56
= 42

2n  2  2n 

1 
2
1
2n 2  2  

= lim
n 2 n 

2
2
32.(4)
The only subsets of R which are both closed aswell
as open sets are R and  . So if A is closed subset
98
i 2 i4
i194
 Z  1, e 98 , e 98 , ..., e 98
31.(2)
lim
1
i.e., 98th roots of unity
As codomain is bounded so the function is
bounded, but limit at no point exist.
= n 
2
S  {zC | z 98  1 & z n 1, 0  n  98}
n
so some eigenvalues of A are 0.
  ,   : R 2  , 
4
38.(2)
ˆ
Aˆ  Aˆ , as  is union of A and all relatively compact
connected components of X \ A.
of R and A  , A  R then A will not be open
39.(3)
33.(2)
R[X] i.e., set of all polynomials in x with real coefficieents is unique factorization domain but not a princip
al ideal domain.
2
x3  y 2   xy   1 implies that group G containing
x and y will contain {1, x, x 2 , y, xy, x 2 y} . So, order
of G is 6
40.(2)

f (z)   n log n zn a  n log n
n
34.(1)
n1
T ( z) 

az b
(ax b)i(ay)

cz  d
(cx  d )i(cy)
 (axb)i(ay)(cx d )i(cy)
(cx  d )2  (cy) 2
n  a  n2  n 1 n  a 1 n  ( n 2 ) 1 n
n
n
2
1
 lim n n
 1n 
1
  lim an n  lim  n 
u 2 ( x  y u ) ( y  x u)  0
1
1
 1  lim an n  1  lim an n  1
n2
 e
g(z)  
n 1 n
zn
n2
,
 n2
1
e
e
an 
 an n  
n
 n

1
en
lim
n lim
n   an n   1  
n n
R
1
 n




....(1)
u( x, 0)  1 in eq. 1 gives L.H .S = 1( x  0 1) (0  x 1)
 x 1 x 1  0  R.H .S .
45.(3)
f ( x)  ax 100  f '( x)  a
So, radius of convergence
xn 1  f (xn ) will converge if | f '( x)|  1  |a|  1
So, a = 0.1 is possible from given options
1
0

46.(4)
41.(2)


F  5iˆ 2 ˆj 3kˆ , r  2iˆ  ˆj  2kˆ
u(x, y) 

, Torque = r̂  F
iˆ ˆj kˆ
2 1 2
=
= iˆ16 ˆj 9kˆ
5 2 3
x3
 C where C is any real number can
6
become particular integral so there is more than
one particular integral.
47.(1)
y '( x)  f ( y( x))
f ( y(x))  y '(x)
42.(2)
F (x, y, y ') 
 2 y  Sinx 
y2  y' 2  2 y
 f ( y ( x)  f ( y( x))   y '( x)
 y '( x)   f ( y( x))
d
(2 y ')  0  2( y " y)  Sin x
dx
1
 y " y   Sin x . C.F. is y  c e x  c e  x
1
2
2
P.I. is y 
 f ( y( x))  y '( x)
f d  f 
 
Sin x
  0
y dx  y ' 
 y( x)
is not a solution but  y ( x) is a solution
48.(4)
1
sin x So, Extremal is
2
 dy1 
 dt 
 1 1   y1 
dY
 AY  
  
  y2 
dt
 dy2 
 0 1  
 dt 
1
y  C1e x  C2e  x  Sin x
2
43.(2)
 r
R( x, t , )    kr (t )
r 0

dy2
 dt  ln y2  t  k
y2
k0 (t )  1 ; k1 (t )  tx zdz  ( x  t )
k2 (t )  tx ( z t )dz 
( x t ) 2
( x t ) r
 kr (t ) 
2
r!

 R( x, t, )    r
r 0


( x t ) r
r!
r 0
( x t )r
r!
 y2  c1e t ; y2 (0)  1  c1  1  y2 (t)  e t
Now,
dy1
  y1  e t
dt
A.E. is m + 1 = 0 => m = - 1
C.F. is y1 (t )  C2 e t
 e
( x  t )
44.(2)
A.E. is
dx
dy
du


x y
y  x u
u

dy1
dy
  y1  y2 & 2   y2
dt
dt
dx  dy  du
du

0
u
1
t
P.I. is y1 (t )  ( D 1) e  tet
 y1(t )  C2 et te t
y1 (0)  0  C2  0
y1' (t )  te t
y1' (t )  (1t ) e t  0 , if t  1
 dx  dy  du  0  (x  y  u)  c1
So, y1(t ) is monotonically decreasing if t > 1.
u( x, 0)  1,  c1  x 1
y2' (t)  et  0 if t > 1 , So y2 (t ) is also
From these conditions solution of IVP satisfies
monotonically decreasing if t > 1.
49.(4)
62.(1,3)
23!
12
1

Required probability = (6)3 
216 18
50.(3)
1
F ( x)  F (  x )  P ( X  x)  P ( X   x)  1 P( X  x)
(0, 0, 0)
51.(4)
Mode will remain same
52.(2)
UE ( 2 ) 
1 2 2
e1  e2
2


x 2  y 2  z 2  1 represents points on unit sphere
53.(2)
54.(3)
55.(2)

which is closed as well as bounded. Also,  An
n 1
(0, 5)
(0, 2)
with product topology where An with two element
0 and 1 has discrete topology for n = 1, 2, 3, ...
admits finite subcover for every cover, so it is
compact.
63.(1, 2, 3, 4)
x1 x2  1
5x1  3 x2  15
 5 12 
 7, 7 


Sup
xR
 45 , 20 
 19 19 


(0, 1)
i.e f '( x) is bounded so f ( x) is uniformly
continuous .Also, under uniformly continuous
function f
(i)
bounded sequence is mapped onto
bounded sequence
(ii)
Convergent sequence is mapped onto
convergent sequence
(iii)
Cauchy sequence is mapped onto cauchy
sequence
2 x1 5x2  10
3, 0
 5 12 
5, 0
 45 20 
All the points lying on  7 , 7  and  19 , 19  are




solutions so there are infinitely many solutions.
56.(4)
P( X 1  X 2 ... X n  n log n) converges to 0.
57.(1)
58.(2)
59.(4)
| f '( x)|   implies that modulus of f '( x)
64,(1,3)
Given series is
60.(2)
n k 2 xk y
 y
lim

 y R 
 ey
n 

!
 k 0

1
0
i.e., the series is convergent. But if |x| < 1 then only
PART ‘C’
k
 x will be convergent iff |x| < 1 i.e, x(1, 1) so
61.(1,2)
for (x, y) = (-1, 1)  (0,  ) and (-1, 1)  (-1, 1) series
is convergent
lim P ( )  2 lt a   ltb
n n 0
0
n 0 n
lim P ( )   2 lt a  lt b
n n 1
1
n 1
n
20 lt an 0 lt bn  l1
2 lt a  lt b  l
1
n 1
n
2
65.(3, 4)
| f ( x)  f ( y)|  |cos x cos y|
.....(1)
 | f ( x)  f ( y)|  | x  y| . so f(x) is Lipschitz function
.....(2)
so, f(x) is uniformly continuous on (0, 1) so also
limit at either end points must exist. So,
Equation (1)  1 and equation (2)   2 gives
10 (0  1) lt an  1l1  0l1
lim f ( x) exist
x0
66.(2, 4)
1l1  0l2
 lt an 
1 0 ( 0 1) (i.e exist)
In U if x > 0 or x < 0 which also is the sign of y,
then there will be one- to- one correspondence
between U and R 2 and also if principal value is
taken so, options 2 & 4 are correct
lt an exist and  02 lt an  0 lt bn exist so lt bn exist
x 2 lt a n  x ltbn which exist for
Thus lim
n  Pn ( x) =
all xR . Also,
Pn' ( x)
 2an bn
lim P ' ( x ) 2 x lim
a lim
b
n n
n   n n   n which also exist
for all xR .
67.(2, 4)


1
1 
S   m 
, n
|m, n, p, qZ 

4| p|
4|q| 


 
and 0 f ( x)dx  0 , So, 0 f n ( x)  0 f ( x)dx

 S 1 Derived set of S  m n : m n  Z .Also,
S c has polygonal path between any two points in
71.
Jacobian of transformation  t  n
it so it is path connected
68.(1, 2, 3)
 
n
B f (tx)dx  B f ( x )t dt
a
ta

A  x, y  R2 : x  y  1
x 
 y
f (x, y)  
,
1

x

y
1

x
 y  ,Jacobian of

and also if y  (0,0,...,0)
R n f ( x  y)dx  R n f ( x)dx
72.(1, 2)
f 1 (G1 G2 )  f 1(G1)  f 1(G2 )
trasformation
f 1 (G1C )  f 1(R 2 \G1 )
y
1 x



2
2 
1

x

y
1

x

y






J  

1 y
x



2
2
 1x y
1 x y  

 f 1(R 2 )\ f 1(G1)
 R 2 \ f 1(G1) 
Now.
Now.
x  ( x1, x2 ,..., xn ) tx  (tx1, tx2 , ..., txn )
f
1
(G1)

c
Let f ( x, y)  ( x 2 , y 2 )
1
Det (J) = |J| = 
which does not vanish on
1x y
G1  (1, 1) ; G2  {(1, 1)}
A. Also on A f (x, y) is infinitely differentiable

f ( x, y ) 
a, b 

gives

f (G1 G2 )  f ()  
f (G1 ) 
 
f (G 2 )  { 1, 1 }
and infact
f (G1 G2 )  f (G1) f (G2 ) . Also, G1   & G2  
y  a(1 x  y) & x  b(1 x  y)
 (1 a) y  ax  a
implies G1 is open and G2 is closed then G1 G2
is both open and closed
b y  (1b) x  b
73.(1, 3)
1a a  y   a 
 
    
 b 1b  x   b 
. Determinant of coefficient matrix is 1-a-b . So, if a
+ b  1 system has unique solution . So, f is oneto-one on A . For a + b = 1 there cannot be (x, y) A
such that f (x, y )  (a, b) so f ( A) is not equal to R 2
then A1 ( AB) A  BA so, AB and BA are similar
matrices
74.(1, 2, 3, 4)
dimV  rank (T )  nullity(T ) &
69.(3,4)
| xy| is differentiable at (0, 0) so f ( x, y ) is
3 2
differentiable at (0, 0) , Df (0, 0)   2 3  ,


dimV  rank (T 2 )  nullity(T 2 )
Rank (T) = Rank (T 2 )
 Nullity (T) = Nullity (T 2 )
| Df (0, 0)|  5  0 So, At (0, 0) Df(0, 0) is also
As T ()  0 T 2 ()  0
invertiable
 Null space of T  Null space of T 2 But
70.(3, 4)
If { f n} is real valued continuous functions in
[0, ) and converges to f uniformly on [0, ) ,
if
nullity (T) = Nullity (T 2 )
 Null space of T = Null sapce of T 2
So, T 2 ()  0  T ()  0
So, kernel (T) and Rank (T) are disjoint
then f is continuous on [0, )
Now,
If A and B are square matrices of same order then
AB and BA have same eigen values. Also A1 exist
x
f ( x)
;
n
n2
x n
So, also kernel (T 2 ) and Rank (T 2 ) are disjoint
75.(1, 2, 3, 4)
0
o therw ise
T ( p( x))  0  an  an 1 x ... a0 x n  0
and x[0, ) Then f ( x)  0 . Also f n (x) onverges
ges
uniformly to f ( x ) but
n

0 fn (x)dx  0
1
x
dx 
n2
2
 a0  a1  ... an  0
So, Nullity (T) = 0 and Rank (T) = n+1 . Hence T is
one- to -one & So T is also invertible Transformation
matrix T is
T 





0
0

1
0 0 0  0 1
1, 4 5C 4C (41)!  30 way






0 1 0 0 0  0
1 0 0 0 0  0

0 0 0 0 0  0
1
0 0 0  1 0
4
2, 3 5C 3C (21)!(31)!  20 way
2
3
 a5  50
Total 50 ways
For n = 4 disjoint cycles will be 1,
det (T) = one or minus one = 1
76.(1, 3, 4)
G.M. of eigen value2 of A is 3Rank (A-2I) = 3  2 = 1
As A.M.  G.M. for eigen value 2, hence A is not
 a4  11
82, (1, 2, 4)
2
X 3 contains factors X , X and X 3 , So, there are 3
Q[ X }
X3
diagonalisable So, C ( x)  ( x  2)2 ( x 3) and
distinct proper ideals in
m( x)  ( x 2) 2 ( x 3) , so its Jordan canomicalform
unique. A is not an integral domain
2 1 0
0 2 0 
B

is

 . As eigen values of A are from Q,
0 0 3
, but prime ideal is
 
 
f  g  0  f(0) g(0) = 0 as . f  g  0 at least
one of them must be zero
83.(3,4)
So A is similar to B over Q.
W  Cos
77.(2, 4)
If u and v are solutions of Ax = b then Au = b & Av
2
2  
 i Sin  e 5
10
10
2
 i 2 
i4 
W2  e 5   e 5




= b  A(u  (1)v)   b(1)b  b so u  (1)v
is also solution of Ax = b for R . If Rank A =n
then then Ax = b has unique solution which also
means at most one solution .
Also w5  1 Set {1, w, w2 , w3, w4} and
78.(1, 3)
{1, w2 , w4 , w6 , w8} are same so, these set over Q
every non-zero vectors are eigen vectors of A,
will span L  Q(w) & K= Q(w2 ) respectively
HenceL = K
means AX X (AI)X 0 has n linearl independent
solutions, so rank of AI must be0 ;hence AI
null matrix  A  I for some C  All eigen
values of A are equal
84.(1)
Number of sylow -5 subgroups = 1+ 5k where
1+ 5k divides 60  1+ 5k = 1 or 6 . But as G is a
simple group so, 1 cannot be the possibility hence
G has six sylow - 5 subgroups
79.(3, 4)
f : R  R2 givenby f ( x)  (sin x, cos x)
is continuous map and also


f ( R)  ( x, y)R 2 | x 2  y 2  1 . Also,
f :  0, 1   2, 3 {0, 1}
85.(3,4)
Q[ x]
X 2  X 1 is irreducible over Q , so x 2  x 1 is field


given by
f ( x)  0
if x[0, 1
1
if x[2, 3]
is a continuous map from [0, 1] U [2, 3] to {0, 1}
80.(2, 4)
X  2(mod 5) & X 3(mod 7)
 X  17(mod35)
F2[ X }
also X 2  X 1 is irreducible over F2 so  X 2  X 1 is
also a field
86.(1, 2, 3, 4)
1
an   lim an  0 .So f(z) is constant
n
1
a  (1)n 1   lim a  0 . So, again f(z) is
n
n
n
X  17 (mod 35) & X  4 (mod11)
 X 367(mod 385)
 X  367, 752, 1137, 1522, ...
constant an 
1
 lim an  0 .So, f(z) is again
2n
1
System has a solution in [601,800] & [1001,1200]
constant an  n if 4 does not divide n ,  is 4
n
For n = 5 disjoint cycles will be
1
divides n , has subsequence a4n 
such that
81.(1, 4)
4n
lima  0,
4n
so, f ( z) is again constant
87.(1, 4)
u(t )  t 4  u '(t )  4t 3
 Z 
3
4u 4 (t )  4t 3 So, u(t )  t 4 is also a solution
163 is prime number so, 
 is cyclic also,
 163Z 

u(t )  0 ;0  t  1

group of it’s units is also cyclic units of Z 10Z =
{1 , 3, 7, 9} which is cyclic group generated by {3}
or {7}
88.(1, 2, 3, 4)
f (z) 
Now
 (t 1) 4 ;t 1
 u '(t )  0 ;0  t  1
 4(t 1)3 ; t 1
1
; ez  1
e z 1
0 ;t 1
e z 1  0  e z  1
So, this u(t) also satisfioes the given differential
equation. similarly
 z  ln1  ln1 ei2n
u(t)  0
; 0  t  10
has
4
 (t 10) ; t 10
 z = i2n  ; n = 0,  1  2,...
Also
u '(t )  0 ; 0  t  10
z i2n 0
[ Case ]
e z 1 0
lim
Z i2n
 4(t 10)3 ; t  10
1
lim
z i2n z [ By L’ Hospital rule} = 1  0 so f ( x)
e
So this u(t) also satisfies given differential equation
94.(1,3)
95.(3)
has simple pole at 2ni . As, f(z) has only
singularities as pole, so f(z) is homomorphic
function Also all the poles are on imaginary axis
 2u  2u
For elliptic equation x2  y 2  0 , solution u( x, y)
89.(2, 3)
such that
1
1
f ( z)  on annulus A  {z  C : | z | 2} has
z
2
laurent series expansion about the centre of the
annulu whose pole is 0 lying out side A. Such
approximatin sequence cannot be polynomial but
it will be rational function only.
lim u( x,
y 
y)  0 and u( x, y)  y 0  sin x

will be u( x, y )   an sin(nx  bn )e ny
n 1
96.
4
f ( x)  x 3
90.(4)
x0  0 ; xn 1  f ( xn )
K = Compact subset of C with non-empty interior

 x1  3 ; x2  3 3
K will be norm on vector sapce of entire function
but not on vector sapce of continuous functions.
91.(4)
97.
y(t )  y(0)  0t y(s)ds
...(1)
u "( x)  2u( x)
Squaring both sides we get

y 2 (t )  y 2 (0)  0t y(s)ds
2

 u "(x)  2u(x)  0
2 y(0)0t y(s)ds
 ( D2 2 )u  0
 u ( x)  A cos x  B sin x
Differentiating both sides of equation (1) we get
y '(t )  y(t )  y(t )
Now.
2
 cet
 y(t )  y(0)
et
u(0)  0  A  0
u(1) 0 A0
  y(s)ds   2 y(0) y(s)ds
t
0

y2 (0)(et
t
0
1)2  2 y2 (0)(et
 y 2 (0)(e2t 1)2  2 y 2 (0)(
....point of convergence is 3 3 3 ...
1,2
 u ( x)  B sin x
u '( x)B cos x
1)
u '(0)  B 
e 2t 1
)  2 t y 2 (s)ds.
0
2
So, y2 (t)  y2 (0)  20t y 2 (s)ds
So, u '2 ( x)  2u2 (x)  2b2  u '2 (0)
Now,
92.(3,4)
1
2 2
0  B cos 2x dx
2
Both H and L are quadratic in V and also H = L
93.(1,2,3,4)
3
u '(t )  4u 4 (t )
....(1)
u(t )  0  u '(t)  0 equation - (1) is satisfied
 01 / 2 2 B cos 2x dx = 0
 01 u '2 ( x)dx 2 01 u 2 ( x)dx  0
98.(1,2,3,4)
For existence of a minimizer it is sufficient to have
101.(2,3)
a convergent subsequence of {n ) and J is
continuous
.....(option 3)
Rest option 1, 2 & 4 are also sufficient because
they imply option 3.
99.(1,2,3)
2  x2 2
1
xn 1   xn    n
xn 
2
2xn
 x  f ( x) where f ( x) 
x2 2
for fixed point
2x
iteration
2
For the given hyperbolic equation
u( x, 0)  ei x
by
Also, f (x)  x2  2  f '( x)  2 x so, for .Newton’ss
method iterative formula is
2
u u

with
t 2 x 2
separtion
of
xn 1  xn 
function u( x, t )  eix eit
u( x, t )  eix ei t
 eit  eit
ix
u
(
x
,
t
)
e

as well as

2






102.
2
Satisfies the given equation but for u( x, t )  t 
u( x, 0) 
x
2
x2
so it does not satisfy the given
2
condition
100.(3)
u  px  qy  p2  q2 . It is quadratic in p so solution
will be parabolic in u- x plane
103.
106.
109.
112.
115.
118.
f ( xn )
x 2 2
 xn  n
f '( xn )
2xn
2xn2  xn2  2 xn2  2

2 xn
2xn
1, 3
Characteristic numbers are distinct for the given
homogeneous integral equation and the
corresponding eigen functions are orthogonal to
each so, 1  2 and 01 f1( x) f 2 ( x)dx0
1,2,3,4
104.
1,4
105.
3
107.
1,2
108.
2,4
110.
2,4
111.
4
113.
1,3,4
114.
2,3,4
116.
1,3
117.
1,2,3
119.
3
120.
2
1,4
4
1,3
2,4
1,3,4
************************************* Pi AIM************************************