CSIR-UGC/NETMATHEMATICAL

SCIENCES

SOLVED PAPER DEC-2013

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21. Let ]1,0[]2,1[: nf be given by nn xxf )2()( for all non-negative

integers n. Let )(lim)( xfxf nn for 21 x . Then which of the following

is true?(a) f is a continuous function on [1, 2](b) nf converges uniformly to f on [1, 2] as n

(c) 2 2

1 1lim ( ) ( )n nf x dx f x dx

(d) for any )2,1(a we have )()(lim afafnn

Exp: Given :[1,2, ] [0,1]nf be given by ( ) (2 ) nnf x x V 0n

( ) lim ( )

x nnf f x

If 1 2;0 2 1 x x lim (2 ) 0

nn

x

( ) 0, 1 2 1, 1f x x

x

2

1

( ) 0 f x dx

2 2 1

1 1

(2 ) 1( ) (2 )1 1

nn

nxf x x dx

n n

2

1

( ) 0nnLim f x dx

Ans: (c)

22. For a fixed positive integer ,3n let A be the nn matrix defined as

1A I Jn

, where I is the identity matrix and J is the nn matrix with all

entries equal to 1. Which of the following statements is NOT true?

(a) AAk for every positive integer k.

(b) Trace 1)( nA

(c) Rank )(A + Rank nA )1( .

(d) A is invertible.

5

Exp: 1

A I Jn Suppose take 3 × 3 matrix

1 0 0 1 1 110 1 0 1 1 13

0 0 1 1 1 1

A

2 1 13 3 31 2 1

3 3 31 1 2

3 3 3

2 4 1 1 2 1 1 1 2det( )3 9 9 3 9 9 3 9 9

A = 0

A is not invertible.Ans:(d)

23. Let A be a 5 × 4 matrix with real entries such that 0xA if and only if 0xwhere x is a 4 × 1 vector and 0 is a null vector. Then, the rank of A is(a) 4 (b) 5 (c) 2 (d) 1Exp: A be a 5 × 4 matrix

0 iff 0 Ax x . If there are no free variables ( ) rark A n column . Thenonly solution is x = 0. ( ) 4rank A Ans:(a)

24. Letsin

if 0( )

1 if 0

xx

f x xx

. Then f is

(a) discontinuous (b) continuous but not differentiable(c) differentiable only once (d) differentiable more than once

Exp:sin( ) xf x

x if 0x = 1, x = 0

The Taylor series for sin x is 2 2

0

sin ( 1)(2 1)!

nn

n

xxn

2

0

sin ( 1)( )(2 1)!

n n

n

x xf xx n

Putting 0x in the above series we get 1 which agrees with the definition of f.

Let2( 1)( )

(2 1)!

n n

nxf x

nThen easily we can show that nf converges uniformly to f on [1, –1] and

nf also converges uniformly on [–1, 1] and nf also converges uniformlyon [–1, 1] and so on. Hence f is infinitely times differentiable.Ans:(d)

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25. Consider the following row vectors : )0,1,0,0,1,1(),0,0,1,0,1,1( 21 aa ,

)1,0,0,0,1,1(3 a 4 5(1,0,1,1,0,0), (1,0,1,0,1,0)a a , 6 (1,0,1,0,0,1)a

The dimension of the vector space spanned by these row vectors is(a) 6 (b) 5 (c) 4 (d) 3

Exp:The vectors 1 2 3 4, , , are linearly independent

1 1 0 0 01 1 0 1 01 1 0 0 11 0 1 0 01 0 1 0 01 0 1 0 1

Also 4 1 (0, 1,1,0,0, 0)

Satisfies 5 1 6 3&

Thus 1 2 3 4 1 2 3 4 5 6( , , , ) ( , , , , , )vector vector & 1 2 3 4( , , , ) is abasis. The space has dimension 4.Ans:(c)

26. Let (( )), 3n n ijA a n , where 2 2( )ij i ja b b , , 1 , 2 , . . . ,i j n

for some distinct real numbers ...., 21 nbbb Then ...., 21 nbbb Then )det(A is

(a) )( jiji bb (b) )( jiji bb (c) 0 (d) 1

Exp: ( ) 3 n n ijA a nSuppose n = 3

2 2 2 2 2 211 1 2 12 1 2 13 1 3a b b a b b a b b

2 2 2 2 2 221 2 1 22 2 2 23 2 3a b b a b b a b b

2 2 2 2 2 231 3 1 32 3 2 33 3 3a b b a b b a b b

2 2 2 21 2 1 3

2 2 2 22 1 2 32 2 2 23 1 3 2

0

0

0

b b b b

A b b b b

b b b b

2 2 2 2 2 2 2 2 2 2 2 21 2 3 1 2 3 1 3 2 1 3 2det( ) ( )( ( ))( ) ( )( )( )A b b b b b b b b b b b b

2 2 2 2 2 2 2 2 2 2 2 21 2 3 1 2 3 1 3 2 1 3 2( ) ( ) ( ) ( ) ( ) ( ) b b b b b b b b b b b b =0

Ans:(c)

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27. Let }{},{ nn ba be sequences of real numbers satisfying nn ba for all 1n .Then(a) nna converges whenever nnb converges

(b) nna converges absolutely whenever nnb converges absolutely

(c) nnb converges whenever n na converges.

(d) nnb converges absolutely whenever nna converges absolutelyAns:(b)

28. If

1n na is absolutely convergent, then which of the following is NOTtrue?

(a)

0mm n

a as n (b) 1 sinnna n

is convergent

(c) 1na

ne

is divergent (d) 21n

na

is divergent

Exp: Given na converges absolutely 2 na is not divergent.

na , we have lim 0na

Hence 1nk N a V n k

2n na a V n k

2n n

n k n k

a a

2

1n

k

a

2

1

nn

a is divergent.

Ans: (d)29. Let A be an nn matrix with real entries. Which of the following is correct?

(a) If ,02 A then A is diagonalisable overcomplex numbers

(b) If 2 ,A I then A is diagonalisable over realnumbers

(c) If ,2 AA then A is diagonalisable only overcomplex numbers

(d) The only matrix of size n satisfying thecharacteristic polynomial of A is AExp: (a) It is not true in general

For example0 10 0

A

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(b) True, if a matrix is annihilated by a polynomial with single roots,then it is diagonalizable and its eigenvalues are roots of thepolynomial.

(c) They are diagonalizable as the polynomial 1nx has only single

roots. But not necessarily over complex numbers (ie.)1 00 1

A

(d) Not true0 1 0 0

&0 0 0 0

A B satisfy 2( ) p x x which is theircharacteristic polynomials.

Ans: (b)

30. Let ]1,0[]1,0[: f be any twice differentiable function satisfying

)()1()())1(( yfaxafyaaxf for all ]1,0[, yx and any ]1,0[a .

Then for all )1,0(x

(a) 0)( xf (b) 0)( xf (c) 0)( xf (d) 0)( xf

Exp:Given :[0,1] [0,1]f be twice differentiable.

( (1 ) ) ( ) (1 ) ( )f ax a y a f x a f y V , [0,1]x y V [0,1]a

f is a convex function, we know that f is convex iff ( ) 0 f x .

Ans:(b)31. Let A be a 4 × 4 invertible real matrix. Which of the following is NOT

necessarily true?(a) The rows of A form a basis of R4

(b) Null space of A contains only the 0 vector(c) A has 4 distinct eigenvalues

(d) Image of the linear transformation Axx on 4R is 4RExp: Given A to be a 4 × 4 invertible matrix then (c) is not correct as if we

take A to be

1 0 0 00 1 0 00 0 1 00 0 0 1

. Then eigen values of A are 1, 1, 1, 1

Ans: (c)

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32. The partial differential equation uxu

tu

2

2 can be transformed to

2

2

xv

tv

. For

(a) uev t (b) uev t (c) tuv (d) tuv

Exp: tV e u tu e vt tv ue u e

t t

tv ue

x x

2 2

2 2

tv uex x

2

2

t t tu ue u e et x

2

2u uut x

2

2

u u ut x

Ans: (a)

33. The integral equation 1

0)(),()()( dyyyxKxfx For 2),( xyyxK has

a solution(a) )()( xfx (b) ),()( xxKx

(c) 3)( xx (d) 1

02 )(

34)()( dxxfxxxfx

Exp:

12

0

( ) ( ) ( ) x f x xy y dy =

12

0

( ) ( ) f x x y y dy

( ) ( )x f x xc

where

12

0

( ) c y y dy1

2 3

0

( ( ) ) c y f y y c dy

1 12 3

0 0

( ) y f y dy y cdy1

2

0

( )4

cc y f y dy

12

0

( )4

cc y f y dy

12

0

4 ( )3

c y f y dy

12

0

4( ) ( ) ( )3

xx f x y f y dy

Ans:(d)

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34. For any integers ba, let baN , denote the number of positive integers

1000x satisfying )27(modax and )37(modbx . Then,

(a) there exist ba, such that 0, baN .

(b) for all 1,, , baNba

(c) for all ,, , 1a ba b N

(d) there exist ba, such that 1, baN , andthere exist ba, such that

2, baN

Exp: Given ,a bN is the number of positive integers 1000x ; satisfying

(mod 27)(mod 27)

x ax b

(27,37) 1 gcd

By Chinese - remainder theorem. There exist a integer0 0

0

(mod 27)

(mod37)

x x a

x b

and all the solutions are given by 0 37.27x k 0 999 x k where k Z .

Hence ,a bN = 1Ans:(b)

35. Let 1 be the product (standard) topology on R2 generated by the base.

tsvutsB :),(),{(1 , vu where },,, Rvuts ( 1B is the collectionof product of open intervals) Given RRr , with Rr 0 and

221 ),( Raaa ,

Let }),{(),,( 221 RxxRraC 2 2 2 2

1 1 2 2( ) ( ) } r x a x a R

Let }0,,,);,,({ 22 RrRRrRaRraCB

Let 2 be the topology generated by the base 2B . Then

(a) 1 2 1 2, (b) 2 1 1 2,

(c) 1 2 (d) 1 2 2 1,

Exp: 1 is generated by open rectangles and 2 is generated by open annuluses.

Consider any open rectangle say

11

Take any point inside it x

Then we can get an open annulus inside it. x easily,xCentre of annulus

Then we can get an open rectangle arrow inside it xCentre of annulus

Then we can get an open rectangle arrow x inside the open rectanglex

Centre of annulus

Ans:(c)

36. The number of group homomorphisms from the symmetric group 3S to theadditive group ZZ 6/ is(a) 1 (b) 2 (c) 3 (d) 0Exp: Let 3 6: S Z be a homomorphism. Then ker is a normal subgroupof 3S . The possible normal subgroups of 3S are 3, ,{ , (1,2,3), (1,3,2)}e S e . If

( ) kernel e , then by first isomorphism theorem, S3 and Z6 will beisomorphic which is not true. If 3ker S then 3( ) 0S . It is a

homomorphism. If ker { , (1, 2,3), (1,3, 2)} e then ( (1,2,3)) ((1,2)) 2O O .Hence ( (1,2))O = 1 or 2.

If it is 1, then (1,2) = 0. Since S3 is generated by (1,2 ) & (1,2,3)0 .

Hence ( (1,2)) 2 (1,2) 3O .

Similarly ((1,23)) ((3,1)) 3 We can easily check that this is a homomorphismAns:(b)

37. If )1,0(]1,0[: f is a continuous mapping then which of the following isNOT true?

(a) ]1,0[F is a closed set implies )(Ff is closed in R.

(b) If )1()0( ff then ])1,0([f must be equal to )]1(),0([ ff .

(c) There must exist )1,0(x such that xxf )( .

(d) )1,0(]1,0([: f .

12

Exp: If (0) (1)f f then ([0,1])f may not be [ (0), (1)]f f consider f given

by, 1 1(0) , (1)3 2

f f ½

3

1

Ans:(b)38. How many normal subgroups does a non-abelian group G of order 21 have

other than the identity subgroup }{e and G?(a) 0 (b) 1 (c) 3 (d) 7

Exp: 21 3.7 GNo. of sylow 3 subgroups 1 + 3k dividing 7 is 1 or 7No. of sylow 7 subgroups 1 + 7k dividing 3 is 1So we have a unique sylow 7 subgroup and so it is normal. Sylow 3 subgroupsis not normal.G has only one non trivial normal subgroup.Ans:(b)

39. Suppose 1 2, ....., nX X X are independent and identically distributed randomvariables each having an exponential distribution with parameter 0 .Let (1) (2) ( )... nX X X be the corresponding order statistics. Then theprobability distribution of ( ) ( 1) (1)( ) /n nX X nX is(a) Chi-square with 1 degree of freedom(b) Beta with parameters 2 and 1(c) F with parameters 2 and 2(d) F with parameters 2 and 1.Ans:(c)

40. A population contains three units 1 2,u u and 3u . For 1, 2, 3i let iY be thevalue of a study variable for iu . A simple random sample of size two isdrawn from the population without replacement. Let 1T denote the usualsample mean and let 2T and 3T be two other estimators, defined as follows :

1 2 1 2

2 1 3 1 3

2 3 2 3

1 ( ) if , are in the sample2

1 2( ) if , are in the sample2 31 1 if , are in the sample2 3

Y Y u u

T Y Y u u

Y Y u u

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1 2 1 2

3 1 3 1 3

2 3 2 3

1 ( ) if , are in the sample2

1 if , are in the sample2

1 1 if , are in the sample2 2

Y Y u u

T Y Y u u

Y Y u u

IfY

is the population mean, then which of the following statements is true ?

(a) All the three estimators 321 ,, TTT areunbiased for Y

.

(b) 2T and 3T are biased estimator for Y

but 1T is not.

(c) 1T and 2T are unbiased for Y

but 3T is not.

(d) 1T and 3T are unbiased for Y

but 2T is not.Ans: (c)

41. Let nXXX ,...,, 21 be a random sample from ),( 2N where 02 is known.

Suppose has the Cauchy prior with density..

121 1 ,

with and known. Then with reference to the posterior distribution of(a) the posterior mean does not exist and theposterior variance is(b) the posterior mean exists but the posteriorvariance is(c) the posterior mean exists and the posteriorvariance is finite(d) the posterior variance is finite but theposterior mean does not existAns:(c)

42. In a 2 × 2 contingency table if we multiply a particular column by an integer( 1)k , then the odds ratio

(a) will increase(b) will decrease(c) remains same

(d) will increase if 2k and will decrease if 2k .Ans:(c)

43. A popular car comes in both a petrol and diesel version. Each of these isfurther available in 3 models, L, V and Z. Among all owners of the petrol

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version of this car, 50% have model V and 20% have model Z. Among dieselcar customers, 50% have model L and 20% model V, 60% of all customershave bought diesel cars. If a randomly chosen customer has model V, what isthe probability that the car is a diesel car?(a) 3/8 (b) 3/5 (c) 1/5 (d) 2/3Ans:(a)

44. Let 1 2, , ..., nX X X be a random sample from uniform 1 1,2 2

.

Consider the problem of testing 01:2

H against 21:1 H . Define

min)1( X 1 2{ , ,...,X X }nX . Consider the following test : Reject 0H if

(1) 0X , accept otherwise. Which of the following is true?

(a) power of the test = 0, size of the test = 0.(b) power of the test = 0, size of the test = 1.(c) power of the test = 1, size of the test = 0.(d) power of the test = 1, size of the test = 1.Ans:(c)

45. Suppose the cumulative distribution function of failure timeT of a component

is 1 exp( ), 0, 1, 0ct t c Then the hazard rate of )(t is

(a) constant(b) non-constant monotonically increasing in t.(c) non-constant monotonically decreasing in t.(d) not a monotone function in t.Ans: (b)

46. Let ,..., 21 XX be a Markov chain with state space {1, 2, 3, 4}. Let the

transition probability matrix P be given be

1/ 3 0 0 2 / 31/ 4 1/ 4 1/ 4 1/ 4

0 0 1 02 / 3 0 0 1/ 3

P

Which of the following is a stationary distribution for the Markov chain?

(a) 1/ 4 1/ 4 1/ 4 1/ 4 (b) 1 / 3 0 0 2 / 3

(c) 0 1/ 4 1 / 2 1 / 4 (d) 1/ 3 0 1 / 3 1 / 3

Ans: (d)

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47. A factorial experiment involving 4 factors, 1 2 3, ,F F F and 4F each of 2 levels,0 and 1, is planned in 4 blocks each of size 4. One of these blocks has the

followiong contents :

1 2 3 4

0 0 0 00 1 0 11 0 1 11 1 1 0

F F F F

The confounded factorial effect are

(a) 1 2 1 3 2 3, ,F F F F F F (b) 1 3 1 2 4 2 3 4, ,F F F F F F F F

(c) 1 4 1 2 3 1 2 4, ,F F F F F F F F (d) 1 4 2 3 1 2 3 4, ,F F F F F F F FAns: (b)

48. Let 0 1 2 3, , ,X and 4 be independent and identically distributed normal

random variables with mean 0 and variance .02 Define

21 11i i iX X where 0 1, for ,0i 1, 2, 3. Let .ij k

denote the partial correlation between iX given jX given kX . Then 2.14p =

(a) 3 (b) 2 (c) (d) 0Ans: (d)

49. Consider the following probability mass function )(21, xP where the

parameters ),( 21 take values in the parameter space

1 1 1 1,3 , , 2 , 2, , 3,3 2 2 3

;

1 2 1 1 1 13 2 2 33 2 2 3

1 1 11 1 7 1 8 1 92 1 11 1 14 1 16 1 93 8 11 5 7 3 4 2 34 1 11 1 14 1 17 1 9

( , ), , , ,

/ / / // / / // / / // / / /

x

Let X be a random observation from the distribution. If the observed value ofX is 3, then(a) MLE of ,3/11 MLE of 32

(b) MLE of 1 1/ 2, MLE of 2 2

(c) MLE of 1 2, MLE of 2 1/ 2

(d) MLE of 1 3, MLE of 2 1/3 .Ans: (c)

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