CSIR UGC NET_ MATHEMATICS_ FREE SOLVED PAPER vpm class.pdf

C SIR NET, GA GA TE, IIT-JA IIT-JA M, UGC UGC NET , TI FR, IISc , JE JEST , JNU, BHU BHU , ISM , IBPS IBPS , CSAT, CSAT, SLET, NIM CET, CET, CT ET

CSIR CSI R NET - MA MATHEMA THEMATICAL TICAL SCIENCE MOCK MO CK TEST TE ST PAPER



This paper contains 60 Multiple Choice Questions



part A 15, 15, part B 25 and a nd part C 20



Each Ea ch quest questii on in Part 'A 'A'' carries two marks



Part Pa rt 'B' carries carries 3 marks



Pa rt 'C' ca Part carries rries 4 .75 marks respectively. respectively. Part Pa rt C has more than one correct correct options and there is no negative marking in Part C



Th ere Ther e wi wi ll be be negative marking @ 25% Part A, 0.75 marks in Part B for each wrong answer.



P attern attern of questions : MCQs

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Total mark marks s

: 200



Dur ura at ion of t est

: 3 Hours

For IIT II T-JAM, JNU, GATE, NET, NIMCET and Other Entrance Entr ance Exa!

1-C-8, Sheela Chowdhary Road, Talwandi, Kota (Raj.) Tel No. 0744-2429714

Web Site ww www w.vpmcl .vpmclass asses.c es.com om E-mail E-m ail-vpmclas -vpmclass [email protected]

Phone: 0744-2429714 Mobile: 9001 90012 2 9711 97111, 1, 9829 98295 5 67114, 90012 9 00129 9 7243 Website: www.vpml!sses.om www.vpml!sses.om "-M!il: vpm l l!sse !sse s#$!hoo.om s#$!hoo.om % % in in&& o#v pml!sses. ml!sses.o om ' (()ess: (()ess: 1-* 1-*-8, -8, +hee +hee l! *h * how(h!)$ o! o!( (, ++, ' /W' /W', , ' , ''+' ''+' , 324 3240 0 05

Page 1


PART A (1-15) 1 T wenty four cl cl erk can can cl ear 180 files in 15 days. Number of cl cl erk require require d to clea r 24 0 files files i n 12 days day s is (1) 38 (2) 39 (3) 40 (4) 42 2.

In the the gi ven fi gu re, RA RA = SA = 9cm and and QA = 7cm. If PQ i s the diam eter, eter, the then n radius i s

P

A

R

S

Q 65 cm (1) 7 130 cm (2) 7 (3) 8 cm (4) None 3.

If the ci rcles are are drawn with wi th radii radii R1, R2, R3 wi th cen ce ntre at the verti ces of a triang triangle le as sho wn i n

figure. Side of triangle is a, b, c respectively, then R1 + R2 + R3 is equal to


Page 2


R3 R1

3(a + b + c)

1 (a + b + c) 3 1 (a + b + c) 2 (4) 2 (a + b + c)

4.

Study St udy the followin owin g graph and an swer the question given below it

Production in a Tool Factory

) 0 0 0' n i ( s l o o T f o . o N

50

20

45

18

40

16

35

14

30

12

25

10

20

8

15

6

10

4

) s e r o r c s R n i ( s l o o t f o e u l a v l a t o T

2 1984

1985

1986

1987

1988

1989

Years –– Number Number of Tools

----- Value

What was the value of each tool in 1985?


Page 3


5 (1) Rs

1 3 thousand

(2) Rs 50 t housan housand d (3) Rs 5 , 103 10 3

5 (4)

5 9

5.

T he tot al adults i n a ci ty i s 60000. T he vari vari ous ous sect i ons of the them m are i ndic dicated at ed bel ow i n the circle

II

I II III IV

108° V 54°

18°

V

→ employees in the public sector → employees i n the private secto sectorr → employees i n the corporate sector → self employed → unemployed

IV IIII II

What

percentage of the employed persons is self employed?

5 (1)

5 19

19 (2)

1 5

(3) 20 (4) 5

6.

Lo ok at this thi s seri es: 14, 28 , 20, 20, 40, 32, 64, ... Wha t numbe numbe r should come ne xt?

(1) 52 (2) 56 (3) 96


Page 4


(4) 128 7.

A car owne ownerr buys petrol at Rs.7 .50, Rs Rs. 8 and Rs. Rs. 8.50 8.50 pe r liter for fo r three three success successii ve yea yea rs. rs. Wha Wha t

approximately is the average cost per liter of petrol if he spends Rs. 4000 each year? (1) Rs. 7.98 (2) Rs. 8 (3) Rs. 8.50 (4) Rs. 9

8.

In a certain store, to re, the the pro fit i s 320 320% % of the cost. ost. If the cost i ncreas ncreases es by 25% 25 % bu t the selli selli ng pr p ri ce

rem ains cons consta ta nt, nt , approxima approxima tel tel y what percentage of the selling pr p ri ce i s the p rofit? ofi t? (1) 30% (2) 70% (3) 100% (4) 250%

T oday is F riday afte r 62 days, days, it will be :

9.

(1) T hursday (2) Friday (3) Wednesday (4) Tuesday T uesday

10 . A car car travelling with of i ts act ual ual spe spe ed covers covers 42 km km in 1 hr 40 mi mi n 48 sec. Fi Fi nd the actual speed peed o f the the car.

17 (1)

6 km /h / hr 7

(2) 25 km/hr (3) 30 km/hr (4) 35 km/hr 11 .

P is i s a working working and Q i s a sl sl eepin eepin g partne partne r. P puts i n Rs. 34 00 and Q puts Rs.650 Rs.650 0. P receiv recei ves 20 %

of the profits for managing. The rest is distributed in proportion to their capitals. Out of a total profit of Rs.990, ho w much did P ge t ?


Page 5


(1) 460 (2) 470 (3) 450 (4) 480 12 .

A lawn i s the for fo rm o f a rectan rectangle gle having having i ts si de i n the the rratio atio 2:3 2:3 T he area area of the the lawn lawn i s 1/6 hecta hectares res..

Fi nd the length and breadth of the law l awn. n. (1) 25m (2) 50m (3) 75m (4) 100 m 13 .

An aeroplane covers a certai n distan ce at a spe ed of of 240 kmph kmph i n 5 hou hours rs.. To cove r the same

distance in 1 hours, it must travel at a speed of: (1) 300 kmph (2) 360 kmph (3) 600 kmph (4) 720 kmph

14 .

Find out the missi issi ng number of the given given questi questi on: 2

7

4

5

2

3

1

?

6

10

42

72

(1) 2 (2) 4 (3) 5 (4) 3 15 .

All of the followi followi ng are are the th e same in a manne r. Fi nd out out the the one one which i s dif ferent ferent among them:

(1) BFJQ (2) RUZG (3) GJOV (4) ILQX


Page 6


PART B (16 -40) 16.

T he deg deg ree of exten sion Q  1 6 + 3



7

8  over the field Q is



(1) 8 (2) 7 (3) 6 (4) 5

17

T he rand andom om var va ri abl abl e X ha s a t-distributi t-distribution on with with v degrees degrees of freedom. freed om. Then the the probabili probabili ty distri distri bu tion tion of X 2 is (1) Chi-square distribution with 1 degree of freedom (2) Chi-square distribution with v degrees of freedom (3) F-di F-di stri butio bution n wi th (1, (1, v) degree degree s of of free free dom dom (4) F- distribution with (v, 1) degrees of freedom

18.

Let T : Rn → R m be a linear transformation and Am×n be its matrix representation then choose the correct statement. (1) Col Col umns of A are LI ⇒ –T –T i s onto on to (2) Columns of A span Rm ⇒ T is onto (3) Col Col umns of A are LI ⇒ T i s one one one (4) T is one one one ⇒ columns of A are L I

19.

Let byx an and d bxy denote the regression coefficient of Y on X and of X on Y respectively are equal, The (1) σy = σx (2) ρ = 1 (3) σ = 0 (4) None of the above


Page 7


20 .

Wha t probabi probabi lity li ty model model is appropriate appropria te to describ describ e a situati situation on where 100 mis mi spri pri nts are distrib uted uted rando ran doml ml y throughout throughou t the 100 pag page e of a book? bo ok? (1) Neces Nece ssaril y P oisson (2) Necessarily Exponential (3) Necessarily Normal (4) Could n ot decide decide

21.

Which one of the following be true for the function

 1 f ( x ) = x 2 si s in   , if x ≠ 0, f ( 0 ) = 0 x (1) Function f is not continuous on [0, 1] (2) Function f is not bounded variation on [0, 1] (3) Function Func tion f does not exi e xi sts (4) Function f is of bounded variation on [0, 1]

22.

If [φ ψ ] be the Poisson bracket Then

∂ [φ, ψ] = ∂t

∂φ ∂ψ  (1)  ,  ∂t ∂t  ∂ φ   ∂ ψ  (2 )  , ψ  +  φ,  ∂t   ∂t  ∂ φ (3 )  ,  ∂t

 

 

φ +  ψ ,

∂ψ  ∂t 

 ∂φ ∂ψ  + [ φ, ψ ] (4)  ,  ∂t ∂t  23.

A re al com ompl plet ete e ma tri tri x of o f order order ‘n ’ hasn has n mutually independent real eigenvectors. then (1) All E.V. are orthogonal (2) All E.V. are orthonormal (3) All E.V. form orthonormal basis. (4) None of these


Page 8


24.

Let I = {1} ∪{ 2} ∪{ 3} ⊂ ℝ for x ∈ ℝ Let φ(x) = [x] + [1 – x] Then (1) φ is discontinuous somewhere on ℝ (2) φ is continuous on ℝ but not differentiable only at x = 1 (3) φ is continuous on I but not differentiable at 1 (4) f is continuous on ℝ but not differentiable at I

25 .

1 1 T he radius of conve rgen genc ce of of the power power seri es of the functi on f(z) = about z = i s 1− z 4 (1) 1 (2)

1 4

(3)

3 4

(4) 0 26.

Let A be a 2 × 2 matrix for which there is a constant ‘k’ such that the sum of entries in each row and each column is k which of the following must be an eigenvector of A

1  (I)   0

0  (II)   1 

1 (III)   1

(1) I only (2) II onl y (3) III only (4) I and II only 27 .

In the Lau Laurent rent seri es exp ansion ansion of f(z) f(z) =

1 1 − valid i n the the region 1 < | z | < 2, the coe coe ffi ffi ci ent of of z −1 z −2

1 is z2 (1) –1 (2) 0 (3) 1 (4) 2


Page 9


28.

Let A and B be (n x n) matrices with the same minimal polynomial. The (1) A is similar to B (2) A i s di di ago agonali nali zable zable if i f B is i s diagno sab le

(3) A-B is sin gula r (4) A and B commute 29.

The image of the infinite strip 0 < y < 1/2c under the map w =

1 is z

(1) A half plane (2) Exterior of the circle (3) Exterior of an ellipse (4) Inter I nterii or of an elli pse pse

30 .

If AB be the arc arc α ≤ θ ≤ β of the th e circle circle |Z| = R and (1) Lim

∫

(2) Lim

∫

f ( z ) dz = i ( β − α ) k

(3) Lim

∫

f ( z ) dz = ( β − α ) k

z →∞

R→∞

R→∞

(4) Lim R →∞

31 .

AB

z →∞

then –

f ( z ) dz = i ( β − α ) k

AB

AB

∫

Li m zf ( z ) = k

AB

f ( z) dz = (β − α ) k

Let σ and τ be the permutations defined by

1 2 3 4 5 6 7 8 9   1 2 3 4 5 6 7 8 9 and τ =    1 3 5 7 9 6 4 8 2   7 8 3 4 9 6 5 2 1

σ=

Then (1) σ and τ genera genera te the grou g roup p of per p erm m utation utati ons s on {1, 2, 3 , 4, 5, 6, 7, 8, 9} (2) σ is co nta ntaii ned in the the g roup gen gen erate erated d by τ (3) τ i s con tained tained in the group generat generated by σ (4) σ and τ are in the same conjugancy class

32.

The number of characteristics curves of the PDE x 2u xx – 2xy uxy + y2 uyy – xux + 3yu y = 8y/x


Page 10


(1) 0 (2) 1 (3) 2 (4) 3 33.

Let R be a ring with unity If 1 is of additive order n then Charracte ristic of R is Cha (1) 1 (2) 0 (3) n (4) ∞

34 .

Whi ch of the following following are subgroups of (Z21, X 21) (1) H = {[x] 21 / x ≡ 1 (mod 3)}

(2) K = {[x] 21 | x ≡ (mod (mod 7)}

(1) Only 1 (2) Only 2 (3) Both 1 and 2 (4) Non Non e of these these

35.

Le t f(x) f(x) = X TAX be a ‘+ve ’ definit definite e quadrati c form form then– then– (1) Zero may be the Eigen value of A 2

(2) a ij < ai jajij i

∀i ≠ j

2

(3) a ij > a ij a ji (4) The diagonal elements of A are +ve 36.

T he maximum step step si ze h such such that the the err e rror or in in linear i nterpol nterpolation ation for the funct functii on y = si si n x in in [0,π ] i s le ss tha than n 5 x 10 -5 is (1) 0.02 (2) 0.002 (3) 0.04 (4) 0.06


Page 11


37.

Comment on the following values of regression coefficients: b xy = 3.2 and byx = 0.8 (1) These coeffi coeffi cien t are correct orrect (2) These coefficient are totally incorrect (3) These coefficient are correct if b yx = 1 .8 (4) These coefficients are correct if b YX = 0

38 .

Con si de r the followin foll owing g Linea r Program Pro gramm m ing Problem Problem : Maximize 3x1 + 8x 2 Su bject bject to 2 x1 + 5 x2 ≤ 10 6 x1 + x 2 ≤ 6 x1, x2 ≥ 0 T he op tim tim al value value of the object objectii ve functi function on is (1) 0 (2) 3 (3)

112 7

(4) 16 39.

If a particle moves under the influence of gravity on the frictionless inner surface of the elliptical paraboloid bx2 + cy2 = 9z where a, b, c ∈ R + Then which of the following is equation of motion of itɺɺ = λ 2bx (1 ) mx ɺɺ + m g = 0 (2 ) mz ɺɺ = 2cy λ (3 ) my ɺɺ = 2a bλ (4 ) mz

40.

Con si de r an example exampl e from a maintenance sho p. The inte r-arrival r-arrival times times at tool crib are expo ex ponen nentia tiall wit h an ave rage tim e of 10 minute s. T he length length of the service ervice time is a ssumed ssumed to be exponen exponen ti ally ally di stributed, wi th mea me an 6 minu mi nutes tes,, Fi nd:


Page 12


Estimate th e fraction fraction of the day that tool ool crib ope ope rato r will will be idle. (1) 40% (2) 50% (3) 60% (4) 70%

PART C (41 -60)

41 .

α

α

T he fun ction f such that that f(x) = x on (0, ∞) to R is continuous and Dx = αx

–1 α –1

for for x ∈ (0, (0, ∞ ) then

(1) α > 0 (2) 0 < α < 1 (3) α ∈ ℂ (4) α < –1 ∞

42 .

If

∑a

is series of real numbers and if f is continuous function on R then power series given by

n

n= 0

fn(x) =

∑a

n

zn with radius 1 then –

(1) {f n(z)} z)} ten tend ds to f( f(1) as z → 1 (2) f n(z) converges for z < 1 (3)

1− z

(1− | z | )

rem ain s bou bou nde nded d

∞

(4)

∑a

n

Converges Conve rges to zero

n=0

43 .

A compa compa ny distribute distributes s its product s by by t rucks l oa ded at its only lo adi adi ng stat ion ion both company’s trucks an d contra contractor’s ctor’struck truc ks are used used for fo r this pu rpose rpose It was foun fo un d tha tha t an average average o f 5 minu tes one truck arrived arri ved and average l oading time wa s three th ree mi mi nute s 50% 50% of the tru cks cks belo ng to the contracto ontra ctorr T hen (1) The probability that a truck has to wait is ρ = 0.6 (2) The waiting time of truck = 7.5 minutes (3) Expected waiting time of contractor per day is 10.8 hrs (4) idle time is 2.2 hrs


Page 13


44.

−4

Let f(z) = e− z ( z ≠ 0 ) be a function defined on complex plane Then– (1) if f(0) = 0 ten f(z) is not analytic at z = 0 (2) f(z) satisfies Cauchy Riemann equations (3) f(z) does not exists for finite values of z (4) f(z) is a continuous function

45.

2

Le t f(z) f(z) = log z g(z) = x + 1 Then (1) f(z) has a branch cut at z = 0 (2)

∫

∞

0

f (x ) g( x )

dx = 0 2

3 f ( x ) π (3) ∫ dx = 0

∞

g(x )

8

nx

(4) f(z) has essential singularity at e wh ere n being being natu ral numbe numbe r 46.

The equation of surface satisfying 4yz p + q + 2y = 0 and passing through y 2 + z2 = 1, x + z = 2 (1) Lies on x-y plane (2) Is y2 + z2 + x + z – 3 = 0 (3) Lies on z-axis (4) Is x2 + y2 + x + y – 3 = 0

47.

1/3

T he ini ini ti al value alue p rob robl em y = 2x , y (0) = 0 in an interval around t = 0 has (1) No sol ution (2) A unique solution (3) Fini tely tel y man man y linearl linearl y independent solu tion tion (4) Infinitely many linearly independent solution

48.

An extremal of the functional

I [y (x)] =

∫

b

a

F ( x y y' y ') dx y (a) = y 1, y(b) = y 2 satisfies Eular’s equation which in general

(1) Admit a unique solution satisfying the conditions y(a) = y1 , y(b) y(b ) = y2 (2) May Ma y not admi admi t a solu solu tion sat sat i sfying the condi condi tions y(a) = y1 , y(b) = y2 (3) Is a second order linearly differential equation Phone: 0744-2429714 Mobile: 9001 90012 2 9711 97111, 1, 9829 98295 5 67114, 90012 9 00129 9 7243 Website: www.vpml!sses.om www.vpml!sses.om "-M!il: vpm l l!sse !sse s#$!hoo.om s#$!hoo.om % % in in&& o#v pml!sses. ml!sses.o om ' (()ess: (()ess: 1-* 1-*-8, -8, +hee +hee l! *h * how(h!)$ o! o!( (, ++, ' /W' /W', , ' , ''+' ''+' , 324 3240 0 05

Page 14


(4) Do not have any non-linear ODE of any order

49.

Cons Con si de r A boundary value problem

d 2y = f ( x ) with y(0) = α, y'(1) = β 2 dx

(1) The BVP has infinitely many solutions (2) The BVP has unique solutions for α = β (3) The Green function G(x,ξ) correspond ing to BV P is

− x

0≤x ≤s s ≤x ≤1

G ( x, ξ) = 

 −s

(4) Green function corresponding to BVP does not exists

50.

If X 1, X2….X k are i ndependent ndependent γ vari ants ants wi th parameters parameters λ1 , λ2 ……. λk then (1) For large value of parameters γ variants vari ants foll follows ows standa rd norm norm al di stri bution bution (2) For large values of parameters γ variants variants follows normal distribution (3) X1 + X 2 + …. X k is also γ variants variants with parameter λ1 + λ2 + …. λk λi

(4) MGF of γ vari vari ant Xi is ( 1 − t )

51.

For approximating a polynomial some of iterative scheme are given as (a)

xn +1 =

1  a  xn  1 + 2  2  xn 

(b)

xn +1 =

1 xn 2

(c)

3a x2n   xn +1 = xn  6 + 2 −  8  xn a

x2n   − 3   a 

1

(1) (a) (a) an d (b) (b) both con verges to the som e limit limit

a

(2) (c) diverges (3) The order of convergence of (a) (b) (c) is 2 (4) The order of convergence of (c) is 3 52.

If G if a g rou p of order order 30


Page 15


Then (1) G has 10 3-SSG (2) Both 3-SSG, 3-SSG, 5 -SSG are normal (3) G has a normal subgroup of order 15 (4) G has 20 element of order 3 53.

The solution of integral equation.

φ(x) = x +

1

∫ xtφ ( t) dt 0

sati at i sfi es

 2 (1) φ(0) + φ  = 1  3  1 (2) φ   + φ (1) = 1  2 (3) φ (2) (2) + φ(4) = 9 1 2

(4) φ (1) (1) + φ(0) =

54.

T he sho sho rtest rtest path from the poi nt A –2, (–2, 3) to the the poi nt B(2, B(2, 3) locat located ed in the region region i s (1) x2 (2) 2x – 1 (3) –2x + 1 (4) –2x – 1

55.

We have I =

Z3 [ x] 3

< x + 2x + 1 >

(1) I is a field with 12 invertiable elements (2) I is a field with 27 elements 2

2

(3) Inverse of x + 1 in I i s x – 1 (4) x2 + 1 is an invertiable element 56.

n-5

Let A be be an (n (n × n) mat matrrix n ≥ 5 with characteristic polynomial x n

(1) A = A

5

(x -1) Then

n-5

(2) Rank A is 5 (3) Rank of A is at least 5


Page 16


(4) There exist non-zero vectors x and y such that A(x + y) = x – y

57.

–1

If A = [aij] n×n n×n ; [aij] = [a ji] if λ1 λ2….. λn be the eigen values of A and P AP = d(λ1λ2….. λn) then (1) A is symmetric matrix (2) A and P has orthonormal vectors (3) P is an orthogonal matrix (4) P is singular

58.

If V be be a 7-di 7-dimens mension iona al vecto vectorr space space over R and Let T : V → V be a linear operator operator with m inima inimall 2

3

polynomi polynomi al m(t) m(t) = (t – 2t + 5 ) (t – 3) then – (1) There are only two possibilities of characteristic polynomial (2) There are 3 possible canonical forms (3) There must not any subspace of order 3 (4) There does not exist any possible Jordan canonical form. 59.

Let V = W1 ⊕ W2 ⊕ . . .. . ⊕ Wr , for each k suppose S k is a linearly independent subset of Wk Then (1) S = ∪ Sk is linearly independent in V K

(2) If S k is basis of Wk then

∪ Sk

is basis of V

K

(3) dim V =

∑ dim W

k

K

(4) dim V = r 60.

If X1 ~ N (µ 1,σ1

2

)

an d X2 ~ N( µ 2 , σ2

2

(1) X1 + X 2 ~ N( µ1 + µ2 , σ1 + σ2

2

)

(2) X1 ± X 2 ~ N( µ1 ± µ2 , σ1 + σ2

2

)

2

2

(3) X1 ± X 2 ~ N( µ1 + µ2 , σ1 ± σ2 2

(4) X1 ± X 2 ~ N( µ1 ± µ2 , σ1 ± σ2 2

2

2

)

then

)

)


Page 17


Answer key Que. 1

Ans. 3

Que. 16

Ans. 2

Que. 31

Ans. 4

Que. 46

Ans. 2

2 3

1 3

17 18

3 1

32 33

1,2 3

47 48

3 2,3

4 5 6

4 1 2 1 2 4 4

19 20 21

1 1 4 2 1 1 1

34 35 36

3 4 1 2 3 1,3 1

49 50 51

1,3 2,3 1,4 1,2,3,4 1,3 1,2,4 2,4

7 8 9 10 11

2

22 23 24 25 26

3

37 38 39 40 41

1,2,3

52 53 54 55 56

12 13

2 4

27 28

3 2

42 43

1,3 1,2,3

57 58

1,3 1,2

14 15

4 1

29 30

2 1

44 45

1,2,4 1,2 ,3

59 60

1,2,3 1,2

1,2

HINTS AND SOLUTION S OLUTION PART A (1-15) PART A (1-15)

1.(3)

m1D1 m2 D2 = w1 w2 24 × 15 m2 × 12 = 1 80 24 m2 = 40

2.(1)

RA × SA QA

= PA ⇒

9×9 7

= PA

Diameter = PA + AQ


Page 18


81 13 0 +7= 7 7 Radius =

3.(3)

65 Diameter ∴ Radius = 7 2

R1 + R2 = a R2 + R3 = b R3 + R1 = c R1 + R2 + R2 + R3 + R3 + R1 = a + b + c

⇒

a+b+c

R 1 + R2 + R3 =

2

alue of each tool in 1985 4. (4) V alue =

10 × 107 18 × 10

= 5

5.(1)

7

[Since 1 crore = 10 ]

3

5 9

T housan housand d

The required required perc entage

=

(sin ce total ot al employ employ ed

18 × 100 ( 360 –18 )

= 3 60 – unem une mp loyed) loye d) =

18 3 42

× 10 0 = 5

5 19

%

6.(2)

This is an alternating alternating m ultiplication ltiplication and a nd subtractin subtracting g s eries: eries: Firs t, multiply multiply b y 2 and then subtra sub tract ct 8.

7.(1)

Total quantity of petrol = 

 4000 4000 40 00  + +  litres 8 8.50   7.50

consumed in 3 years

 2 1 2  + +  li ters  15 8 17 

4000 

 76700   litres  51 

=


Page 19


Total amount amount spent s pent = Rs. (3 x 4000) = Rs. Rs. 12000.

6120  12000 × 51 = Rs.7.98  = Rs. 767  76700 

Average cost = 

8.(2)

Let C. C. P.= Rs. 100. Then, Pr of it = Rs. Rs. 320 320,, S. P. = Rs. Rs. 420. 420. New C.P. = 125% of Rs. 100 = Rs. 125 New S.P. S.P. = Rs. 420. Prof it = Rs. (420 - 125) = Rs. 295.

 29 5  1 47 5 Requ Req u ired pe p ercentage rc entage =  % = 70% (a pproximately) × 1 00  = 21  42 0 % A student multiplied a number by

9.(4)

3 5 instead instead of 5 3

Each day of the w eek is re r epeated af af ter 7 days. So, af af ter 63 6 3 days, day s, it w ill ill be Friday . Hence enc e af ter 63 days , it w ill be Thu Thurs rs day. day . Theref Theref ore the the requi r equired red day is Thursday .

4 51 12 6 hrs. 10 .(4) 40 mi n = 1 hr s = 5 75 75 Time taken = 1 hr 40 min 48 sec = 1 hr Let the actual speed be x km/hr. 5 1 26 x× = 42 7 75

Then,

 42 × 7 × 75  35km / hr. hr.  = 35km  5 × 126 

x =

pr of it = Rs. 990 11.(2) Given, Total prof Ration of their capitals = 34 : 65. Now , prof it am amoun ountt got go t by P = 20% 20% of total total profit + P’s share sha re in in balance 80% prof it f or his capital c apital 34    0.2 + 0. 8 × 3 4 + 6 5  = 470  


Page 20


1000)s q m = 5000/3 5000/3 sq m 12.(2) Now area = (1/6 × 1000)s 2x × 3x = 5000/3 =>x × x = 2500 / 9 x = 50/3 50/3 length length = 2x = 100/3 m and breadth = 3x = 3× (50/3) = 50m

13. (4) Distance = (240 x 5) = 1200 km. Speed = Distance/Time Speed Speed = 1 200/(5/3 200/(5/3)) km/hr.

[We can w rite rite 1 hours as 5/3 hou hourrs]

Requ Req u ired speed = 1 12 2 00 x 3 km/hr km/hr = 720 km/hr.

14.(4) A s,

2 × 5 × 1 = 20

and

4 × 3 × 6 = 72

Similarly, Similarly,

7 × 2 × ? = 42 ?

42 =

=

3

14 ∴ 15.(1) According to question,

Therefore, B F J Q is odd.

PART-B(16-40)


Page 21


16.(2)

Let u = 7 16 + 3

7

8 sin sin ce u =

(

7

2+3

Q by Einstein’s criterion we have  Q



17.(3)

3

)( 2) 7

it follows that u ∈ Q

( 2 ) since x – 2 is irreducible over 7

7

( 2 ) : Q = 7 7

X ~ t(v) If ξ ~ N(0, 1) and p ~ χ (2n) Then X =

ξ p/ v

~ t ( n) 2

⇒

X =

ξ2 ξ2 /1 = p/ v p/v

Being the ratio of two linearly independent chisquare var vari ates di vi ded ded by the theii r respective Degree s of freedom freed om i s F(1, v) 2

t ~ F(1, v)

n

m

18.(1) Since T : R → R be a linear transformation and [A]m×n be its matrix representation Then we can check that Columns of A are LI ⇔ T is one one Columns of A span Rm ⇔ t is onto an d

–T will be onto onto whenever T i s onto

⇔ Columns of A span Rm ⇔ –T i s on to

19. (1)

b YX = r

σ Y σX

b XY = r

σ X σY

since b XY = bXY ⇒

σX2 = σX2

⇒

σY = σX


Page 22


20.(1)

Here the random variable X representing the number of misprints in a page follows the Poisson distribution with parameter m = Average number of misprints per page = 1

21.(4)

1  2 x sin  The Given function f ( x) =  x  0 Lf '( 0 ) = L im

;x ≠0 :x =0

f ( 0 − h) − f ( 0 ) h

h →0

  ( − h) sin   − 0  −h  1

2

= Lim h→ 0

−h

 1 = L im − h sin  −  h→ 0  h  1 im h si si n  = L im h →0 n

= 0 × sin ∞ = 0 Rf’(0) Rf’(0) = Lim

f (0 + h ) − f ( 0) h

h→ 0

 1 h 2 sin   − 0  h = Lim h →0 h

=0 Lf ' (0 ) = Rf ' ( 0 ) = 0

Function is differentiable at x = 0 i.e. in [0, 1]  1 1  ,... . .. .. .. 1 And we can find a partition P = 0, ,  n n−1 

Of [0, 1] Let ∆fr = f ( xr ) − f ( x r−1 ) V ([ 0, 1] , P, f ) =

n

∑ Σ f (r) r =1

P([0, 1], f) = Sup V ([a, b], b], P, f)


Page 23


Where supremum being taken over all partition of [0, 1] ⇒

22.(2)

f is of bound boun ded variation variation on [0, 1]

From the definition of Poisson bracket

 ∂φ ∂ψ

[ φ, ψ ] = ∑  k

∂ [φ, ψ] = ∂t

 ∂qk ∂pk

−

∂φ ∂ψ ∂ψ   ∂pk ∂qk 

 ∂  ∂φ  ∂φ ∂φ ∂  ∂ψ  ∂  ∂φ ∂φ ∂φ ∂  ∂ψ    ∂t  ∂p + ∂p ∂p  ∂t  − ∂p  ∂t  ∂q − ∂ q ∂q  ∂t    k  k k  k  k k   k  k

∑  ∂q k

 ∂  ∂φ  ∂φ  ∂φ ∂  ∂ψ ∂φ ∂  ∂ψ    ∂φ ∂φ  ∂φ ∂φ  ∂ψ − + ∑           ∂qk  ∂t  ∂pk ∂p k  ∂t  ∂qk k  ∂qk ∂pk  ∂t  ∂pk ∂qk  ∂t   

= ∑ k

 ∂φ , ψ +  φ, ∂ψ   ∂t   ∂t 

=

23.(1)

If a real comp comp lete matri matr i x of order n has n mutu mut ually indep independe endent nt real eigenv ectors cto rs t hen all eigenvectors are orthogonal.

24.(1)

φ ( x) = [ x ] + 1 − x

−1 ≤ x ≤ 3


Page 24


⇒

 −1 + 1 − x −1 ≤ x < 0  0 +1− x 0 ≤ x < 1  φ ( x) =   1+ x −1 1 ≤ x < 2  2 + x − 1 2 ≤ x ≤ 3

⇒

 −x 1 − x  φ ( x) =   x 1 + x

−1 ≤ x < 0 0≤ x<1 1≤ x < 2 2 ≤x <3

which could be shown as Cle arl arl y from above figu figu re y i s no t cont conti nuous an d n ot di di fferentiabl fferentiable e at x = {0, 1, 2}

⇒

25.(1)

A i s correct option

If f ( z) =

1 1− z

First we will determine the power series of function f ⇒

f (z ) =

1 1− z ∞

f (z ) =

|z| < 1

∑ (z )

m

Provi Pr ovided ded |z| < 1

m =0

So the th e radius radius of conver converge gence nce of power po wer series series is 1

26.(3)

If k be the sum of each row and column

 1 Then we get   colum colum n vector vector w.r.to eig envalue envalue k.  1

27.(3)

f (z) =

1 1 − z − 1 z −2

at 1 < |z| < 2 f (z) =

1

 

z  1−

1 z 

1 1 = 1 −  z z

+

−1

1

 

2 1 − 1

z 2  z

−1

+ 1 −  2 2


Page 25


 1 1 1  1 z  z  + + + + + + + 1 . . . . . . . 1 . . . . . .   2    2 z  z z  2  2   2

=

Coefficient of

28.(2)

1 is 1 2 z

If A and B be n × n matrices with same minimal polynomial i.e. the eigen values of A and B are same [in case their multiplici t ies can can b e different] different] If A is diagonalizable then B must be diagonalizable i.e. if A have linearly independent eigenvectors then B also must have linearly Independent eigenvectors to all of eigen values But A may be similar to B or may be not.

29. (2)

Since

and

0
⇒

0<−

⇒

v<0

1 ⇒ 2c

y<

−

v 2

u +v

2

⇒

v u + v2 2

2

<

1 2c

2

u + v = 2cv > 0

Hence Hence the t he image of infinit infinit e st rip 0 < y <

1 2 2 2 is giv given en by by v < 0 and and u + (v + c) > c 2c

The image region lies below the v-axis in the w-plane and this is the exterior of the circle With centre (0, - c) and radius c

30.(1)

For given ε > 0 we can find positive real numbers R such that |Z f(z) – k| < e when ever |Z| > R or fo rmall mal l y Zf(z) = k + e where ε → 0 as a s z → ∞ Thus for sufficiently large R

∫

AB

f ( z ) dz =

∫

k +ε dz AB Z


Page 26


β

= ∫ ( k + ε) i d θ = z = Rei θ α = i (β – α) +

∫

AB

f ( z ) dz = i (β − α ) k ≤

∫

β

α

∫

β

α

iε dθ

i ε dθ < ε ( β − α )

31.(4) σ and τ are the permutations with 9 symbols so we can find some i nvertiable element x ∈ Sq

such ⇒

32.(1) (2) here so

⇒ ⇒

33.(3)

that

−1

τ = x σx

τ and σ are conjugate to each other or τ or σ belongs to the same conjugacy class.

Given Pde x2u xx – 2xy uxy + y2 uyy – xux + 3yu y = 8y/x R = x2 S = –2xy T = y2 S 2 – 4 RT = 0 T he gi ven pde pde i s pa raboli raboli c everywhere everywhere It must must onl y one charac characte terri stic curve

Let additiv addit ivee order of 1 be n (i.e. order of 1 in the group (R, +) is n) Then n.1 = 0 and n is the least ‘+ve’ integer Now for any x ∈ R nx = x + x + ………… + x = 1.x. + 1.x + ……. 1.x = (1 + 1 + …. 1)x = 0.x ⇒

34.(3)

Ch R = n

=0

The subset H is finite and nonempty at [1] 21 ∈ H so it is enough to show that H is closed under m ul ti pli pli cation cation i f [x]21 and [y]21 belong to H then x ≡ 1(mo m od 3 ) and y ≡ 1(mo m od 3 ) so i t follo follo ws that that xy ≡ 1(mo mo d 3 ) therefore


Page 27


[x] [x]21 [y ]21 = [xy]21 ∈ H Similar argument shows that K is subgroup of (Z21, X21 )

35.(4) A posi posi ti ve def defi nite mat rix can have only p osit ositi ve greater greater th an zero zero eigenvalu eigenvalues es but zero could could n ot be an eigenvalu eigenvalue e of i t. We can take an example The quadratic form 6x12 + 3x 22 + 3x 32 – 4x1x2 – 2x2x3 + 4x 3x1 is positive definite. m atr at rix co rrespon ponding ding to to it i t is

 6 −2 2    A =  −2 3 −1  2 − 1 3  2

2

he re we can check a ij >/ a ija j i

36. (1)

We have E 2 ( f ; ( x) ) ≤

h

or a ij
3

9 3

M3

for f(x) = sin x we obtain f ''' ( x ) = − co s x M3 = max cos x = 1 0 ≤x ≤

π 4

Hence the step size h is given by h

3

9 3

or

≤ 5 × 10

−5

h ≈ 0 .0 2

37. (2) If bXY = 3.2

bYX = 0.8

But bXY bYX ≤ 1 But But it is not not p ossi oss ible for bXY = 3.2 or 1.8 ⇒ These coefficients are totally incorrect.

38.(3)

Given LPP

M aximize aximize

3x1 + 8x2

s.to

2x1 + 5 x2 ≤ 1 0


Page 28


6x1 + x2 ≤ 6

and

x1x2 ≥ 0

By Graphical method

(0, 2)

 6 24   ,   7 14 

(1, 0)

6 12 The Th e boundary boundary p oints oints of criti criti cal region are (0, 2) (1, 0)  ,  7 7  6 12 by  ,  gives maximum value 7 7 

which is

112 112 7

39.(1,3) The equation of the paraboloid is bx2 + cy2 = az.

∴ The equation of the constraint is given by 2b x dz + 2cy dy – a dz = 0

... (1) (1)

The coefficients in the constraints equation. (1) are given by A x = 2bx, A y = 2 cy and Az = –a

... .. . (2)

The Lagrangian of the particle is given by L=

1 m ( xɺ 2 + ɺy2 + zɺ 2 ) − mg z, 2


Page 29


which gives

       ∂L ∂L = mzɺ , = −mg  ∂ɺz ∂z  ∂L = mxɺ , ∂xɺ ∂L = myɺ , ∂yɺ

∂L = 0, ∂x ∂L =0 ∂y

... .. . (3)

T he Lag Lag range’s range’s eq uatio uation n are a re d  ∂L  ∂L = A x λ,  − d t  ∂ɺx  ∂x d  ∂L  ∂L = A y λ,  − d t  ∂ɺy  ∂y and

d  ∂L  ∂L = A z λ,  − d t  ∂ɺz  ∂z

or

mɺɺ x = 2bx λ, mɺɺ y = 2cy λ mɺɺ z + mg = −a λ

and

40. (1) here

.... .. .. (4 ) from (2) and (3).

λ = 60/10 = 6 p er ho hour µ = 60/6 = 10 per hour

A p erson wi w ill have to wait if the service service is not idle idle Probabili Probability ty that the th e service service facilit facility y is idle = p robabil robability ity of no n o customer in t he sy sy stem (P 0) Prob Pro bability of wait ing

= 1 – P0 = 1 – (1 – ρ) = ρ =

λ = 0.6 µ

P 0 = 1 – ρ = 0.4 ⇒ 40% of the time of tool crib operator is idle.

PART-C(41-60)


Page 30


41. (1, 2, 3)

α

+

Let α ∈ ℝ Then the function f defined on R such that f(x) = x is continuous and α

α-1

differentiable and Dx = αx

for x ∈ (0, ∞ )

Then By chain rule Dx α = D e α lnx = e α lnx D( α lnx) lnx) α x

α

=x ⋅

α-1

= αx

for x ∈ (0, ∞ )

if α > 1 the power function strictly increasing on (0, ∞ ) to

ℝ and

if α < 0 the function

α

f(x) = x is strictly decre d ecreas asing ing Thus α ∈ ℝ is possible value

42. (1,3) S tatement. atement. If

∑

∞

a conver converg ges, es, t he the p ower series series f(z) f(z ) = n= 0 n

∑

∞ n =0

n

a n z with w ith R = 1 tends to

f(1) as z → 1, p rovided rovided |1 – z|/(1 – |z|) rema rema ins bounded. Proof. If we assume

∑

∞ n =0

a n = 0. This can be done by adding a constant to a0. So, f(1) = 0.

and we can can write 2

Sn(z) = a0 + a1z + a2z + … anz

n

= S0 + (S1 – S0)z + (S2 – S1)z2 + ….. … .. + (Sn –S –Sn–1)z n = (1 – z) (S0 + S1z + …… + Sn–1z

n–1

n

) + Snz .

But S nzn → 0 AS n → ∞ (f(1) = 0), z ≤ 1 ) so that we can write f ( z ) = (1 − z )

∞

∑S z . n

n

n =0

Since |1 – z|/(1 – |z|) remains bounded, there exists a constant K such that 1− z 1− | z |

≤ K.

Again, for given ε > 0, t here here is a pos itive int integer eger m such that n ≥ m implies S n ≤ ε. Finally Finally,, we say that the remainde remainderr Σn≥ m Sn zn , is the th e dominated dominated by the th e geo geometric metric ser ies


Page 31


∞

ε∑ z = n

n ≥m

εz

m

<

1− z

ε . 1− z

Hence, m −1

f (z ) ≤ 1− z

∑S z n

n

+ Kε.

n =0

The first term on the right be made arbitrarily small by choosing z suf ficiently suf ficiently close to 1, and we conclude that f ( z ) → 0 when z → 1, su s ubject bject to the t he stat stat ed condition. condition. This comp completes letes the th e proof. 43.(1,2,3) Here Here we a re g i ven : Average arrival rate of trucks, λ =

60 rucks / hr. = 12 trucks 5

Ave rage rage service rate of t rucks, µ =

60 = 20 trucks/hr. 3

(i) Probability that a truck has to wait is given by :

ρ=

λ 12 = = 0.6 µ 20

(ii ) T he waiting time of a tru ck that wai wai ts is given given by : Ws =

1

µ −λ

=

1 2 0 −1 2

=

1 8

hour 7.5 minutes.

(iii ) The expected wai wai ting tin g time of contrac contractor’s tor’s truck truc k per da y

(assum (assum i ng 24 hrs. shift) hif t)

= (No. of trucks per day) × (Contractor’s percentage) × (Expected waiting time of a truck)

= 12× 24×

= 288 ×

44.(1,2,4)

p ut

λ 50 × 100 µ ( µ − λ )

1 12 54 × = or 10.8 10.8 h rs. rs. 2 20 × 8 5

w = f(z) = e − z w = u + iv, u + iv = e

−(x +iy )

−4

z = x + iy 4 −


Page 32


 4   ( x +i y)   4   (x 2 + y 2 )   

u + iv = e

−

=e −

1

(x

4

+ y2 )

 x4 + y2 − 6 x2 y2 − 4ixy ( x2 − y 2 )  

x4 + y 4 −6 x2 y 2

(x

u=e

4

−

2

2

4

+ y2 )

4

2

x + y − 6x y

(x

v=e

2

+y

2

)

2

4

 4 xy ( x 2 − y2 )2   ⋅ sin   x2 + y 2 4  )   ( u ( x, 0 ) − ( 0, 0 ) ∂u = Lim ∂x x →0 x

At z = 0 = Lim

 4xy ( x 2 − y2 )2   cos   x2 + y2 4  )   (

e

−x

−4

x

x→ 0

=0 u ( 0, y ) − u ( 0, 0 ) ∂u = Lim ∂y y →0 y

= Lim y →∞

e

−y

−4

y

=0 v ( x,0 ) − v ( 0, 0 ) 0 ∂v = L im = Lim = 0 x → 0 x → 0 x x ∂x v ( 0, y ) − v ( 0, 0 ) ∂v 0 = L im = Lim = 0 y → 0 y → 0 ∂y y y

Hence C-R equations are satisfied at z = 0 But

f ' ( 0 ) = Li m

f ( z ) − f (0 ) z

z→ 0

= Lim z →0

1 i (π /4 )

re

= Lim z→ 0

1 exp ( −r

−4

)

1 1/ z 4

ze

Taking z = re i(π / 4)


Page 33


= Lim r →∞

1 i (π / 4 )

re

⋅

1 exp ( − 1/ r

4

)

=∞

showing that f '( z ) does does not not exist existss at z = 0 and and hence f(z) is not analytic at z = 0

45. (1,2,3)

f(z) f(z ) = log z has a branch branch cut cut at at z = 0

we have ℓn z = ℓnr + iθ suppose that we start at some point z1 ≠ 0 in the complex plane for which r = r1 θ = θ 1 so that ℓn z1 = ℓn r1 + iθ1 Then after making one complete circuit about the origin in the positive or counter clockwise direction we find on returning to z 1 t hat r = r θ = θ1 + 2 π

so that

ℓnz 1 = ℓnr1 + i(θ1 + 2π)

Thus we are on another branch of the function so z = 0 is the branch point at z = 0 Consider 2

∫

(logz ) C

2

x +1

dz,

where C = [ε, R] ∪ CR ∪ [ −R,− ε] ∪ C ε is the contour depicted in Fig, and take the branch |z| > 0, -

π /2 < ar ar g z < 3π /2.

i

CR C

ε

−ε

ε

We have Phone: 0744-2429714 Mobile: 9001 90012 2 9711 97111, 1, 9829 98295 5 67114, 90012 9 00129 9 7243 Website: www.vpml!sses.om www.vpml!sses.om "-M!il: vpm l l!sse !sse s#$!hoo.om s#$!hoo.om % % in in&& o#v pml!sses. ml!sses.o om ' (()ess: (()ess: 1-* 1-*-8, -8, +hee +hee l! *h * how(h!)$ o! o!( (, ++, ' /W' /W', , ' , ''+' ''+' , 324 3240 0 05

Page 34


( ℓn x)

R

∫

2

dx +

2

x +1

ε

2

2

∫

x2 + 1

−R

−ε

∫

dz +

2

z +1

CR

log x ) ( log

−ε

∫

(lo g z )

−R

2

( lo g x ) 2

x +1

2

∫

dx +

Cε

2

(l og z ) 2

z +1

2

dz +

∫ C

(lo g z ) 2

z +1

d z.

2

R l og ( − x ) lo g ( − x)  d x. = dx = − ∫ d x 2 ∫ 2

ε

x +1

R

x +1

ε

Thus, 2

  ℓn ( − x ) + + d x d x ∫+∫ 2 ∫ε x 2 + 1 C C x +1 

( log x)

R

∫ ε

2

R

R

2  ( log z)2 (l og z ) dz = 2πi Re sz =i 2 .  2  z +1 z 1 + 

ε

on taking limits ε → 0 and R → ∞. T hus the above above equation simp simp lifies, lifies, and ∞

2

∫

2

( ℓ nx )

x2 + 1

0

∞

d x + 2i π

∞

ℓn x

∫x

2

0

+1

dx − π2

∫x

0

dx 2

+1

=−

π3 . 4

But ∞

2

π

∫ 0

∞ π3 1 2 −1   = π = d x t a n x .  0 x2 + 1 2

Hence, we have ∞

2

∫

0

2

( ℓ nx )

x2 + 1

∞

ℓ nx

∫x

d x + 2i π

2

+1

0

dx =

π3 , 4

Where Where up on, on, by b y t aking aking the real and ima ginary p art s ∞

∫ 0

46.(2)

( ℓn x)

2

3

π dx = , 2 x +1 8

∞

ℓn x

∫x

0

2

+1

d x = 0.

Gi ven 4yzp 4y zp + q = –2y. –2y. ... (1) (1) 2 2 Gi ven curve is y + z = 1, x + z = 2. ... (2) (2) The Lagrange’s auxiliary equations of (1) are dx dy dz = = . .... .. .. (3) 4yz 1 − 2y Taking the first and third fraction of (3), we have dx + 2zdz = 0 so that x + z2 = c1. ... (4) T aking aking the las last two t wo fraction fraction s of of (3), (3 ), we we have dz + 2ydy = 0 so that z + y2 = c2. .... .. .. (5) Adding (4) and (5), (y2 + z2) + (x + z) = c 1 + c2 or 1 + 2 = c1 + c2, u sing in g (2) .... .. .. (6)


Page 35


Pu tting tting the val ues of c1 and c2 from (4) and (5) in (6), the equation of the required surface is given by 3 = x + z2 + z + y2 or y2 + z2 + x + z – 3 = 0.

1/ 3 47. (3) IVP is y ' = 2x , y ( 0 ) = 0

⇒

dy 1/ 3 = 2x dx

⇒

dy = 2x dx dx

1/3

on integrating y=

3 4 /3 x +c 2

where c is constant of integration now if y(0) = 0 but we can not determine a particular value of const ant c th ere exist infinitely infinitely many indep endent endent solu solution ⇒ there But there exists finitely many linearly independent solution in an interval around t = 0

48. (2,3) The variational problem is given by I y (x )  =

∫

b

a

F( x, x, y, y ') dx dx

y (a) = y 1

y (b) = y 2

Obviously the solution may not admit admit a solution satisfying satisfying the condition condition y (a) = y , y ( b ) = F is independent of y ' so t hat Then the eular’s equation this th is reduces reduces t o

1 if 2

∂F =0 ∂y '

∂ F d  ∂F  −  =0 ∂ y d x  ∂y ' 

∂F =0 ∂y

which is a finite equation and not a differential equation the solution of

∂F = 0 does not ∂y


Page 36


contain contain any arbitrary arbitrary constants constants and and therefore therefore generally generally speaking does not satis sat isffy the th e boundary boundary conditions conditions y(a 1) = y 1 and y(b 2) = y 2 Hence in general there does not exist a solution of this variational problem only in exceptional cases when the curve

∂F = 0 p asses t hroug hrough h ∂y

the th e boundar boundary y p oints (a1, y 1) and (b2, y 2) does there exist a curve on which an a n extremum can be attained when the function function F is linearly linearly dependent any such such t hat F(x, F(x, y , y ' ) = M (x, (x, y) + N(x, N(x, y ) y ' ∂F ∂M ∂N = + y' ∂y ∂y ∂y

∂F = N ( x, y ) ∂y '

d  ∂F  ∂N( x, y)   = dx d x  ∂y ' 

Hence the Eular’s equation becomes

∂F d  ∂F  −  =0 ∂ y d x  ∂y ' 

∂M ∂N ∂N y '− + ( x, y) = 0 ∂y ∂y ∂x

or

∂M ∂N  ∂N ∂N dy  y' + +  =0 ∂y ∂y  ∂x ∂y dx 

so that

∂M ∂N − =0 ∂y ∂x

whic wh ich h is a finit finite e equation, and not a differential differentia l equat equation ion so t hat hat curve cur ve given by by (1) does not in general general sa s at isfy t he given given bound ary condition condition y(a) = y , y (b) = y 2 Hence the variational problem in general does not possess a solution in the class of continuous function.

t he solu so lution tion of BVP is 49.(1,3) Let the y (x) (x) = y 1(x) + y 2(x)

… (1)

such that y 1(x) is a solution satisfying the homogeneous boundary conditions


Page 37


y 1(0) = 0

y ' (1) = 0

… (2)

and y 2(x) is a solution of the homogeneous differential equation subject to the in homogeneous boundary conditions y 2(0) = α y '2 (1) = β

…. (3)

The solution of differential equation y '' = 0 is

y 2(x) = Ax + B

using t he boundary conditions conditions we get get y 2(x) = α + β x

… (4)

now we w e dete det ermine as as t he so s olution of y '' = f ( x ) under the boundary condition (2) Here we have have p (x) = 1 , q(x) q(x) = 0

a = 0, b = 1

and the boundary conditions y(0) = y ’(1) = 0 now we introduce the functions u(x) and v(x) which satisfies the homogeneous differential equation y ''( x) = 0 under the corresponding boundary conditions y (0) = 0

v ' (1) = 0

… (5)

u(x) u(x) = Ax Ax v(x) v(x) = B where A and B are const const ant ant s. T he Wronskian Wronskian W will be W = –AB Hence Hence the t he Green’s Green’s function is given given by − x

G ( x, s) = 

 −s

and

y1 ( x ) =

∫

1 0

y (x ) =

…. (6)

G ( x, s) f ( s) ds ds

The Th e complete complete solution is or

0≤ x ≤ s s ≤ x ≤1

y = y 1(x) + y 2(x)

1

∫ G ( x s) f ( s) ds + α + βx 0

for a, b being a constant BVP has infinitly many solutions.


Page 38


50. (2,3) A random variable X is said to have a γ distribution distribution with parameter λ > 0 if its its p df is is given  e −x xɺ λ −1 ; λ > 0, 0 < x < δ  f ( x) =  λ  0 o th erw ise 

by

2

we know that if x ~ γ ( λ ) then E(X) = λ = µ(say) Var(X) = λ = σ (say ) The Th e standard standard γ variate variate is given by Z=

X−µ X− λ = σ λ

M z(t) = exp(– exp(– µt/ σ)M x(t / σ) = exp(–µt/ σ) 1 − t   σ  

−λ

K z ( t) =

=e

−tλ / λ

t   1 −  λ 

−λ





t 





λ 

λ ⋅  t − λ log  1 −



 t  t2 t3 = − λ t − λ + + + . . . . . .  3 /2  λ 2λ 3 + t  = − λ t + λt +

t2 + 0 ( λ−1/ 2 ) 2

1/2 1/2

1/2

where 0( λ- ) are terms containing λ and and h igher igher p owers of λ in the t he denomin denominator ator L im k Z ( t ) = λ→∞

t2 2

⇒

 t2   2

L imMz ( t) = exp  λ →∞

which is the mgf of a standard normal variate hence by uniqueness theorem of mgf standard γ variate tends to standard normal variate as λ → ∞ In other words γ distribu distributt ion ion tends t o normal normal distributio distribut ion n for larg large valu v aluee of par p aramet amet ers ers The Th e sum of o f indep indepen endent dent γ variate variate is also a γ varia vari at e if X 1, X 2,…….. Xk are ind ep endent endent γ variates variates with w ith p arameters arameters l1 λ2 …… λ k resp resp ectively ectively Then X 1 + X2 + …… X k is also a γ variat var iatee with wit h parame parameter ter λ1 + λ2 + … + λ k


Page 39


51. (1,4)

1



a 



xn 

(a)

xn +1 =

(b)

2  xn  xn +1 = xn 3 −  2  a

(c)

xn +1 =

2

xn  1 +

2



1

1 xn 9

2  3a xn  + − 6   2 a xn 

Taking the limits as n → ∞ and nothing that Ln→im∞ xn = ξ Ln→im∞ xn +1 = ξ where ξ is the exact root 2

we obtain from all the three methods ξ = a Thus all the three methods determine

a where a

is any any p ositi osit ive real number number. Substituting

xn = ξ + εn xn+1 = ξ + εn+1

(a)

2

and

a = ξ , we get get

ξ + εn +1

  ξ2 1 = ( ξ + εn ) 1 +  2  (ξ + εn )2  



  ε  1 = (ξ + εn ) 1 + 1 + n  2 ξ    =

Therefore

−2

 1  2 ε 3 ε2   = ( ξ + εn ) 2 − n + 2n ... . .. .. ξ ξ  2  

1 ε2n  2 2 2 3 2 ε + − ε + − +. .. .. . .. ( ) ( ) n n  2 ξ 

ξn+1 =

1 2 3 ξn + 0 ( ξn ) 2ξ

… (1)

Hence the method has second order convergence with the error constant c = (b)

ξ + εn +1 =

1 2

ξ

1 1  2 ( ξ + εn ) 3 − 2 ( ξ + εn )  2  ξ 



= ( ξ + εn ) 1 −



εn ε2  − n2  ξ 2ξ 


Page 40


εn +1 = −

3 2 ε n + o ( εn3 ) 2ξ

…. (2)

Hence the method has second order convergence with the error constant c* = −

3 2ξ

Therefore the magnitude of the error in the first formula is about one-third of t hat in t he second formula (c)

If

3εn +1 =

3 2 3 εn + 3o ( εn ) 2ξ

By (1)

and

εn +1 = −

3 2 3 ε n + o ( εn ) 2ξ

By (2)

⇒

εn +1 = o ( εn3 )

Thus the order of convergence of (c) is 3

52. (1,2,3,4)

o(G) = 30 = 2 × 3 × 5

T he number of Sy Sy low low 3-su 3-s ubgroup s is 1 + 3k and (1 + 3k) 3k)

| 10 ⇒ k = 0 or 3

If k = 0 , then Sylow 3-subgroup is normal. Let k ≠ 0, t hen k = 3. T his gives gives 10 Sy Sy low 3-subgroup 3-subgroup s H i each of order 3 and so we have 20 element of order 3. [Notice (for i ≠ j ) o (Hi ∩ Hj ) | o (Hi ) = 3 ⇒ o (Hi ∩ Hj ) = 1 only and so these 20 element are different. Each Hi has one element e of order 1 and other two of order 3. a ∈ H i ⇒ o(a)|

o(H i) = 3 ⇒ o(a) = 1, 3]. The Th e number of Sy low low 5-subgroup 5-subgroup s is 1 + 5k’ and (1 + 5k ') | 6 ⇒ k ' = 0 or 1. If k ' = 0. The Sylow 5-subgroup is normal. Let k ' ≠ 0. Then k ' = 1 . This gives 6 Sylow 5 subgroups each of order 5 and we get 24 elements of order 5. But we have already counted 20 elements of order 3. Thus we have more than 44 elements in G, a contradiction. So,. either k = 0 or k ' = 0. i.e. either Sylow 3-subgroup or Sylow 5-subgroup is normal in G.


Page 41


Let H be a Sylow 3-subgroup of order 3 and K, a Sylow 5-subgroup of order 5. By (i), either H is normal in G or K is normal in G. In any any case, case, HK ≤ G, o(HK) = 15 as as o (H ∩ K ) divides o(H) = 3 and o(K) = 5 ⇒ o (H ∩ K ) = 1 . Sinc Sincee index of HK H K in G is 2, HK H K is normal normal in G. Suppose, H is normal normal in G, G , K is not not normal normal in G. By (i) G has 6 Sylow 5-su bgroups and so 24 element element s of order order 5. But But o(HK) = 15 ⇒ HK is cyclic ⇒ HK has ϕ(15) = 8 elements of order 15. Th T hus G has 24 + 8 = 32 eleme elements, nts, a contr contradic adiction. tion.

∴ K is normal in G. If H is not normal in G, they by (i), G has 10 Sylow 3-subgroups and so 20 elements of order 3. From above HK has 8 elements of order 15 and K has 4 elements of order 5. This gives 20 + 8 + 4= 32 element element s in G, a contradict contradictio ion. n.

∴ 53. (1,3)

H is is normal in G. So So both bot h H and K are norm normal al in G.

The integral equation 1

φ ( x) = x + ∫ xt φ( t ) d t 0

1

φ ( x) = x + x ∫ t φ ( t ) dt 0

φ( x) = x + cx

… (1)

1

where c = ∫ t φ ( t ) d t 0 c=

1

∫ t ( t + ct ) d t 0

1

 t3 ct 3  = +  3 0 3 c=

1 c + 3 3

2c 1 = 3 3


Page 42


c=

⇒

1 2

φ ( x) =

3 x 2

1 3 3 3 φ   + φ( 1) = + = 4 2 4 2

2  φ (0 ) + φ   = 0 + 1 = 1 3 

φ(2) + φ(4) = 3 + 6 = 9 φ (1) = φ(0) =

3 2

54.(1,2,4) The problem problem is t o find the extremum extremum o f the functional functional

I [y] =

2

2 ∫−2 1 + y ' ( x )

Subject to the conditions y ≤ x2 ,

1/ 2

dx y(–2) y(– 2) = 3,

y(2) y(2 ) = 3

Cle arl arl y, th e extremals extremals of I [y] are the strai strai ght line line y = C1 + C2x.

(

3/ 2

)

If F i s the i nte ntegrand grand in I [y], then Fy ' y ' = 1 + y '2 ( x ) ≠ 0. The desired extremal will consist of portions of the straight line AP and QB both tangent to the parabola y = x2 a nd o f the the portion portion PO Q of the pa rabol rabol a.


Page 43


Fig–1 Extremal Extremal through through tw o give give n points outside outside a parabolic region. Le t the the ab sciss cissa of P and Q be – x and x, respectivel respec tively. y. T hen t he con con dition of tang tan gen cy of AP and BQ at P and Q demands C1 + C2 x = x 2 , C2 = 2x ... .. . (1) Sin ce the tangen tangen t QB pa sse s throu th rou gh (2, (2, 3), 3 ), C2 + 2C 2 = 3. ... .. . (2) Sol uti uti on of (1) (1) and (2) gi gi ves two values for vi z., and the the second valu e i s clear clea rl y in admi admis ssi ble and so x1 = 1. This gives from (1) C1 = –1, C2 = 2 . T hus the req uired uired e xtremal tremal i s y = –2 x – 1 i f − 2 ≤ x ≤ − 1, x2 i f − 1 ≤ x ≤ 1, 2x – 1 i f 1 ≤ x ≤ 2. This obviously minimizes the functional.

55. (2,4)

I=

Z3 [x ] x 3 + 2 x +1

Since Z 3[x] = {0, 1, 2} 3

and x + 2x + 1 is irreducible in Z 3 [x] ⇒ I is a field

and no. of elements of I is 3 3 = 27 2

x + 1 is an inv inv ertiable ertiable element element as it it s inverse inverse in I ex e xists ist s Phone: 0744-2429714 Mobile: 9001 90012 2 9711 97111, 1, 9829 98295 5 67114, 90012 9 00129 9 7243 Website: www.vpml!sses.om www.vpml!sses.om "-M!il: vpm l l!sse !sse s#$!hoo.om s#$!hoo.om % % in in&& o#v pml!sses. ml!sses.o om ' (()ess: (()ess: 1-* 1-*-8, -8, +hee +hee l! *h * how(h!)$ o! o!( (, ++, ' /W' /W', , ' , ''+' ''+' , 324 3240 0 05

Page 44


A be an (n × n) matrix n ≥ 5

56. (1,2)

n-5

5

Its characteristic polynomial is x (x – 1) By ∈H t heorem heorem A n–5(A 5 – 1) = A n – A n–5 = 0 n

⇒ A = A

n–5

The rank rank of A is 5

A = [aij ]n×n

57. (1,3)

S.t.

aij = a ji ⇒ A is sy mmetric matrix

if

λ1 λ2……….. λn be eigenvalues eigenvalues of A

and if we can determine a non singular matrix p –1

s.t. P AP = d( d(λ 1λ2…..λn) ⇒

P is an ort hogonal hogonal matrix matr ix

58. (1,2) Since dim V = 7 there are only two possible characteristic polynomials 2

2

∆1(y) = (t – 2t + 5) (t – 3)

3

or

2

∆1(t) = (t – 2t + 5) (t – 3)

5

The Th e sum of o f the orders of the companion companion matr matrice icess must add up to 7 T hus M must must be one of following following b lock diag dia gonal matrice matr icess  0 0 2 7     0 − 5 0 −5    dia g   , , 1 0 −2 7        1 2   1 2  0 1 9        0 − 5 dia g   , 1 2    

0 0 2 7    1 0 −2 7 ,  0 −9      1 6     0 1 9   

   0 − 5 dia g   , 1 2   

 0 0 2 7   1 0 −2 7 , 3 , 3    [ ] [ ]   0 1 9  


Page 45


59. (1,2,3)

(a)

Suppose a1u1 + …. + mum + b 1w 1 + …. + bnw n = 0, whe w here re a1, b j are scalars.

Then (a1u1 + …… am um ) + (b1w 1 + …. + bnw n) = 0 = 0 + 0 where 0, 1u 1 + …. + am um ∈ U and 0, b 1w1 + ….. + b nwn ∈ W. Since such a sum for 0 is unique, this leads to a1u1 + … + am um = 0 and

b1w1 + …… bnw n = 0

Since S1 is linearly independent, each a i = 0, and sin ce S2 is linearly linearly indep independe endent nt,, each b j = 0. Thus S = S1 ∪ S2 is linearly independent. (b)

By part(a), S = S1 ∪ S2 is linearly independent, and S = S1 ∪ S 2 spans V = U + W. Thus S = S1 ∪ S2 is a basis of V.

(c)

T his follows direct direct ly from p art(b). We can generlise these results for r subsets

60. (1,2)

If

X1 ~ N ( µ, σ1

2

X 2 ~ N ( µ 2 , σ2

2

)

)

X1 2 ± X2N( µ1 ± µ2, σ1 + σ2 2

2

)


Page 46

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