No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical, photocopying, or otherwise without the prior permission of the author.GATE SOLVED PAPERElectrical EngineeringENGINEERING MATHEMATICSCopyright By NODIA & COMPANYInformation contained in this book has been obtained by authors, from sources believes to be reliable.However, neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Nodia and its authors are supplying information but are not attempting to render engineeringor other professional services.NODI A AND COMPANYB-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, J aipur 302039Ph :+91 - 141 - 2101150www.nodia.co.inemail : enquiry@nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount*GATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inYEAR 2013ONE MARKQ. 1Given a vector field F y xa yza x ax y z2 2= - -vv v v, the line integralF dl :v v# evaluated along a segment on thex-axis from x 1 =to x 2 =is(A). 233 - (B) 0(C). 233(D) 7Q. 2The equation xx21210012--= > > > H H H has(A) no solution(B) only one solution xx0012= > > H H(C) non-zero unique solution(D) multiple solutionsQ. 3Square roots ofi - , where i 1 = - , are(A) i ,i -(B)cos sin i4 4p p- + - d d n n, cos sin i4343 p p+ b b l l(C)cos sin i4 43 p p+ d b n l, cos sin i434p p+ b d l n(D) cos sin i4343 p p+ - b b l l, cos sin i4343 p p- + b b l lQ. 4The curl of the gradient of the scalar field defined by V x y y z z x 2 3 42 2 2= + +is(A) 4 6 8 xya yza zxax y z+ +v v v(B)4 6 8 a a ax y z+ +v v v(C)xy za x yz a y zx a 4 4 2 6 3 8x y z2 2 2+ + + + +v v v^ ^ ^ h h h(D) 0Q. 5A continuous random variableXhas a probability density functionf x ex=-^ h , x 0< < 3. Then PX 1 > " , is(A) 0.368(B) 0.5(C) 0.632(D) 1.0YEAR 2013TWO MARKSQ. 6WhentheNewton-Raphsonmethodisappliedtosolvetheequation f x x x 2 1 03= + - = ^ h , the solution at the end of the first iteration with the initial value as. x 120=is(A). 082 - (B). 049(C). 0705(D). 169Q. 7A function y x x 5 102= +is defined over an open interval, x 1 2 =^ h. Atleast at one point in this interval,/ dy dx is exactly(A) 20(B) 25(C) 30(D) 35GATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 8 zzdz4422+-# evaluated anticlockwise around the circlez i 2 - = , where i 1 = -, is(A)4p - (B) 0(C) 2 p + (D)i 2 2 +Q. 9A Matrix has eigenvalues1 -and2 - . The corresponding eigenvectors are 11 -> H and 12 -> H respectively. The matrix is(A) 1112 - -> H(B) 1224 - -> H(C) 1002--> H(D) 0213 - -> HYEAR 2012ONE MARKQ. 10TwoindependentrandomvariablesX and Y areuniformlydistributedinthe interval, 1 1 -6 @. The probability that, max X Y6 @ is less than/ 1 2 is(A)/ 3 4(B)/ 9 16(C)/ 1 4(D)/ 2 3Q. 11If, x 1 = -then the value of xx is(A) e/ 2 p -(B) e/ 2 p(C) x(D) 1Q. 12Given( ) f zz z 1132=+-+. If Cis a counter clockwise path in the z-plane such thatz 1 1 + = , the value of( )jf z dz21Cp# is(A)2 - (B)1 -(C) 1(D) 2Q. 13With initial condition( ) . x 1 05 = , the solution of the differential equation tdtdxx t + = , is(A) x t21= - (B) x t21 2= -(C) xt22= (D) xt2=YEAR 2012TWO MARKSQ. 14Given thatand A I52301001=- -= > > H H, the value of A3 is(A) 15 12 A I + (B) 19 30 A I +(C) 17 15 A I + (D) 17 21 A I +Q. 15The maximum value of( ) f x x x x 9 24 53 2= - + +in the interval [ , ] 1 6 is(A) 21(B) 25(C) 41(D) 46GATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 16A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is(A)/ 1 3(B)/ 1 2(C)/ 2 3(D)/ 3 4Q. 17ThedirectionofvectorA isradiallyoutwardfromtheorigin,withkr An= . where r x y z2 2 2 2= + +and k is a constant. The value of n for whichA 0 : d =is(A)2 - (B) 2(C) 1(D) 0Q. 18Consider the differential equation ( ) ( )( )dtd y tdtdy ty t 222+ + ( ) t d =with( ) 2 0 and y tdtdytt00=- ===--The numerical value of dtdyt 0 =+ is(A)2 - (B)1 -(C) 0(D) 1YEAR 2011ONE MARKQ. 19Roots of the algebraic equation x x x 1 03 2+ + + =are(A) ( , , ) j j 1 + + - (B) ( , , ) 1 1 1 + - +(C) ( , , ) 0 0 0 (D) ( , , ) j j 1 - + -Q. 20A point Z has been plotted in the complex plane, as shown in figure below.The plot of the complex number YZ1=isGATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 21With Kas a constant, the possible solution for the first order differential equation dxdyex 3=- is(A)e Kx31 3- +-(B)e Kx31 3- +(C)e Kx31 3- +-(D)e K 3x- +-YEAR 2011TWO MARKSQ. 22Solution of the variables x1 and x2 for the following equations is to be obtained by employing the Newton-Raphson iterative method.Equation (1). sin x x 10 08 02 1- =Equation (2). cos x x x 10 10 06 0222 1- - =Assuming the initial values are. x 001=and. x 102= , the jacobian matrix is(A) ..1000806--> H(B) 100010> H(C) ..0100806--> H(D) 1010010 -> HQ. 23The function( ) f x x x x 2 32 3= - - +has(A) a maxima at x 1 =and minimum at x 5 =(B) a maxima at x 1 =and minimum at x 5 =-(C) only maxima at x 1 =and(D) only a minimum at x 5 =Q. 24A zero mean random signal is uniformly distributed between limitsa -anda +and its mean square value is equal to its variance. Then the r.m.s value of the signal is(A) a3(B) a2(C) a 2(D) a 3Q. 25The matrix [ ] A2411=-> H is decomposed into a product of a lower triangularmatrix [ ] Land an upper triangular matrix [ ] U . The properly decomposed [ ] Land [ ] Umatrices respectively are(A) 1401 -> H and 1012 -> H(B) 2401 -> H and 1011> H(C) 1401> H and 2011 -> H(D) 2403 -> H and . 10151> HQ. 26The two vectors [1,1,1] and [1, , ] aa2 where a j2123=- +_ i, are(A) Orthonormal(B) Orthogonal(C) Parallel(D) CollinearYEAR 2010ONE MARKQ. 27The value of the quantity P , where P xe dxx01= # , is equal to(A) 0(B) 1(C) e(D) 1/ eGATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 28Divergence of the three-dimensional radial vector fieldr is(A) 3(B)/ r 1(C)i j k + +t t t(D) 3( ) i j k + +t t tYEAR 2010TWO MARKSQ. 29A box contains 4 white balls and 3 red balls. In succession, two balls are randomly andremovedformthebox.Giventhatthefirstremovedballiswhite,the probability that the second removed ball is red is(A) 1/ 3(B) 3/ 7(C) 1/ 2(D) 4/ 7Q. 30An eigenvector of P100120023=JLKKKNPOOO is(A)1 1 1T-8 B(B)1 2 1T8 B(C)1 1 2T-8 B(D)2 1 1T-8 BQ. 31For the differential equation dtd xdtdxx 6 8 022+ + =with initial conditions( ) x 0 1 =and dtdx0t 0==, the solution is(A)( ) 2 x t e et t 6 2= -- -(B)( ) 2 x t e et t 2 4= -- -(C)( ) 2 x t e et t 6 4=- +- -(D)( ) 2 x t e et t 2 4= +- -Q. 32Forthesetofequations,x x x x 2 4 21 2 3 4+ + + = and3 6 3 12 6 x x x x1 2 3 4+ + + = . The following statement is true.(A) Only the trivial solution0 x x x x1 2 3 4= = = =exists(B) There are no solutions(C) A unique non-trivial solution exists(D) Multiple non-trivial solutions existQ. 33At t 0 = , the function( )sinf tt t=has(A) a minimum(B) a discontinuity(C) a point of inflection(D) a maximumYEAR 2009ONE MARKQ. 34Thetraceanddeterminantofa2 2#matrixareknowntobe2 - and35 -respectively. Its eigen values are(A)30 -and5 - (B)37 -and1 -(C)7 -and 5(D) 17.5 and2 -YEAR 2009TWO MARKSQ. 35( , ) f x yis a continuous function defined over ( , ) [ , ] [ , ] x y 0 1 0 1#! . Given the two constraints, x y >2 and y x >2, the volume under( , ) f x yis(A)( , ) f x y dxdyx yx yy y012====# # (B)( , ) f x y dxdyx yxy xy 1 12 2====# #(C)( , ) f x y dxdyx xy y0101====# # (D)( , ) f x y dxdyx x yy y x0 0 ====# #GATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 36Assume for simplicity that N people, all born in April (a month of 30 days), are collected in a room. Consider the event of at least two people in the room being born on the same date of the month, even if in different years, e.g. 1980 and 1985. What is the smallest N so that the probability of this event exceeds 0.5 ?(A) 20(B) 7(C) 15(D) 16Q. 37A cubic polynomial with real coefficients(A) Can possibly have no extrema and no zero crossings(B) May have up to three extrema and upto 2 zero crossings(C) Cannot have more than two extrema and more than three zero crossings(D) Will always have an equal number of extrema and zero crossingsQ. 38Letx 117 02- = .TheiterativestepsforthesolutionusingNewton-Raphons method is given by(A) x xx 21 117k kk1= ++ b l(B) x xx117k kk1= -+(C) x xx117k kk1= -+(D) x x xx 21 117k k kk1= - ++ b lQ. 39( , ) ( ) ( ) x y x xy y xy F a ax y2 2= + + +t t.Itslineintegraloverthestraightlinefrom ( , ) ( , ) x y 0 2 =to ( , ) ( , ) x y 2 0 =evaluates to(A)8 - (B) 4(C) 8(D) 0YEAR 2008ONE MARKSQ. 40Xis a uniformly distributed random variable that takes values between 0 and 1. The value of{ } EX3 will be(A) 0(B) 1/ 8(C) 1/ 4(D) 1/ 2Q. 41The characteristic equation of a (3 3 # ) matrix Pis defined as( ) a I P 2 1 03 2l l l l l = - = + + + =If Idenotes identity matrix, then the inverse ofmatrix Pwill be(A) ( ) P P I 22+ +(B) ( ) P P I2+ +(C)( ) P P I2- + +(D)( ) P P I 22- + +Q. 42If the rank of a ( ) 5 6 #matrix Q is 4, then which one of the following statement is correct ?(A) Q will have four linearly independent rows and four linearly independent columns(B) Q will have four linearly independent rows and five linearly independent columns(C) QQT will be invertible(D) QQT will be invertibleGATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inYEAR 2008TWO MARKSQ. 43Consider function( ) ( 4) f x x2 2= -wherex is a real number. Then the function has(A) only one minimum(B) only tow minima(C) three minima(D) three maximaQ. 44Equatione 1 0x- = isrequiredtobesolvedusingNewtonsmethodwithan initial guess x 10=- . Then, after one step of Newtons method, estimate x1 of the solution will be given by(A) 0.71828(B) 0.36784(C) 0.20587(D) 0.00000Q. 45Aism n # fullrankmatrixwithm n > andI isidentitymatrix.Letmatrix ' ( ) A AA A1 T T=-, Then, which one of the following statement is FALSE ?(A)' AA A A = (B) ( ') AA2(C)' A A I = (D)' ' AA A A =Q. 46Adifferentialequation/ ( ) dx dt e u tt 2=-,hastobesolvedusingtrapezoidalrule ofintegrationwithastepsize. h 001 = s.Function( ) u t indicatesaunitstep function. If( ) x 0 0 =-, then value of x at. t 001 =s will be given by(A) 0.00099(B) 0.00495(C) 0.0099(D) 0.0198Q. 47Let Pbe a 2 2 #real orthogonal matrix and x is a real vector [ ] x,x1 2T with length ( ) x x x/1222 1 2= + . Then, which one of the following statements is correct ?(A)Px x #where at least one vector satisfiesPx x (D) No relationship can be established betweenxandPxYEAR 2007ONE MARKQ. 48xx x xn 1 2Tg =8 B is an n-tuple nonzero vector. The n n #matrixV xxT=(A) has rank zero(B) has rank 1(C) is orthogonal(D) has rank nYEAR 2007TWO MARKSQ. 49The differential equation dtdx x 1=t- is discretised using Eulers numerical integration method with a time stepT 0 > 3 . What is the maximum permissible value ofT 3to ensure stability of the solution of the corresponding discrete time equation ?(A) 1(B)/ 2 t(C) t(D) 2tQ. 50The value of ( ) zdz1C2+# where Cis the contour1 zi2- =is(A)i 2p (B) p(C) tanz1 -(D)tan i z1p-GATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 51The integral( ) sin cos t d2102pt tt -p# equals(A) sin cos t t(B) 0(C)cost21(D)sint21Q. 52A loaded dice has following probability distribution of occurrencesDice Value 1 2 3 4 5 6Probability 1/ 4 1/ 8 1/ 8 1/ 8 1/ 8 1/ 4If three identical dice as the above are thrown, the probability of occurrence of values 1, 5 and 6 on the three dice is(A) same as that of occurrence of 3, 4, 5 (B) same as that of occurrence of 1, 2, 5(C) 1/ 128(D) 5/ 8Q. 53Letx andy be two vectors in a 3 dimensional space andx,y < > denote their dot product. Then the determinantdetx,xy,xx,yy,y< >< >< >< >= G(A) is zero when x and y are linearly independent(B) is positive when x and y are linearly independent(C) is non-zero for all non-zero x and y(D) is zero only when either x or y is zeroQ. 54Thelinearoperation( ) L x isdefinedbythecrossproductL(x) b x # = ,where b 0 1 0T=8 B andxx x x1 2 3T=8 B are three dimensional vectors. The 3 3 #matrix M of this operations satisfies( ) M xxxL x123=RTSSSSVXWWWWThen the eigenvalues of M are(A), , 0 1 1 + - (B), , 1 1 1 -(C), , i i 1 - (D), , i i 0 -Statement for Linked Answer Question 55 and 56Cayley-HamiltonTheoremstatesthatasquarematrixsatisfiesitsown characteristic equation. Consider a matrixA3220=--= GQ. 55A satisfies the relation(A) A I A 3 2 01+ + =-(B)2 2 0 A A I2+ + =(C) ( )( ) A I A I 2 + + (D)( ) 0 exp A =Q. 56A9 equals(A) 511 510 A I + (B) 309 104 A I +(C) 154 155 A I + (D)( ) exp A 9GATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inYEAR 2006TWO MARKSQ. 57The expression( / ) V R h H dh 1H2 20p = -# for thevolume of a cone is equal to(A)( / ) R h H dr 1R2 20p -#(B)( / ) R h H dh 1R2 20p -#(C)( / ) rH rRdh 2 1H0p -# (D)rHRrdr 2 1R20p -` j#Q. 58Asurface( , ) S x y x y 2 5 3 = + - isintegratedonceoverapathconsistingofthe points that satisfy ( ) ( ) x y 1 2 1 2 2 + + - = . The integral evaluates to(A) 17 2(B) 17 2(C)/ 2 17(D) 0Q. 59Two fair dice are rolled and the sum rof the numbers turned up is considered(A)( 6) Pr r >61=(B) Pr ( / r 3 is an integer) 65=(C) Pr ( / r r 8 4 ; =is an integer) 95=(D)( 6 / Prr r 5 ; =is an integer) 161=Statement for Linked Answer Question 60 and 61, , P Q R10132592712T T T=-=-- =-RTSSSSRTSSSSRTSSSSVXWWWWVXWWWWVXWWWW are three vectors.Q. 60An orthogonal set of vectors having a span that contains P,Q,Ris(A) 636423----RTSSSSRTSSSSVXWWWWVXWWWW(B) 4245711823-- -RTSSSSRTSSSSRTSSSSVXWWWWVXWWWWVXWWWW(C) 671322394 --- -RTSSSSRTSSSSRTSSSSVXWWWWVXWWWWVXWWWW(D) 43111313534RTSSSSRTSSSSRTSSSSVXWWWWVXWWWWVXWWWWQ. 61Thefollowingvectorislinearlydependentuponthesolutiontotheprevious problem(A) 893RTSSSSVXWWWW(B) 21730--RTSSSSVXWWWW(C) 445RTSSSSVXWWWW(D) 1323 -RTSSSSVXWWWWYEAR 2005ONE MARKQ. 62In the matrix equationx q P = , which of the following is a necessary condition for the existence of at least on solution for the unknown vector x(A) Augmented matrix [ ] q Pmust have the same rank as matrix P(B) Vector q must have only non-zero elements(C) Matrix Pmust be singular(D) Matrix Pmust be squareGATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 63If Pand Q are two random events, then the following is TRUE(A) Independence of P and Q implies that probability ( ) P Q 0 + =(B) Probability ( ) P Q , $ Probability (P) + Probability (Q)(C) If Pand Q are mutually exclusive, then they must be independent(D) Probability ( ) P Q + # Probability (P)Q. 64If S xdx31=3-#, then S has the value(A) 31- (B) 41(C) 21(D) 1Q. 65The solution of the first order DE'( ) ( ) xt x t 3 =- ,(0) x x0=is(A)( ) x t x et03=- (B)( ) x t x e03=-(C)( ) x t x e/01 3=- (D)( ) x t x e01=-YEAR 2005TWO MARKSQ. 66For the matrix p300220211=--RTSSSSVXWWWW, one of the eigen values is equal to2 -Which of the following is an eigen vector ?(A) 321-RTSSSSVXWWWW(B) 321--RTSSSSVXWWWW(C) 123-RTSSSSVXWWWW(D) 250RTSSSSVXWWWWQ. 67If R122013112=--RTSSSSVXWWWW, then top row of R1 - is(A)5 6 48 B(B)5 3 1 -8 B(C)2 0 1 -8 B(D)/ 2 1 1 2 -8 BQ. 68A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is(A) 1/ 6(B) 1/ 2(C) 3/ 6(D) 3/ 4Q. 69For the function( ) f x x ex 2=-, the maximum occurs when x is equal to(A) 2(B) 1(C) 0(D)1 -GATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 70For the scalar field uxy2 322= + , magnitude of the gradient at the point (1, 3) is(A) 913(B) 29(C)5(D) 29Q. 71For the equation''( ) '( ) ( ) x t xt x t 3 2 5 + + = ,the solution( ) x tapproaches which of the following values as t " 3 ?(A) 0(B) 25(C) 5(D) 10***********GATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inSOLUTI ONSol . 1Option (B) is correct.Given, the field vectorFvy xa yza x ax y z2 2= - -v v vFor the line segment along x-axis, we havedlvdxax=vSo,F dl :v vy xdx2=^ ^ h hSince, on x-axis y 0 =so,F dl :v v0 =or,F dl :v v#0 =Sol . 2Option (D) is correct.Given the equations in matrix form as xx212112--> > H H 00=> HSo,itisahomogenoussetoflinearequation.Ithaseitheratrivialsolution x x 01 2= = ^ h or an infinite no. of solution. Since, for the matrixA 2121=--> Hwe have the determinant A0 =Hence, it will have multiple solutionsSol . 3Option (B) is correct.We know thati e/ i 2=p(In phasor form)or,i - e/ i 2=p -So,i - e e/// i i 21 24! ! = =p p - -^ hcos sin i4 4!p p= - d d n n= Gcos sin i4 4p p= - d d n n ;cos sin i4 4p p- + d d n n i212= -; i12- +This is equivalent to the given option (B) only.Sol . 4Option (D) is correct.Given the scalar fieldV x y y z z x 2 3 42 2 2= + +Its gradient is given byV d xy za x yz a y zx a 4 4 2 6 3 8x y z2 2 2= + + + + +v v v^ _ ^ h i hSo, the curl of the gradient is obtained asGATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inv#d d ^ h axy zaz yzay zx 4 4 2 6 3 8xxyyzz2 2 2=+ + +222222v v v8 8 a y y a z z a x x 6 6 4 4x y z= - - - + -v v v^ ^ ^ h h h0 =Note : From the properties of curl, we know that curl of gradient of any scalar field is always zero. So, there is no need to solve the curl and gradient.Sol . 5Option (A) is correct.Given, the PdF of random variable x asf x ^ hex=-x 0< < 3So, P x 1 > ^ he dxx1=3-# e1x1=-3 -: De1=-. 0368 =Sol . 6Option (C) is correct.Given, the equationf x ^ hx x 2 1 03= + - =as initial condition is x0. 12 =so, from N R -method we obtainxn 1 +xf xf xnnn= -l^^hhHerex0. 12 =f x0 ^h. 2 . 1 3.128 12 123= + - =^ ^ h hAlso,f x l^ hx 3 22= +So,f x0l^h. . 3 12 2 6322= + = ^ hHence, 1st iterative value isx1xf xf x000= -l^^hh...126323128= -. 0705 =Sol . 7Option (B) is correct.Given the functiony5 10 f x x x2= = +^ h in the internal, x 1 2 =^ hSince, function y is continuous in the interval, 1 2 ^ h as well as its is differentiable at each point so, from Lagranges mean value theorem there exist at least a point wheref c l^ h b af b f a=-- ^ ^ h hHere, we havea 1 =, b 2 =So, for x a 1 = = , we obtainy5 f a f 1 1 10 1 152= = = + = ^ ^ ^ ^ h h h hand for x b 2 = =y5 f b f 2 2 10 2 402= = = + = ^ ^ ^ ^ h h h hTherefore,f c l^ h 2 140 1525 =--=GATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inSol . 8Option (A) is correct.Given the contour integral zzdz4422+-#It has two poles given aszi 2 ! =Now, the contour is defined by circlez i 2 - =which is shown in the figure belowSo, it can be observed that the given contour enclosed z i 2 =while z i 2 =-is out of the contour. So, we obtain the residue at z i 2 =only as residue z iz24z i22=+-= i iiii2 22 44822=+-=-=^ hHence, contour integral is given as zzdz4422+-#2 i p =(sum of residues)2 i i 2 p = ^ h4p =-Sol . 9Option (D) is correct.We know that the characteristic equation is given byA X 6 6 @ @X l =6 @whereA 6 @ is the matrix as l is the scalar which gives eigen values. Now, we consider the matrixA 6 @ ac bd=> Hmatrix 2 2# ^ hFor eigen value1 -as eigen vector is 11 -> H, so, we have ac bd11 -> > H H111=--> Hor,a b - 1 =- ....(1)c d - 1 = ....(2)Similarly, for eigen value2 -with eigen vector 12 -> H, we obtain ac bd12 -> > H H212=--> Hor,a b 2 - 2 =- ....(3)c d 2 - 4 = ....(4)GATE SOLVED PAPER - EEENGI NEERI NG MATHEMATI CS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inSolving Eqs. (1) and (3), we obtaina,b 0 1 = =and solving Eqs. (2) and (4), we obtainc,d 2 3 =- =Thus, the required matrix is ac bd> H 0213=- -> HSol . 10Option (B) is correct.Probability density function of uniformly distributed variables Xand Yis shown as [ ( , )] max P x y21