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GATE SOLVED PAPERElectronics & Communication Engineering Mathematics
Copyright By NODIA & COMPANY
Information 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 engineering or other professional services.
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GATE SOLVED PAPER - ECENGINEERING MATHEMATICS
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2013 ONE MARK
Q. 1 The maximum value of q until which the approximation sin .q q holds to within 10% error is(A) 10c (B) 18c
(C) 50c (D) 90c
Q. 2 The minimum eigen value of the following matrix is352
5127
275
R
T
SSSS
V
X
WWWW
(A) 0 (B) 1
(C) 2 (D) 3
Q. 3 A polynomial ( )f x a x a x a x a x a4 4 3 3 2 2 1 0= + + + - with all coefficients positive has(A) no real roots
(B) no negative real root
(C) odd number of real roots
(D) at least one positive and one negative real root
2013 TWO MARKS
Q. 4 Let A be an m n# matrix and B an n m# matrix. It is given that determinant I ABm+ =^ h determinant I BAn+^ h, where Ik is the k k# identity matrix. Using
the above property, the determinant of the matrix given below is2111
1211
1121
1112
R
T
SSSSSS
V
X
WWWWWW
(A) 2 (B) 5
(C) 8 (D) 16
2012 ONE MARK
Q. 5 With initial condition ( ) .x 1 0 5= , the solution of the differential equation t dtdx x t+ = , is
(A) x t 21= - (B) x t 2
12= -
(C) x t22= (D) x t2=
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GATE SOLVED PAPER - EC ENGINEERING MATHEMATICS
Q. 6 Given ( )f z z z11
32= + - + .
If C is a counter clockwise path in the z -plane such that z 1 1+ = , the value of ( )j f z dz2
1Cp # is
(A) 2- (B) 1-(C) 1 (D) 2
Q. 7 If ,x 1= - then the value of xx is(A) e /2p- (B) e /2p
(C) x (D) 1
2012 TWO MARKS
Q. 8 Consider the differential equation
( ) ( )
( )dt
d y tdt
dy ty t22
2
+ + ( )td= with ( ) 2 0andy t dtdy
tt
00
=- == =- -The numerical value of dt
dyt 0= +
is(A) 2- (B) 1-(C) 0 (D) 1
Q. 9 The direction of vector A is radially outward from the origin, with krA n= . where r x y z2 2 2 2= + + and k is a constant. The value of n for which A 0:d = is(A) 2- (B) 2(C) 1 (D) 0
Q. 10 A 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 4
Q. 11 The maximum value of ( )f x x x x9 24 53 2= - + + in the interval [ , ]1 6 is(A) 21 (B) 25
(C) 41 (D) 46
Q. 12 Given that
andA I52
30
10
01=
- - => >H H, the value of A3 is(A) 15 12A I+ (B) 19 30A I+(C) 17 15A I+ (D) 17 21A I+
2011 ONE MARK
Q. 13 Consider a closed surface S surrounding volume V . If rv is the position vector of a point inside S , with nt the unit normal on S , the value of the integral r n dS5
S$v t##
is
(A) V3 (B) V5
(C) V10 (D) V15
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Q. 14 The solution of the differential equation , (0)dxdy ky y c= = is
(A) x ce ky= - (B) x kecy=(C) y cekx= (D) y ce kx= -
Q. 15 The value of the integral ( )z z
z dz4 5
3 4
c2+ +- +# where c is the circle z 1= is given
by(A) 0 (B) 1/10
(C) 4/5 (D) 1
2011 TWO MARKS
Q. 16 A numerical solution of the equation ( )f x x 3 0+ - = can be obtained using Newton- Raphson method. If the starting value is x 2= for the iteration, the value of x that is to be used in the next step is(A) 0.306 (B) 0.739
(C) 1.694 (D) 2.306
Q. 17 The system of equations
4 6 20
4
x y z
x y y
x y z
6
ml
+ + =+ + =+ + =
has NO solution for values of l and given by(A) 6, 20ml = = (B) 6, 20ml = =Y(C) 6, 20ml = =Y (D) 6, 20ml = =Y
Q. 18 A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is(A) 2/36 (B) 2/6
(C) 5/12 (D) 1/2
2010 ONE MARKS
Q. 19 The eigen values of a skew-symmetric matrix are(A) always zero (B) always pure imaginary
(C) either zero or pure imaginary (D) always real
Q. 20 The trigonometric Fourier series for the waveform ( )f t shown below contains
(A) only cosine terms and zero values for the dc components
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(B) only cosine terms and a positive value for the dc components
(C) only cosine terms and a negative value for the dc components
(D) only sine terms and a negative value for the dc components
Q. 21 A function ( )n x satisfied the differential equation ( ) ( )
dxd n x
Ln x
022
2- =where L is a constant. The boundary conditions are : (0)n K= and ( )n 03 = . The solution to this equation is(A) ( ) ( / )expn x K x L= (B) ( ) ( / )expn x K x L= -(C) ( ) ( / )expn x K x L2= - (D) ( ) ( / )expn x K x L= -
2010 TWO MARKS
Q. 22 If e x /y x1= , then y has a(A) maximum at x e= (B) minimum at x e=(C) maximum at x e 1= - (D) minimum at x e 1= -
Q. 23 A fair coin is tossed independently four times. The probability of the event the number of time heads shown up is more than the number of times tail shown up(A) 1/16 (B) 1/3
(C) 1/4 (D) 5/16
Q. 24 If A xya x ax y2= +v t t , then A dlC
$v v# over the path shown in the figure is
(A) 0 (B) 3
2
(C) 1 (D) 2 3
Q. 25 The residues of a complex function
( )( )( )
x zz z z
z1 2
1 2= - --
at its poles are
(A) ,21
21- and 1 (B) ,2
121- and 1-
(C) , 121 and 2
3- (D) ,21 1- and 2
3
Q. 26 Consider differential equation ( )
( )dxdy x
y x x- = , with the initialcondition ( )y 0 0= . Using Eulers first order method with a step size of 0.1, the value of ( . )y 0 3 is(A) 0.01 (B) 0.031
(C) 0.0631 (D) 0.1
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Q. 27 Given ( )( )
f t Ls s k s
s4 3
3 113 2= + + -
+- ; E. If ( ) 1lim f tt ="3 , then the value of k is(A) 1 (B) 2
(C) 3 (D) 4
2009 ONE MARK
Q. 28 The order of the differential equation dtd y
dtdy y e t2
2 34+ + = -c m is
(A) 1 (B) 2
(C) 3 (D) 4
Q. 29 A fair coin is tossed 10 times. What is the probability that only the first two tosses will yield heads?
(A) 21 2c m (B) C 2110 2
2b l(C) 2
1 10c m (D) C 2110 210b l
Q. 30 If ( )f z c c z0 1 1= + - , then ( )zf z
dz1
unit circle
+# is given by(A) c2 1p (B) ( )c2 1 0p +(C) jc2 1p (D) ( )c2 1 0p +
2009 TWO MARKS
Q. 31 The Taylor series expansion of sinx
xp- at x p= is given by
(A) !
( )...
x1
3
2p+ - + (B) !
( )...
x1
3
2p- - - +
(C) !
( )...
x1
3
2p- - + (D) !
( )...
x1
3
2p- + - +
Q. 32 Match each differential equation in Group I to its family of solution curves from Group IIGroup I Group II
A. dxdy
xy= 1. Circles
B. dxdy
xy=- 2. Straight lines
C. dxdy
yx= 3. Hyperbolas
D. dxdy
yx=-
(A) A 2, B 3, C 3, D 1- - - -(B) A 1, B 3, C 2, D 1- - - -(C) A 2, B 1, C 3, D 3- - - -(D) A 3, B 2, C 1, D 2- - - -
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Q. 33 The Eigen values of following matrix are
130
310
563
-- -R
T
SSSS
V
X
WWWW
(A) 3, 3 5 , 6j j+ - (B) 6 5 , 3 , 3j j j- + + -(C) 3 , 3 , 5j j j+ - + (D) 3, 1 3 , 1 3j j- + - -
2008 ONE MARKS
Q. 34 All the four entries of the 2 2# matrix Ppp
pp
11
21
12
22= = G are nonzero,
and one of its eigenvalue is zero. Which of the following statements is true?(A) p p p p 111 12 12 21- = (B) p p p p 111 22 12 21- =-(C) p p p p 011 22 12 21- = (D) p p p p 011 22 12 21+ =
Q. 35 The system of linear equations
x y4 2+ 7= x y2 + 6= has(A) a unique solution
(B) no solution
(C) an infinite number of solutions
(D) exactly two distinct solutions
Q. 36 The equation ( )sin z 10= has(A) no real or complex solution(B) exactly two distinct complex solutions(C) a unique solution(D) an infinite number of complex solutions
Q. 37 For real values of x , the minimum value of the function( ) ( ) ( )exp expf x x x= + - is
(A) 2 (B) 1
(C) 0.5 (D) 0
Q. 38 Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x 0= ?(A) ( )sin x3 (B) ( )sin x2
(C) ( )cos x3 (D) ( )cos x2
Q. 39 Which of the following is a solution to the differential equation( )
( )dt
dx tx t3 0+ = ?
(A) ( )x t e3 t= - (B) ( )x t e2 t3= -(C) ( )x t t2
3 2=- (D) ( )x t t3 2=
2008 TWO MARKS
Q. 40 The recursion relation to solve x e x= - using Newton - Raphson method is(A) x en x1 n=+ - (B) x x en n x1 n= -+ -
(C) (1 )x xe
e1n n x
x
1 n
n= + ++ --
(D) ( )
xx e
x e x1 1n
nx
nx
n1
2
n
n
= -- - -
+ --
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Q. 41 The residue of the function ( )f z ( ) ( )z z2 2
12 2= + - at z 2= is
(A) 321- (B)
161-
(C) 161 (D)
321
Q. 42 Consider the matrix P02
13= - -= G. The value of ep is
(A) e ee e
e ee e
2 32 2 5
2 1
2 1
1 2
2 1
--
--
- -- -
- -- -> H
(B) e ee e
e ee e2 423 2
1 1
1 2
2 1
1 2
+-
-+
- --
- -- -> H
(C) e ee e
e ee
52 6
34 6
2 1
2 1
1 2
2 1
--
-+
- -- -
- -- -> H
(D) e ee e
e ee e
22 2 2
1 2
1 2
1 2
1 2
-- +
-- +
- -- -
- -- -> H
Q. 43 In the Taylor series expansion of ( ) ( )exp sinx x+ about the point x p= , the coefficient of ( )x 2p- is(A) ( )exp p (B) . ( )exp0 5 p(C) ( )exp 1p + (D) ( )exp 1p -
Q. 44 The value of the integral of the function ( , )g x y x y4 103 4= + along the straight line segment from the point ( , )0 0 to the point ( , )1 2 in the x y- plane is(A) 33
(B) 35
(C) 40
(D) 56
Q. 45 Consider points P and Q in the x y- plane, with ( , )P 1 0= and ( , )Q 0 1= . The line integral 2 ( )xdx ydy
P
Q +# along the semicircle with the line segment PQ as its diameter(A) is 1-(B) is 0
(C) is 1
(D) depends on the direction (clockwise or anit-clockwise) of the semicircle
2007 ONE MARK
Q. 46 The following plot shows a function which varies linearly with x . The value of the
integral I ydx1
2= # is
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(A) 1.0 (B) 2.5
(C) 4.0 (D) 5.0
Q. 47 For x 1
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(C) ( )f t2 and ( )f t3 are orthogonal D) ( )f t1 and ( )f t2 are orthonormal
Q. 54 If the semi-circular control D of radius 2 is as shown in the figure, then the value
of the integral ( )s
ds1
1
D2-# is
(A) jp (B) jp-(C) p- (D) p
Q. 55 It is given that , ...X X XM1 2 at M non-zero, orthogonal vectors. The dimension of the vector space spanned by the M2 vectors , ,... , , ,...X X X X X XM M1 2 1 2- - - is(A) M2 (B) M 1+(C) M
(D) dependent on the choice of , ,...X X XM1 2
Q. 56 Consider the function ( )f x x x 22= - - . The maximum value of ( )f x in the closed interval [ , ]4 4- is(A) 18 (B) 10
(C) 225- (D) indeterminateQ. 57 An examination consists of two papers, Paper 1 and Paper 2. The probability
of failing in Paper 1 is 0.3 and that in Paper 2 is 0.2. Given that a student has failed in Paper 2, the probability of failing in Paper 1 is 0.6. The probability of a student failing in both the papers is(A) 0.5 (B) 0.18
(C) 0.12 (D) 0.06
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2006 ONE MARK
Q. 58 The rank of the matrix 111
111
101
-R
T
SSSS
V
X
WWWW is
(A) 0 (B) 1
(C) 2 (D) 3
Q. 59 P4#4# , where P is a vector, is equal to(A) P P P2#4# 4- (B) ( )P P24 4 4#+(C) P P24 4#+ (D) ( )P P24 4$ 4-
Q. 60 ( )P ds4# $# # , where P is a vector, is equal to(A) P dl$# (B) P dl4 4 $# ##(C) P dl4 $## (D) Pdv4$# # #
Q. 61 A probability density function is of the form ( ) , ( , )p x Ke xx 3 3!= -a- . The value of K is(A) 0.5 (B) 1
(C) .0 5a (D) aQ. 62 A solution for the differential equation ( ) ( ) ( )x t x t t2 d+ =o with initial condition
(0 ) 0x =- is(A) ( )e u tt2- (B) ( )e u tt2
(C) ( )e u tt- (D) ( )e u tt
2006 TWO MARKS
Q. 63 The eigenvalue and the corresponding eigenvector of 2 2# matrix are given byEigenvalue Eigenvector
81l = v 11
1 = = G42l = v 1
12 = -= G
The matrix is
(A) 62
26= G (B) 46 64= G
(C) 24
42= G (D) 48 84= G
Q. 64 For the function of a complex variable lnW Z= (where, W u jv= + and Z x jy= +, the u = constant lines get mapped in Z-plane as(A) set of radial straight lines (B) set of concentric circles
(C) set of confocal hyperbolas (D) set of confocal ellipses
Q. 65 The value of the constant integral z 4
1
z j
2
2+- =
# dz is positive sense is(A) j
2p (B)
2p-
(C) j2p- (D)
2p
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Q. 66 The integral sin d30
q qp# is given by(A)
21 (B)
32
(C) 34 (D)
38
Q. 67 Three companies ,X Y and Z supply computers to a university. The percentage of computers supplied by them and the probability of those being defective are tabulated below
Company % of Computer Supplied Probability of being supplied defective
X %60 .0 01
Y %30 .0 02
Z %10 .0 03
Given that a computer is defective, the probability that was supplied by Y is (A) 0.1 (B) 0.2
(C) 0.3 (D) 0.4
Q. 68 For the matrix 42
24= G the eigenvalue corresponding to the eigenvector 101101= G is
(A) 2
(B) 4
(C) 6
(D) 8
Q. 69 For the differential equation dxd y k y 02
22+ = the boundary conditions are
(i) y 0= for x 0= and (ii) y 0= for x a=The form of non-zero solutions of y (where m varies over all integers) are
(A) siny Aa
m xm
m
p=/ (B) cosy A am x
mm
p=/(C) y A xm a
m
m
= p/ (D) y A em am xm
= p-/
Q. 70 As x increased from 3- to 3, the function ( )f xe
e1 x
x= +(A) monotonically increases
(B) monotonically decreases
(C) increases to a maximum value and then decreases
(D) decreases to a minimum value and then increases
2005 ONE MARK
Q. 71 The following differential equation has
dtd y
dtdy y3 4 22
2 3 2+ + +c cm m x=(A) degree 2= , order 1= (B) degree 1= , order 2=(C) degree 4= , order 3= (D) degree 2= , order 3=
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Q. 72 A fair dice is rolled twice. The probability that an odd number will follow an even number is(A) 1/2 (B) 1/6
(C) 1/3 (D) 1/4
Q. 73 A solution of the following differential equation is given by dxd y
dxdy y5 6 02
2
- + =(A) y e ex x2 3= + - (B) y e ex x2 3= +(C) 3y e x x2 3= +- (D) y e ex x2 3= +- -
2005 TWO MARKS
Q. 74 In what range should ( )Re s remain so that the Laplace transform of the function e( )a t2 5+ + exits.(A) ( ) 2Re s a> + (B) ( )Re s a 7> +(C) ( )Re s 2< (D) ( )Re s a 5> +
Q. 75 The derivative of the symmetric function drawn in given figure will look like
Q. 76 Match the following and choose the correct combination:Group I Group 2E. Newton-Raphson method 1. Solving nonlinear equationsF. Runge-kutta method 2. Solving linear simultaneous equationsG. Simpsons Rule 3. Solving ordinary differential equationsH. Gauss elimination 4. Numerical integration 5. Interpolation 6. Calculation of Eigenvalues(A) E 6, F 1, G 5, H 3- - - - (B) E 1, F 6, G 4, H 3- - - -(C) E 1, F 3, G 4, H 2- - - - (D) E 5, F 3, G 4, H 1- - - -
Q. 77 Given the matrix 44
23
-= G, the eigenvector is(A)
32= G (B) 43= G
(C) 21-= G (D) 12-= G
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Q. 78 Let, .
A20
0 13=
-= G and A ab01 21
=- = G. Then ( )a b+ =(A) 7/20 (B) 3/20
(C) 19/60 (D) 11/20
Q. 79 The value of the integral expI x dx21
8
2
0p= -3 c m# is
(A) 1 (B) p(C) 2 (D) 2p
Q. 80 Given an orthogonal matrix
A
1110
1110
1101
1101
= -- -
R
T
SSSSS
V
X
WWWWW
AAT 1-6 @ is
(A) 000
0
00
00
0
000
41
41
21
21
R
T
SSSSSS
V
X
WWWWWW
(B) 000
0
00
00
0
000
21
21
21
21
R
T
SSSSSS
V
X
WWWWWW
(C)
1000
0100
0010
0001
R
T
SSSSS
V
X
WWWWW
(D) 000
0
00
00
0
000
41
41
41
41
R
T
SSSSSS
V
X
WWWWWW
***********
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SOLUTIONS
Sol. 1 Option (B) is correct.Here, as we know sinLim
0q
"q 0.
but for %10 error, we can check option (B) first,
q 18 .18180
0 314#c c cp= = =
sinq .sin18 0 309c= = % error .
. . % . %0 3090 314 0 309 100 0 49#= - =
Now, we check it for 50cq = q .50 50
1800 873#c c c
p= = = sinq .sin50 0 77c= = % error .
0.77 0.873 . %0 873 12 25=- =-
so, the error is more than %10 . Hence, for error less than 10%, 18cq = can have the approximation sinq . q
Sol. 2 Option (A) is correct.For, a given matrix A6 @ the eigen value is calculated as A Il- 0=where l gives the eigen values of matrix. Here, the minimum eigen value among the given options is l 0=We check the characteristic equation of matrix for this eigen value A Il- A= (for 0l = )
352
5127
275
=
3 5 260 49 25 14 35 24= - - - + -^ ^ ^h h h 33 55 22= - + 0=Hence, it satisfied the characteristic equation and so, the minimum eigen value is l 0=
Sol. 3 Option (D) is correct.Given, the polynomial
f x^ h a x a x a x a x a4 4 3 3 2 2 1 0= + + + -Since, all the coefficients are positive so, the roots of equation is given by f x^ h 0=It will have at least one pole in right hand plane as there will be least one sign change from a1^ h to a0^ h in the Routh matrix 1st column. Also, there will be a
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corresponding pole in left hand planei.e.; at least one positive root (in R.H.P)and at least one negative root (in L.H.P)Rest of the roots will be either on imaginary axis or in L.H.P
Sol. 4 Option (B) is correct.Consider the given matrix be
I ABm+ 2111
1211
1121
1112
=
R
T
SSSSSS
V
X
WWWWWW
where m 4= so, we obtain
AB
2111
1211
1121
1112
1000
0100
0010
0001
= -
R
T
SSSSSS
R
T
SSSSSS
V
X
WWWWWW
V
X
WWWWWW
1111
1111
1111
1111
=
R
T
SSSSSS
V
X
WWWWWW
1111
=
R
T
SSSSSS
V
X
WWWWWW
1 1 1 16 @
Hence, we get
A
1111
=
R
T
SSSSSS
V
X
WWWWWW
, B 1 1 1 1= 8 B
Therefore, BA 1 1 1 1= 8 B 1111
R
T
SSSSSS
V
X
WWWWWW
4=From the given property Det I ABm+^ h Det I BAm= +^ h
& Det
2111
1211
1121
1112
R
T
SSSSSS
V
X
WWWWWW
Det
1000
0100
0010
0001
4= +
R
T
SSSSSS
V
X
WWWWWW
Z
[
\
]]
]]
_
`
a
bb
bb
1 4= + 5=Note : Determinant of identity matrix is always 1.
Sol. 5 Option (D) is correct.
t dtdx x+ t=
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dtdx
tx+ 1=
dtdx Px+ Q= (General form)
Integrating factor, IF e e e tlntPdt t dt1= = = =# #
Solution has the form,
x IF# Q IF dt C#= +^ h# x t# ( )( )t dt C1= +# xt t C2
2= +Taking the initial condition,
( )x 1 .0 5= 0.5 C2
1= + & C 0=
So, xt t22= x t2& =
Sol. 6 Option (C) is correct.
( )f z z z11
32= + - +
( )j f z dz21
Cp # = sum of the residues of the poles which lie inside the given closed region.
C z 1 1& + =Only pole z 1=- inside the circle, so residue at z 1=- is. ( )f z
( )( )z zz1 3
1= + +- +
( )( )( )( )
limz zz z
1 31 1
22 1
z 1= + +
+ - + = ="-
So ( )j f z dz21
Cp # 1=Sol. 7 Option (A) is correct.
x i1= - = cos sini2 2p p= +
So, x ei 2= p
xx eix
2= p^ h & ei i2p^ h e 2= p-Sol. 8 Option (D) is correct.
( ) ( )
( )dt
d y tdtdy t
y t2
2
2
+ + ( )td=By taking Laplace transform with initial conditions
( ) ( ) [ ( ) ( )] ( )s Y s sy dtdy sy s y Y s0 2 0
t
2
0- - + - +
=; E 1=
& ( ) 2 ( ) ( )s Y s s sY s Y s2 0 22 + - + + +6 6@ @ 1= ( ) [ ]Y s s s2 12+ + s1 2 4= - - ( )Y s
s ss2 1
2 32= + +- -
We know that, If, ( )y t ( )Y sL
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then, ( )dt
dy t ( ) ( )sY s y 0L -
So, ( ) ( )sY s y 0- ( )( )s s
s s2 1
2 322= + +
- - +
( )s s
s s s s2 1
2 3 2 4 22
2 2= + +- - + + +
( ) ( )sY s y 0- ( ) ( ) ( )ss
ss
s12
11
11
2 2 2= ++ = +
+ + +
( )s s11
11
2= + + +Taking inverse Laplace transform
( )dt
dy t ( ) ( )e u t te u tt t= +- -
At t 0= +, dtdy
t 0= + e 0 10= + =
Sol. 9 Option (A) is correct.Divergence of A in spherical coordinates is given as
A:d ( )r r
r A1 r22
22= ( )
r rkr1 n2
2
22= +
( )rk n r2 n2
1= + +
( )k n r2 0n 1= + =- (given) n 2+ 0= & n 2=-
Sol. 10 Option (C) is correct.Probability of appearing a head is /1 2. If the number of required tosses is odd, we have following sequence of events.
,H ,TTH , ...........TTTTH
Probability P .....21
21
213 5= + + +b bl l
P 1 3
241
21
= - =
Sol. 11 Option (B) is correct.
( )f x x x x9 24 53 2= - + +
( )dxdf x
x x3 18 24 02= - + =
& ( )dxdf x
6 8 0x x2= - + = x 4= , x 2=
( )
dxd f x
2
2
x6 18= -
For ,x 2= ( )dx
d f x12 18 6 0
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Sol. 12 Option (B) is correct.Characteristic equation.
A Il- 0=
52
3ll
- - -- 0=
5 62l l+ + 0= 5 62l l+ + 0=Since characteristic equation satisfies its own matrix, so
5 6A A2+ + 0= 5 6A A I2& =- -Multiplying with A 5 6A A A3 2+ + 0= 5( 5 6 ) 6A A I A3+ - - + 0= A3 19 30A I= +
Sol. 13 Option (D) is correct.
From Divergence theorem, we have
Adv4$v v### A n dss$= v t#
The position vector
rv u x u y u zx y z= + +t t t^ hHere, 5A r=v v, thus A4$ v u
xu
yu
zu x u y u zx y z x y z:2
222
22= + + + +t t t t t tc ^m h
5dxdx
dydy
dzdz= + +c m 3 5#= 15=
So, 5r n dss
$v t## 15dv V15= =###
Sol. 14 Option (C) is correct.
We have dxdy ky=
Integrating ydy# k dx A= +#
or lny kx A= +Since ( )y 0 c= thus lnc A=So, we get, lny lnkx c= +or lny ln lne ckx= +or y cekx=
Sol. 15 Option (A) is correct.
C R Integrals is z z
z dz4 5
3 4
C2+ +- +# where C is circle z 1=
( )f z dzC# 0= if poles are outside C.
Now z z4 52+ + 0= ( )z 2 12+ + 0=Thus z ,1 2 j z2 1>,1 2&!=-So poles are outside the unit circle.
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Sol. 16 Option (C) is correct.
We have ( )f x x x 3 0= + - = ( )f xl
x1
21= +
Substituting x0 2= we get ( )f x0l .1 35355= and ( ) .f x 2 2 3 0 4140 = + - =Newton Raphson Method
x1 ( )( )
xf xf x
00
0= -l
Substituting all values we have
x 1 ..2 1 3535
0 414= - .1 694=Sol. 17 Option (B) is correct.
Writing :A B we have
:::
111
144
16
620
l m
R
T
SSSS
V
X
WWWW
Apply R R R3 3 2" -
::: 20
110
140
16
6
620
l m- -
R
T
SSSS
V
X
WWWW
For equation to have solution, rank of A and :A B must be same. Thus for no solution; 6, 20!ml =
Sol. 18 Option (C) is correct.Total outcome are 36 out of which favorable outcomes are :(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6);(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) which are 15.
Thus ( )P E ..No of total outcomes
No of favourable outcomes3615
125= = =
Sol. 19 Option (C) is correct.Eigen value of a Skew-symmetric matrix are either zero or pure imaginary in conjugate pairs.
Sol. 20 Option (C) is correct.For a function ( )x t trigonometric fourier series is
( )x t [ ]cos sinA A n t B n to n nn 1
w w= + +3=/
Where, Ao ( )T x t dt1
T0 0
= # T0" fundamental period
An ( )cosT x t n t dt2
T0 0
w= #
Bn ( )sinT x t n t dt2
T0 0
w= #For an even function ( ),x t B 0n =Since given function is even function so coefficient B 0n = , only cosine and constant
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terms are present in its fourier series representation.Constant term :
A0 ( )T x t dt1
/
/
T
T
4
3 4= -#
T Adt Adt1 2
/
/
/
/
T
T
T
T
4
4
4
3 4= + --: D# # T
TA AT1 2 2 2= -: D A2=-Constant term is negative.
Sol. 21 Option (D) is correct.
Given differential equation
( ) ( )
dxd n x
Ln x
2
2
2- 0=Let ( )n x Ae x= lSo, A e
LAex x2
2l -ll
0=
L12
2l - L01& !l= =
Boundary condition, ( )n 03 = so take L1l =-
( )n x Ae Lx= -
( )n 0 Ae K A K0 &= = =So, ( )n x Ke ( / )x L= -
Sol. 22 Option (A) is correct.
Given that ey x x1=
or lney lnx x1=
or y lnx x1=
Now dxdy lnx x x x
1 1x1
2= + - -^ h lnx x1
2 2= -For maxima and minima :
dxdy (1 ) 0ln
xx12= - =
lnx 1= " x e 1=Now
dxd y
2
2
lnx
xx x x
2 2 1 13 3 2=- - - -b bl l
lnx x
xx
2 2 12 3 3=- + -
dyd x
at x e2
2
1=
e e e2 2 1 0
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Sol. 24 Option (C) is correct.
Av xya x ax y2= +t t dlv dxa dyax y= +t t A dl
C
:v v# ( ) ( )xya x a dxa dyax yC
x y2 := + +t t t t#
( )xydx x dyC
2= +#
xdx xdx dy dy3 34
31
/
/
/
/
2 3
1 3
1
3
3
1
1 3
2 3= + + +# # ##
[ ] [ ]21
34
31
23
31
34
34 3 1 3
1 1 3= - + - + - + -: :D D 1=
Sol. 25 Option (C) is correct.Given function
( )X z ( )( )z z z
z1 2
1 2= - --
Poles are located at 0, 1, 2andz z z= = =At Z 0= residues is R0 ( )z X z
Z 0:= = ( )( )0 1 0 2
1 2 0#= - -- 2
1=at z 1= , R1 ( 1) ( )Z X Z
Z 1:= - =
( )1 1 2
1 2 1 1#= -- =
At z 2= , R2 ( ) ( )z X z2z 2
:= - =
( )2 2 11 2 2
23#= -
- =-Sol. 26 Option (B) is correct.
Taking step size h 0.1= , ( )y 0 0=x y
dxdy x y= + y y hdx
dyi i1 = ++
0 0 0 . ( )y 0 0 1 0 01 = + =0.1 0 0.1 . ( . ) .y 0 0 1 0 1 0 012 = + =0.2 0.01 0.21 . . . .y 0 01 0 21 0 1 0 0313 #= + =0.3 0.031
From table, at . , ( . ) .x y x0 3 0 3 0 031= = =Sol. 27 Option (D) is correct.
Given that
( )f t ( )s s K s
s4 3
3 1L 1 3 2= + + -+- ; E
( )lim f tt"3
1=By final value theorem
( )lim f tt"3
( )lim sF s 1s 0
= ="
or ( )
( )lim
s s K ss s4 3
3 1s 0 3 2
:
+ + -+
" 1=
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or [ ( )]
( )lim
s s s Ks s4 33 1
s 0 2+ + -+
" 1=
K 31- 1=
or K 4=Sol. 28 Option (B) is correct.
The highest derivative terms present in DE is of 2nd order.
Sol. 29 Option (C) is correct.Number of elements in sample space is 210. Only one element
, , , , , , , , ,H H T T T T T T T T" , is event. Thus probability is 2110
Sol. 30 Option (C) is correct.We have
( )f z c c z0 1 1= + -
( )f z1 ( )
zf z
zc c z1 1 0 1 1= + = + + - ( )
zz c c1
20 1= + +
Since ( )f z1 has double pole at z 0= , the residue at z 0= is Res ( )f z z1 0= . ( )lim z f z
z 0
21=
" .
( )lim z
zz c c1
z 0
220 1= + +
" c m c1=Hence
( )f z dz1unit circle
# [ ( )]zf z
dz1
unit circle
= +# j2p= [Residue at z 0= ]
jc2 1p=Sol. 31 Option (D) is correct.
We have ( )f x sinx
xp= -
Substituting x p- y= ,we get ( )f y p+ ( )sin sin
yy
yyp= + =- ( )sin
yy1= -
! !
...yy y y1
3 5
3 5
= - - + -c mor ( )f y p+
! !...y y1
3 5
2 4
=- + - +Substituting x yp- = we get ( )f x
!( )
!( )
...x x
13 5
2 4p p=- + - - - +Sol. 32 Option (A) is correct.
(A) dxdy
xy=
or ydy# x
dx= #or logy log logx c= +or y cx= Straight LineThus option (A) and (C) may be correct.
(B) dxdy
xy=-
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or ydy# x
dx=- #or logy log logx c=- +or logy log log
xc1= +
or y xc= Hyperbola
Sol. 33 Option (D) is correct.Sum of the principal diagonal element of matrix is equal to the sum of Eigen values. Sum of the diagonal element is 1 1 3 1- - + = .In only option (D), the sum of Eigen values is 1.
Sol. 34 Option (C) is correct.The product of Eigen value is equal to the determinant of the matrix. Since one of the Eigen value is zero, the product of Eigen value is zero, thus determinant of the matrix is zero.
Thus p p p p11 22 12 21- 0=Sol. 35 Option (B) is correct.
The given system is
xy
42
21= =G G 76= = G
We have A 42
21= = G
and A 42
21 0= = Rank of matrix ( )A 2 , e 1>x and 0 1e<
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Sol. 40 Option (C) is correct.
We have x e x= -or ( )f x x e x= - - '( )f x e1 x= + -The Newton-Raphson iterative formula is
xn 1+ '( )( )
xf xf x
nn
n= -Now ( )f xn x en xn= - - '( )f xn e1 xn= + -
Thus xn 1+ xe
x e1
n xn
x
n
n= - +-
-
-
( )ex e
11
xn
x
n
n= ++
-
-
Sol. 41 Option (A) is correct.
Res ( )f z z a= ( )!( ) ( )
n dzd z a f z
11
n
nn
z a1
1= - ---
=6 @Here we have n 2= and a 2=Thus Res ( )f z z 2= ( )!
( )( ) ( )dz
d zz z2 1
1 22 2
1
z a
22 2= - - - + =; E
( )dz
dz 2
1
z a2= + =; E ( )z 2
2
z a3= +
-=
; E 64
2=- 321=-
Sol. 42 Option (D) is correct.
eP ( )L sI A1 1= -- -6 @ L
ss00 0
213
11
= - - --
-e o= =G G L
ss2
13
11
= -+-
-e o= G L ( )( )
( )( )
( )( )
( )( )
s ss
s s
s s
s ss
1 1 23
1 22
1 21
1 2= - + +
+
+ +-
+ +
+ +f p> H
e ee e
e ee e
22 2 2
1 2
1 2
1 2
1 2= -- +-
- +- -
- -
- -
- -= GSol. 43 Option (B) is correct.
Taylor series is given as
( )f x ( )!
'( )!
( )"( ) ...f a x a f a
x af a
1 2
2
= + - + - +For x p= we haveThus ( )f x ( )
!'( )
!( )
"( )...f x fx
f x1 2
2
p p p p= + - + -
Now ( )f x sine xx= + '( )f x cose xx= + "( )f x sine xx= - "( )f p sine ep= - =p p
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Thus the coefficient of ( )x 2p- is !
"( )f2p
Sol. 44 Option (A) is correct.The equation of straight line from ( , )0 0 to ( , )1 2 is y x2= .Now ( , )g x y x y4 103 4= +or, ( , )g x x2 x x4 1603 4= +Now ( , )g x x2
0
1# ( )x x dx4 1603 40
1= +# [ ]x x32 334 5 01= + =
Sol. 45 Option (B) is correct.
I ( )xdx ydy2P
Q= +# xdx ydy2 2
P
Q
P
Q= + ## xdx ydy2 2 0
0
1
1
0= + =##Sol. 46 Option (B) is correct.
The given plot is straight line whose equation is
x y1 1- + 1=
or y x 1= +Now I ydx
1
2= # ( )x dx11
2= +#
( )x21 2 2= +; E .29 24 2 5= - =
Sol. 47 Option (C) is correct.
cothx sinhcosh
xx=
as 1, 1coshx x
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( )x e1 2 2= - - -6 @ Neglecting higher powers ( )x e3 2= - -
Sol. 51 Option (D) is correct.
We have kdxd y2
2
2
y y2= -
or dxd y
ky
2
2
2- ky
22=-
A.E. Dk122- 0=
or D k1!=
C.F. C e C e1 2kx
kx= +-
P.I. D k
y y1
k2 1 2
22
22
= -- =c m
Thus solution is
y C e C e y1 2 2kx
kx= + +-
From ( )y y0 1= we get C C1 2+ y y1 2= -From ( )y y23 = we get that C1 must be zero.Thus C2 y y1 2= - y ( )y y e y1 2 2k
x= - +-Sol. 52 Option (B) is correct.
We have
( )f x x x x4 43 2= - + - '( )f x x x3 2 42= - +Taking x 20 = in Newton-Raphosn method x1 '( )
( )x
f xf x
00
0= - ( ) ( )
( )2
3 2 2 2 42 2 4 2 4
2
3 2
= - - +- + -
34=
Sol. 53 Option (C) is correct.For two orthogonal signal ( )f x and ( )g x
( ) ( )f x g x dx3
3
-
+# 0=i.e. common area between ( )f x and ( )g x is zero.
Sol. 54 Option (A) is correct.We know that
s
ds1
1
D2-# j2p= [sum of residues]
Singular points are at s 1!= but only s 1=+ lies inside the given contour, Thus Residue at s 1=+ is ( ) ( )lim s f s1
s 1-
" ( )lim s
s1
11
21
s 1 2= - - ="
s
ds1
1
D2-# j j2 2
1p p= =` j
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Sol. 55 Option (C) is correct.For two orthogonal vectors, we require two dimensions to define them and similarly for three orthogonal vector we require three dimensions to define them. M2 vectors are basically M orthogonal vector and we require M dimensions to
define them.
Sol. 56 Option (A) is correct.We have
( )f x x x 22= - + '( )f x x2 1 0= - = x
21" =
"( )f x 2=Since "( )f x 2 0>= , thus x
21= is minimum point. The maximum value in
closed interval ,4 4-6 @ will be at x 4=- or x 4=Now maximum value
[ ( 4), (4)]max f f= - ( , )max 18 10= 18=
Sol. 57 Option (C) is correct.Probability of failing in paper 1 is ( ) .P A 0 3=Possibility of failing in Paper 2 is ( ) .P B 0 2=Probability of failing in paper 1, when student has failed in paper 2 is 0.6P B
A =^ hWe know that
P BAb l ( )
( )P BP B+=
or ( )P A B+ ( )P B P BA= b l . . .0 6 0 2 0 12#= =
Sol. 58 Option (C) is correct.We have
A 111
111
101
110
110
100
+= - -R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW R R3 1-
Since one full row is zero, ( )A 3
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We know ( )p x dx3
3
-# 1=
Thus Ke dxx3
3 a--# 1=
or Ke dx Ke dxx x0
0 + 33
a a--
## 1=or
( )K e k ex x0
0a a+ -33a a
--6 6@ @ 1=
or K Ka a+ 1=or K
2a=
Sol. 62 Option (A) is correct.
We have ( ) ( )x t x t2+o ( )s t=Taking Laplace transform both sides
( ) ( ) ( )sX s x X s0 2- + 1=or ( ) ( )sX s X s2+ 1= Since ( )x 0 0=- ( )X s
s 21= +
Now taking inverse Laplace transform we have
( )x t ( )e u tt2= -Sol. 63 Option (A) is correct.
Sum of the Eigen values must be equal to the sum of element of principal diagonal of matrix.
Only matrix 62
26= G satisfy this condition.
Sol. 64 Option (B) is correct.
We have W ln z= u jv+ ( )ln x jy= +or eu jv+ x jy= +or e eu jv x jy= + ( )cos sine v j vu + x jy= +Now cosx e vu= and siny e vu=Thus x y2 2+ e u2= Equation of circle
Sol. 65 Option (D) is correct.We have
z
dz4
1
z j2
2+
- =
# ( )( )z i z i dz2 21
z j 2
= + -- =
#
( , )P 0 2 lies inside the circle z j 2- = and ( , )P 0 2- does not lie.Thus By cauchys integral formula
I ( )( )( )
limi z iz i z i
2 22 2
1z i2
p= - + -" i ii
2 22
2C
p p= + =#Sol. 66 Option (C) is correct.
I sin d30
q q= p#
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sin sin d4
3 30
q q q= -p ` j# sin sin sin3 3 4 3q q q= - cos s4
3123
0 0q w q= - =p p: :D D 43 43 121 121 34= + - + =8 8B B
Sol. 67 Option (D) is correct.Let d " defective and y " supply by Y
p dya k ( )
( )P d
P y d+= ( )P y d+ . . .0 3 0 02 0 006#= = ( )P d . . . . . . .0 6 0 1 0 3 0 02 0 1 0 03 0 015# # #= + + = P d
ya k .. .
0 0150 006 0 4= =
Sol. 68 Option (C) is correct.
We have A 42
24= = G
Now [ ]A I Xl-6 @ 0=or
42
24
101101
ll
--= =G G 00= = G
or ( )( ) ( )101 4 2 101l- + 0=or l 6=
Sol. 69 Option (A) is correct.
We have dxd y k y2
22+ 0=
or D y k y2 2+ 0=The AE is m k2 2+ 0=The solution of AE is m ik!=Thus sin cosy A kx B kx= +From x 0= , y 0= we get B 0= and ,x a y 0= = we get sinA ka 0=or sinka 0= k
am xp=
Thus y sinAa
m xm
m
p= ` j/Sol. 70 Option (A) is correct.
We have ( )f x e
e1 x
x= +For x " 3, the value of ( )f x monotonically increases.
Sol. 71 Option (B) is correct.Order is the highest derivative term present in the equation and degree is the power of highest derivative term.Order 2= , degree 1=
Sol. 72 Option (D) is correct.Probability of coming odd number is 2
1 and the probability of coming even number is 2
1 . Both the events are independent to each other, thus probability of coming
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odd number after an even number is 21
21
41# = .
Sol. 73 Option (B) is correct.
We have dxd y
dxdy y5 62
2
- + 0=The . .A E is m m5 62- + 0= m ,3 2=The CF is yc C e C ex x1 3 2 2= +Since Q 0= , thus y C e C ex x1 3 2 2= +Thus only (B) may be correct.
Sol. 74 Option (A) is correct.
We have ( )f t e( )a t2 5= + + .e e( )a t5 2= +Taking Laplace transform we get
( )F s ( )
es a 2
15= - +; E Thus ( ) ( )Re s a 2> +Sol. 75 Option (C) is correct.
For x 0> the slope of given curve is negative. Only (C) satisfy this condition.Sol. 76 Option (C) is correct.
Newton - Raphson " Method-Solving nonlinear eq.
Runge - kutta Method " Solving ordinary differential eq.
Simpsons Rule " Numerical Integration
Gauss elimination " Solving linear simultaneous eq.
Sol. 77 Option (C) is correct.
We have A 44
23=
-= GCharacteristic equation is
A Il- 0=
or 4
42
3l
l-
- 0=or ( 4 )(3 ) 8l l- - - - 0=or 12 82l l- + + - 0=or 202l l+ - 0=or l ,5 4=- Eigen valuesEigen vector for 5l =- ( )A I Xil- 0=
( ) xx
1 54
28 4
1
2
- --= =G G 00= = G
xx
10
20
1
2= =G G 00= = G R R42 1-
x x21 2+ 0=Let x x2 11 2&- = =- ,Thus X
21= -= G Eigen vector
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Sol. 78 Option (A) is correct.We have
A .2
00 13=
-= G and A ab01 21
=- = GNow AA 1- I=
or . a
b20
0 13 0
21-= =G G 10 01= = G
or .a b
b10
2 0 13-= G 10 01= = G
or .a2 0 1- 0= and b3 1=Thus solving above we have b
31= and a
601=
Therefore a b+ 31
601
207= + =
Sol. 79 Option (A) is correct.Gaussian PDF is
( )f x e dx21 ( )x
2 2
2
p s= 33 -
-sm-# for x3 3# #-
and ( )f x dx3
3
-# 1=
Substituting 0m = and 2s = in above we get e dx
2 21 x
8
2
p 33 -
-# 1=
or 2 e dx2 21
0
x8
2
p3 -# 1=
or e dx21
0
x8
2
p3 -# 1=
Sol. 80 Option (C) is correct.From orthogonal matrix
[ ]AAT I=Since the inverse of I is I , thus
[ ]AAT 1- I I1= =-
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