1

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GATE SOLVED PAPER
Electronics & Communication
Engineering Mathematics
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GATE SOLVED PAPER - EC
ENGINEERING MATHEMATICS
2013 ONE MARK
Q. 1 The maximum value of q until which the approximation sinq . 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 is
R
3
5
2
5
12
7
2
7
5
SSSS
WWWW
© www.nodia.co.in
T
V
X
(A) 0 (B) 1
(C) 2 (D) 3
4
Q. 3 A polynomial f (x) a4x a 3
x a x a x a
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
^Im + ABh = determinant ^In + BAh, where Ik is the k#k identity matrix. Using
the above property, the determinant of the matrix given below is
2
1
1
1
1
2
1
1
1
1
2
1
1
1
1
2
R
SSSSS
S
T
V
WWWWW
W
X
(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 dx dt
+ x = t , is
(A) x t 2
= - 1 (B) x = t 2- 1
2
= (D) x t2
(C) x t
2
2
=

GATE SOLVED PAPER - EC ENGINEERING MATHEMATICS
= 1
2 + - + .
Q. 6 Given f (z) z 1 z
3
If C is a counter clockwise path in the z -plane such that z + 1 = 1, the value of
1
2 j f (z)dz
C p # is
(A) -2 (B) -1
(C) 1 (D) 2
Q. 7 If x = -1, then the value of xx is
(A) e-p/2 (B) ep/2
(C) x (D) 1
2012 TWO MARKS
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Q. 8 Consider the differential equation
d y t
dy t
2 y t 2
( ) ( )
( )
dt
dt
2
dy
+ + = d(t) with y(t) =- 2 and dt =
0
t
t
0
0
=
=
-
-
dy
The numerical value of dt
t=0+
is
(A) -2 (B) -1
(C) 0 (D) 1
Q. 9 The direction of vector A is radially outward from the origin, with A = krn .
where r2 = x2+ y2+ z2 and k is a constant. The value of n for which d:A = 0 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) = x3- 9x2+ 24x + 5 in the interval [1,6] is
(A) 21 (B) 25
(C) 41 (D) 46
Q. 12 Given that
- -
> H = > H, the value of A3 is
5
2
3
0
A and I
1
0
0
= 1
(A) 15A+ 12I (B) 19A+ 30I
(C) 17A+ 15I (D) 17A+ 21I
2011 ONE MARK
Q. 13 Consider a closed surface S surrounding volume V . If rv is the position vector of a
## v $ t
point inside S , with nt the unit normal on S , the value of the integral 5r n dS
S
is
(A) 3V (B) 5V
(C) 10V (D) 15V

dy = ky y = c is
Q. 14 The solution of the differential equation dx , (0)
(A) x = ce-ky (B) x = kecy
(C) y = cekx (D) y = ce-kx
Q. 15 The value of the integral
# - + where c is the circle z = 1 is given
z dz
4 5
3 4
2+ +
(z z )
by
c
(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
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Q. 17 The system of equations
+ + =
+ + =
+ + =
x y z
x y y
x y z
6
4 6 20
4
l m
has NO solution for values of l and μ given by
(A) l = 6, m = 20 (B) l = 6, m =Y 20
(C) l =Y 6, m = 20 (D) l =Y 6, m = 20
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

(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
2
d n x
n x
( ) ( )
dx
2 - =
0 2
L
where L is a constant. The boundary conditions are :n(0) = K and n(3) = 0.
The solution to this equation is
(A) n(x) = Kexp(x/L) (B) n(x) = Kexp(- x/ L)
(C) n(x) = K2exp(- x/L) (D) n(x) = Kexp(- x/L)
2010 TWO MARKS
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Q. 22 If ey = x1/x , 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
v = t + 2 t , then A dl
Q. 24 If A xyax x ay
# v $ v over the path shown in the figure is
C
(A) 0 (B)
2
3
(C) 1 (D) 2 3
Q. 25 The residues of a complex function
( )
-
( )( )
x z
z z z z
1 2
= 1 2 - -
at its poles are
(A) 1
2,
- 1 2
and 1 (B) 1
2
, - 1 and -1
2
1 and 2
(C) 2, 1
-3 (D) , 2
1 - 1 and 2
3
Q. 26 Consider differential equation
dy x
( )
dx ( )
- y x = x , with the initial
condition y(0) = 0. Using Euler’s 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

Q. 27 Given ( )
- 1 3 s
+ 1
; E. If limf (t) 1
3 2 =
+ + -
s 4 s k 3
s
( )
f t L
t
=
"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
2 3
+ + 4 = - c m is
d y
dt
dy y et
dt
2
(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) 2
10 1
c 1 m 2 (B) C 2
2
2 b l
1 10 c m (D) C 2
(C) 2
10 1
2
co.in
nodia.www.© 10 b l
= + -1, then
Q. 30 If f (z) c0 c1z
# + f z
is given by
( )
z
dz
1
unit circle
(A) 2pc1 (B) 2p(1 + c0)
(C) 2pjc1 (D) 2p(1 + c0)
2009 TWO MARKS
Q. 31 The Taylor series expansion of sin
x
x
- p at x = p is given by
(A)
+ ( - p )
2 + (B)
!
...
x
1
3
- - ( - p )
2 +
!
...
x
1
3
(C)
- ( - p )
2 + (D)
!
...
x
1
3
- + ( - p )
2 +
!
...
x
1
3
Q. 32 Match each differential equation in Group I to its family of solution curves from
Group II
Group I Group II
A.
dy
dx
= y 1. Circles
x
B.
dy
dx
=-y 2. Straight lines
x
C.
dy
dx
y x
= 3. Hyperbolas
D.
dy
dx
y x
=-
(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

Q. 33 The Eigen values of following matrix are
R
-
- -
1
3
0
3
1
0
5
6
3
SSSS
T
V
WWWW
X
(A) 3, 3 + 5j, 6 - j (B) -6 + 5j, 3 + j, 3 - j
(C) 3 + j, 3 - j, 5 + j (D) 3, - 1 + 3j, - 1 - 3j
2008 ONE MARKS
Q. 34 All the four entries of the 2 # 2 matrix P
p
p
p
p
11
21
12
22
= = G are nonzero,
and one of its eigenvalue is zero. Which of the following statements is true?
(A) p11p12- p12p21 = 1 (B) p11p22- p12p21 =- 1
(C) p11p22- p12p21 = 0 (D) p11p22+ p12p21 = 0
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Q. 35 The system of linear equations
4x + 2y = 7 2x + y = 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
f (x) = exp(x) + exp(- 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
dx t
( )
( )
dt
+ 3x t = 0 ?
(A) x(t) = 3e-t (B) x(t) = 2e-3t
(C) x(t) =- 3 t 2 (D) x(t) = 3t2
2
2008 TWO MARKS
Q. 40 The recursion relation to solve x = e-x using Newton - Raphson method is
= n +
(A) xn e
x
1
- (B) x x e n n
x
= - n +
1
-
-
x
= +
e
n n n 1 x
(C) x (1 x )
+ + -
1 n
e
(D)
- - -
x e ( 1 x )
1
x
+ -
x e
n
n
x
n
x
n
1
2
n
n
=
-
-

Q. 41 The residue of the function f (z)
1
2 2 =
+ -
(z 2) (z 2)
at z = 2 is
(A)
- 1 (B)
32
- 1
16
(C)
1 (D)
16
1
32
Q. 42 Consider the matrix P
0
2
1
= =- -3G. The value of ep is
(A)
- -
- -
2 1
2 1
-
-
e e
e e
- -
1 2
2 1
-
-
e e
e e
2 3
2 2 5
> - - H
(B)
- -
-
1 1
1 2
+
-
e e
e e
- -
2 1
1 2
-
+
e e
2
3 2
> - - H
2 4 e e
(C)
- -
- -
2 1
2 1
-
-
e e
e e
- -
1 2
2 1
-
+
e e
e
5
2 6
3
4 6
> - - H
(D)
- -
- -
1 -
2
1 2
e e
e e
- -
1 -
2
1 2
e e
e e
2
2 2 2
> - - H
- +
- +
© www.nodia.co.in
Q. 43 In the Taylor series expansion of exp(x) + sin(x) about the point x = p, the
coefficient of (x - p)2 is
(A) exp(p) (B) 0.5 exp(p)
(C) exp(p) + 1 (D) exp(p) - 1
Q. 44 The value of the integral of the function g(x,y) = 4x3+ 10y4 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
Q
# + along the semicircle with the line segment PQ as
line integral 2 (xdx ydy)
P
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 ydx
1
2
= # is

(A) 1.0 (B) 2.5
(C) 4.0 (D) 5.0
Q. 47 For x << 1, coth (x) can be approximated as
(A) x (B) x2
(C)
1 (D)
x
1
2
x
q
b l is
sin 2
0 q
Q. 48 lim
q"
(A) 0.5 (B) 1
(C) 2 (D) not defined
Q. 49 Which one of following functions is strictly bounded?
(A) 1/x2 (B) ex
in
(C) x2 co.(D) e - x2
nodia.www.© Q. 50 For the function e-x , the linear approximation around x = 2 is
(A) (3 - x)e-2
(B) 1 - x
(C) 63 + 3 2 - (1 - 2)x @e-2
(D) e-2
2007 TWO MARKS
Q. 51 The solution of the differential equation k
2 d y y y
dx
2
2
= - 2 under the boundary
conditions
(i) y = y1 at x = 0 and
(ii) y = y2 at x = 3, where k,y1 and y2 are constants, is
x y = 1- 2 a- 2 k+ 2
(A) y (y y )exp
k
x y = 2- 1 a- k+ 1
(B) y (y y )exp k
x y = ^ 1- 2h a k+ 1
(C) y y y sinh k
x y = ^ 1- 2h a- k+ 2
(D) y y y exp k
Q. 52 The equation x3- x2+ 4x - 4 = 0 is to be solved using the Newton - Raphson
method. If x = 2 is taken as the initial approximation of the solution, then next
approximation using this method will be
(A) 2/3 (B) 4/3
(C) 1 (D) 3/2
Q. 53 Three functions f1(t), f2(t) and f3(t) which are zero outside the interval [0, T] are
shown in the figure. Which of the following statements is correct?

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(A) f1(t) and f2(t) are orthogonal (B) f1(t) and f3(t) are orthogonal
(C) f2(t) and f3(t) are orthogonal D) f1(t) and f2(t) 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
# 1
ds
is
(s 2 - 1
)
D
(A) jp (B) -jp
(C) -p (D) p
Q. 55 It is given that , ... X X XM
1 2 at M non-zero, orthogonal vectors. The dimension of
the vector space spanned by the 2M vectors X1,X2,...XM, - X1, - X2,... - XM is
(A) 2M (B) M+ 1
(C) M
(D) dependent on the choice of , ,... X X XM
1 2
Q. 56 Consider the function f (x) = x2- x - 2. The maximum value of f (x) in the closed
interval [- 4, 4] is
(A) 18 (B) 10
(C) -225 (D) indeterminate
Q. 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

2006 ONE MARK
Q. 58 The rank of the matrix
R
1
1
1
1
1
1
1
0
1
-
SSSS
T
V
WWWW
X
is
(A) 0 (B) 1
(C) 2 (D) 3
Q. 59 4#4# P, where P is a vector, is equal to
(A) P P 2P #4# -4 (B) 42P +4 (4# P)
(C) 42P +4# P (D) 4(4$ P) -42P
Q. 60 ## (4# P) $ ds , where P is a vector, is equal to
(A) # P $ dl (B) # 4#4# P $ dl
(C) # 4# P $ dl (D) ### 4$ Pdv
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Q. 61 A probability density function is of the form p(x) = Ke-a x ,x ! (-3,3). The
value of K is
(A) 0.5 (B) 1
(C) 0.5a (D) a
Q. 62 A solution for the differential equation xo (t) + 2x(t) = d(t) with initial condition
x(0-) = 0 is
(A) e-2tu(t) (B) e2tu(t)
(C) e-tu(t) (D) etu(t)
2006 TWO MARKS
Q. 63 The eigenvalue and the corresponding eigenvector of 2 # 2 matrix are given by
Eigenvalue Eigenvector
l1 = 8 v
1
1
1 == G
l2 = 4 v
1
1
2 =
= G
-
The matrix is
(A)
6
2
2
= 6G (B)
4
6
6
= 4G
(C)
2
4
4
= 2G (D)
4
8
8
= 4G
Q. 64 For the function of a complex variable W = lnZ (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
1
# dz is positive sense is
z 4
z j
2
2
+
- =
(A) j
p (B)
2
2 p
-
- p (D)
(C) j
2
2 p

q q p # is given by
Q. 66 The integral sin3 d
0
(A)
1 (B)
2
2
3
(C)
3 4
(D)
8
3
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
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Q. 68 For the matrix
4
2
2
= 4G the eigenvalue corresponding to the eigenvector
101
=101G is
(A) 2
(B) 4
(C) 6
(D) 8
Q. 69 For the differential equation
2
d y k y 0 2
dx
+ 2 = 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) y A sin
=/ m p x
(B) y A cos
a
m
m
=/ p
m x
a
m
m
= / m
p (D) y Ame a
(C) y Amx a
m
= / - p
m x
m
Q. 70 As x increased from -3 to 3, the function f (x)
e
1 e
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
2 3 2 c m+ c m + + = x
3 d y
4 dy y 2 2
dt
dt
(A) degree = 2, order = 1 (B) degree = 1, order = 2
(C) degree = 4, order = 3 (D) degree = 2, order = 3

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
© www.nodia.co.in
d y
dx
5 dy 6y 0 2
dx
2
- + =
(A) y = e2x + e-3x (B) y = e2x + e3x
(C) y = e-2x + 33x (D) y = e-2x + e-3x
2005 TWO MARKS
Q. 74 In what range should Re(s) remain so that the Laplace transform of the function
e(a+2)t+5 exits.
(A) Re(s) > a + 2 (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 2
E. Newton-Raphson method 1. Solving nonlinear equations
F. Runge-kutta method 2. Solving linear simultaneous
equations
G. Simpson’s Rule 3. Solving ordinary differential
equations
H. 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
-
= G, the eigenvector is
4
4
2
3
(A)
3
=2G (B)
4
=3G
(C)
2
=-1G (D)
-
= G
1
2

Q. 78 Let,
0 1
.
A
2
0
-
= G and A
= 3
1 2 1
a
0 b
- = = G. Then (a + b) =
(A) 7/20 (B) 3/20
(C) 19/60 (D) 11/20
2
# c m is
Q. 79 The value of the integral I = 1
3
exp - x dx
2
8
p 0
(A) 1 (B) p
(C) 2 (D) 2p
Q. 80 Given an orthogonal matrix
A
R
1
1
1
0
1
1
1
0
1
1
0
1
1
1
0
1
SSSSS
= -
- -
T
V
WWWWW
X
6AAT @-1 is
1
SSSSSS
WWWWWW
SSSSSS
WWWWWW
0
(A)
0
in
0
co.nodia.www.© 0
1
0
0
0
0
0
0
0
0
4
4
2 1
2 1
R
T
V
X
(B)
0
0
0
0
0
0
0
0
0
0
0
0
2 1
2 1
2 1
2 1
R
T
V
X
(C)
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
R
SSSSS
T
V
WWWWW
X
(D)
1
SSSSSS
0
0
0
0
1
0
0
0
0
1
0
0
0
0
4
4
4
1
4
R
T
V
WWWWWW
X
***********

SOLUTIONS
Sol. 1 Option (B) is correct.
Here, as we know
Limsin
0
q
q"
. 0
but for 10% error, we can check option (B) first,
= = p =
q 18 18 .
c c# 0 314
180
c
sinq = sin18c = 0.309
% error = 0 . 314 - 0 . 309 0 .
309
#100 % = 0 . 49
% Now, we check it for q = 50c
in
q = 50 c = 50 c# p =
0 .
873
180
co.c
sinq nodia.= sin50c = 0.77
% error = 0.77 - 0.873 0 .
873 =-
12 . 25 % www.© so, the error is more than 10%. Hence, for error less than 10%, q = 18c can have
the approximation
sin q . q
Sol. 2 Option (A) is correct.
For, a given matrix A 6 @ the eigen value is calculated as
A - lI = 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 - lI = A (for l = 0)
3
5
2
5
12
7
2
7
5
=
= 3^60 - 49h- 5^25 - 14h+ 2^35 - 24h
= 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^xh a4x 4
a x 3
a x 2
3
2
a x a
= + + + 1 - 0
Since, all the coefficients are positive so, the roots of equation is given by
f^xh = 0
It will have at least one pole in right hand plane as there will be least one sign
change from ^a1h to ^a0h in the Routh matrix 1st column. Also, there will be a

corresponding pole in left hand plane
i.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
Consider the given matrix be
Im + AB
R
2
1
1
1
1
2
1
1
1
1
2
1
V
1
1
1
2
=
SSSSS
S
T
WWWWW
W
X
where m = 4 so, we obtain
AB
R
2
1
1
1
1
2
1
1
1
1
2
1
V
1
1
1
2
R
1
0
0
0
0
1
0
0
0
0
1
0
V
0
0
0
1
SSSSS
WWWWW
= -
S
T
SSSSS
S
T
W
X
WWWWW
W
X
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
SSSSS
WWWWW
=
co.in
nodia.www.© R
S
T
V
W
X
1
1
1
1
=
R
SSSSS
S
T
V
WWWWW
W
X
61 1 1 1@
Hence, we get
A
1
1
1
1
=
R
SSSSS
S
T
V
WWWWW
W
X
, B = 81 1 1 1B
R
Therefore, BA = 81 1 1 1B 1
SSSSS
WWWWW
1
1
1
S
T
V
W
X
= 4
From the given property
Det ^Im + ABh = Det^Im + BAh
& Det
2
1
1
1
1
2
1
1
1
1
2
1
1
1
1
2
R
SSSSS
S
T
V
WWWWW
W
X
Det
R
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
SSSSS
WWWWW
]]
]]
= + 4
S
T
V
W
X
Z
[

_
bb
bb
`
a
= 1 + 4
= 5
Note : Determinant of identity matrix is always 1.
dx + x = t
t dt

dx
dt
t x
+ = 1
dx + Px = Q (General form)
dt
1dt = # = # = =
Integrating factor, IF e Pdt e t elnt t
Solution has the form,
x#IF = # ^Q#IFhdt + C
x#t = # (1) (t)dt + C
2
xt = t2 +
C
Taking the initial condition,
x(1) = 0.5
t2
0.5 = 1 2 co.+ C
& in
C = 0
t
2
So, nodia.xt = & 2
x =
www.© Sol. 6 Option (C) is correct.
1
f (z) z 1 z
3
= 2 + - +
1
C p # = sum of the residues of the poles which lie
2 j f (z)dz
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
1 3
= 1 + +
(z ) (z )
z z
1 1
( )( )
= + +
( )( )
lim
z z
1 3
2 2
1
z 1
+ - + = =
"-
1
C p # = 1
So 2 j f (z)dz
= p + p
x = -1 = i cos 2 i sin 2
So, x = ei 2p
= 2p ^ h & ei i
xx ei x
^ h = e 2-p
2 p
d y t
dy t
( ) ( )
( )
dt
dt
y t
2
2
2
+ + = d(t)
By taking Laplace transform with initial conditions
0 dy 2 sy s y 0 Y s
- - + - +
; E = 1
s Y(s) sy( ) dt [ ( ) ( )] ( )
t
2
0
=
& 6s2Y(s) + 2s - 0@+ 26sY(s) + 2@+ Y(s) = 1
Y(s) [s2+ 2s + 1] = 1 - 2s - 4
Y(s)
- -
s
2 1
2 3
2 =
+ +
s s
We know that, If, y(t) Y(s) L

then,
dy t
( )
dt
sY(s) y(0) L -
So, sY(s) - y(0)
- - +
s s
2 1
2 3
( )
s s
2 2 =
+ +
( )
2 2
- s - s + s + s
+
2 3 2 4 2
2
+ +
(s 2 s 1
)
=
sY(s) - y(0)
s
s
+ 2
=
+ 1
+
1
2 2 2 =
+
+
+
(s 1 ) ( s
1
) ( s
1
)
1
1
2 = + +
s 1 (s +
1
)
Taking inverse Laplace transform
dy t
( )
dt
= e-tu(t) + te-tu(t)
dy
At t = 0+, dt
t=0+
= e0+ 0 = 1
Divergence of A in spherical coordinates is given as
1 1 in
r A
krn
2 co.r
2
nodia.www.© = 2 ( )
d:A ( )
r r
2
2
= 2 +
r r
2
2
k n 2 rn
2
= ( + )
+1
r
= k(n + 2)rn-1 = 0 (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, ...........
1
1
3 1 5
= +b l +b l +
Probability P 2 2
2
.....
P
2
1
1 1
3
4
2
=
-
=
f (x) = x3- 9x2+ 24x + 5
( )
dx
df x
= 3x2- 18x + 24 = 0
&
df x
( )
dx
= x2- 6x + 8 = 0 x = 4, x = 2
d f x
( )
dx
2
2
= 6x - 18
For x = 2,
d f x
( )
dx
12 18 6 < 0 2
2
= - =-
So at x = 2, f (x) will be maximum
f (x)
max
= (2)3- 9(2)2+ 24(2) + 5
= 8 - 36 + 48 + 5 = 25

Characteristic equation.
A- lI = 0
5
2
- - -
l 3
- = 0
l
5l + l2+ 6 = 0
l2+ 5l + 6 = 0
Since characteristic equation satisfies its own matrix, so
A2+ 5A+ 6 = 0 & A2 =- 5A- 6I
Multiplying with A
A3+ 5A2+ 6A = 0
A3+ 5(- 5A- 6I) + 6A = 0
A3 = 19A+ 30I
co.in
nodia.www.© Sol. 13 Option (D) is correct.
From Divergence theorem, we have
###4v $ Avdv A n ds
= # v $ t
s
The position vector
rv = ^utx x + utyy + utz zh
Here, Av = 5rv, thus
= ct + t + t 2 m ^t + t + t h
4$ Av u
2
x y z ux x uyy uz z :
2
x
u
y
u
z
2
2
2
= c dx
+ + dz m = 3#5 = 15
dy
dx dy
dz
5
## v $ t = ### 15dv = 15V
So, 5r nds
s
dy = ky
We have dx
# dy = # k dx + A
Integrating y
or lny = kx + A
Since y(0) = c thus lnc = A
So, we get, lny = kx + lnc
or lny = lnekx + lnc
or y = cekx
C R Integrals is
# - + where C is circle z = 1
z dz
4 5
3 4
z z
C
2+ +
f (z)dz
C #
= 0 if poles are outside C.
Now z 2+ 4z + 5 = 0
(z + 2)2+ 1 = 0
Thus z1,2 2 j z1,2 > 1 =- ! &
So poles are outside the unit circle.

We have f (x) = x + x - 3 = 0
fl(x)
x
= 1
+ 1
2
Substituting x0 = 2 we get
fl(x0) = 1.35355 and f (x0) = 2 + 2 - 3 = 0.414
Newton Raphson Method
x1
f x
( )
= - 0
f x
( )
x
0
0
l
Substituting all values we have
= - 0 414 = 1.694
2 . 1 3535
x 1 .
Writing A:B we have
:
:
:
1
1
1
1
4
4
1
6
6
20
SSSS
WWWW
l m
co.in
nodia.www.© R
T
V
X
Apply R3 R3 R2 " -
:
:
: 20
1
1
0
1
4
0
1
6
6
6
20
l - m -
R
SSSS
T
V
WWWW
X
For equation to have solution, rank of A and A:B must be same. Thus for no
solution; l = 6, m ! 20
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 favourable outcomes
= = 5
.
No of total outcomes
15
36
12
Eigen value of a Skew-symmetric matrix are either zero or pure imaginary in
conjugate pairs.
For a function x(t) trigonometric fourier series is
3
= /
= + w + w
x(t) Ao [An cosn t Bn sinn t]
n 1
Where, Ao T1 x(t)dt
= # T0" fundamental period
0T
0
An = T2 # x(t)cosn w
t dt
0T
0
Bn = T2 # x(t) sinn w
t dt
0T
0
For an even function x(t),Bn = 0
Since given function is even function so coefficient Bn = 0, only cosine and constant

terms are present in its fourier series representation.
Constant term :
3 T
4 A0 =
T1 x(t)dt
4
- #
T
/
/
T1 T
Adt 2Adt
= /
+ 3 4 -
-: # # D
T
/
T
/
T
/
4
4
4
1 TA AT
= : 2 - 2 2 D A2
T
=-
Constant term is negative.
Given differential equation
2
d n x
n x
( ) ( )
dx
2 - = 0
L
2
Let n(x) = Aelx
l
x
in
So, A l 2
e
l x -
Ae
= 0
L
2 l 2 - 1
= l =!co.0 & L1
L
2 Boundary condition, nodia.n ( 3 ) = 0 so take l =-
n ( x ) = Ae -
n(0) = Ae0 = K & A = K
So, www.n(x) = Ke-(x/L)
L x
L © 1
1 =
Given that ey x x
1 =
or ln ey ln x x
or y = x1 lnx
Now dx
dy ln x x x x 1 1
= + ^- - 1
ln
x
2h 1
2 2 = -
x x
For maxima and minima :
dy (1 ln ) 0
dx
1 x
2 = - =
x
lnx = 1 " x = e 1
Now
d y
dx
2
2
=- 2 - ln
x
x
3 b- 2 l- 1 1
x 3 x 2 b x
l
=- 2 + 2 ln
-
1
2 3 3 x x
x
x
d x
dy
atx e
2
2
= 1
2 2 1 < 0 2 3 3 = - + -
e e e
So, y has a maximum at x = e1
According to given condition head should comes 3 times or 4 times
4 1
1
P(Heads comes 3 times or 4 times) C 2 C
1
2
2
4
4
4
3
3
= b l + b l b l
1 16
= : + : : = 5
1 4 1
8
1
2
16

A v
= t + 2 t
dvl = dxatx + dyaty
xyax x ay
# A v : dl
v (xyax x ay) (dxa dya )
C
= # t + 2 t : t + t
C
x y
= # + 2
(xydx x dy)
C
xdx xdx dy dy 3 3 4
3
= # 2 3 + # 1 3
+ # + #
= 1
: - D+ : - D+ - + 1 1 - 3
= 1
3 4
1
3
3
-
= 1 2 - -
© www.nodia.co.in
1
/
/
/
/
2 3
1
3
3
1
1 3
2 3
2
[ ] [ ]
1
3 4
3 4
3 1 3
Given function
X(z)
z ( z z z
1 ) ( 2
) Poles are located at z = 0, z = 1, and z = 2
At Z = 0 residues is
= :
R0 z Z
zX ( ) =
0
- 2
= 1 2#0 - -
(0 1) (0 2)
= 1
Z
at z = 1 , R1 = ( Z - 1) :
X ( ) Z=
1
- =
= 1 2#1 1 -
1(1 2)
At z = 2, R2 (z 2) X(z)
z 2
= - :
=
= 1 2 # 2
3 -
2(2 1)
2
- =-
Taking step size h = 0.1, y(0) = 0
x y
dy = x + y y y hdx
dx
dy
i+1 = i +
0 0 0 y1 = 0 + 0.1(0) = 0
0.1 0 0.1 y2 = 0 + 0.1(0.1) = 0.01
0.2 0.01 0.21 y3 = 0.01 + 0.21#0.1 = 0.031
0.3 0.031
From table, at x = 0.3,y(x = 0.3) = 0.031
Given that
f (t)
- s
+ ; E
L 1 3 1
3 2 =
+ + -
s 4 s (K 3
)s
limf (t)
t"3
= 1
By final value theorem
limf (t)
t"3
= =
limsF(s) 1
s "
0
or
:
+
s ( 3 s
1
)
4 3
+ + -
( )
lim
s 0 3 2
s s K s
"
= 1

or
+
s ( 3 s
1
)
4 3
s 0 s s 2+ s + K
-
[ ( )]
lim
"
= 1
1-
K 3
= 1
or K = 4
The highest derivative terms present in DE is of 2nd order.
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
1
10
2
We have
f (z) = c0 + c1z
-1
in
1 + f ( z
)
-
1
f1(z)
= = 1 + co.c0 + c1z
z
z
nodia.www.© = z ( 1 + c 0 )
+ c
1
z
2
Since f1(z) has double pole at z = 0, the residue at z = 0 is
Res f1(z)z=0 limz .f (z)
z 0
2
= 1
"
= + 0 + 1
.
c m = c1
" z ( 1 c )
c
lim z
z
z 0
2
2
Hence
f1(z)dz
#
unit circle
= # [ 1
+ f ()]
z
dz
= 2pj [Residue at z = 0]
z
unit circle
= 2pjc1
We have f (x) sin
x
x
p
=
-
Substituting x - p = y ,we get
f (y + p)
= sin( y
+ p ) =- sin
y (sin )
y
y
= -1 y
y
3 5
1 y y y
= - c - + - m
! !
...
y
3 5
or f (y + p)
2 4
1 y y ...
=- + - +
3 5
! !
Substituting x - p = y we get
f (x)
=- + ( - p )
2 - - p 4 +
!
( )
!
...
x x
1
3 5
(A)
dy
dx
= y
x
or
# dy
y
= # dx
x
or log y = log x + log c
or y = cx Straight Line
Thus option (A) and (C) may be correct.
(B)
dy
dx
=-y
x

or
# dy
y
=- # dx
x
or log y =- log x + log c
or log y = log 1 + log
c
x
or y
x c
= Hyperbola
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.
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 p11p22- p12p21 = 0
© www.nodia.co.in
The given system is
x
y
4
2
2
= 1G= G
7
= =6G
We have A
4
2
2
= = 1G
and A
4
2
2
= 1 = 0 Rank of matrix r(A) < 2
Now C
4
2
2
1
7
= = 6G Rank of matrix r(C) = 2
Since r(A) ! r(C) there is no solution.
sinz can have value between -1 to +1. Thus no solution.
We have f (x) = ex + e-x
For x > 0, ex > 1 and 0 < e-x < 1
For x < 0, 0 < ex < 1 and e-x > 1
Thus f (x) have minimum values at x = 0 and that is e0+ e-0 = 2.
sin x
3 5
x x x ...
= + + +
3 5
! !
cos x
2 4
1 x x ...
= + + +
2 4
! !
Thus only sin(x3) will have odd power of x .
We have
dx t
( )
( )
dt
+ 3x t = 0
or (D + 3)x(t) = 0
Since m =- 3, x(t) = Ce-3t Thus only (B) may be solution.

We have x = e-x
or f (x) = x - e-x
f'(x) = 1 + e-x
The Newton-Raphson iterative formula is
xn+1
f x
( )
= - n
f x
'( )
x
n
n
x= - - n
Now f (xn) xn e
f'(xn) 1 e x= + - n
x
-
-
x e
1 n x
n = -
Thus xn+1 x
e
n
n
+
-
x e
1
+
( )
n
e
1
x
x
n
n
=
+
-
-
1
d n
-
1
Res f (z)( z in
a ) n
z=a
( n 1
)!
co.f ()
z
dz
n
nodia.www.© 1 z a
=
-
- -
= 6 @
Here we have n = 2 and a = 2
Thus Res f (z)z=2
1 2
( )!
d z
( )
1
; E
dz ( ) ( )
2 1 z z
2 2
z a
2
2 2 =
-
-
- + =
d
1
; E
dz ( z 2
)
z a
2 =
+ =
2
-
(z 2)
z a
3 =
+
=
; E
=- 2 32
64
=- 1
eP = L-1 (sI -A)-1 6 @
-
L
s
0 s
0 0
2
1
3
1
1
= - - -
-
e= G = Go
L
s
2 s
1
3
1
1
=
-
+
-
-
e= G o
s
3
+
L ( )( )
2
-
( )( )
1
( )( )
( )( )
s s
s s
s s
s
s s
1 1 2
1 2
1 2
1 2
= - + +
+ +
+ +
+ +
f> Hp
1 -
2
1 2
e e
e e
1 2
e e
e e
2
2 2 2
= 1 2
- +
-
- +
- -
- -
- -
= - - G
Taylor series is given as
x a
f a x a f a "( ) ...
f (x) ( )
!
'( )
( )
!
f a
1 2
2
= + - + - +
For x = p we have
Thus f (x) ( )
f x f "( )...
!
'( )
( )
!
x
f x
1 2
2
= p + - p p + - p
Now f (x) = ex + sinx
f'(x) = ex + cos x
f"(x) = ex - sinx
f"(p) = ep - sinp = ep

Thus the coefficient of (x - p)2 is
p
f"( )
2
!
The equation of straight line from (0,0) to (1,2) is y = 2x .
Now g(x,y) = 4x3+ 10y4
or, g(x,2x) = 4x3+ 160x4
1
Now g(x,2x)
1 # (4x3 160x4)dx
0
= # +
0
= [x4 + 32x5] 1 =
33
0
Q
= # +
2 xdx 2 ydy
I 2 (xdx ydy)
P
Q
= # + #
P
2 xdx 2 ydy 0
Q
P
= # + # =
0
1
1
0
© www.nodia.co.in
The given plot is straight line whose equation is
x y
-1 1
+ = 1
or y = x + 1
Now I ydx
1
2
= # (x 1)dx
1
2
= # +
(x )
2
1 2 2
= +
; E .
= 9
- 4 = 2 5
2
2
coth x
= cosh
x
sinh
x
as x << 1, coshx . 1 and sinhx . x
Thus coth x
. 1
x
sin
0
lim
^ q h
2
q" q
q
^
sin
0 2 2
^
h
= lim
2
q" q
h
= lim
2
sin
1
2
q" q
0 2
q
^
^
h
h
= 1 = 0 .
5
2
We have, lim
1
x"0 2
x
= 3
lim x
x
2
"3
= 3
lim e
x
x
"3
- = 3
lim e
x
x2
"3
- = 0
lim e
x
x
0
2
"
- = 1 Thus e x- 2 is strictly bounded.
We have f (x) = e-x = e-(x - 2) - 2 = e-(x - 2)e-2
x ...
( )
( )
!
x
1 2 e
2
2 2
= - - + - -2 ; E

= 1 - (x - 2) e-2 6 @ Neglecting higher powers
= (3 - x)e-2
We have k
2
2 d y
dx
2
= y - y2
or
2
d y
dx
y
2 -
k
2
y
2
=- 2
k
A.E. D
2 1
2 - = 0
k
or D
=!1
k
C.F. = C1e - x
x k
+
C2e k
P.I.
=
1 - y 2
2 c 2
m
= in
y
D -
1 k
2
2
k
2
Thus solution is
co.y = C1e - x
x k
+ C2e k
+
y2 From y(0) = y1 we get
C1+ nodia.C2 = y1- y2
From y(3) = y2 we get that C1 must be zero.
Thus www.C2 = y1- y2
y = (y1 - y2)e - x k
+
y2 © Sol. 52 Option (B) is correct.
We have
f (x) = x3- x2+ 4x - 4
f'(x) = 3x2- 2x + 4
Taking x0 = 2 in Newton-Raphosn method
x1
f x
( )
= - 0
'( )
x
f x
0
0
3 2
- + -
2 2 42 4
( )
2
( ) ( )
2
3 2 2 2 4
= -
- +
3 4
=
For two orthogonal signal f (x) and g(x)
# + f (x)g(x)dx
= 0
3
3
-
i.e. common area between f (x) and g(x) is zero.
We know that
# 1
ds
= 2pj [sum of residues]
s
2 - 1
D
Singular points are at s =!1 but only s =+ 1 lies inside the given contour,
Thus Residue at s =+ 1 is
-
lim(s 1) f (s)
s "
1
lim(s )
s
1
1
1
1
2
s 1 2 = -
-
=
"
# 1
ds
2 j j
s
2 - 1
D
= p ` 1 j = p
2

For two orthogonal vectors, we require two dimensions to define them and
similarly for three orthogonal vector we require three dimensions to define them.
2M vectors are basically M orthogonal vector and we require M dimensions to
define them.
We have
f (x) = x2- x + 2
f'(x) = 2x - 1 = 0 x
" = 1
2
f"(x) = 2
Since f"(x) = 2 > 0, thus x
= 1 is minimum point. The maximum value in
2
closed interval 6-4,4@ will be at x =- 4 or x = 4
Now maximum value
= max[f (- 4), f (4)]
= max(18,10)
= 18
© www.nodia.co.in
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 P 0.6 B
^A h =
We know that
A b l
P B
P + B =
( )
P B
( )
= bAl = 0.6 # 0.2 = 0.12
or P(A + B) P(B)P B
We have
A
R
1
1
1
1
1
1
1
0
1
SSSS V
1
1
0
1
1
0
1
0
0
SSSS
WWWW
= - + -
T
R
T
X
V
WWWW
X
R3- R1
Since one full row is zero, r(A) < 3
Now
1
1
1
-1 =- 2 ! 0, thus r(A) = 2
The vector Triple Product is
A# (B #C) = B(A$ C) -C(A$ B)
Thus 4#4#P =4 (4$ P) - P(4$4) =4 (4$ P) -42P
The Stokes theorem is
## (4# F) $ ds = # A $ dl

3
-# = 1
We know p(x)dx
3
3 -a
-# = 1
Thus Ke x dx
3
0
# a + # 3
-a
= 1
- or Ke x dx Ke x dx
0
3
or
K e x 0 k e x
a a 0 3+ -
6 a @ 6 - a @ 3
= 1
( )
-
or K K
a + a = 1
or K
2 a
=
We have xo (t) + 2x(t) = s(t)
Taking Laplace transform both sides
sX(s) - x(0) + 2X(s) = 1
co.in
nodia.www.© or sX(s) + 2X(s) = 1 Since x(0-) = 0
X(s)
= 1
+
s 2
Now taking inverse Laplace transform we have
x(t) = e-2tu(t)
Sum of the Eigen values must be equal to the sum of element of principal diagonal
of matrix.
Only matrix
6
2
2
= 6G satisfy this condition.
We have W = lnz
u + jv = ln(x + jy)
or eu+jv = x + jy
or eu ejv = x + jy
eu (cos v + j sinv) = x + jy
Now x = eu cos v and y = eu sinv
Thus x2+ y2 = e2u Equation of circle
We have
z
dz
4
1
z j
2
#
2 + - =
1
(z i) (z i)
dz
2 2
#
z j 2
=
+ -
- =
P(0,2) lies inside the circle z - j = 2 and P(0, - 2) does not lie.
Thus By cauchy’s integral formula
= p -
I ( )
1
" + -
( )( )
i lim z i
z i z i
2 2
2 2
z 2i
= p p
# 2
i
=
i i
2 2
2
C
+
= q q p #
I sin3 d
0

= # p ` 3 sin q - sin 3
q j d
q
sin3q = 3 sinq - 4 sin3q
4
0
= - q = w q
p p
3
s
4
cos 3
: D : D
12
0 0
4 3
4 3
1
12
1
12
y a k
P y+d
=
( )
P d
= 0 006 = 0 4
. .
0 015
© www.nodia.co.in
3 4
= 8 + B-8 + B=
Let d " defective and y " supply by Y
p 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
y a k
.
We have A
4
2
2
= = 4G
Now 6A - lI @[X] = 0
or
-
= - G= G
4
2
2
4
101
101
l
l
0
= =0G
or (101)(4 - l) + 2(101) = 0
or l = 6
We have
2
d y k y 2
dx
+ 2 = 0
or D2y + k2y = 0
The AE is m2+ k2 = 0
The solution of AE is m =!ik
Thus y = Asinkx + Bcos kx
From x = 0, y = 0 we get B = 0 and x = a,y = 0 we get
Asinka = 0
or sinka = 0
k
= mpx
a
=/ ` p j
Thus y A sin
m x
a
m
m
We have f (x)
e
1 e
x
x
=
+
For x " 3, the value of f (x) monotonically increases.
Order is the highest derivative term present in the equation and degree is the
power of highest derivative term.
Order = 2, degree = 1
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

odd number after an even number is 2 1
# = 1 .
2 1
4
We have
2
d y
dx
5 dy 6y 2
- + = 0
dx
The A.E. is m2- 5m + 6 = 0
m = 3,2
The CF is yc = C 3
x 2
x
1
e + C 2
e Since Q = 0, thus y = C e x + C e 2
x
1
3
2
Thus only (B) may be correct.
We have f (t) = e(a+2)t+5 = e5.e(a+2)t
Taking Laplace transform we get
F(s)
= 5 1
co.in
nodia.www.© ; - + E Thus Re(s) > (a +
2) ( )
e
s a 2
For x > 0 the slope of given curve is negative. Only (C) satisfy this condition.
Newton - Raphson " Method-Solving nonlinear eq.
Runge - kutta Method " Solving ordinary differential eq.
Simpson’s Rule " Numerical Integration
Gauss elimination " Solving linear simultaneous eq.
We have A
-
= G
4
4
2
= 3
Characteristic equation is
A - lI = 0
or
4
4
2
3
l
l
-
- = 0
or (- 4 - l)(3 - l) - 8 = 0
or -12 + l + l2- 8 = 0
or l2+ l - 20 = 0
or l =- 5,4 Eigen values
Eigen vector for l =- 5
- l = 0
( ) A I Xi
1 - ( -
5
) 2
x
= 4
8 - 4
G= x
G
1
2
0
= =0G
x
x
1
0
2
0
1
2
= =G G
0
= =0G R2- 4R1
x1+ 2x2 = 0
Let -x1 = 2 & x2 =- 1,
Thus X
2
= =-1G Eigen vector

We have
A
-
2 .
0
0 1
= G and A
= 3
1 2 1
a
0 b
- = = G
Now AA-1 = I
or
0 1
3 0 2 1
. a
= =G G
b
2
0
-
1
0
0
= = 1G
or
-
a . b
2 01
= G
b
1
0
3
1
0
0
= = 1G
or 2a - 0.1 = 0 and 3b = 1
Thus solving above we have b
= 1 and a
3
= 1
60
Therefore a + b
= 1
+ = 7
3
1
60
20
Gaussian PDF is
1 (x co.)
in
nodia.www.© -m # for -3 # x # 3
3 -
-
f (x) e dx
2
2 2
2
p s
=
3
s
3
-# = 1
and f (x)dx
3
Substituting m = 0 and s = 2 in above we get
1 x
-# 3 -
2
e dx
= 1
2 2
8
p 3
1
2
or 2 e dx
2 2
0
x
8
p
# 3 - = 1
1
2
or e dx
2
0
x
8
p
# 3 - = 1
From orthogonal matrix
[AAT] = I
Since the inverse of I is I , thus
[AAT]-1 = I-1 = I
*********

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