Engineering Mathematics

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UNIT 10

ENGINEERING MATHEMATICS

2013 ONE MARK

10.1 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

10.2 The minimum eigen value of the following matrix is

3 5 2 5 12 7 2 7 5 R T S S SS V X W W WW (A) 0 (B) 1 (C) 2 (D) 3

10.3 A polynomial f x( )=a x₄ 4+a x₃ 3+a x₂ 2+a x₁ -a₀ 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

10.4 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 T S S S S SS V X W W W W WW (A) 2 (B) 5 (C) 8 (D) 16 2012 ONE MARK

10.5 With initial condition x 1( )=0 5. , the solution of the differential

equation t dtdx + =x t, is (A) x= -t ₂1 (B) x ₌t2_-21 (C) x=t₂2 (D) x =₂t 10.6 Given f z( ) z 1 1 z 23 = + - + .

If C is a counter clockwise path in the z-plane such that

z+1 =1, the value of ₂1_j f z dz( )

C

p

#

is

(A) -2 (B) -1

(C) 1 (D) 2

10.7 If x= -1, then the value of xx is

(A) e-p/2 _{(B) e}p/2

(C) x (D) 1

2012 TWO MARKS

10.8 Consider the differential equation ( ) ( ) ( )

dt d y t dt dy t y t 2 2 2 + + =d( )t with y t( ) 2 and dy_dt 0 t t 0 0 =- = = - ₌

-The numerical value of dy_dt

t=0+

is

(A) -2 (B) -1

(C) 0 (D) 1

10.9 The direction of vector A is radially outward from the origin, with

kr

A ₌ n_{. where r}2_=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

10.10 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/

10.11 The maximum value of f x( ) =x3-9x2+24x+5 in the interval [ , ]1 6

is (A) 21 (B) 25 (C) 41 (D) 46 10.12 Given that and A=>- -₂5 ₀3H I=>1₀ 0₁H, the value of A3 _is (A) 15A+12I (B) 19A+30I (C) 17A+15I (D) 17A+21I 2011 ONE MARK

10.13 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 5r n dS

S v t$

##

is

(A) 3V (B) 5V

(C) 10V (D) 15V

10.14 The solution of the differential equation , (0)

dx

dy _{=ky y =c is}

(A) x₌ce-ky _{(B) x =ke}cy

(C) y₌cekx _{(D) y=ce}-kx

10.15 The value of the integral

(z 3z4z 45)dz

c

2_{+ +}

- +

#

where c is the circle

z =1 is given by

(A) 0 (B) 1/10

(C) 4/5 (D) 1

2011 TWO MARKS

10.16 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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4 6 20 4 x y z x y y x y z 6 m l + + = + + = + + =

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

10.18 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

10.19 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

10.20 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

10.21 A function n x( ) satisfied the differential equation

( ) ( ) dx d n x L n x ₀ 2 2 2 - =

where L is a constant. The boundary conditions are : (0)n =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

10.22 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

10.23 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 10.24 If Av =xyat_x+x a2t_y, then A dl C $ v v

#

over the path shown in the figure is

(A) 0 (B)

3 2

(C) 1 (D) 2 3

10.25 The residues of a complex function

( ) _{( )( )}

x z = _{z z-}1-₁ 2_zz_-2

at its poles are

(A) ₂1,-₂1 and 1 (B) 1₂,-₂1 and -1

(C) 1₂, 1 and -₂3 (D) ₂1,-1 and ₂3

10.26 Consider differential equation ( ) ( )

dx dy x

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 10.27 Given ( ) ( ) f t L s s k s s 4 3 3 1 1 3 2 = + ++ -; E. If lim f t( ) 1

t"₃ = , then the value of k is

(A) 1 (B) 2

(C) 3 (D) 4

2009 ONE MARK

10.28 The order of the differential equation

dt d y dt dy _{y e} t 2 2 3 4 + + = -c m is (A) 1 (B) 2 (C) 3 (D) 4

10.29 A fair coin is tossed 10 times. What is the probability that only the

first two tosses will yield heads?

(A) 2c m1 2 (B) 10_{C 2}1 2 2 b l (C) 2c m1 10 (D) 10_{C 2}1 2 10 b l 10.30 If ( )f z =c₀+c z₁ -1, then ( ) z f z _dz 1 unit circle +

#

is given by (A) 2p (B) c1 2 1p( +c0) (C) 2p (D) jc1 2 1p( +c0) 2009 TWO MARKS

10.31 The Taylor series expansion of sin

x-xp at x= is given byp (A) ! ( ) ... x 1 3 2 p + - + (B) ! ( ) ... x 1 3 2 p - - - + (C) ! ( ) ... x 1 3 2 p - - + (D) ! ( ) ... x 1 3 2 p - + - +

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curves from Group II

Group I Group II A. dx dy x y = 1. Circles B. dx dy x y =- 2. Straight lines C. dx dy y x = 3. Hyperbolas D. dx dy 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

10.33 The Eigen values of following matrix are

1 3 0 3 1 0 5 6 3 -R T S S SS V X W W WW (A) 3, 3+5 , 6j - (B) j - +6 5 , 3j +j, 3-j (C) 3+j, 3-j, 5+ (D) j 3, - +1 3 ,j - -1 3j 2008 ONE MARKS

10.34 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) p p11 12-p p12 21= (B) 1 p p11 22-p p12 21=-1

(C) p p11 22-p p12 21= (D) 0 p p11 22+p p12 21=0

10.35 The system of linear equations

x y

4 +2 7=

x y

2 + 6= has (A) a unique solution

(B) no solution

(C) an infinite number of solutions (D) exactly two distinct solutions

10.36 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

10.37 For real values of x , the minimum value of the function

( ) exp( ) exp( )

f x = x + - isx

(A) 2 (B) 1

(C) 0.5 (D) 0

10.38 Which of the following functions would have only odd powers of x

in its Taylor series expansion about the point x= ?0

(A) sin x( )3 _{(B) sin x( )}2

(C) cos x( )3 _{(D) cos x( )}2

10.39 Which of the following is a solution to the differential equation

( ) ( ) dt dx t x t 3 0 + = ? (A) ( )x t ₌3e-t _{(B) x t( )=2e}-3t (C) ( )x t _=-23 2t _{(D) x t( )=3t}2 2008 TWO MARKS

10.40 The recursion relation to solve x=e-x using Newton - Raphson

method is (A) xn+1=e-xn (B) xn+1=xn-e-xn (C) x (1 x ) e e 1 n n x x 1 n n = + ₊ + (D) x ( ) x e x e 1 x 1 n n x n x n 1 2 n n = - _- - -+

-10.41 The residue of the function ( )f z

(z 2) (z 2) 1 2 2 = + - at z= is2 (A) 321 - (B) 161 -(C) 161 (D) 321

10.42 Consider the matrix P 0

2 1 3 = - -= G. The value of ep _is (A) e e e e e e e e 2 3 2 2 5 2 1 2 1 1 2 2 1 -- -- -- -- -> H (B) e e e e e e e e 2 4 2 3 2 1 1 1 2 2 1 1 2 + -+ - -- -- -> H (C) e e e e e e e 5 2 6 3 4 6 2 1 2 1 1 2 2 1 -+ - -- -- -- -> H (D) e e e e e e e e 2 2 2 2 1 2 1 2 1 2 1 2 -- + -- + - -- -- -- -> H

10.43 In the Taylor series expansion of exp( )x +sin( )x about the point

x = , the coefficient of (p x_-p)2 _is

(A) exp p (B) ( ) 0 5. exp( )p

(C) exp( )p + (D) 1 exp( )p -1

10.44 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- plane isy

(A) 33 (B) 35

(C) 40 (D) 56

10.45 Consider points P and Q in the x- plane, with y 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

10.46 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

10.47 For x << , coth ( )1 x can be approximated as

(A) x (B) x2 (C) x1 (D) _x12 10.48 limsin 2 0 q q " q b l is

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(A) 0.5 (B) 1

(C) 2 (D) not defined

10.49 Which one of following functions is strictly bounded?

(A) /x1 2 _{(B) e}x

(C) x2 _{(D) e-}x2

10.50 For the function e-x, the linear approximation around x= is2

(A) (3_-x e) -2 _{(B) 1 -x}

(C) 3₊3 2_-(1_- 2)x e-2

6 @ (D) e-2

2007 TWO MARKS

10.51 The solution of the differential equation k

dx d y _{y y} 2 2 2 2 = - under the boundary conditions (i) y= at xy1 = and0

(ii) y= at xy2 = , where ,3 k y1 and y2 are constants, is

(A) y (y y)exp

kx y

1 2 2 2

= - _a- _k+ (B) (y= y2-y1)expa-_kxk+y1 (C) y=^y1-y2hsinhax_kk+y1 (D) y=^y1-y2hexpa-_kxk+y2

10.52 The equation x3-x2+4x- = is to be solved using the Newton 4 0

- Raphson method. If x= is taken as the initial approximation of 2 the solution, then next approximation using this method will be

(A) 2/3 (B) 4/3

(C) 1 (D) 3/2

10.53 Three functions ( ), ( )f t f t_{1 2} and ( )f t₃ which are zero outside the interval

[ , ]0 T are shown in the figure. Which of the following statements is correct?

(A) ( )f t1 and ( )f t2 are orthogonal

(B) ( )f t1 and ( )f t3 are orthogonal

(C) ( )f t2 and ( )f t3 are orthogonal

D) ( )f t1 and ( )f t2 are orthonormal

10.54 If the semi-circular control D of radius 2 is as shown in the figure,

then the value of the integral

(s 1)ds 1 D 2

-#

is (A) jp (B) -jp (C) - (D) p p

10.55 It is given that ,X X_{1 2}...X_M at M non-zero, orthogonal vectors.

The dimension of the vector space spanned by the M2 vectors , ,... , , ,...

X X1 2 XM -X1 -X2 -XM is

(A) M2 (B) M+1

(C) M

(D) dependent on the choice of ,X X1 2,...XM

10.56 Consider the function ( )f x =x2- - . The maximum value of ( )x 2 f x

in the closed interval [-4 4, ] is

(A) 18 (B) 10

(C) 225- (D) indeterminate

10.57 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

10.58 The rank of the matrix

1 1 1 1 1 1 1 0 1 -R T S S SS V X W W WW is (A) 0 (B) 1 (C) 2 (D) 3

10.59 4#4# , where P is a vector, is equal toP

(A) P P 2P #4# -4 (B) ₄2P ( P) 4 4# + (C) ₄2P P 4# + (D) (_{4 4$}P)_-42P

10.60

##

(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

10.61 A probability density function is of the form

( ) , ( , ) p x Ke x x 3 3 ! = -a -The value of K is (A) 0.5 (B) 1 (C) .0 5a (D) a

10.62 A solution for the differential equation ( )x to +2x t( )=d( )t with

initial condition (0 )x - =0 _is

(A) e u t-2t ( ) _{(B) e u t}2t _{( )}

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2006 TWO MARKS

10.63 The eigenvalue and the corresponding eigenvector of 2 2_{# matrix}

are given by Eigenvalue Eigenvector 8 1 l = v 1 1 1== G 4 2 l = v 1 1 2= -= G The matrix is (A) =6₂ 2₆G (B) =4₆ 6₄G (C) =2₄ 4₂G (D) =4₈ 8₄G

10.64 For the function of a complex variable W=lnZ (where, W= + u jv

and Z= + , the u = constant lines get mapped in Z -plane asx jy

(A) set of radial straight lines (B) set of concentric circles (C) set of confocal hyperbolas (D) set of confocal ellipses

10.65 The value of the constant integral

z 1 4 z j 2 2 + - =

#

dz is positive sense is (A) j 2 p _(B) 2 p -(C) j 2 p - (D) 2 p

10.66 The integral sin d3

0 q q p

#

is given by (A) 2 1 (B) 3 2 (C) 3 4 (D) 3 8

10.67 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 Sup-plied 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 sup-plied by Y is

(A) 0.1 (B) 0.2

(C) 0.3 (D) 0.4

10.68 For the matrix 4

2 2 4

= G the eigenvalue corresponding to the eigenvector 101

101 = G is

(A) 2 (B) 4

(C) 6 (D) 8

10.69 For the differential equation

dx

d y _{k y 0}

2 2

2

+ = the boundary conditions are

(i) y= for x0 = and 0 (ii) y= for x0 =a

The form of non-zero solutions of y (where m varies over all inte-gers) are (A) y A sin a m x m m p =

_/

(B) cosy A a m x m m p =

_/

(C) y A xm a m m =

_/

p (D) y A em a m x m = - p

/

10.70 As x increased from 3- to 3, the function ( )f x

e e 1 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

10.71 The following differential equation has

dt d y dt dy _y 3 2_{2 +}4 3₊ 2₊2 c m c m x=

(A) degree 2= , order 1= (B) degree = , order 21 = (C) degree 4= , order 3= (D) degree = , order 32 =

10.72 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

10.73 A solution of the following differential equation is given by

dx d y dx dy _y 5 6 0 2 2 - + = (A) y₌e2x₊e-3x (B) y₌e2x₊e3x (C) y₌e-2x₊33x (D) y₌e-2x₊e-3x 2005 TWO MARKS

10.74 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

10.75 The derivative of the symmetric function drawn in given figure will

look like

10.76 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

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6. Calculation of Eigenvalues (A) E-6, F-1, G-5, H- (B) 3 E-1, F-6, G-4, H-3 (C) E-1, F-3, G-4, H- (D) 2 E-5, F-3, G-4, H-1

10.77 Given the matrix 4

4 2 3 -= G, the eigenvector is (A) = G (B) ₂3 = G4₃ (C) = G (D) _-2₁ = G-₂1 10.78 Let, A 2 . 0 0 1 3 =₌ - _{G and A} a b 0 1₌ 21 - _{= G. Then (a+ =b)} (A) 7/20 (B) 3/20 (C) 19/60 (D) 11/20

10.79 The value of the integral I exp x dx

2 1 8 2 0 p =

_#

3 _c- _m is (A) 1 (B) p (C) 2 (D) 2p

10.80 Given an orthogonal matrix

A 1 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 = _- -R T S S S S S V X W W W W W AAT 1 -6 @ is (A) 0₀ 0 0 0 0 0 0 0 0 0 0 4 1 4 1 2 1 2 1 R T S S S S SS V X W W W W WW (B) 0₀ 0 0 0 0 0 0 0 0 0 0 2 1 2 1 2 1 2 1 R T S S S S SS V X W W W W WW (C) 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 R T S S S S S V X W W W W W (D) 0₀ 0 0 0 0 0 0 0 0 0 0 4 1 4 1 4 1 4 1 R T S S S S SS V X W W W W WW

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SOLUTIONS

10.1 Option (B) is correct. Here, as we know sin Lim 0 q " q .0

but for 10% error, we can check option (B) first,

q 18 18 . 180 0 314 # c c c p = = = sin q =sin 18c=0 309. % error = 0 314. _{0 309.}-0 309. #100%=0 49. %

Now, we check it for q = 50c

q 50c 50c#₁₈₀ 0 873. c p = = = sin q =sin 50c=0 77. % error = 0.77_{0 873.}-0.873=-12 25. %

so, the error is more than 10%. Hence, for error less than 10%,

18c

q = can have the approximation

sin q .q

10.2 Option (A) is correct.

For, a given matrix 6 @A 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 49 5 25 14 2 35 24 = ^ - h- ^ - h+ ^ - h 33 55 22 = - + 0 =

Hence, it satisfied the characteristic equation and so, the minimum eigen value is

l =0

10.3 Option (D) is correct.

Given, the polynomial

f x^ h =a x4 4+a x3 3+a x2 2+a x1 -a0

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 ^ ha1 to ^ ha0 in the Routh matrix 1st col-umn. 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

10.4 Option (B) is correct.

Consider the given matrix be

Im+AB 2 1 1 1 1 2 1 1 1 1 2 1 1 1 1 2 = R T S S S S SS V X W W W W WW

where m=4 so, we obtain

AB 2 1 1 1 1 2 1 1 1 1 2 1 1 1 1 2 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 = -R T S S S S SS R T S S S S SS V X W W W W WW V X W W W W WW 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 = R T S S S S SS V X W W W W WW 1 1 1 1 = R T S S S S SS V X W W W W WW 1 1 1 1 6 @ Hence, we get A 1 1 1 1 = R T S S S S SS V X W W W W WW , B=81 1 1 1B Therefore, BA =81 1 1 1B 1 1 1 1 R T S S S S SS V X W W W W WW 4 =

From the given property

Det ^Im+ABh =Det I^ m+BAh & Det 2 1 1 1 1 2 1 1 1 1 2 1 1 1 1 2 R T S S S S SS V X W W W W WW Det 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 4 = + R T S S S S SS V X W W W W WW Z [ \ ] ] ] ] _ ` a b b b b 1 4 = + 5 =

Note : Determinant of identity matrix is always 1.

10.5 Option (D) is correct. t dtdx +x =t dt dx t x + =1 dt dx _{+Px =Q (General form)} Integrating factor, IF =e#Pdt=e#1tdt=elnt =t

Solution has the form,

x#IF =

#

^Q#IF dth +C

x#t =

#

( )( )1 t dt+C xt =t₂2+C

Taking the initial condition,

( ) x 1 =0 5. 0.5 = +₂1 C & C =0 So, xt =t₂2 &x = t₂ 10.6 Option (C) is correct. ( ) f z _{= + - +z} 1_{1 z} 2 ₃

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( )

j f z dz

21p

#

C = 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 1z)(z1 3) = ₊- +₊ ( )( ) ( )( ) lim z z z z 1 3 1 1 2 2 ₁ z 1 = +₊ - +₊ = = " -So ₂1_j f z dz( ) C p

#

=1

10.7 Option (A) is correct.

x = - =1 i =cosp₂+isinp₂ So, x ₌eip₂ xx ₌^ h_eip₂ x _& ^ h_eip₂ i _=e-p₂ 10.8 Option (D) is correct. ( ) ( ) ( ) dt d y t dt dy t y t 2 2 2 + + =d( )t

By taking Laplace transform with initial conditions

( ) ( ) [ ( ) ( )] ( ) s Y s sy 0 dy_dt 2 sy s y 0 Y s t 2 0 - - + - + = ; E =1 & 6s Y s2 ( )+2s-0@+26sY s( )+2@+Y s( ) =1 ( )[ ] Y s s2₊2s₊1 _{= -1 2s-4} ( ) Y s s s s 2 1 2 3 2 = + + -

-We know that, If, y t( ) L Y s( )

then, dy t_dt( ) L sY s( )-y 0( ) So, sY s( )-y 0( ) ( ) ( ) s s s s 2 1 2 3 2 2 = + + - - ₊ (s s ) s s s s 2 1 2 3 2 4 2 2 2 2 = + + - - + + + ( ) ( ) sY s -y 0 (s ) ( ) ( ) s s s s 1 2 1 1 1 1 2 2 2 = ++ = ++ + + ( ) s 11 _s 1₁ 2 = + + +

Taking inverse Laplace transform

( ) dt dy t ( ) ( ) e u tt te u tt = - + -At t=0+_, dt dy t=0+ e0 0 1 = + =

10.9 Option (A) is correct.

Divergence of A in spherical coordinates is given as A : d ( ) r12₂2r r A2 r = ( ) r12₂2r krn 2 = + ( ) rk n 2 r n 2 1 = + + ( ) k n 2 rn 1 0 = + - = _(given) n+2 =0 & n =-2 10.10 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 b₂1l3+b1₂l5+... P 1 2 ₄1 23 1 = _- = 10.11 Option (B) is correct. ( ) f x ₌x3_-9x2₊24x₊5 ( ) dx df x x x 3 2 18 24 0 = - + = & ( ) dx df x 6 8 0 x2 x = - + = x =4, x =2 ( ) dx d f x 2 2 x 6 18 = -For x=2, ( ) dx d f x 12 18 6<0 2 2 = - =-So at x=2, f x( ) will be maximum ( ) f x max ( )2 9 2( ) 24 2( ) 5 3 2 = - + + 8 36 48 5 = - + + =25 10.12 Option (B) is correct. Characteristic equation. A-lI =0 5 2 3 l l - - -- =0 5_l+l2₊6 ₌₀ 5 6 2 l + l+ =0

Since characteristic equation satisfies its own matrix, so

5 6 A2₊ A_{+ =0 & A}2_=-5A-6I Multiplying with A 5 6 A3₊ A2₊ A ₌₀ 5( 5 6 ) 6 A3_{+ -} A_- I ₊ A ₌₀ A3 _=19A+30I 10.13 Option (D) is correct.

From Divergence theorem, we have

Adv

4$v v

###

A n ds

s $

=

_#

v _t

The position vector

rv =^u xtx +u yty +u ztz h Here, Av=5r_{v, thus} A 4$ v u x u y u z u x u y u z x y z : x y z 2 2 2 2 2 2 =_ct + t + t _{m ^}t + t + t _h 5 dx dx dy dy dz dz =_c + + _m =3#5 =15 So, 5r n ds s v t$

##

=

###

15dv=15V 10.14 Option (C) is correct. We have _dxdy =ky Integrating

#

dy_y =

#

k dx+A or ln y =kx+A Since y 0( ) =c thus ln c =A So, we get, ln y =kx+lnc or ln y ₌lnekx₊lnc or y ₌cekx

10.15 Option (A) is correct.

C R Integrals is z 3z4z 45dz C 2_{+ +} - +

#

where C is circle z =1 ( ) f z dz

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Now z2₊4z₊5 ₌₀ (z₊2)2₊1 ₌₀ Thus z1 2, =-2!j& z1 2, >1 So poles are outside the unit circle.

10.16 Option (C) is correct. We have f x( ) = +x x - =3 0 ( ) f xl x 1 2 1 = + Substituting x0 =2 we get ( ) f xl 0 =1 35355. and f x( )0 = +2 2- =3 0 414. Newton Raphson Method

x1 ( ) ( ) x f x f x 0 0 0 = -l

Substituting all values we have

x1 = -2 _{1 3535}0 414_.. =1 694. 10.17 Option (B) is correct. Writing A B: we have : : : 1 1 1 1 4 4 1 6 6 20 l m R T S S S S V X W W W W Apply R3"R3-R2 : : : 20 1 1 0 1 4 0 1 6 6 6 20 l- m -R T S S S S V X W W W W

For equation to have solution, rank of A and A B: must be same. Thus for no solution; l = 6,m !20

10.18 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 favourable outcomes._{No of total outcomes.} =15₃₆ =₁₂5

10.19 Option (C) is correct.

Eigen value of a Skew-symmetric matrix are either zero or pure imaginary in conjugate pairs.

10.20 Option (C) is correct.

For a function x t( ) trigonometric fourier series is

( ) x t Ao [Ancosn t Bnsinn t] n 1 w w = + 3 + =

/

Where, Ao _T1 x t dt( ) T 0 0 =

#

T0" fundamental period An _T2 x t( )cosn t dt T 0 0 w =

#

Bn _T2 x t( )sinn t dt T 0 0 w =

#

For an even function x t B( ), n=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 : A0 _T1 x t dt( ) / / T T 4 3 4 =

-#

_T1 Adt 2Adt / / / / T T T T 4 4 4 3 4 = + -:

#

#

D T1 TA2 2 2AT = : - D =-A₂

Constant term is negative.

10.21 Option (D) is correct.

Given differential equation

( ) ( ) dx d n x L n x 2 2 2 - =0 Let n x( ) ₌Aelx So, A e L Ae x x 2 2 l l - l ₌₀ L1 2 2 l - =0&l=!_L1

Boundary condition, n( )3 =0 so take l =-_L1 ( )

n x =Ae-Lx

( )

n 0 ₌Ae0₌K_&A₌K

So, n x( ) ₌Ke-( / )x L

10.22 Option (A) is correct.

Given that ey _=xx1 or ln ey _{=ln xx}1 or y = _x1lnx Now _dxdy _{x x}1 1 lnx x x 1 2 = + _^- - _h ln x12 x2 =

-For maxima and minima :

dx dy _{(1 ln ) 0} x12 x = - = ln x =1"x =e1 Now dx d y 2 2 ln x23 x x23 x1 12 x =- - _b- l- b l ln x x x x 2 2 1 2 3 3 =- + -dy d x at x e 2 2 1 = e e e 2 2 1 _<0 2 3 3 = - +

-So, y has a maximum at x₌e1

10.23 Option (D) is correct.

According to given condition head should comes 3 times or 4 times

(Heads comes times or times)

P 3 4 4C 1₂ C ₂1 ₂1 4 4 4 3 3 = _{b l} + _{b l b l} 1:₁₆1 4:₈1:₂1 ₁₆5 = + = 10.24 Option (C) is correct. Av =xyatx+x a2ty dlv =dxatx+dyaty A dl C : v v

#

(xyax x ay) (dxa dya )

C x y 2 : =

#

t + t t + t (xydx x dy) C 2 =

#

+ xdx 3xdx ₃4dy ₃1dy / / / / 2 3 1 3 1 3 3 1 1 3 2 3 =

#

+

#

+

#

+

#

[ ] [ ] 2 1 3 4 3 1 2 3 3 1 3 4 3 4 3 1 ₃1 1 3 = : - D+ : - D+ - + -1 = 10.25 Option (C) is correct. Given function ( ) X z ( )( ) z z 11 2zz 2 = _--

-Poles are located at z=0,z=1,andz=2

At Z=0 residues is R0 z X z( ) Z 0 : = _{= =(0}1 _{1 0)(}2#0₂₎ -- - = 21

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at z=1, R1 =(Z-1):X Z( )_Z=1 ( ) 1 1 2 1 2#1 ₁ = - _- = At z=2, R2 =(z-2):X z( )_z=2 ( ) 2 2 1 1 2 2 2 3 # = - _- =-10.26 Option (B) is correct.

Taking step size h =0.1, y 0( ) =0

x y dx dy _{= +x y y y} hdxdy 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. 10.27 Option (D) is correct. Given that ( ) f t ( ) s s K s s 4 3 3 1 L 1 3 2 = + ++ -; E ( ) lim f t t " 3 =1

By final value theorem

( ) lim f t t " 3 =lim sF ss"0 ( ) =1 or ( ) ( ) lim s s K s s s 4 3 3 1 s 0 3 2 : + + -+ " =1 or [ ( )] ( ) lim s s s K s s 4 3 3 1 s 0 2+ + -+ " =1 K1-3 =1 or K =4 10.28 Option (B) is correct.

The highest derivative terms present in DE is of 2nd order.

10.29 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

2 1 10 10.30 Option (C) is correct. We have ( ) f z c= 0+c z1 -1 ( ) f z1 ( ) z f z z c c z 1 1 0 1 1 = + = + + - ( ) z z 1 c c 2 0 1 = + +

Since ( )f z1 has double pole at z= , the residue at z0 = is0

Res f z1( )z=0 lim z f z. ( ) z 0 2 1 = " . ( ) lim z z z 1 c c z 0 2 2 0 1 = + + " c m c= 1 Hence ( ) f z dz1 unit circle

#

[1 _zf z( )]dz unit circle =

_#

+ =2pj [Residue at z= ]0 jc 2p 1 = 10.31 Option (D) is correct. We have f x ( ) sin x xp = -Substituting x- yp = ,we get

( ) f y+ p sin( ) sin y y y y p = + =- (sin ) y1 y = -! ! ... y y y y 1 3 5 3 5 = - _c - + - _m or f y( + p) ! ! ... y y 1 3 5 2 4 =- + - + Substituting x- = we getp y ( ) f x ! ( ) ! ( ) ... x x 1 3 5 2 4 p p =- + - - - +

10.32 Option (A) is correct.

(A) dx dy x y = or y dy

#

=

_#

dx_x

or log y log= x+logc

or y cx= Straight Line

Thus option (A) and (C) may be correct. (B) dx dy x y =-or y dy

#

=-

_#

dx_x

or log y =-logx+logc

or log y log log

x1 c = + or y x c = Hyperbola 10.33 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- - + = .In 1 3 1 only option (D), the sum of Eigen values is 1.

10.34 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 p11 22-p p12 21 0=

10.35 Option (B) is correct.

The given system is

x y 4 2 2 1 = =G G =_{= G}7₆ We have A =₌4₂ 2_1G

and A = 4₂ 2₁ = 0 Rank of matrix ( )r A <2 Now C =₌4₂ 2₁ 7_6G Rank of matrix ( )r C =2 Since ( )r A !r( )C there is no solution.

10.36 Option (A) is correct.

sin z can have value between 1- to 1+ . Thus no solution.

10.37 Option (A) is correct.

We have f x e( ) ₌ x₊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 = and that is 0 e0₊e-0₌2_.

10.38 Option (A) is correct.

sin x ! ! ... x x x 3 5 3 5 = + + + cos x ! ! ... x x 1 2 4 2 4 = + + +

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Thus only sin x( )3 _{will have odd power of x .}

10.39 Option (B) is correct. We have ( ) ( ) dt dx t x t 3 + 0= or (D+3) ( )x t 0=

Since m=- , 3 x t Ce( ) ₌ -3t _{Thus only (B) may be}

solution. 10.40 Option (C) is correct. We have x e₌ -x or f x x( ) _{= -}e-x '( ) f x _{= +}1 e-x

The Newton-Raphson iterative formula is

xn 1+ xn _{f x}f x_{'( )}( ) n n = -Now f x( )n x= n-e-xn '( ) f xn = +1 e-xn Thus xn 1+ x e x e 1 n n _x x n n = -+ -- _{( )} e x e 1 1 x n x n n = + +

-10.41 Option (A) is correct.

Res f z( )z=a _{(n )!} ( ) ( ) dz d _{z a f z} 1 1 n n _n z a 1 1 = _- -_{- 6} - _@= Here we have n= and a2 =2

Thus Res ( )f z z=2 _{( )!} ( ) ( ) ( ) dzd z _{z z} 2 11 2 2 ₂ 21 ₂ 2 _{z a} = _- -- + ₌ ; E ( ) dzd _z 1₂ 2 _{z a} = + ₌ ; E _(z 2₂₎ z a 3 = + -= ; E 642 =- =-₃₂1 10.42 Option (D) is correct. eP _=L-1 _(sI-A)-1 6 @ L₌ 1 s_{0 s}0 0₂ 1₃ 1 -- -e= G = Go L₌ 1 s_{2 s}-1₃ 1 + - _{e= G}- _o L ( )( ) ( )( ) ( )( ) ( )( ) s s s s s s s s ss 1 1 3 2 12 2 11 2 1 2 = - + + + + - + + + + +

f

>

H

p

e e e e e e e e 2 2 2 2 1 2 1 2 1 2 1 2 = _- -_{+ -} -₊ - -- -- -- -= G 10.43 Option (B) is correct.

Taylor series is given as ( ) f x ( ) ! '( ) ! ( ) "( ) ... f a x a f a x a f a 1 2 2 = + - + - + For x= we havep Thus f x ( ) ( ) ! '( ) ! ( ) "( )... f x f x f x 1 2 2 p p p p = + - + -Now f x ( ) ₌ex₊sinx '( ) f x ₌ex₊cosx "( ) f x ₌ex_-sinx "( ) f p =ep-sinp =ep

Thus the coefficient of (x_-p)2 _is

! "( )

f

2

p

10.44 Option (A) is correct.

The equation of straight line from ( , )0 0 to ( , )1 2 is y=2x.

Now g x y ( , ) ₌4x3₊10y4 or, g x x( , )2 ₌4x3₊160x4 Now g x x( , )2 0 1

#

(4x3 160x dx4) 0 1 =

_#

+ [x4 32x5] 33 0 1 = + = 10.45 Option (B) is correct. I 2 (xdx ydy) P Q =

_#

+ 2 xdx 2 ydy P Q P Q =

_#

+

_#

xdx ydy 2 2 0 0 1 1 0 =

_#

+

_#

= 10.46 Option (B) is correct.

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 . 2 9 2 4 _{2 5} = - = 10.47 Option (C) is correct. coth x sinh cosh xx =

as x <<1,coshx. and sinh x1 .x

Thus coth x

x1

.

10.48 Option (A) is correct.

limsin 0 2 q " q q ^ h _limsin 2 0 ₂ 2 = " q q q ^^hh lim sin 2 1 0 ₂ 2 = " q q q ^^hh = 21 =0 5. 10.49 Option (D) is correct. We have, lim x 1 x"0 2 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-}x2 is strictly bounded.

10.50 Option (A) is correct.

We have f x e( ) ₌ -x _e= - -(x 2 2)- _e= - -(x 2)_e-2 ( ) ! ( ) _... x x e 1 2 2 2 2 2 = - - + - -; E (x )e 1 2 2 = - -

-6 @ Neglecting higher powers (= 3-x e) -2 10.51 Option (D) is correct. We have k dx d y 2 2 2 y= -y2 or dx d y k y 2 2 2 -k y 2 2 =-A.E. D k 1 2 2 - 0= or D k1 ! = C.F. C e1 k C e2 x k x = - + P.I. D k k y _y 1 1 2 2 2 22 2 = -- ₌ c m Thus solution is

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y C e1 k C e2 y2 x k x = - + + From ( )y 0 = we gety1 C1+C2 y= 1-y2

From ( )y 3 =y2 we get that C1 must be zero.

Thus C2 y= 1-y2 y (y1 y e2) k y2 x = - - + 10.52 Option (B) is correct. We have ( ) f x x₌ 3_-x2₊4x_-4 '( ) f x ₌3x2_-2x₊4 Taking x0= in Newton-Raphosn method2

x1 x0 _{f x}f x_{'( )}( ) 0 0 = -( ) ( ) ( ) 2 3 2 2 2 4 2 2 4 2 4 2 3 2 = -- + - + _- 3 4 = 10.53 Option (C) is correct.

For two orthogonal signal ( )f x and ( )g x

( ) ( ) f x g x dx 3 3 -+

#

0=

i.e. common area between ( )f x and ( )g x is zero.

10.54 Option (A) is correct.

We know that s 1ds 1 D 2

-#

=2pj [sum of residues] Singular points are at s=!1 but only s=+ lies inside the given 1 contour, Thus Residue at s=+ is1

( ) ( ) lim s 1 f s s"1 - lim s( 1)s 1 1 2 1 s 1 2 = -- = " s 1ds 1 D 2

-#

2 j j 2 1 p p = _{` j}= 10.55 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.

10.56 Option (A) is correct.

We have ( ) f x x₌ 2_{- +}x 2 '( ) f x =2x- = x1 0 2 1 " = "( ) f x 2= Since "( )f x =2>0, thus x 2 1

= is minimum point. The maximum value in closed interval ₆-4 4, _@ will be at x=- or x4 =4

Now maximum value

[ ( 4), (4)] max f f = -( , ) max 18 10 = 18= 10.57 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 P_{^ hB}A ₌0.6 We know that P Bb lA = (P B_{P B( )}+ ) or P A B( + ) =_{P B P B}( ) _{b l}A =0 6 0 2. # . =0 12. 10.58 Option (C) is correct. We have A 1 1 1 1 1 1 1 0 1 1 1 0 1 1 0 1 0 0 + = - -R T S S SS R T S S SS V X W W WW V X W W WW R3-R1 Since one full row is zero, ( )r A <3

Now 1_{1 -} 1₁ =-2!0, thus ( )r A =2

10.59 Option (D) is correct.

The vector Triple Product is ( ) A# B C# =B A C( $ )-C A B( $ ) Thus 4#4#P =4 4$( P)-P(4$4) ( P) 2P 4 4$ 4 =

-10.60 Option (A) is correct.

The Stokes theorem is

(4#F ds)$

##

=

#

A dl$ 10.61 Option (C) is correct. We know p x dx( ) 3 3

-#

1= Thus Ke x dx 3 3 _-a

-#

1= or Ke dxx Ke xdx 0 0 + 3 3 a -a -

#

#

1= or ( ) K e x 0 k _e x 0 a 3+ -a 3 a a -6 @ 6 @ 1= or K_a +K_a 1= or K 2 a =

10.62 Option (A) is correct.

We have x to( )+2x t( ) =s t( ) Taking Laplace transform both sides

( ) ( ) ( ) sX s -x 0 +2X s 1= or sX s( )+2X s( ) 1= Since x 0( )- =0 ( ) X s s 12 = ₊ Now taking inverse Laplace transform we have

( )

x t ₌e u t-2t ( )

10.63 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 2_{6G satisfy this condition.}

10.64 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 sin ) eu v+j v x= +jy

Now x =eucosv and y=eusinv

Thus x2+ ey2 ₌ 2u _{Equation of circle}

10.65 Option (D) is correct. We have z 4dz 1 z j 2 2 + - =

#

(z i z)( i) dz 2 1 2 z j 2 = ₊ -- =

#

( , )

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Thus By cauchy’s integral formula

I ( ) ( )( ) lim i z i z i z i 2 2 2 1 2 z 2i p = -+ -" i i i 22 2 2 C p p = + =

#

10.66 Option (C) is correct. I sin d3 0 q q =

_#

p sin sin _d 4 3 3 0 q _{q q}

=

_#

p_` - _j sin3_q=3sin_q-4sin3_q

cos s 43 q ₀ w q123 ₀

= -: Dp=: D p =₈₄3 + ₄3_B-₈₁₂1 +₁₂1 _B= 4₃

10.67 Option (D) is correct.

Let d " defective and y " supply by Y

p da ky = P y(_{P d( )}+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 da ky _.. . 0 015 0 006 _{0 4} = = 10.68 Option (C) is correct. We have A =₌4₂ 2_4G Now ₆A-lI X_@[ ] 0= or =4-₂l _4-2_l =G G ₁₀₁101 =_{= G}0₀ or (101 4)( -l)+2 101( ) 0= or l 6=

10.69 Option (A) is correct.

We have dx d y _{k y} 2 2 2 + 0= or D y2 +k y2 0= The AE is m2+ 0k2 = The solution of AE is m=!ik

Thus y=Asinkx+Bcoskx

From x= , y0 = we get B0 = and 0 x=a y, = we get0 sin A ka 0= or sin ka 0= k a m xp = Thus y A sin a m x m m p =

_/

_{` j}

10.70 Option (A) is correct.

We have f x ( ) e e 1 x x = +

For x " 3, the value of ( )f x monotonically increases.

10.71 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=

10.72 Option (D) is correct.

Probability of coming odd number is ₂1 _{and the probability of}

coming even number is ₂1_{. Both the events are independent to each}

other, thus probability of coming odd number after an even number is ₂1 2 1 4 1 # = . 10.73 Option (B) is correct. We have dx d y dx dy _y 5 6 2 2 - + 0= The . .A E is m2_-5m_{+ 0}6 ₌ m =3 2, The CF is yc C e= 1 3x+C e2 2x Since Q= , thus 0 y C e x C e x 1 3 2 2 = +

Thus only (B) may be correct.

10.74 Option (A) is correct.

We have f t e( ) ₌ (a+2)t+5 _{=e e}5_. (a+2)t

Taking Laplace transform we get ( ) F s ( ) e s a1 2 5 = _{; - + E} Thus Re s( )>(a+2) 10.75 Option (C) is correct.

For x> the slope of given curve is negative. Only (C) satisfy this 0 condition.

10.76 Option (C) is correct.

Newton - Raphson " Method-Solving nonlinear eq. Runge - kutta Method " Solving ordinary differential eq. Simpson’s Rule " Numerical Integration

Gauss elimination " Solving linear simultaneous eq.

10.77 Option (C) is correct. We have A =₌-4₄ 2_3G Characteristic equation is A-lI 0= or 4-₄l _{3- 0}2_l = or ( 4- -l)(3 -l)- 08 = or _-12_{+ +l l}2_{- 0}8 ₌ or _l2_{+ -l} 20 ₀₌ or l =-5 4, Eigen values Eigen vector for l =-5

(A-lI X) i 0= ( ) x x 1 5 4 2 8 4 1 2 -= =G G =_{= G}0₀ x x 1 0 2 0 1 2 = =G G =_{= G} 0₀ R2-4R1 x1+2x2 0= Let x- =1 2&x2=- ,1

Thus X _{= -= 1}2_{G Eigen} vector

10.78 Option (A) is correct.

We have A =₌₀2 -₃0 1. _{G and A}-1₌₌₀21 a_bG Now AA-1 _I= or ₌2₀ -₃0 1. _=G021 a_bG 1 0 0 1 =_{= G} or =₀1 2a-_3b0 1. bG =₌1₀ 0_1G or 2a-0 1. 0= and b3 =1 Thus solving above we have b

3 1 = and a 601 = Therefore a+ b 3 1 601 207 = + =

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10.79 Option (A) is correct.

Gaussian PDF is ( ) f x e dx 2 1 (x ) 22 2 p s = 3 3 -s m

-#

for -3# #x 3 and f x dx( ) 3 3

-#

1=

Substituting m = and 0 s = in above we get2

e dx 2 2 1 x 8 2 p 3 3

-#

1= or 2 e dx 2 2 1 0 x 8 2 p 3

-#

1= or e dx 2 1 0 x 8 2 p 3

-#

1= 10.80 Option (C) is correct.

From orthogonal matrix [AAT] I=

Since the inverse of I is I , thus [AAT]-1 _I= -1_=I