SOLUTION
SOL 2.1 Option (D) is correct.
Numerical Integration Scheme Order of Fitting Polynomial P. Simpson’s 3/8 Rule 3. Third order
Q. Trapezoidal Rule 1. First order R. Simpson’s 1/3 Rule 2. Second order
SOL 2.2 Option (C) is correct.
Let a square matrix A x
y y
= > xH
The characteristic equation for the eigen value is given by A-lI =0
So, eigen values are real if matrix is real and symmetric.
SOL 2.3 Option (D) is correct.
We have
Order is determined by the orders of the highest derivative present in it. So, it is a second order partial differential equation.
It is also a non-linear equation because in linear equation, the product of u with
x u 2
2 is not allow. Therefore, it is a second order, non-linear partial differential equation.
SOL 2.4 Option (C) is correct.
We know
cos x2 =2cos x2 -1 cos x2 = -1 2sin x2
cos x2 = -1 sin x2 The linear equation is given by
y =mx+c
This equation satisfy the above three equations, so that cos x2 , sin x2 , cos x2 are linearly dependent.
SOL 2.5 Option (C) is correct.
We have dt
d f22 +f =0 D2+1 f
^ h =0 The auxiliary equation is m2+1 =0
m =!1
Thus the solution of this equation becomes f t^ h =C1cosx+C2sinx
Hence, the laplace transform is L f t^ h =4L6sinx@
We know, the Gauss divergence theorem is F n dA
Thus the gauss theorem transformed surface integral to volume integral.
F
Hence the given integral becomes dA
SOL 2.7 Option (C) is correct.
Let I e x ln x dx
1
=
#
^ hFrom ILATE, consider ln x^ h as first and x as second function.
I ln x e x dx e dxd ln x e x dx dx
SOL 2.8 Option (B) is correct.
We have
dx
d u kdxdu
2
2 - =0
GATE MCQ Mechanical Engineering (4-volumes)
or ^D2-kD uh =0 The auxiliary equation is
m2-km =0 m m^ -kh =0
or m =0, k
Thus the complete solution is
u C e1 0x C ekx
= + 2
or u =C1+C e2 kx
From the given condition
u 0^ h=0: 0 =C1+C2
C1+C2 =0 ....(i)
and
u L^ h=U: U =C1+C e2 kL ....(ii) Subtracting equation (i) from (ii), we get
U =C e2^ kL-1h
or C2
e U
kL 1
= ^ - h From equation (i), we have
C1 C
Substitute these values in the expression for u, we get
u e
SOL 2.9 Option (D) is correct.
Let A be the event when student knows the answer and B be the event when student guesses the answer. Therefore
P A^ h =P A^ +Bh= 32
where 32 is the probability of correct answer and 31 is the probability that student does not know the answer. So guesses the answer and probability of correct guess is 41. Therefore total probability of correct answer
32 31
41 129
= + # =
Conditional probability that student knows the correct answer /
SOL 2.10 Option (A) is correct.
For y =x straight line and
y =x2 parabola, curve is as given. The shaded region is the area, which is bounded by the both curves (common area).
We solve given equation as follows to gett the intersection points : In y=x2 putting y=x we have x =x2 or
x2-x =0 & x x( -1) =0 & x =0 1,
Then from y=x, for x =0 & y=0 and x =1 & y=1 Curve y=x2 and y=x intersects at point ( , )0 0 and ( , )1 1 So, the area bounded by both the curves is
A dydx
SOL 2.11 Option (C) is correct.
Given f x( )= x (in -1#x#1)
For this function the plot is as given below.
At x = 0, function is continuous but not differentiable because.
For x >0 and x<0 Therefore it is not differentiable.
SOL 2.12 Option (B) is correct.
Let y ( )
It forms : D condition. Hence by L00 -Hospital rule
y ( )
Still these gives : D condition, so again applying L00 -Hospital rule
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SOL 2.13 Option (D) is correct.
We have f x( ) =x3+1
SOL 2.14 Option (A) is correct.
Given : x2+y2+z2 =1
This is a equation of sphere with radius r=1
The unit normal vector at point , ,
SOL 2.15 Option (D) is correct.
First using the partial fraction : ( )
Comparing the coefficients both the sides,
(A+B) =0 and A =1, B =-1
-SOL 2.16 Option (B) is correct.
Given A 3
5 1
=> 3H
For finding eigen values, we write the characteristic equation as A-lI =0
& (5-l)(3-l)-3 =0 8 12
l2- l+ =0 & l =2 6, Now from characteristic equation for eigen vector.
x
Normalized eigen vector 2 1
SOL 2.17 Option (D) is correct.
Given : No. of Red balls =4 No. of Black ball =6
3 balls are selected randomly one after another, without replacement.
1 red and 2 black balls are will be selected as following
Manners Probability for these sequence R B B
Hence Total probability of selecting 1 red and 2 black ball is P 61
SOL 2.18 Option (A) is correct.
We have x
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-Now substitute in equation (i)
[ (D D-1)+D-4]y =0
Substitute C1&C2 in equation (ii), the required solution be y =x2
SOL 2.19 Option (C) is correct.
For given equation matrix form is as follows A The augmented matrix is
: Which is less than the number of unknowns (3)
r6 @A =r8A : BB=2<3 Hence, this gives infinite No. of solutions.
SOL 2.20 Option (B) is correct.
sin q ...
3 5 7
3 5 7
q q q q
= - + - +
SOL 2.21 Option (D) is correct.
Let y lim sin
0 qq
=q"
( )
q Applying L-Hospital rule cos10
= =1
SOL 2.22 Option (C) is correct Let a square matrix
A x y
y
=> xH
We know that the characteristic equation for the eigen values is given by A-lI =0
So, eigen values are real if matrix is real and symmetric.
SOL 2.23 Option (A) is correct.
Let, z1=(1+i), z2=(2-5i)
z =z1#z2 =(1+i)(2-5i) i i i
2 5 2 5 2
= - + - = -2 3i+5 = -7 3i i2=-1
SOL 2.24 Option (D) is correct.
For a function, whose limits bounded between -a to a and a is a positive real number. The solution is given by
( )
SOL 2.25 Option (C) is correct.
Let, f x( ) x dx1
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[1.33 2] . 31 3 3 3333
= + = =1.111
SOL 2.26 Option (D) is correct.
Given : dxdy =(1+y x2)
( y) dy
1+ 2 =xdx
Integrating both the sides, we get y
SOL 2.27 Option (D) is correct.
The probability of getting head p= 21 And the probability of getting tail q 1 21
21
= - = The probability of getting at least one head is
( 1)
SOL 2.28 Option (C) is correct.
Given system of equations are, x x x
2 1+ 2+ 3 =0 ...(i)
x2-x3 =0 ...(ii)
x1+x2 =0 ...(iii)
Adding the equation (i) and (ii) we have
x x
2 1+2 2 =0
x1+x2 =0 ...(iv)
We see that the equation (iii) and (iv) is same and they will meet at infinite points. Hence this system of equations have infinite number of solutions.
SOL 2.29 Option (D) is correct.
The volume of a solid generated by revolution about x-axis bounded by the function f x( ) and limits between a to b is given by
SOL 2.30 Option (B) is correct.
Given:
Order is determined by the order of the highest derivation present in it. So, It is third order equation but it is a nonlinear equation because in linear equation, the product of f with d f d2 / h2 is not allow.
Therefore, it is a third order non-linear ordinary differential equation.
SOL 2.31 Option (D) is correct.
Let I
SOL 2.32 Option (B) is correct.
Let, z ii
SOL 2.33 Option (C) is correct.
( )
Checking the continuity of the function.
At x =23, Lf x( ) lim f 32 h Now checking the differentiability :
( )
SOL 2.34 Option (A) is correct.
Let, A 3
2 1
=> 2H
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And l1 and l2 are the eigen values of the matrix A.
The characteristic equation is written as A-lI =0 So, the eigen vector is
K
Since option A 2 -1
> H is in the same ratio of x1 and x2. Therefore option (A) is an eigen vector.
SOL 2.35 Option (A) is correct.
( )
f t is the inverse Laplace
So, f t( )
Compare the coefficients of s s2, and constant terms and we get A+C =0; A+B =0 and B =1
SOL 2.36 Option (C) is correct.
The box contains : Number of washers = 2
Number of nuts = 3
Number of bolts = 4
Total objects = 2 + 3 + 4 = 9
First two washers are drawn from the box which contain 9 items. So the probability of drawing 2 washers is,
P1
After this box contains only 7 objects and then 3 nuts drawn from it. So the probability of drawing 3 nuts from the remaining objects is,
P2
After this box contain only 4 objects, probability of drawing 4 bolts from the box, P3 CC
11 1
4 4 4 4
= = =
Therefore the required probability is, P =P P P1 2 3 361
351 1 12601
# #
= =
SOL 2.37 Option (B) is correct.
Given : h =60c- =0 60c
h =60#180p = p3 =1 047. radians From the table, we have
y0=0, y1=1066, y2=-323, y3=0, y4=323, y5=-355 and y6=0 From the Simpson’s 1/3rd rule the flywheel Energy is,
E =h y y36( 0+ 6)+4(y1+y3+y5)+2(y2+y4)@ Substitute the values, we get
E =1 047 0 0.3 6( + )+4 1066( + -0 355)+2(-323+323)@
.3 ( )
1 047 4 711 2 0#
= 6 + @ =993 Nm rad (Joules/cycle)
SOL 2.38 Option (A) is correct.
Given : M x5
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Comparing both sides a12 element,25 x
12+53 =0 " x 1225 35
54
=- #
=-SOL 2.39 Option (C) is correct.
Let, V =3xzi+2xyj-yz2k
SOL 2.40 Option (C) is correct.
Let f s( )
s s L 1 21+
= - ; E
First, take the function s s
SOL 2.41 Option (D) is correct.
Total number of cases =23=8
& Possible cases when coins are tossed simultaneously.
H
From these cases we can see that out of total 8 cases 7 cases contain at least one head. So, the probability of come at least one head is = 87
SOL 2.42 Option (C) is correct.
Given : z = +x iy is a analytic function ( )
f z =u x y( , )+iv x y( , )
u =xy ..(i)
Analytic function satisfies the Cauchy-Riemann equation.
xu So from equation (i),
xu 2
2 =y &
yv y 22 = uy
2
2 =x & xv x 22 =-Let v x y( , ) be the conjugate function of u x y( , )
dv xv dx
yv dy 2
2
2
= +2 = -( x dx) +( )y dy Integrating both the sides,
#
dv =-#
xdx+#
ydy v x2 y2 k2 2
=- + + = 21(y2-x2)+k
SOL 2.43 Option (A) is correct.
Given xdxdy +y =x4 dxdy
x y1
+b l =x3 ...(i)
It is a single order differential equation. Compare this with dxdy +Py=Q and we get
P =x1 Q=x3 Its solution will be
( . .)
y I F =
#
Q I F dx( . .) +C . .I F =e#Pdx=e #x dx1 =elog xe =x Complete solution is given by,
yx =
#
x3#xdx+C =#
x dx4 +C x5 C= 5+ ...(ii) and y 1( )= 56 at x=1 & y= 56 From equation (ii),
56#1 = +51 C & C 56
51 1
= - = Then, from equation (ii), we get
yx x 55 1
= + & y x x 54 1
= +
SOL 2.44 Option (B) is correct.
The equation of circle with unit radius and centre at origin is given by, x2+y2=1
Finding the integration of (x+y)2 on path AB traversed in counter-clockwise sense So using the polar form
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Integrating above equation, we getcos22 /
SOL 2.45 Option (A) is correct.
The given equation of surface is
z2 = +1 xy ...(i)
From equation (i) and (ii), we get d2-x2-y2 = +1 xy
d2 =x2+y2+xy+1
Let f x y( , ) =d2=x2+y2+xy+1 ...(iii) The f x y( , ) be the maximum or minimum according to d2 maximum or minimum.
Differentiating equation (iii) w.r.t x and y respectively, we get x Applying maxima minima principle and putting
x So, the nearest point is
z2 = +1 xy= +1 0
& z =!1
SOL 2.46 Option (A) is correct.
Given : y2=4x and x2=4y draw the curves from the given equations,
The shaded area shows the common area. Now finding the intersection points of the curves. So the enclosed area is given by
A (y y dx)
Integrating the equation, we get
A x x
SOL 2.47 Option (A) is correct.
The cumulative distribution function
( )
GATE MCQ Mechanical Engineering (4-volumes)
So, standard deviation
12 1 0
12
= - = 1
SOL 2.48 Option (C) is correct.
Taylor’s series expansion of f x( ) is given by,
SOL 2.49 Option (D) is correct.
Given : xp+3x =0 and x 0( )=1
(D2+3)x =0 D=dtd
The auxiliary Equation is written as m2+3 =0
m =! 3i=0! 3i Here the roots are imaginary
m1 =0 and m2= 3 Solution is given by
x =em t1 (Acosm t2 +Bsinm t2 ) Substituting in equation (i),
1 =[Acos 3 0( )+Bsin 3 0( )]=A+0 A =1
Differentiateing equation (i) w.r.t. t,
xo = 3[-Asin 3t+Bcos 3t] ...(ii) Given x 0o( ) =0 at t=0, x =o 0
Substituting in equation (ii), we get
0 = 3[-Asin0+Bcos0] B =0
Substituting A&B in equation (i) x =cos 3t ( )
x 1 =cos 3 =0 99.
SOL 2.50 Option (B) is correct.
Let f x( ) Substitute the limits, we get
( )
f x = 31(8)-2 3/ = 31(2 )3 -2 3/ 4 1 3 121
= # =
SOL 2.51 Option (A) is correct.
In a coin probability of getting Head
p ..
No of Total cases No of Possible cases 21
= = Probability of getting tail
q 1 21 21
= - =
So the probability of getting Heads exactly three times, when coin is tossed 4 times is
SOL 2.52 Option (C) is correct.
Let, A
Sum of eigen values of matrix = Sum of the diagonal element of matrix A
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1 2 3
l +l +l = + +1 0 p
2 3
l +l = + -1 p l1= + -1 p 3 = -p 2
SOL 2.53 Option (D) is correct.
We know that the divergence is defined as 4:V
Let V =(x-y)i+(y-x)j+(x+ +y z)k
SOL 2.54 Option (A) is correct.
Given :
The equation of line in intercept form is given by x y
Again Integrating and substituting the limits, we get xydxdy
SOL 2.55 Option (B) is correct.
Direction derivative of a function f along a vector P is given by a =grad f: aa
-(2xi 4yj k) (3i 4 )j
-At point P(1, 1, 2) the direction derivative is a 6 1 516 1
105 2
# #
= - =-
=-SOL 2.56 Option (B) is correct.
Given : 2x+3y =4 x+ +y z =4 x+2y-z =a
It is a set of non-homogenous equation, so the augmented matrix of this system is :
So, for a unique solution of the system of equations, it must have the condition [ : ]A B
r =r[ ]A So, when putting a=0 We get r[ : ]A B =r[ ]A
SOL 2.57 Option (D) is correct.
Here we check all the four options for unbounded condition.
(A) / tan xdx Both log 0 and log (–1) undefined so it is unbounded.
SOL 2.58 Option (A) is correct.
Let I =
#
f z dz( ) and f z( )= cosz zThen I cos
z z dz
=
#
=#
zcos-z dz0 ...(i)Given that z =1 for unit circle. From the Cauchy Integral formula
GATE MCQ Mechanical Engineering (4-volumes)
( ) zf z-adz
#
=2pi f a( ) ...(ii)Compare equation (i) and (ii), we can say that, a =0 and f z( )=cosz Or, f a( ) =f(0)=cos0=1 Now from equation (ii) we get
zf z( ) 0dz
#
- =2pi#1=2pi a=0SOL 2.59 Option (D) is correct.
Given y = 32x3 2/ ...(i)
Differentiate equation(i) w.r.t. x dx
SOL 2.60 Option (B) is correct.
Let A 2
2 1
=>0 H l1 and l2 is the eigen values of the matrix.
For eigen values characteristic matrix is, A-lI =0
SOL 2.61 Option (C) is correct.
Given f x y( , ) =yx
First partially differentiate the function w.r.t. y yf
2
2 =xyx 1
-Again differentiate. it w.r.t. x x y
2f 2 2
2 =yx-1( )1 +x y^ x-1logyh =yx 1- ^xlogy+1h At : x=2, y =1
x y
2f 2 2
2 =( )1 2 1- (2log1+1) =1 2( #0+1)=1
SOL 2.62 Option (A) is correct.
Given : ym+2yl+y =0
(D2+2D+1)y =0 where D=d dx/ The auxiliary equation is
m2+2m+1 =0
(m+1)2 =0, m =- -1, 1
The roots of auxiliary equation are equal and hence the general solution of the given differential equation is,
y =(C1+C x e2 ) m x1 (C C x e2 ) x
= 1+ - ..(i)
Given y 0( )=0 at x =0, & y=0 Substitute in equation (i), we get
0 (C1 C2 0)e 0
= + #
0 =C1#1&C1=0 Again y 1( )=0, at x =1 &y=0
Substitute in equation (i), we get
0 [C1 C2 ( )]1 e 1
= + # - [C C e]1
1 2
= +
C1+C2 =0 &C2=0
Substitute C1 and C2 in equation (i), we get y =(0+0x e) -x =0 And y 0 5( . ) =0
SOL 2.63 Option (B) is correct.
Given : y =x2 ...(i)
and interval [1, 5]
At x=1 &y =1
And at x=5 y =( )5 2=25
Here the interval is bounded between 1 and 5 So, the minimum value at this interval is 1.
SOL 2.64 Option (A) is correct Let square matrix
A x y
y
=> xH
The characteristic equation for the eigen values is given by A-lI =0
x y
y x l
l
-- =0 (x-l)2-y2 =0 (x-l)2 =y2
GATE MCQ Mechanical Engineering (4-volumes)
x-l =!y l =x!y
So, eigen values are real if matrix is real and symmetric.
SOL 2.65 Option (B) is correct.
The Cauchy-Reimann equation, the necessary condition for a function f z( ) to be analytic is
2x 2j
2y
=2y
2y 2j
2x
=-2y when
2x 2j,
2y 2j,
2y 2y,
2x
2y exist.
SOL 2.66 Option (A) is correct.
Given :
x2 y x y 0
2
2 2
2 2
2 2
2 2
2 j + j + j + 2j =
Order is determined by the order of the highest derivative present in it.
Degree is determined by the degree of the highest order derivative present in it after the differential equation is cleared of radicals and fractions.
So, degree = 1 and order = 2
SOL 2.67 Option (B) is correct.
Given y = +x x+ x+ x+...3 ...(i) y-x = x+ x+ x+....3
Squaring both the sides,
(y-x)2 = +x x+ x+...3
(y-x)2 =y From equation (i)
y2+x2-2xy =y ...(ii)
We have to find y 2( ), put x=2 in equation (ii), y2+ -4 4y =y
y2-5y+4 =0 (y-4)(y-1) =0
y =1 4, From Equation (i) we see that
For y 2( ) y = +2 2+ 2+ 2+...3 >2 Therefore, y =4
SOL 2.68 Option (B) is correct.
Vector area of TABC,
A 21BC BA
= # 21(c b) (a b)
= - #
-[c a c b b a b b]
-SOL 2.69 Option (C) is correct.
Given : dxdy =y2 or y dy
2 =dx
Integrating both the sides y
SOL 2.70 Option (A) is correct.
Let f( )t tf t dt( )
SOL 2.71 Option (A) is correct.
From the Trapezoidal Method ( )
GATE MCQ Mechanical Engineering (4-volumes)
( ) sin
f x = x 0 0.707 1 0.707 0 -0 707. -1 -0 707. 0 Now from equation(i),
sin xdx
0
#
2p = p8[0+2(0.707+ +1 0.707+ -0 0.707- -1 0.0707+0)]8 #0 0
= p =
SOL 2.72 Option (D) is correct.
The X and Y be two independent random variables.
So, E XY( ) =E X E Y( ) ( ) (i)
& covariance is defined as ( , )
For two independent random variables
( )
Var X+Y =Var( )X +Var( )Y and E X Y( 2 2) =E X E Y( 2) ( 2) So, option (D) is incorrect.
SOL 2.73 Option (B) is correct.
Let, f x( ) lim
SOL 2.74 Option (B) is correct.
Let, A 2
0 1
=> 2H
Let l is the eigen value of the given matrix then characteristic matrix is
A-lI =0 Here I 1 So, only one eigen vector.
SOL 2.75 Option (D) is correct.
Column I
P. Gauss-Seidel method 4. Linear algebraic equation Q. Forward Newton-Gauss method 1. Interpolation
R. Runge-Kutta method 2. Non-linear differential equation S. Trapezoidal Rule 3. Numerical integration
So, correct pairs are, P-4, Q-1, R-2, S-3
SOL 2.76 Option (B) is correct.
Given : dxdy +2xy =e-x2 and y 0( )=1
It is the first order linear differential equation so its solution is ( . .)
y I F =
#
Q I F dx( . .) +C So, I F. . =e#Pdx=e#2xdx( ) compare with dxdy +P y =Q e2 xdx
= # =e2#x22=ex2 The complete solution is,
yex2 =
#
e-x2#e dxx2 +C dx C=
#
+ = +x Cy e
x c
x2
= + ...(i)
Given y(0) =1
At x =0 &y=1 Substitute in equation (i), we get
1 =C1 &C=1
Then y ( )
e
x x21 x 1 e x
= + = + - 2
SOL 2.77 Option (C) is correct.
The incorrect statement is, S={ :x x!Aandx!B} represents the union of set A and set B.
The above symbol ( )! denotes intersection of set A and set B. Therefore this statement is incorrect.
SOL 2.78 Option (D) is correct.
Total number of items = 100 Number of defective items = 20 Number of Non-defective items = 80
Then the probability that both items are defective, when 2 items are selected at random is,
P C CC
100 2 20 280
= 0
100!! !! !! 98 2 18 220
=
20 19 10022 99
#
#
= = 49519
Alternate Method :
Here two items are selected without replacement.
Probability of first item being defective is P1 10020
15
= =
After drawing one defective item from box, there are 19 defective items in the 99 remaining items.
Probability that second item is defective, P2 = 89919
then probability that both are defective
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P =P1#P2 51 9919
49519
= # =
SOL 2.79 Option (A) is correct.
Given : S 3
2 2
=> 3H
Eigen values of this matrix is 5 and 1. We can say l =1 1 l =2 5 Then the eigen value of the matrix
S2 =S S is l12, l22
SOL 2.80 Option (B) is correct.
Given f x( ) =(x-8)2 3/ +1
The equation of line normal to the function is
(y-y1) =m x2( -x1) ...(i) Slope of tangent at point (0, 5) is
( )
=-We know the slope of two perpendicular curves is -1. m m1 2 =-1
m2 m1 / 1 31 3
=- 1 = -- = The equation of line, from equation (i) is
(y-5) =3(x-0) y =3x+5
SOL 2.81 Option (A) is correct.
Let f x( ) / e dtit
SOL 2.82 Option (B) is correct.
Given f x( )
-" Applying L – Hospital rule
Substitute the limit, we get
( )
SOL 2.83 Option (A) is correct.
(P) Singular Matrix " Determinant is zero A =0
(Q) Non-square matrix " An m#n matrix for which m!n, is called non-square matrix. Its determinant is not defined (R) Real Symmetric Matrix " Eigen values are always real.
(S) Orthogonal Matrix " A square matrix A is said to be orthogonal if AAT=I Its determinant is always one.
SOL 2.84 Option (B) is correct.
Given : The auxiliary Equation is,
m2+4m+3 =0 & m =-1,-3
SOL 2.85 Option (C) is correct.
Given EF =G where G= = Identity matrixI
We know that the multiplication of a matrix and its inverse be a identity matrix AA-1 =I
SOL 2.86 Option (B) is correct.
The probability density function is, ( )
-For standard deviation first we have to find the mean and variance of the function.
( )
GATE MCQ Mechanical Engineering (4-volumes)
SOL 2.87 Option (A) is correct.
The Stokes theorem is, dr
Here we can see that the line integral F dr
C :
#
and surface integral (CurlF) dsS :
##
is related to the stokes theorem.
SOL 2.88 Option (B) is correct.
Let, P = defective items Q = non-defective items
10% items are defective, then probability of defective items P =0 1.
Probability of non-defective item Q = -1 0 1. =0 9.
The Probability that exactly 2 of the chosen items are defective is ( ) ( )
SOL 2.89 Option (A) is correct.
Let f x( ) (sin x sin x dx)
a
a 6 7
= +
#
-sin xdx sin xdx
a
SOL 2.90 Option (C) is correct.
We know, from the Echelon form the rank of any matrix is equal to the Number of non zero rows.
Here order of matrix is 3#4, then, we can say that the Highest possible rank of this matrix is 3.
SOL 2.91 Option (A) is correct.
We can draw the graph from the limits of the integration, the limit of y is from y x
Comparing the limits and get r=0, s=2, p=0, q=4y
SOL 2.92 Option (A) is correct.
Let, A The characteristic equation for eigen values is given by,
A-lI =0
Solving this, we get
(5-l)(5-l)[(2-l)(1-l)-3] =0
GATE MCQ Mechanical Engineering (4-volumes)
SOL 2.93 Option (C) is correct.
Given : x+y =2 ...(i)
. x . y
1 01 +0 99 =b, db =1 unit ...(ii) We have to find the change in x in the solution of the system. So reduce y From the equation (i) and (ii).
Multiply equation (i) by 0.99 and subtract from equation (ii) 1.01x+0.99y-(0.99x+0.99 )y = -b 1 98.
. x . x
1 01 -0 99 = -b 1 98. . x
0 02 = -b 1 98. Differentiating both the sides, we get
. dx 0 02 =db
dx = 0 02.1 =50 unit db = 1
SOL 2.94 Option (A) is correct.
Given, x u v( , ) =uv
SOL 2.95 Option (D) is correct.
Given : Radius of sphere r =1 Let, Radius of cone =R
Height of the cone =H
Finding the relation between the volume and Height of the cone From DOBD, OB2 =OD2+BD2
1 =(H-1)2+R2 =H2+ -1 2H+R2 R2+H2-2H =0
R2 =2H-H2 ...(i)
Volume of the cone, V = 31pR H2
Substitute the value of R2 from equation (i), we get
V 31 (2H H H) (2H H ) 31
2 2 3
p p
= - =
-Differentiate V w.r.t to H
dVdH = 31p[4H-3H2] Again differentiate
dH d V
2
2 = 31p[4-6H]
For minimum and maximum value, using the principal of minima and maxima.
Put dVdH =0
[ H H ] 31p 4 -3 2 =0
[ ]
H 4-3H =0 & H=0 and H = 34 At H = 34,
dH d V2
2 31 4 6 34 [ ]
31 4 8 34 <0
p # p p
= : - D= - =- (Maxima)
And at H=0, dH d V2
2 = 31p[4-0]= 34p>0 (Minima) So, for the largest volume of cone, the value of H should be 4 3/
SOL 2.96 Option (D) is correct.
Given : x dx2dy +2xy ln( ) x x
= 2
dx
dy +2xy ln( ) x
x 2
= 3
Comparing this equation with the differential equation dxdy +P y( )=Q we have P= x2 and ln( )
Q x
x 2
= 3
GATE MCQ Mechanical Engineering (4-volumes)
The integrating factor is,
I.F.= e#Pdx =e#x dx2 e2lnx =elnx2=x2 Complete solution is written as,
( . .)
Substitute the value from equation (iii) in equation (i), ( ) So from equation (iv), we get
x y2 =(ln x)2
SOL 2.97 Option (A) is correct.
Potential function of v=x yz2 at P 1 1 1( , , ) is =12#1#1=1 and at origin
SOL 2.98 Option (C) is correct.
Let, f x( ) =x3+3x-7 From the Newton Rapson’s method
xn 1+
( )
f xl =3x2+3 ( )
f xl 0 =3#( )1 2+ =3 6
Then, x1 = -1 (-63)= + = +1 63 1 21=23 =1 5.
SOL 2.99 Option (D) is correct.
We know a die has 6 faces and 6 numbers so the total number of ways 6#6 36
= =
And total ways in which sum is either 8 or 9 is 9, i.e.
( , ),( , )( , )( , )( , )( , )( , )( , )( , )2 6 3 6 3 5 4 4 4 5 5 4 5 3 6 2 6 3
Total number of tosses when both the 8 or 9 numbers are not come 36 9 27
= - =
Then probability of not coming sum 8 or 9 is, = 3627 = 43
SOL 2.100 Option (C) is correct.
Given :
dx
d y22 +pdxdy +qy =0
The solution of this equation is given by, y c e1 mx c enx
= + 2 ...(i)
Here m & n are the roots of ordinary differential equation
Given solution is, y =c e1 -x+c e2 -3x ...(ii)
SOL 2.101 Option (C) is correct.
Given : ( 1)
The auxilliary equation of equation (i) is written as m2+4m+4 =0 & m =-2,-2
Here the roots of auxiliary equation are same then the solution is y =(c1+c x e2 ) mx =xe-2x Let c
SOL 2.102 Option (C) is correct.
Given : x =a q( +sinq), y =a 1( -cosq) First differentiate x w.r.t. q,
ddx
q =a 1[ +cosq] And differentiate y w.r.t. q
d
GATE MCQ Mechanical Engineering (4-volumes)
Substitute the values of dyd
q and dxd q dx
dy
[ ]
sin cos sincos
a a 1 1
SOL 2.103 Option (C) is correct.
Given : P 0 866 0 500 0( . , . , ), so we can write P =0.866i+0.5j+0k ( . , . , )
Q= 0 259 0 966 0 , so we can write
Q =0.259i+0.966j+0k For the coplanar vectors
P:Q = P Q cosq
SOL 2.104 Option (B) is correct.
Let A
We know that the sum of the eigen value of a matrix is equal to the sum of the diagonal elements of the matrix
So, the sum of eigen values is,
1+5+1 =7
SOL 2.105 Option (D) is correct.
Given : Total number of cards = 52 and two cards are drawn at random.
Number of kings in playing cards = 4
So the probability that both cards will be king is given by, P CC
-SOL 2.106 Option (B) is correct.
Given : U t( -a) 0,
From the definition of Laplace Transform [ ( )]F t
-( ) ( )
-SOL 2.107 Option (D) is correct.
First we have to make the table from the given data
Take x0=0 and h=1 Then P = -x hx0 =x From Newton’s forward Formula
( )
SOL 2.108 Option (A) is correct.
SOL 2.108 Option (A) is correct.