Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on Euler’s Theorem and Answers

Engineering Mathematics Multiple Choice Questions on “Euler’s Theorem – 1”.

1. f(x, y) = x3 + xy2 + 901 satisfies the Euler’s theorem.
a) True
b) False
Answer: b
Explanation: The function is not homogenous and hence does not satisfy the condition posed by euler’s theorem.

2. f(x, y)=(frac{x^3+y^3}{x^{99}+y^{98}x+y^{99}}) find the value of fy at (x,y) = (0,1).
a) 101
b) -96
c) 210
d) 0
Answer: b
Explanation: Using Euler theorem
xfx + yfy = n f(x, y)
Substituting x = 0; n=-96 and y = 1 we have
fy = -96. f(0, 1) = -96.(1⁄1)
= – 96.

3. A non-polynomial function can never agree with euler’s theorem.
a) True
b) false
Answer: b
Explanation: Counter example is the function
(f(x, y)=x^9.y^8sin(frac{x^2+y^2}{xy})+cos(frac{x^3}{x^2y+yx^2})x^{11}.y^6).

4. (f(x, y)=x^9.y^8sin(frac{x^2+y^2}{xy})+cos(frac{x^3}{x^2y+yx^2})x^{11}.y^6) Find the value of fx at (1,0).
a) 23
b) 16
c) 17(sin(2) + cos(1⁄2))
d) 90
Answer: c
Explanation: Using Eulers theorem we have
xfx + yfy = nf(x, y)
Substituting (x,y)=(1,0) we have
fx = 17f(1, 0)
17 (sin(2) + cos(1⁄2)).

5. For a homogeneous function if critical points exist the value at critical points is?
a) 1
b) equal to its degree
c) 0
d) -1
Answer: c
Explanation: Using Euler theorem we have
xfx + yfy = nf(x, y)
At critical points fx = fy = 0
f(a, b) = 0(a, b) → critical points.

6. For homogeneous function with no saddle points we must have the minimum value as _____________
a) 90
b) 1
c) equal to degree
d) 0
Answer: d
Explanation: Substituting fx = fy = 0 At critical points in euler theorem we have
nf(a, b) = 0 ⇒ f(a, b) = 0(a, b) → critical points.

7. For homogeneous function the linear combination of rates of independent change along x and y axes is __________
a) Integral multiple of function value
b) no relation to function value
c) real multiple of function value
d) depends if the function is a polynomial
Answer: c
Explanation: Euler’s theorem is nothing but the linear combination asked here, The degree of the homogeneous function can be a real number. Hence, the value is integral multiple of real number.

8. A foil is to be put as shield over a cake (circular) in a shape such that the heat is even along any diameter of the cake.
Given that the heat on cake is proportional to the height of foil over cake, the shape of the foil is given by
a) f(x, y) = sin(y/x)x2 + xy
b) f(x, y) = x2 + y3
c) f(x, y) = x2y2 + x3y3
d) not possible by any analytical function
Answer: b
Explanation:Given that the heat is same along lines we need to choose a homogeneous function.
Checking options we get that only option satisfies condition for homogeneity.

9. f(x, y) = sin(y/x)x3 + x2y find the value of fx + fy at (x,y)=(4,4).
a) 0
b) 78
c) 42 . 3(sin(1) + 1)
d) -12
Answer: c
Explanation: Using Euler theorem we have
xfx + yfy = nf(x, y)
Substituting (x,y)=(4,4) we have
4fx + 4fy = 3f(4, 4) = 3⁄4(43 . sin(1) + 43)
= 42 . 3(sin(1) + 1).

250+ TOP MCQs on Improper Integrals and Answers

Engineering Mathematics Multiple Choice Questions on “Improper Integrals – 1”.

1. Integration of function is same as the ___________
a) Joining many small entities to create a large entity
b) Indefinitely small difference of a function
c) Multiplication of two function with very small change in value
d) Point where function neither have maximum value nor minimum value
Answer: a
Explanation: Integration of function is same as the Joining many small entities to create a large entity.

2. Integration of (Sin(x) + Cos(x))ex is______________
a) ex Cos(x)
b) ex Sin(x)
c) ex Tan(x)
d) ex (Sin(x)+Cos(x))
Answer: b
Explanation: Let f(x) = ex Sin(x)
∫ ex Sin(x)dx = ex Sin(x) – ∫ ex Cos(x)dx
∫ ex Sin(x)dx + ∫ ex Cos(x)dx = ∫ ex [Cos(x)+Sin(x)]dx = ex Sin(x).

3. Integration of (Sin(x) – Cos(x))ex is ___________
a) -ex Cos(x)
b) ex Cos(x)
c) -ex Sin(x)
d) ex Sin(x)
Answer: a
Explanation: Add constant automatically
Let f(x) = ex Sin(x)
∫ ex Sin(x)dx = -ex Cos(x) + ∫ ex Cos(x)dx
∫ ex Sin(x)d-∫ ex Cos(x)dx = ∫ ex [Sin(x)-Cos(x)]dx = -ex Cos(x).

4. Value of ∫ Cos2 (x) Sin2 (x)dx.
a) (frac{1}{8} [x-frac{Cos(2x)}{2}])
b) (frac{1}{4} [x-frac{Cos(2x)}{2}])
c) (frac{1}{8} [x-frac{Sin(2x)}{2}])
d) (frac{1}{4} [x-frac{Sin(2x)}{2}])
Answer: c
Explanation: Add constant automatically
Given,f(x)=(int Cos^2 (x) Sin^2 (x)dx=frac{1}{4} int Sin^2 (2x) dx=frac{1}{4} int frac{[1-Cos(2x)]}{2} dx=frac{1}{8} [x-frac{Sin(2x)}{2}])

5. If differentiation of any function is zero at any point and constant at other points then it means?
a) Function is parallel to x-axis at that point
b) Function is parallel to y-axis at that point
c) Function is constant
d) Function is discontinuous at that point
Answer: a
Explanation: Since slope of a function is given by dy⁄dx at that point. Hence, when dy⁄dx = 0 means slope of a function is zero i.e, parallel to x axis.
Function is not a constant function since it has finite value at other points.

6. If differentiation of any function is infinite at any point and constant at other points then it means ___________
a) Function is parallel to x-axis at that point
b) Function is parallel to y-axis at that point
c) Function is constant
d) Function is discontinuous at that point
Answer: a
Explanation: Since slope of a function is given by dy⁄dx at that point.Hence,when dy⁄dx = ∞ means slope of a function is 90 degree i.e,parallel to y axis.

7. Integration of function y = f(x) from limit x1 < x < x2 , y1 < y < y2, gives ___________
a) Area of f(x) within x1 < x < x2
b) Volume of f(x) within x1 < x < x2
c) Slope of f(x) within x1 < x < x2
d) Maximum value of f(x) within x1 < x < x2
Answer: a
Explanation: Integration of function y=f(x) from limit x1 < x < x2 , y1 < y < y2, gives area of f(x) within x1 < x < x2.

8. Find the value of ∫ ln⁡(x)⁄x dx.
a) 3a2
b) a2
c) a
d) 1
Answer: a
Explanation: Add constant automatically
Given, f(x)=(int frac{ln⁡(x)}{x} dx)
Let, z=ln⁡(x)=>dz=(frac{dx}{x}=>f(x)=int zdz=z^2/2=frac{ln^2⁡(x)}{2})

9. Find the value of ∫t⁄(t+3)(t+2) dt, is?
a) 2 ln⁡(t+3)-3 ln⁡(t+2)
b) 2 ln⁡(t+3)+3 ln⁡(t+2)
c) 3 ln⁡(t+3)-2 ln⁡(t+2)
d) 3 ln⁡(t+3)+2ln⁡(t+2)
Answer: c
Explanation: Add constant automatically
Given, et = x => dx = et dt,
Given, f(x)=(int frac{ln⁡(x)}{x} dx)
Let, z=ln⁡(x)=>dz=(frac{dx}{x}=>f(x)=int zdz=frac{z^2}{2}=frac{ln^2⁡(x)}{2})

10. Find the value of ∫ cot3(x) cosec4 (x).
a) –([frac{cot^4⁡(x)}{4}+frac{cosec^6⁡(x)}{6}])
b) –([frac{cosec^4⁡(x)}{4}+frac{cosec^6⁡(x)}{6}])
c) –([frac{cot^4⁡(x)}{4}+frac{cot^6⁡(x)}{6}])
d) –([frac{cosec^4⁡(x)}{4}+frac{cot^6⁡(x)}{6}])
Answer: c
Explanation: Add constant automatically
Given, (int cot^3⁡(x)cosec^4 (x)dx=-int cot^3⁡(x)cosec^2 (x)dcot(x))
=-(int t^3 (1+t^2)dt=-[frac{t^4}{4}+frac{t^6}{6}]=-[frac{cot^4⁡(x)}{4}+frac{cot^6⁡(x)}{6}])

11. Find the value of (int frac{sec^4⁡(x)}{sqrt{tan⁡(x)}} dx).
a) (frac{2}{5}sqrt{tan⁡(x)}[5+sec^2⁡(x)])
b) (frac{2}{5}sqrt{sec⁡(x)}[5+tan^2⁡(x)])
c) (frac{2}{5}sqrt{tan⁡(x)}[6+tan^2⁡(x)])
d) (frac{2}{5}sqrt{tan⁡(x)}[5+tan^2⁡(x)])
Answer: d
Explanation: Add constant automatically
Given, (int frac{sec^4⁡(x)}{sqrt{tan⁡(x)}} dx)
=(int frac{sec^2⁡(x) sec^2⁡(x)}{sqrt{tan⁡(x)}} dx)
=(int frac{1+t^2}{sqrt{t}} dt)
=(int [frac{1}{sqrt{t}}+t^{3/2}]dt)
=(2sqrt{t}+frac{2}{5} t^{5/2})
=(frac{2}{5}sqrt{tan⁡(x)}[5+tan^2⁡(x)])

12. Find the value of (int frac{1}{4x^2+4x+5} dx).
a) 1⁄8 sin(-1)⁡(x + 1⁄2)
b)1⁄4 tan(-1)⁡(x + 1⁄2)
c) 1⁄8 sec(-1)⁡(x + 1⁄2)
d) 1⁄4 cos(-1)⁡(x + 1⁄2)
Answer: b
Explanation: Add constant automatically
Given, (int frac{1}{4x^2+4x+5} dx)
=(int frac{1}{4 (x^2+x+frac{5}{4}+frac{1}{4}+frac{1}{4})} dx =int frac{1}{4[(x+frac{1}{2})^2+1^2])}dx=frac{1}{4} tan^{-1}(x+frac{1}{2}))

13. Find the value of (int sqrt{4x^2+4x+5} dx).
a) (2left [frac{1}{2} (x+frac{1}{2}) sqrt{{(x+frac{1}{2})^2+1)}}right ]+ln⁡left [(x+frac{1}{2})+sqrt{(x+frac{1}{2})^2+1} right ])
b) (2left [frac{1}{2} sqrt{(x+frac{1}{2})^2+1)}right ]+frac{1}{2} ln⁡left [(x+frac{1}{2})+sqrt{(x+frac{1}{2})^2+1} right ])
c) (2left [frac{1}{2} (x+frac{1}{2}) sqrt{(x+frac{1}{2})^2+1)}right ]+frac{1}{2} ln⁡left [(x+frac{1}{2})+sqrt{(x+frac{1}{2})^2+1} right ])
d) (2left [(x+frac{1}{2}) sqrt{{(x+frac{1}{2})^2+1)}}right ]+frac{1}{2} ln⁡left [(x+frac{1}{2})+sqrt{(x+frac{1}{2})^2+1} right ])
Answer: c
Explanation: Add constant automatically
Given, (int sqrt{4x^2+4x+5} dx=int 2sqrt{(x+frac{1}{2})^2+1^2} dx)
=(int 2sqrt{t^2+1^2} dt=2left [frac{1}{2} tsqrt{t^2+1}right ]+frac{1}{2} ln⁡[t+sqrt{t^2+1}])
=(2left [frac{1}{2} (x+frac{1}{2}) sqrt{(x+frac{1}{2})^2+1)} right ]+frac{1}{2} ln⁡left [(x+frac{1}{2})+sqrt{(x+1/2)^2+1}right ])

250+ TOP MCQs on Clairaut’s and Lagrange Equations and Answers

Ordinary Differential Equations Multiple Choice Questions on “Clairaut’s and Lagrange Equations”.

1. Singular solution for the Clairaut’s equation (y = y’x+frac{a}{y’}) is given by _______
a) (frac{x^2}{a^2} + frac{y^2}{a^2} = 1)
b) y2=-4ax
c) y2=4ax
d) x2=-2ay
Answer: c
Explanation: Let ( p = y’ = frac{dy}{dx})
Given equation is of the form y = px + f(p), whose general solution is y = cx + f(c)
thus the general solution is (y = cx + frac{a}{c}) …………..(1), to find the value of c
we differentiate (1) partially w.r.t ‘c’ i.e ( 0 = x-frac{a}{c^2} rightarrow c^2 = frac{a}{x} rightarrow c = sqrt{frac{a}{x}})
hence (1) becomes (y = sqrt{frac{a}{x}} x + a sqrt{frac{x}{a}} rightarrow y=2sqrt{ax})
y2=4ax is the singular solution.

2. Obtain the general solution for the equation xp2+px-py+1-y=0 where p=(frac{dy}{dx}).
a) y=cx+(frac{1}{c+1})
b) x=cy-(c+1)
c) x=cy-(frac{1}{c+1})
d) y=cx+(c+1)
Answer: a
Explanation: xp2+px-py+1-y=0
xp2+px+1=y(p+1)
(y=frac{xp(p+1)+1}{p+1} ,or, y=px+frac{1}{p+1})……(1) thus (1) is in the Clairaut’s equation form y=px+f(p),
thus general solution is y=cx+(frac{1}{c+1}).

3. Find the general solution for the equation (px-py)(py+x)=2p by reducing into Clairaut’s form by using the substitution X=x2, Y=y2 where p=(frac{dy}{dx}).
a) (y^2 = x + frac{c}{c+1})
b) (y^2 = cx^2 – frac{2c}{c+1})
c) (x^2 = cy^2 – frac{1}{2c+1})
d) (x^2 = y^2 + frac{c}{2c+2})
Answer: b
Explanation: (X=x^2 rightarrow frac{dX}{dx} = 2x)
(Y=y^2 rightarrow frac{dY}{dy} = 2y)
now (p = frac{dy}{dx} = frac{dy}{dY} frac{dY}{dX} frac{dX}{dx} ,and, ,let, P=frac{dY}{dx})
(p=frac{1}{2y} * P * 2x ,or, p=frac{x}{y} ,P ,i.e, p=sqrt{frac{X}{Y}} P)
now consider (px-py)(py+x)=2p substituting the value of p we get
(left(sqrt{frac{X}{Y}} P sqrt{X} – sqrt{Y}right)left(sqrt{frac{X}{Y}} P sqrt{Y} + sqrt{X}right) = 2sqrt{frac{X}{Y}} P)
(frac{(PX-Y)}{sqrt{Y}} (P+1)sqrt{X} = 2sqrt{frac{X}{Y}} P rightarrow (PX-Y)(P+1)=2P ,or, Y(P)=PX-frac{2P}{P+1}) is in the Clairaut’s form
hence general solution is (y^2 = cx^2 – frac{2c}{c+1}).

4. Find the general solution of the D.E 2y-4xy’-log y’=0.
a) (y(p) = frac{2c}{p} – 1 + frac{log⁡p}{2} )
b) (y(p) = frac{c}{2p} – 2 + log⁡p)
c) (x(p) = frac{-1}{p} + frac{c}{p^2} )
d) (x(p) = frac{1}{2p} + frac{c}{p^{1/2}} )
Answer: a
Explanation: Let y’=p and hence given equation is in Lagrange equation form
i.e 2y=4xp+log p …..(1) differentiating both sides of the equation
2dy=4xdp+4pdx+(frac{dp}{p}) and dy=pdx
–> 2pdx=4xdp+4pdx+(frac{dp}{p}) –> -2pdx=4pdx+(frac{dp}{p})
-2p(frac{dx}{dp} = 4x + frac{1}{p} rightarrow frac{dx}{dp} + frac{2}{p} x = frac{-1}{2p^2}) (p≠0)…….this is a linear D.E for the function x(p)
I.F is (e^{int frac{2}{p} ,dp} = e^{log⁡p^2} = p^2) and solution is x(p) (p^2 = int p^2 *frac{-1}{2p^2} ,dp + c)
(x(p) = frac{-1}{2p} + frac{c}{p^2}) substituting back in (1) we get (2y=4pleft(frac{-1}{2p} + frac{c}{p^2}right) + log p)
(y(p) = frac{2c}{p} – 1 + frac{log⁡p}{2} ).

5. Find the general solution of the D.E y = 2xy’ – 3(y’)2.
a) (y(p) = p^{1/2} + frac{c}{2p})
b) (y(p) = p^2 + frac{2c}{p})
c) (x(p) = -cp + frac{c}{p^2})
d) (x(p) = 2p + frac{2c}{p^2} )
Answer: b
Explanation: Let y’=p –> y = 2xp – 3p2 ….(1) is in the Lagrange equation form
now differentiating we get dy=2xdp+2pdx-6pdp and dy=pdx
thus pdx=2xdp+2pdx-6pdp→-pdx=2xdp-6pdp –> (frac{dx}{dp} + frac{2}{p} x – 6=0)…(2)
(2) is a linear D.E whose I.F=(e^{int frac{2}{p} ,dp} = p^2) hence its solution is
(p^2 x(p) = int 6p^2 ,dp + c rightarrow x(p) = 2p + frac{c}{p^2}) ….substituting in (1) we get
(y(p) = 2(2p+frac{c}{p^2})p-3p^2 = p^2 + frac{2c}{p}.)

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250+ TOP MCQs on Laplace Transform by Properties and Answers

Engineering Mathematics Multiple Choice Questions on “Laplace Transform by Properties – 3”.

1. Time domain function of (frac{s}{a^2+s^2}) is given by?
a) Cos(at)
b) Sin(at)
c) Cos(at)Sin(at)
d) Sin(t)
Answer: a
Explanation: L[Cos(at)] = (frac{s}{a^2+s^2})
L-1 ([frac{s}{a^2+s^2}]) = Cos(at).

2. Inverse Laplace transform of (frac{1}{(s+1)(s-1)(s+2)}) is?
a) –1⁄2 et + 1⁄6 e-t + 1⁄3 e2t
b) –1⁄2 e-t + 1⁄6 et + 1⁄3 e-2t
c) 1⁄2 e-t – 1⁄6 et – 1⁄3 e-2
d) –1⁄2 e-t + 1⁄6 e-t + 1⁄3 e-2
Answer: b
Explanation:
Given, (F(s)=frac{1}{(s+1)(s-1)(s+2)}=frac{-1}{2(s+1)} +frac{1}{6(s-1)}+frac{1}{3(s+2)})
Hence, inverse laplace transform is (f(t)=-frac{1}{2} e^{-t}+frac{1}{6} e^t+frac{1}{3} e^{-2})

3. Inverse laplace transform of (frac{1}{(s-1)^2 (s+5)}) is?
a) 1⁄6 e – t – 1⁄36 et + 1⁄36 e-5t
b) 1⁄6 ett – 1⁄36 et + 1⁄36 e-5t
c) 1⁄6 e-tt2 – 1⁄36 e-t + 1⁄36 e5t
d) 1⁄6 e-t t-1⁄36 e-t + 1⁄36 e5t
Answer: a
Explanation:
Given, (F(s)=frac{1}{(s-1)^2 (s+5)}=frac{1}{(s-1)} left [frac{1}{(s-1)(s+5)}right ])
=(frac{1}{(s-1)} left [frac{1}{6(s-1)}-frac{1}{6(s+5)}right ])
=(frac{1}{6} left [frac{1}{(s-1)^2}-frac{1}{(s-1)(s+5)}right ])
=(frac{1}{6} left [
frac{1}{(s-1)^2} – frac{1}{6} left [frac{1}{(s-1)} – frac{1}{(s+5)}right ]right ])
=(frac{1}{6(s-1)^2}-frac{1}{36(s-1)}+frac{1}{36(s+5)})
Inverse Laplace transform is (f(t)=frac{1}{6} e^t t-frac{1}{36} e^t+frac{1}{36} e^{-5t})

4. Find the inverse laplace transform of (frac{1}{(s^2+1)(s – 1)(s + 5)}).
a) 1⁄12 et – 1⁄13 Cos(-t) – 1⁄12 Sin(-t) – 1⁄156 e-5t
b) 1⁄12 e-t – 1⁄13 Cos(t) – 1⁄12 Sin(t) – 1⁄156 e5t
c) 1⁄12 et – 1⁄13 Cos(t) – 1⁄12 Sin(t) – 1⁄156 e-5t
d) 1⁄12 et + 1⁄13 Cos(t) + 1⁄12 Sin(t) + 1⁄156 e-5t
Answer: c
Explanation:
Given , F(s)=(frac{1}{(s^2+1)(s-1)(s+5)})
F(s)=(frac{1}{6(s^2+1)}left [frac{1}{s-1}-frac{1}{s+5}right ]=frac{1}{6(s^2+1)(s-1)}-frac{1}{6(s^2+1)(s+5)})
=(frac{1}{6} left [frac{1}{2*(s – 1)}-frac{1}{2} frac{s+1}{(s^2+ 1)}right ]-frac{1}{6}left [frac{1}{26*(s + 5)}-frac{1}{26} frac{s-5}{(s^2+1)}right ])
=(frac{1}{12(s – 1)}-frac{1}{26} frac{2s+3}{(s^2+ 1)}-frac{1}{156(s + 5)})
=(frac{1}{12(s – 1)}-frac{1}{13} frac{s}{(s^2+ 1)}-frac{1}{12} frac{1}{(s^2+ 1)}-frac{1}{156(s + 5)})
=(frac{1}{12} e^t-frac{1}{13} Cos(t)-frac{1}{12} Sin(t)-frac{1}{156}e^{-5t})

5. Find the inverse laplace transform of (frac{s}{(s^2+ 4)^2}).
a) 1⁄4 sin(2t)
b) t2⁄4 sin(2t)
c) t⁄4 sin(2t)
d) t⁄4 sin(2t2)
Answer: c
Explanation:
Given, (Y(s)=frac{s}{(s^2+ 4)^2})
Inverse Laplace transform of (frac{1}{s^2+4})=sin⁡(2t)
Now, (frac{d}{ds} (frac{1}{s^2+4}))=-tsin(2t)
Inverse lapalce of (frac{-2s}{(s^2+4)^2}=-frac{t}{2} sin(2t))
Inverse lapalce of (frac{s}{(s^2+4)^2}=frac{t}{4} sin(2t))

6. Final value theorem states that _________
a) x(0)=(lim_{xrightarrow ∞} sX(s))
b) x(∞)=(lim_{xrightarrow ∞} sX(s))
c) x(0)=(lim_{xrightarrow 0} sX(s))
d) x(∞)=(lim_{xrightarrow 0} ⁡sX(s))
Answer: d
Explanation: Final value theorem states that
x(∞)=(lim_{xrightarrow 0} ⁡sX(s))

7. Initial value theorem states that ___________
a) x(0)=(lim_{xrightarrow ∞} sX(s))
b) x(∞)=(lim_{xrightarrow ∞} sX(s))
c) x(0)=(lim_{xrightarrow 0} sX(s))
d) x(∞)=(lim_{xrightarrow 0} ⁡sX(s))
Answer: a
Explanation: Initial value theorem states that
x(0)=(lim_{xrightarrow ∞} sX(s))

8. Find the value of x(∞) if (X(s)=frac{2s^2+5s+12/s}{s^3+4s^2+14s+20}).
a) 5
b) 4
c) 12⁄20
d) 2
Answer: c
Explanation:
Given, (X(s)=frac{2s^2+5s+12/s}{s^3+4s^2+14s+20})
Hence, (sX(s)=frac{2s^3+5s^2+12}{s^3+4s^2+14s+20})
Hence, by final value theorem,
(x(∞)=lim_{xrightarrow 0} ⁡sX(s)=frac{12}{20})

9. Find the value of x(0) if (X(s)=frac{2s^2+5s+12/s}{s^3+4s^2+14s+20}).
a) 5
b) 4
c) 12
d) 2
Answer: d
Explanation:
Given, (X(s)=frac{2s^2+5s+12/s}{s^3+4s^2+14s+20})
Hence, (sX(s)=frac{2s^3+5s^2+12}{s^3+4s^2+14s+20})
Hence, by initial value theorem,
(x(0)=lim_{xrightarrow infty} ⁡sX(s)=2)

10. Find the inverse lapace of (frac{(s+1)}{[(s+1)^2+4][(s+1)^2+1]}).
a) 1⁄3 et [Cos(t) – Cos(2t)].
b) 1⁄3 e-t [Cos(t) + Cos(2t)].
c) 1⁄3 et [Cos(t) + Cos(2t)].
d) 1⁄3 e-t [Cos(t) – Cos(2t)].
Answer: d
Explanation:
Given, (Y(s)=frac{(s+1)}{[(s+1)^2+4][(s+1)^2+1]})
=(frac{s+1}{3(s^2+ 2*s + 2)}-frac{s+1}{3(s^2+ 2*s + 5)})
=(frac{s+1}{3[(s+1)^2+1]}-frac{s+1}{3[(s+1^2+4)]})
=(frac{1}{3} [e^{-t} Cos(t)]-frac{1}{3}[e^{-t} Cos(2t)])
=(frac{1}{3} e^{-t} [Cos(t)-Cos(2t)])

11. Find the inverse laplace transform of (Y(s)=frac{2s}{1-s^2}e^{-s}).
a) -e-t + 1 + et – 1
b) -e-t + 1 – et + 1
c) -e-t + 1 + et + 1
d) -e-t + 1 – et – 1
Answer: d
Explanation: Given,
Y(s)=(frac{2s}{1-s^2}e^{-s})
Let,G(s)=(frac{2s}{1-s^2}=-frac{1}{s – 1}-frac{1}{s + 1})
hence,g(t)=(-e^{-t} – e^t)
Since,Y(s)=(e^{-s} G(s)=>y(t)=g(t-1))
hence,y(t)=(-e^{-t+1}-e^{t-1})

12. Find the inverse laplace transform of (frac{1}{s(s-1)(s^2+1)}).
a) 1⁄2 e-t + 1⁄2 Sin(-t) – 1⁄2 Cos(-t)
b) 1⁄2 et + 1⁄2 Sin(t) – 1⁄2 Cos(t)
c) 1⁄2 et + 1⁄2 Sin(t) + 1⁄2 Cos(t)
d) 1⁄2 et – 1⁄2 Sin(t) – 1⁄2 Cos(t)
Answer: b
Explanation: We know that,
Given, Y(s)=(frac{1}{s(s-1)(s^2+1)})
Let, G(s)=(frac{1}{(s-1)(s^2+1)}=frac{1}{2(s^2-1)}-frac{s+1}{2(s^2+1)}=frac{1}{2*(s-1)}-frac{s}{2(s^2+1)}-frac{1}{2(s^2+1)})
Now, g(t)=(frac{1}{2}e^t-frac{1}{2}cos(t)-frac{1}{2}cos(t))
Now, Y(s)=(frac{1}{2}G(s)=>y(t)=int_0^t g(t)dt=frac{1}{2}e^t+frac{1}{2}sin(t)-frac{1}{2}cos(t))

250+ TOP MCQs on Cayley Hamilton Theorem and Answers

Linear Algebra Multiple Choice Questions on “Cayley Hamilton Theorem”.

1. Find the inverse of the given Matrix, using Cayley Hamilton’s Theorem.
A=(begin{bmatrix}1&2&3\2&3&4\3&4&5end{bmatrix})
a) A-1=(frac{1}{16} begin{bmatrix}2&-3&-1\4&-2&-6\-6&9&11end{bmatrix})
b) A-1=(frac{1}{8} begin{bmatrix}2&-3&-1\4&-2&-3\-6&9&11end{bmatrix})
c) A-1=(frac{1}{16} begin{bmatrix}2&-1&-1\4&-2&-6\-6&9&11end{bmatrix})
d) A-1=(frac{1}{8} begin{bmatrix}2&-3&-1\4&-2&-6\-6&9&11end{bmatrix})
Answer: d
Explanation: For the given Matrix,
A=(begin{bmatrix}1&2&3\2&3&4\3&4&5end{bmatrix})
The characteristic polynomial is given by –
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0
α3-7α2+11α-8=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.
Thus,
A3-7A2+11A-8I=0
To find A-1, multiply both the sides of the equation by A-1
A2 A A-1-7AA A-1+11A A-1-8I A-1=0
We know that A A-1=I
A2I-7AI+11I-8IA-1=0
A2-7A+11-8 A-1=0
A2-7A+11=8 A-1

8A-1=(begin{bmatrix}19&18&13\-3&1&1\15&9&7end{bmatrix}-7begin{bmatrix}4&3&2\-1&2&1\3&0&1end{bmatrix}+11begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})

8A-1=(begin{bmatrix}19-28+11&18-21&13-14\-3+7&1-14+11&1-7\15-21&9&7-7+11end{bmatrix})

8A-1=(begin{bmatrix}2&-3&-1\4&-2&-6\-6&9&11end{bmatrix})

A-1=(frac{1}{8} begin{bmatrix}2&-3&-1\4&-2&-6\-6&9&11end{bmatrix}).

2. Find the value of A3 where A=(begin{bmatrix}-1&-1&2\0&1&-1\2&2&1end{bmatrix}).
a) (begin{bmatrix}3&5&-1\-2&-9&2\-2&-4&-5end{bmatrix})
b) (begin{bmatrix}3&5&-1\1&-9&1\-2&-4&-5end{bmatrix})
c) (begin{bmatrix}3&5&-1\-2&-9&1\-2&-4&-5end{bmatrix})
d) (begin{bmatrix}3&5&-1\-1&-9&1\-2&-4&-5end{bmatrix})
Answer: c
Explanation: For the given Matrix,
A=(begin{bmatrix}-1&-1&2\0&1&-1\2&2&1end{bmatrix})
The characteristic polynomial is given by-
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0

α3-α2+3α+5=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,
A3-A2+3A+5I=0
A3=A2-3A-5I
A3=(begin{bmatrix}5&2&5\-2&-1&-2\4&2&3end{bmatrix}-3begin{bmatrix}-1&-1&2\0&1&-1\2&2&1end{bmatrix}-5begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})

A3=(begin{bmatrix}5+3-5&2+3&5-6\-2+0&-1-3-5&-2+3\4-6&2-6&3-3-5end{bmatrix})

A3=(begin{bmatrix}3&5&-1\-2&-9&1\-2&-4&-5end{bmatrix}).

3. Find the value of A3+19A, A=(begin{bmatrix}2&-3&1\2&0&-1\1&4&5end{bmatrix}).
a) (begin{bmatrix}42&-14&70\21&+21&-21\105&119&203end{bmatrix})
b) (begin{bmatrix}42&-7&70\21&-21&-21\105&119&203end{bmatrix})
c) (begin{bmatrix}42&-14&70\21&-21&-21\105&119&203end{bmatrix})
d) (begin{bmatrix}42&-7&70\21&+21&-21\105&119&203end{bmatrix})
Answer: c
Explanation: Explanation: For the given Matrix,
A=(begin{bmatrix}2&-3&1\2&0&-1\1&4&5end{bmatrix})
The characteristic polynomial is given by –
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0

α3-7α2+19α-49=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,
A3-7A2+19A-49I=0
A3+19A=7A2+49I
A3+19A=7(begin{bmatrix}-1&-2&10\3&-10&-3\15&17&22end{bmatrix}+49begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
A3+19A=(begin{bmatrix}-7+49&-17&70\21&-70+49&-21\105&119&154+49end{bmatrix})
A3+19A=(begin{bmatrix}42&-14&70\21&-21&-21\105&119&203end{bmatrix}).

4. Find the value of 2A3+4A2, where = (begin{bmatrix}5&0&-1\1&2&-1\-3&4&1end{bmatrix}).
a) (begin{bmatrix}-200&0&-24\24&-32&-24\-72&96&-56end{bmatrix})
b) (begin{bmatrix}-200&0&-24\24&-32&-12\-72&96&-56end{bmatrix})
c) (begin{bmatrix}-200&0&-24\12&-32&-24\-72&96&-56end{bmatrix})
d) (begin{bmatrix}-100&0&-12\12&-16&-12\-36&48&-28end{bmatrix})
Answer: a
Explanation: Explanation: For the given Matrix,
A=(begin{bmatrix}5&0&-1\1&2&-1\-3&4&1end{bmatrix})

The characteristic polynomial is given by-
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0

α3+2α2-12α-40=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,
A3+2A2-12A+40I=0
A3+2A2=12A-40I
A3+2A2=(12begin{bmatrix}5&0&-1\1&2&-1\-3&4&1end{bmatrix}-40begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})

A3+2A2=(begin{bmatrix}-60-40&0&-12\12&24-40&-12\-36&48&12-40end{bmatrix})

A3+2A2=(begin{bmatrix}-100&0&-12\12&-16&-12\-36&48&-28end{bmatrix})

2A3+4A2=(begin{bmatrix}-200&0&-24\24&-32&-24\-72&96&-56end{bmatrix}).

5. Find the value of A3-3A2-28A, A = (begin{bmatrix}-1&2&8\-2&3&0\-4&5&1end{bmatrix}).
a) (begin{bmatrix}80&-126&-504\126&-172&-63\252&-316&-46end{bmatrix})
b) (begin{bmatrix}80&-126&-504\126&-172&-63\252&-315&-46end{bmatrix})
c) (begin{bmatrix}40&-126&-504\126&-172&-63\252&-315&-46end{bmatrix})
d) (begin{bmatrix}40&-126&-504\126&-172&-63\252&-316&-46end{bmatrix})
Answer: b
Explanation: For the given Matrix,
A=(begin{bmatrix}-1&2&8\-2&3&0\-4&5&1end{bmatrix})

The characteristic polynomial is given by-
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0

α3-3α2+35α-17=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,
A3-3A2+35A-17I=0
On performing long division (α3-3α2+35α-17)/(α2-7α)
Q=α+4 and R=63α-17
Using division properties,
α3-3α2+35α-17=(α2-7α)×(α+4)+(63α-17)
α3-3α2+35α-17=(α3-3α2-28α)+( 63α-17)
0=(α3-3α2-28α)+(63α-17) ————— (From Characteristic Polynomial)
(α3-3α2-28α) = -63α+17
(A3-3A2-28A) = -63A+17I
(A3-3A2-28A) = (-63begin{bmatrix}-1&2&8\-2&3&0\-4&5&1end{bmatrix}+17begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
(A3-3A2-28A)=(begin{bmatrix}63+17&-126&-504\126&17-189&-63\252&-315&17-63end{bmatrix})
(A3-3A2-28A)=(begin{bmatrix}80&-126&-504\126&-172&-63\252&-315&-46end{bmatrix}).

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250+ TOP MCQs on First Order Non-Linear PDE and Answers

Fourier Analysis and Partial Differential Equations Multiple Choice Questions on “First Order Non-Linear PDE”.

1. Which of the following is an example of non-linear differential equation?
a) y=mx+c
b) x+x’=0
c) x+x2=0
d) x”+2x=0
View Answer

Answer: c
Explanation: For a differential equation to be linear the dependent variable should be of first degree. Since in equation x+x2=0, x2 is not a first power, it is not an example of linear differential equation.

2. Which of the following is not a standard method for finding the solutions for differential equations?
a) Variable Separable
b) Homogenous Equation
c) Orthogonal Method
d) Bernoulli’s Equation
View Answer

Answer: c
Explanation: The following are the different standard methods used in finding the solution of a differential equation:

  • Variable Separable
  • Homogenous Equation
  • Non-homogenous Equation reducible to Homogenous Equation
  • Exact Differential Equation
  • Non-exact Differential Equation that can be made exact with the help of integrating factors
  • Linear First Order Equation
  • Bernoulli’s Equation

3. Solution of a differential equation is any function which satisfies the equation.
a) True
b) False
View Answer

Answer: a
Explanation: A solution of a differential equation is any function which satisfies the equation, i.e., reduces it to an identity. A solution is also known as integral or primitive.

4. A solution which does not contain any arbitrary constants is called a general solution.
a) True
b) False
View Answer

Answer: a
Explanation: The solution of a partial differential equation obtained by eliminating the arbitrary constants is called a general solution.

5. Which of the following is a type of Iterative method of solving non-linear equations?
a) Graphical method
b) Interpolation method
c) Trial and Error methods
d) Direct Analytical methods
View Answer

Answer: b
Explanation: There are 2 types of Iterative methods, (i) Interpolation methods (or Bracketing methods) and (ii) Extrapolation methods (or Open-end methods).

6. A particular solution for an equation is derived by substituting particular values to the arbitrary constants in the complete solution.
a) True
b) False
View Answer

Answer: a
Explanation: A solution which does not contain any arbitrary constants is called a general solution whereas a particular solution is derived by substituting particular values to the arbitrary constants in this solution.

7. Singular solution of a differential equation is one that cannot be obtained from the general solution gotten by the usual method of solving the differential equation.
a) True
b) False
View Answer

Answer: a
Explanation: A differential equation is said to have a singular solution if in all points in the domain of the equation the uniqueness of the solution is violated. Hence, this solution cannot be obtained from the general solution.

8. Which of the following equations represents Clairaut’s partial differential equation?
a) z=px+f(p,q)
b) z=f(p,q)
c) z=p+q+f(p,q)
d) z=px+qy+f(p,q)
View Answer

Answer: d
Explanation: Equations of the form, z=px+qy+f(p,q) are known as Clairaut’s partial differential equations, named after the Swiss mathematician, A. C. Clairaut (1713-1765).

9. Which of the following represents Lagrange’s linear equation?
a) P+Q=R
b) Pp+Qq=R
c) p+q=R
d) Pp+Qq=P+Q
View Answer

Answer: b
Explanation: Equations of the form, Pp+Qq=R are known as Lagrange’s linear equations, named after Franco-Italian mathematician, Joseph-Louis Lagrange (1736-1813).

10. A partial differential equation is one in which a dependent variable (say ‘x’) depends on an independent variable (say ’y’).
a) False
b) True
View Answer

Answer: a
Explanation: An ordinary differential equation is divided into two types, ordinary and partial differential equations.
A partial differential equation is one in which a dependent variable depends on one or more independent variables.
Example: (F(x,t,y,frac{∂y}{∂x},frac{∂y}{∂t},……)= 0. )

11. What is the complete solution of the equation, (q= e^frac{-p}{α})?
a) (z=ae^frac{-a}{α}y)
b) (z=x+e^frac{-a}{α}y)
c) (z=ax+e^frac{-a}{α} y+c)
d) (z=e^frac{-a}{α}y)
View Answer

Answer: c
Explanation: Given: (q= e^frac{-p}{α})
The given equation does not contain x, y and z explicitly.
Setting p = a and q = b in the equation, we get (b= e^frac{-a}{α}.)
Hence, a complete solution of the given equation is,
(z=ax+by+c,,with , b= e^frac{-a}{α})
(z=ax+e^frac{-a}{α} y+c.)

12. A particular solution for an equation is derived by eliminating arbitrary constants.
a) True
b) False
View Answer

Answer: b
Explanation: A particular solution for an equation is derived by substituting particular values to the arbitrary constants in the complete solution thereby eliminating any arbitrary constants present in the solution. Such solution represents a particular member of the family of surfaces given by the complete solution.

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