300+ REAL TIME Engineering Mathematics Objective Questions & Answers 2023

250+ TOP MCQs on Change of Order of Integration: Double Integral and Answers

Differential and Integral Calculus Multiple Choice Questions on “Change of Order of Integration: Double Integral”.

1. Which of the following is not a property of double integration?
a) ∬ af(x,y)ds = a∬ f(x,y)ds, where a is a constant
b) ∬ (f(x,y)+g(x,y))ds = ∬f(x,y)ds+ ∬g(x,y)ds
c) (∬_0^Df(x,y)ds = ∬_0^{D1}f(x,y)ds+ ∬_{D1}^{D2}f(x,y)ds,) where D is union of disjoint domains D1 and D2
d) ∬(f(x,y)*g(x,y))ds = ∬f(x,y)ds*∬g(x,y)ds
View Answer

Answer: d
Explanation: The following are the properties of double integration:

∬af(x,y)ds = a∬f(x,y)ds
∬f(x,y)+g(x,y))ds = ∬f(x,y)ds+ ∬g(x,y)ds
(∬_0^Df(x,y)ds = ∬_0^{D1}f(x,y)ds+ ∬_{D1}^{D2}f(x,y)ds)

2. The region bounded by circle is an example of regular domain.
a) False
b) True
View Answer

Answer: b
Explanation: A domain D in the XY plane bounded by a curve c is said to be regular in the Y direction, if straight lines passing through an interior point and parallel to Y axis meets c in two points A and B. Hence, region bounded by circle is an example of regular domain.

3. What is the result of the integration (∫_3^4∫_1^2(x^2+y)dxdy)?
a) (frac{83}{6} )
b) (frac{83}{3} )
c) (frac{82}{6} )
d) (frac{81}{6} )
View Answer

Answer: a
Explanation: Given: (∫_3^4∫_1^2(x^2+y)dxdy)
Integrating with respect to y first, we get,
(∫_3^4(x^2(y)_1^2+(frac{y^2}{2})_1^2)dx= ∫_3^4(x^2+frac{3}{2}) dx)
Next integrating with respect to x, we get,
((frac{x^3}{3})_3^4+frac{3}{2}(x)_3^4= frac{37}{3}+frac{3}{2}=frac{83}{6})

4. Volume of an object expressed in spherical coordinates is given by (V = ∫_0^2π∫_0^frac{π}{3}∫_0^1 r cos∅ ,dr ,d∅ ,dθ.) The value of the integral is _______
a) (frac{√3}{2})
b) (frac{1}{√2} π)
c) (frac{√3}{2}π)
d) (frac{√3}{4} π)
View Answer

Answer: d
Explanation: Given: (V = ∫_0^2π∫_0^frac{π}{3}∫_0^1 r cos∅ ,dr ,d∅ ,dθ.)
( V = ∫_0^2π∫_0^{frac{π}{3}}(frac{r^2}{2})_0^1 cos∅, d∅, dθ)
( V = frac{1}{2} ∫_0^{2π}(sin∅)_0^frac{π}{3} d∅ ,dθ)
(V = frac{1}{2}×frac{√3}{2} ∫_0^2π dθ)
( V = frac{1}{2}×frac{√3}{2} ×2π)
( V = frac{√3}{2} π )

5. Which of the following equation represents Moment of Inertia of a plane region relative to x-axis?
a) ∬x² f(x,y)dxdy
b) ∬xf(x,y)dxdy
c) ∬y² f(x,y)dxdy
d) ∬yf(x,y)dxdy
View Answer

Answer: c
Explanation: Moment of Inertia of a plane region,
Relative to x-axis is given by,
I_xx=∬y² f(x,y)dxdy
Relative to y-axis is given by,
I_yy=∬x² f(x,y)dxdy

6. What is the mass of the region R as shown in the figure?
” alt=”” width=”360″ height=”311″ data-src=”2020/06/integral-calculus-questions-answers-change-order-integration-double-integral-q6″ data-srcset=”2020/06/integral-calculus-questions-answers-change-order-integration-double-integral-q6 360w, 2020/06/integral-calculus-questions-answers-change-order-integration-double-integral-q6-300×259 300w” data-sizes=”(max-width: 360px) 100vw, 360px” />
a) 8
b) 9
c) (frac{9}{2} )
d) (frac{9}{4} )
View Answer

Answer: b
Explanation: From the above figure, we can see that X-axis ranges from 0 to 3 and Y-axis ranges from 0 to 2.
Therefore, the mass of the region is given by,
(M = ∫_0^2∫_0^3xy ,dx,dy)
( = ∫_0^2y(frac{x^2}{2})_0^3 dy = frac{9}{2}(frac{y^2}{2})_0^2 = 9 )

7. The half-interval method in numerical analysis is also known as __________
a) Newton-Raphson method
b) Regula Falsi method
c) Taylor’s method
d) Bisection method
View Answer

Answer: d
Explanation: The Bisection method, also known as binary chopping or half-interval method, is a starting method which is used, where applicable, for few iterations, to obtain a good initial value.

8. Given (∫_0^8x^frac{1}{3}dx,) find the error in approximating the integral using Simpson’s 1/3 Rule with n=4.
a) 1.8
b) 2.9
c) 0.3
d) 0.35
View Answer

Answer: d
Explanation: Given: (∫_0^8x^frac{1}{3}dx,n = 3,)
Let (f(x)= x^frac{1}{3},)
(∆x = frac{b-a}{2}= frac{8-0}{2}=4) ………………since b=8, a=0 (limits of the given integral)
Hence endpoints x_i have coordinates {0, 2, 4, 6, 8}.
Calculating the function values at x_i, we get,
(f(0)= 0^frac{1}{3}=0)
(f(2)= 2^frac{1}{3})
(f(4)= 4^frac{1}{3})
(f(6)= 6^frac{1}{3})
(f(8)= 8^frac{1}{3} =2)
Substituting these values in the formula,
(∫_0^8x^frac{1}{3} dx ≈ frac{∆x}{3} [f(0)+4f(2)+2f(4)+4f(6)+f(8)])
( ≈frac{2}{3}[0+4(2^frac{1}{3})+2(4^frac{1}{3})+ 4(6^frac{1}{3})+2] ≈ 11.65)
Actual integral value,
(∫_0^8x^frac{1}{3} dx= left(frac{x^frac{4}{3}}{frac{4}{3}}right)_0^8=12)
Error in approximating the integral = 12 – 11.65 = 0.35

9. A sphere with the dimensions is shown in the figure. What is the error that can be incorporated in the radius such that the volume will not change more than 4%?
” alt=”” width=”197″ height=”195″ data-src=”2020/05/integral-calculus-questions-answers-change-order-integration-double-integral-q9″ />
a) 0.127%
b) 0.0127%
c) 12.7%
d)1.27%
View Answer

Answer: b
Explanation: We know that volume of the sphere is,
(V = frac{4}{3} πR^3 )
Differentiating the above equation with respect to R we get,
(frac{dV}{dR}= frac{4}{3} π×3R^2=4πR^2)
Since the volume of the sphere should not exceed more than 4%,
(dR=frac{dV}{4πR^2}=frac{0.04}{4π(5)^2}=0.000127)
Error in radius = 0.0127%

10. The x-coordinate of the center of gravity of a plane region is given by, (x_c=frac{1}{M}∬xf(x,y)dxdy.)
a) True
b) False
View Answer

Answer: a
Explanation: The coordinates (x_c,y_c) of the centroid of a plane region with mass M is given by,
(x_c=frac{1}{M} ∬xf(x,y)dxdy)
(y_c=frac{1}{M} ∬yf(x,y)dxdy)

Global Education & Learning Series – Differential and Integral Calculus.

To practice all areas of Differential and Integral Calculus, here is complete set of 1000+ Multiple Choice Questions and Answers.

250+ TOP MCQs on Law of Natural Growth and Decay and Answers

Ordinary Differential Equations Interview Questions and Answers for Experienced people focuses on “Law of Natural Growth and Decay”.

1. The population of the Mysore city of a country was 30 lakhs in the year 2000 and 60 lakhs in the year 2010. What is the estimated population in the year 1990 with the help of the population model?
a) 18 lakhs
b) 15 lakhs
c) 20 lakhs
d) 12 lakhs
Answer: b
Explanation: Here rate of change of population is proportional to the population present at that instant of time i.e (frac{dp}{dt} ∝ P) i.e (frac{dp}{dt}=kP), k is a constant of proportionality.
solving using the variable separable integral form we get (int frac{1}{P} ,dP = int K ,dt + a)
log P=kt+c –> e^kt+a = P or P=ce^kt…where e^a is a constant let the year 2000 be the initial reference year –> P(0)=30=ce^k(0) = c i.e c = 30 & P(10) = ce^k10
=30e^k10 = 60 –> log 60 = log 30 + 10k –> log ⁶⁄₃ =log 2 = 10k –> k = 0.1*0.693 = 0.0693
to find P(-10) = ce^kt = 30*e^-0.693=15 lakhs.

2. Urine culture test is being conducted in the pathology lab from the sample of a patient to prescribe a suitable drug based on bacterial growth. There were 3000 bacteria in the initial sample and it had increased to 3400 in the sample collected after one hour. Based on the exponential growth how long will it take for the bacterial population to reach 10000?
a) 10 hrs
b) 8.5 hrs
c) 9.6 hrs
d) 3.6 hrs
Answer: c
Explanation: As we know according to law of exponential growth, the bacterial population model is
B(t)=ce^kt at t=0 B(0)=ce⁰=3000……given –> c=3000 and B(1)=3400=ce^k=3000*e^k
e^k=(frac{17}{15}) taking log, k=log 17 – log 15 = 0.125, to find t when B(t)=10000 i.e
10000=3000*e^kt = 3000*e^0.125t –> 3.33=e^0.125t –> log 3.33 = 0.125t –> t=9.62 hrs.

3. Uranium disintegrates at a rate proportional to the amount present at any time instant. If a and b grams of uranium are present at times t’ & t’’ respectively. What is the expression for the half life of uranium?
a) (frac{(t”+t’)log2}{log⁡(frac{b}{a})} )
b) (frac{(t”-2t’)log2}{2log(frac{a}{b})} )
c) (frac{(t”+t’)}{2log(frac{a}{b})} )
d) (frac{(t”-t’)log2}{log⁡(frac{b}{a})} )
Answer: d
Explanation: Let P grams of uranium be present at any time t. According to law of decay, (frac{dp}{dt} = -kP…k) is a constant its solution is P=ce^-kt given that at t=t’ P=a
a=ce^{-kt^‘}…..(1) and b=ce^{-kt^‘}….(2) dividing (1) & (2) we get (frac{a}{b}=e^{k(t^{”}-t’)})
taking log on both side (log(frac{a}{b})) = k(t’’-t’)…(3) at t=0 ‘c’ is the initial mass of uranium present in the sample thus at half life i.e t=T ,P=c/2 P = P=ce^-kt becomes (frac{c}{2}) = ce^-kT
log 2=kT substituting the value of k from (3) we get (T = frac{(t”-t’)log2}{log⁡(frac{b}{a})} ) is the equation for half life period.

Global Education & Learning Series – Ordinary Differential Equations.

To practice all areas of Ordinary Differential Equations for Interviews,

250+ TOP MCQs on Solution of DE With Constant Coefficients using the Laplace Transform and Answers

Ordinary Differential Equations Multiple Choice Questions & Answers focuses on “Solution of DE With Constant Coefficients using the Laplace Transform”.

1. While solving the ordinary differential equation using unilateral laplace transform, we consider the initial conditions of the system.
a) True
b) False
Answer: a
Explanation: When bilateral laplace transformation is used in solving differential equations, we don’t consider the initial conditions as the transformation is from -∞ to +∞. But when we consider unilateral laplace transformation, the integral is from 0 to ∞. So, the initial conditions are considered.

2. With the help of _____________________ Mr.Melin gave inverse laplace transformation formula.
a) Theory of calculus
b) Theory of probability
c) Theory of statistics
d) Theory of residues
Answer: d
Explanation: Let f(t) be the function in time. The laplace transformation of the function is L[f(t)] = F(s). So, the inverse laplace transform of F(s) comes out to be the function f(t) in time. The formula for laplace transform is derived using the theory of residues by Mr.Melin.

3. What is the laplce tranform of the first derivative of a function y(t) with respect to t : y’(t)?
a) sy(0) – Y(s)
b) sY(s) – y(0)
c) s² Y(s)-sy(0)-y'(0)
d) s² Y(s)-sy'(0)-y(0)
Answer: b
Explanation: Let (f(t) = y(t) )
(L[f’(t)] = int_0^∞ e^{-st} f'(t)dt )
( = e^{-st} f(t)(from , 0 , to , infty) – int_0^∞ (-s) e^{-st} f(t)dt )
( = -f(0) + sint_0^{infty} e^{-st} f(t)dt )
( = -f(0) + sF(s) )
( = sY(0) -y(0) ).

4. Solve the Ordinary Differential Equation by Laplace Transformation y’’ – 2y’ – 8y = 0 if y(0) = 3 and y’(0) = 6.
a) (3e^t cos(3t)+tsint(3t) )
b) (3e^t cos(3t)+te^{-t} sint(3t) )
c) (2e^{-t} cos(3t)-2 frac{t}{3} sint(3t) )
d) (2e^{-t} cos(3t)-2 frac{te^{-t}}{3} sint(3t) )
Answer: a
Explanation: L[y’’ – 2y’ – 8y ] = 0
s² Y(s) – sy(0) – y'(0) – 2sY(s) + 2y(0) – 8Y(s) = 0
(s² – 2s – 8)Y(s) = 2s
(L[y(t)] = 2 frac{s}{(s^2-2s-8)} )
Therefore, y(t) = 3e^t cos(3t) + tsint(3t).

5. Solve the Ordinary Differential Equation y’’ + 2y’ + 5y = e^-t sin(t) when y(0) = 0 and y’(0) = 1.(Without solving for the constants we get in the partial fractions).
a) (e^t [Acost+A1sint+Bcos(2t)+frac{(B1)}{2} sin(2t)] )
b) (e^{-t} [Acost+A1sint+Bcos(2t)+B1sin(2t)] )
c) (e^{-t} [Acost+A1sint+Bcos(2t)+frac{(B1)}{2} sin(2t)] )
d) (e^t [Acost+A1sint+Bcos(2t)+(B1)sin(2t)] )
Answer: c
Explanation: (L[y’’+2y’ +5y = e^{-t} sin(t)] )
(s^2 Y(s)-sy(0)-y'(0)+ 2sY(s) -2y(0) + 5Y(s) = frac{1}{(s+1)^2+1} )
((s^2+2s+5)Y(s)= frac{1}{(s+1)^2+1}+1 )
((s^2+2s+5)Y(s)= frac{(s^2+2s+3)}{(s^2+2s+2)} )
( Y(s) = frac{(s^2+2s+3)}{(s^2+2s+2)(s^2+2s+5)} )
( = frac{(s+1)^2+2}{((s+1)^2+1)((s+1)^2+4)} )
( y(t) = e^{-t} L^{-1} [frac{(As+A1)}{(s^2+1)}+frac{(Bs+B1)}{(s^2+4)}] )
( = e^{-t} [Acost+A1sint+Bcos(2t)+frac{(B1)}{2} sin(2t)]).

6. Solve the Ordinary Diferential Equation using Laplace Transformation y’’’ – 3y’’ + 3y’ – y = t² e^t when y(0) = 1, y’(0) = 0 and y’’(0) = 2.
a) (2e^t frac{t^5}{720}+e^t+2e^t frac{t}{6}+4e^t frac{t^2}{24} )
b) (e^t frac{t^5}{720}+2e^{-t}+2e^t frac{t}{6}+4e^t frac{t^2}{24} )
c) (e^{-t} frac{t^5}{720}+e^{-t}+2e^{-t} frac{t}{6}+4e^{-t} frac{t^2}{24} )
d) (2e^{-t} frac{t^5}{720}+e^{-t}+2e^{-t} frac{t}{6}+4e^{-t} frac{t^2}{24} )
Answer: a
Explanation: L[y’’’ – 3y’’ + 3y’ – y = t² e^t]
s³ Y(s) – s² y(0) – sy'(0) – y”(0) – 3s² Y(s) + 3sy(0) + 3y'(0) + 3sY(s) – 3y(0) – Y(s) = (frac{2}{(s-1)^3} )
(Y(s) = frac{2}{(s-1)^6} +frac{(s^2+3s+5)}{(s-1)^3} )
(y(t) = 2e^t frac{t^5}{720}+e^t+2e^t frac{t}{6}+4e^t frac{t^2}{24}).

7. Take Laplace Transformation on the Ordinary Differential Equation if y’’’ – 3y’’ + 3y’ – y = t² e^t if y(0) = 1, y’(0) = b and y’’(0) = c.
a) ((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=frac{2}{(s-1)^3} )
b) ((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)+(-3a-c)s)=frac{2}{(s-1)^3} )
c) ((s^3-3s^2+3s)Y(s)+(-as+(3a-b)s+(-3a-c))=frac{2}{(s-1)^3} )
d) ((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=frac{2}{(s-1)^3} )
Answer: a
Explanation: L[y’’’ – 3y’’ + 3y’ – y = t² e^t]
s³ Y(s) – s² y(0) – sy'(0) – y”(0) – 3s² Y(s) + 3sy(0) + 3y'(0) + 3sY(s) – 3y(0) – Y(s) = (frac{2}{(s-1)^3} )
((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=frac{2}{(s-1)^3}.)

8. What is the inverse Laplace Transform of a function y(t) if after solving the Ordinary Differential Equation Y(s) comes out to be (Y(s) = frac{s^2-s+3}{(s+1)(s+2)(s+3)} ) ?
a) (frac{1}{2} e^{-t}+frac{9}{2} e^{-3t}-3e^{-2t} )
b) (frac{-1}{2} e^{-t}+frac{9}{2} e^{-2t}-3e^{-3t} )
c) (frac{1}{2} e^{-t}-frac{3}{2} e^{-2t}-3e^{-3t} )
d) (frac{-1}{2} e^{t}+frac{9}{2} e^{2t}-3e^{3t} )
Answer: b
Explanation: Taking inverse Laplace Transformation for
(Y(s) = frac{(s^2-s+3)}{(s+1)(s+2)(s+3)} )
Solving the partial fractions we get,
(Y(s) = frac{-1}{2} frac{1}{(s+1)}+frac{9}{2} frac{1}{(s+2)}-3 frac{1}{(s+3)} )
Therefore, (y(t) = frac{-1}{2} e^{-t}+frac{9}{2} e^{-2t}-3e^{-3t}. )

9. For the Transient analysis of a circuit with capacitors, inductors, resistors, we use bilateral Laplace Transformation to solve the equation obtained from the Kirchoff’s current/voltage law.
a) True
b) False
Answer: b
Explanation: For the transient analysis of the circuit with capacitors, inductors, resistors, we have to know the initial condition of the components used. So, the unilateral Laplace Transform is used to solve the equations obtained from the Kirchoff’s current/voltage law.

10. While solving an Ordinary Differential Equation using the unilateral Laplace Transform, it is possible to solve if there is no function in the right hand side of the equation in standard form and if the initial conditions are zero.
a) True
b) False
Answer: b
Explanation: It is not possible to solve an equation if the input and the initial conditions are zero becase Y(s) becomes zero where Y(s) is the Laplace Transform of y(t) function.

Global Education & Learning Series – Ordinary Differential Equations.

To practice all areas of Ordinary Differential Equations,

250+ TOP MCQs on Transformation (Reduction) of Quadratic Form to Canonical Form and Answers

Linear Algebra Multiple Choice Questions & Answers focuses on “Transformation (Reduction) of Quadratic Form to Canonical Form”.

1. What is the canonical form of the matrix A = (begin{bmatrix}1 & 0 \ 1 & 1end{bmatrix} )?
a) x+xy+y²
b) x²+xy
c) x²+y²
d) x²+xy+y²
View Answer

Answer: d
Explanation: The quadratic form of the given matrix is,
[x y](begin{bmatrix}1 & 0 \1 & 1 end{bmatrix} begin{bmatrix} x \ y end{bmatrix} = [x + y y]begin{bmatrix} x \ y end{bmatrix}) = x²+xy+y².

2. The solution of the given matrix equation is _____
(begin{bmatrix}3 & 0 & 2\ 6 & 1 & 1\ 2 & 8 & 91end{bmatrix} begin{bmatrix}x_1 \ x_2\ x_3 end{bmatrix} ₌ begin{bmatrix}0 \ 0 \ 0 end{bmatrix} )
a) x₁ = 1, x₂ = 1, x₃ = 2
b) x₁ = 0, x₂ = 0, x₃ = 0
c) x₁ = 3, x₂ = -1, x₃ = -1
d) x₁ = 0, x₂ = -2, x₃ = 4
View Answer

Answer: b
Explanation: Let A = (begin{bmatrix}3 & 0 & 2\ 6 & 1 & 1\ 2 & 8 & 91end{bmatrix} begin{bmatrix}x_1 \ x_2\ x_3 end{bmatrix} ₌ begin{bmatrix}0 \ 0 \ 0 end{bmatrix} )
Hence, the given matrix equation can be written in the form,
AX = B
Multiplying both sides by A^-1, we get
X = A^-1 B
But since B = 0, X = 0 and hence the solution is,
x₁ = 0, x₂ = 0, x₃ = 0.

3. Which one of the following is not a criterion for linearity of an equation?
a) The dependent variable y should be of second order
b) The derivatives of the dependent variable should be of second order
c) Each coefficient does not depend on the independent variable
d) Each coefficient depends only on the independent variable
View Answer

Answer: c
Explanation: The two criterions for linearity of an equation are: The dependent variable y and its derivatives of first degree. Each coefficient depends only on the independent variable.

4. Which among the following does not belong to main types of integrals?
a) Indefinite Integral
b) Proper Definite Integral
c) Improper Definite Integral
d) Real Integral
View Answer

Answer: d
Explanation: There are generally two types of integrals,
1. Definite Integrals: These are further classified as,

Proper Definite Integrals
Improper Definite Integrals

2. Indefinite Integrals

5. Which of the following is true for matrices?
a) (AB)^-1 = B^-1A^-1
b) (A^T) = A
c) AB = BA
d) A*I = I
View Answer

Answer: a
Explanation: The correct forms of the other options are:

(A^T)^T = A
AB ≠ BA
A*I = A

6. Euler’s integral of the first kind, which is a proper integral, is used to define the gamma function.
a) True
b) False
View Answer

Answer: b
Explanation: Euler’s integral of the second kind, which is an improper function, is used to define gamma function for integer x>0.
(Γ(x)= ∫_0^∞ t^{x-1}e^{-t}.dt )

7. Which of the following matrix is not orthogonal?
a) (begin{bmatrix}0.33 & 0.67 & -0.67\ -0.67 & 0.67 & 0.33\ 0.67 & 0.33 & 0.67end{bmatrix} )
b) (begin{bmatrix}cosx &sinx \-sinx & cosxend{bmatrix} )
c) (begin{bmatrix}0.33 & -0.67 & 0.67\0.67 & 0.67 & 0.33\ -0.67 & 0.33 & 0.67end{bmatrix} )
d) (begin{bmatrix}cosx & sinx \-sinx & -cosx end{bmatrix} )
View Answer

Answer: a
Explanation: Out of the given options, (begin{bmatrix}0.33 & 0.67 & -0.67\ -0.67 & 0.67 & 0.33\ 0.67 & 0.33 & 0.67end{bmatrix} ) satisfies the condition for orthogonality, i.e. AA^T = I
(begin{bmatrix}0.33 & 0.67 & -0.67\ -0.67 & 0.67 & 0.33\ 0.67 & 0.33 & 0.67end{bmatrix}
begin{bmatrix}0.33 & -0.67 & 0.67 \ 0.67 & 0.67 & 0.33\-0.67 & 0.33 & 0.67end{bmatrix}= begin{bmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1end{bmatrix} )
Since (begin{bmatrix}0.33 & -0.67 & 0.67\ 0.67 & 0.67 & 0.33\ -0.67 & 0.33 & 0.67end{bmatrix} )is the transpose of A, it is also orthogonal.
Coming to (begin{bmatrix}cosx & sinx\ -sin x &cosxend{bmatrix}, )
(begin{bmatrix}cosx & sinx\-sinx & cosxend{bmatrix} begin{bmatrix}cosx & -sinx \ sinx & cosxend{bmatrix} = begin{bmatrix}1 & 0 \ 0 & 1end{bmatrix} )
The remaining option (begin{bmatrix}cosx & sinx\ -sinx & -cosxend{bmatrix}, )which does not satisfy the condition for orthogonality.

8. The determinant of the matrix whose eigen values are 7, 1, 9 is given by ________
a) 7
b) 63
c) 9
d) 17
View Answer

Answer: b
Explanation: The product of the eigen values of a matrix gives the determinant of the matrix,
Therefore, ∆ = 63.

9. Find the values of x and y in the matrix below if the matrix is a skew symmetric matrix.
P = (begin{bmatrix}0 & x+y & 6 \3 & 0 & 9\ x & 9 & 0end{bmatrix} )
a) x = -6, y = 3
b) x = 3, y = 3
c) x = 6, y = -3
d) x = 0, y = 3
View Answer

Answer: c
Explanation: The general form of a skew symmetric matrix is given by,
(begin{bmatrix}0 & w1 & w2 \ w1 & 0 & w3 \ w2 & w3 & 0end{bmatrix} )

Therefore, from the given matrix,
x = 6,
x+y = 3 → 6+y=3 → y = -3

10. The sum of two symmetric matrices is also a symmetric matrix.
a) False
b) True
View Answer

Answer: b
Explanation: To prove the above statement, let us consider an example,
A = (begin{bmatrix}1 & 3 & 8\ 3 & 0 & 5 \ 8 & 5 & 7 end{bmatrix} )

Therefore, A + A =(begin{bmatrix}1 & 3 & 8 \3 & 0 & 5 \8 & 5 & 7 end{bmatrix}+ begin{bmatrix}1 & 3 & 8\ 3 & 0 & 5\ 8 & 5 & 7 end{bmatrix} = begin{bmatrix}2 & 6 & 16\ 6 & 0 & 10\ 16 & 10 & 14 end{bmatrix} ) which is also a skew-symmetric matrix.

Global Education & Learning Series – Eigen Values and Eigen Vectors

To practice all areas of Linear Algebra, here is complete set of 1000+ Multiple Choice Questions and Answers.

250+ TOP MCQs on Method of Separation of Variables and Answers

Partial Differential Equations Questions and Answers for Entrance exams focuses on “Method of Separation of Variables”.

1. By using the method of separation of variables, the determination of solution to P.D.E. reduces to determination of solution to O.D.E.
a) True
b) False
View Answer

Answer: a
Explanation: A differential equation is an equation involving an unknown function y of one or more independent variables x, t, …… and its derivatives. These are divided into two types, ordinary or partial differential equations.
An ordinary differential equation is a differential equation in which a dependent variable (say ‘y’) is a function of only one independent variable (say ‘x’).
A partial differential equation is one in which a dependent variable depends on one or more independent variables.

2. Separation of variables, in mathematics, is also known as Fourier method.
a) False
b) True
View Answer

Answer: b
Explanation: Separation of variables (or the Fourier method) is a method of solving ordinary and partial differential equations, in which we can rewrite an equation so that each of two variables occur on different sides of the equation.

3. Which of the following equations cannot be solved by using the method of separation of variables?
a) Laplace Equation
b) Helmholtz Equation
c) Alpha Equation
d) Biharmonic Equation
View Answer

Answer: c
Explanation: The method of separation of variables is used to solve a wide range of linear partial differential equations with boundary and initial conditions, such as:

Heat equation
Wave equation
Laplace equation
Helmholtz equation
Biharmonic equation

4. The matrix form of the separation of variables is the Kronecker sum.
a) True
b) False
View Answer

Answer: a
Explanation: If A is n × n, B is m × m and I_k denotes the k × k identity matrix then we can define what is sometimes called the Kronecker sum, ⊕, by,
A⊕B=A⊗I_M+B⊗I_N

5. For a partial differential equation, in a function φ (x, y) and two variables x, y, what is the form obtained after separation of variables is applied?
a) Φ (x, y) = X(x)+Y(y)
b) Φ (x, y) = X(x)-Y(y)
c) Φ (x, y) = X(x)Y(y)
d) Φ (x, y) = X(x)/Y(y)
View Answer

Answer: c
Explanation: The method of separation of variables relies upon the assumption that a function of the form,
Φ (x, y) = X(x)Y(y)
will be a solution to a linear homogeneous partial differential equation in x and y. This is called a product solution and provided the boundary conditions are also linear and homogeneous this will also satisfy the boundary conditions.

6. What is the solution of, (frac{∂^2 u}{∂x^2}=2xe^t,) after applying method of separation of variables ((u(0,t)=t,frac{∂u}{∂x} (0,t)= e^t))?
a) (u=frac{x^3}{3} e^t+xe^t )
b) (u=frac{x^3}{3} e^t+xe^t+t)
c) (u=frac{x^3}{3} e^t+e^t+t)
d) (u=frac{x^2}{2} e^t+xe^t+t)
View Answer

Answer: b
Explanation: Given: (frac{∂^2 u}{∂x^2}=2xe^t,)……………………………………………………………………………… (1)
Integrating (1) with respect to x we get,
(frac{∂u}{∂x}=x^2 e^t+f(t),) where f(t) = arbitrary function ………………………………. (2)
Integrating again with respect to x we get,
(u=frac{x^3}{3} e^t+xf(t)+g(t))………………………………………………………………………………………… (3)
Applying the given initial condition, (frac{∂u}{∂x}(0,t)= e^t) in equation (2), we get, f(t)=e^t.
Applying the initial condition, u(0,t)=t in equation (3), we get, g(t)=t.
Therefore, the solution, (u=frac{x^3}{3} e^t+xe^t+t.)

7. Which of the following is true with respect to formation of differential equation by elimination of arbitrary constants?
a) The given equation should be differentiated with respect to independent variable
b) Elimination of the arbitrary constant by replacing it using derivative
c) If ‘n’ arbitrary constant is present, the given equation should be differentiated ‘n’ number of times
d) To eliminate the arbitrary constants, the given equation must be integrated with respect to the dependent variable
View Answer

Answer: d
Explanation: Consider a general equation, f(x,y,c)=0 ……………………………………… (1)
To form a differential equation by elimination of arbitrary constant, the following steps need to be followed:

Differentiate (1) with respect to x
In case of ‘n’ arbitrary constants, the equation should be differentiated ‘n’ number of times
Eliminate the arbitrary constant using (1) and the derivatives

8. In the formation of differential equation by elimination of arbitrary constants, after differentiating the equation with respect to independent variable, the arbitrary constant gets eliminated.
a) False
b) True
View Answer

Answer: a
Explanation: In the formation of differential equation by elimination of arbitrary constants, the first step is to differentiate the equation with respect to the dependent variable. Sometimes, the arbitrary constant gets eliminated after differentiation.

9. u (x, t) = e − 2π*2t*sin πx is the solution of the two-dimensional Laplace equation.
a) True
b) False
View Answer

Answer: b
Explanation: The solution of the two-dimensional Laplace equation is,
u(x, y) = sin x*cosh y

10. The symbol used for partial derivatives, ∂, was first used in mathematics by Marquis de Condorcet.
a) True
b) False
View Answer

Answer: a
Explanation: Partial derivatives are indicated by the symbol ∂. This was first used in mathematics by Marquis de Condorcet who used it for partial differences.

11. Separation of variables was first used by L’Hospital in 1750.
a) False
b) True
View Answer

Answer: b
Explanation: Guillaume François Antoine, Marquis de L’Hospital (1661 – 2 February 1704), was a French mathematician. His name is firmly associated with L’Hospital’s rule for calculating limits involving indeterminate forms 0/0 and ∞/∞.

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.

Global Education & Learning Series – Fourier Analysis and Partial Differential Equations.

To practice all areas of Partial Differential Equations for Entrance exams, here is complete set of 1000+ Multiple Choice Questions and Answers.

250+ TOP MCQs on Lagrange’s Mean Value Theorem and Answers

Engineering Mathematics Multiple Choice Questions on “Lagrange’s Mean Value Theorem – 1”.

1. For the function f(x) = x² – 2x + 1. We have Rolles point at x = 1. The coordinate axes are then rotated by 45 degrees in anticlockwise sense. What is the position of new Rolles point with respect to the transformed coordinate axes?
a) ³⁄₂
b) ¹⁄₂
c) ⁵⁄₂
d) 1
Answer: a
Explanation:The coordinate axes are rotated by 45 degree then the problem transforms into that of Lagrange mean value theorem where the point in some interval has the slope of tan(45).
Hence differentiating the function and equating to tan(45).
We have
f'(x) = tan(45) = 2x – 2
2x – 2 = 1
x = ³⁄₂.

2. For the function f(x) = x³ + x + 1. We do not have any Rolles point. The coordinate axes are transformed by rotating them by 60 degrees in anti-clockwise sense. The new Rolles point is?
a) (frac{sqrt{3}}{2})
b) The function can never have a Rolles point
c) (3^{frac{1}{2}})
d) (sqrt{frac{sqrt{3}-1}{3}})
Answer: d
Explanation: The question is simply asking us to find if there is some open interval in the original function f(x)
where we have f'(x) = tan(60)
We have
f'(x) = 3x² + 1 = tan(60)
3x² = √3 – 1
x=(sqrt{frac{sqrt{3}-1}{3}})

3. What is the minimum angle by which the coordinate axes have to be rotated in anticlockwise sense (in Degrees), such that the function f(x) = 3x³ + 5x + 1016 has at least one Rolles point
a) ^π⁄₁₈₀ tan^-1(5)
b) tan^-1(5)
c) ¹⁸⁰⁄_π tan^-1(5)
d) -tan^-1(5)
Answer: c
Explanation: For the transformed function to have a Rolles point is equivalent to the existing function having a Lagrange point somewhere in the real number domain, we are finding
the point in the domain of the original function where we have f'(x) = tan(α)
Let the angle to be rotated be α
We have
f'(x) = 9x² + 5 = tan(α)
9x² = tan(α) – 5
For the given function to have a Lagrange point we must have the right hand side be greater than zero, so
tan(α) – 5 > 0
tan(α) > 5
α > tan^-1(5)
In degrees we must have,
α^deg > ¹⁸⁰⁄_π tan^-1(5).

4. For a third degree monic polynomial, it is seen that the sum of roots are zero. What is the relation between the minimum angle to be rotated to have a Rolles point (α in Radians) and the cyclic sum of the roots taken two at a time c
a) α = ^π⁄₁₈₀ * tan^-1(c)
b) Can never have a Rolles point
c) α = ¹⁸⁰⁄_π tan^-1(c)
d) α = tan^-1(c)
Answer: d
Explanation: From Vietas formulas we can deduce that the x² coefficient of the monic polynomial is zero (Sum of roots = zero). Hence, we can rewrite our third degree polynomial as
y = x³ + (0) * x² + c * x + d
Now the question asks us to relate α and c
Where c is indeed the cyclic sum of two roots taken at a time by Vietas formulae
As usual, Rolles point in the rotated domain equals the Lagrange point in the existing domain. Hence, we must have
y ^‘ = tan(α)
3x² + c = tan(α)
To find the minimum angle, we have to find the minimum value of α
such that the equation formed above has real roots when solved for x.
So, we can write
tan(α) – c > 0
tan(α) > c
α > tan^-1(c)
Thus, the minimum required angle is
α = tan^-1(c).

5. For the infinitely defined discontinuous function
(begin{cases}x+sin(2x)& :xin [0,pi] \ x+sin(4x)& :xin (pi,2pi] \ x+sin(6x)& :xin (2pi,3pi] \.\.\x+sin(2nx)& :xin [(n-1)pi,npi)\.\.end{cases})
How many points c∈[0,16x] exist, such that f'(c) = 1
a) 256
b) 512
c) 16
d) 0
Answer: a
Explanation: To find points such that f'(c) = 1
We need to check points on graph where slope remains the same (45 degrees)
In every interval of the form [(n – 1)π, nπ] we must have 2n – 1 points
Because sine curve there has frequency 2n and the graph is going to meet the graph y = x at 2n points.
Hence, in the interval [0, 16π] we have
= 1 + 3 + 5…….(16terms)
=(16)² = 256.

6. Let g(x) be periodic and non-constant, with period τ. Also we have g(nτ) = 0 : n ∈ N. The function f(x) is defined as
f(x)=(begin{cases}2x+g(2x)& :xin [0,tau] \ 2x+g(4x)& :xin (tau,2tau] \2x+g(6x)& :xin (2tau,3tau]\.\.\2x+g(2nx)& :xin [(n-1)tau,ntau)\.\.
end{cases})
How many points c ∈ [0, 18τ] exist such that f'(c) = tan^-1(2)
a) 325
b) 323
c) 324
d) 162
Answer: c
Explanation: To find points such that f ^‘(c) = f ^‘(c) = tan^-1(2)
We need to check points on the graph such that the slope remains the same ( tan^-1(2) radians)
In every interval of the form [(n – 1)τ, nτ] we must have 2n points on the graph because the frequency of periodic function in that interval is 2n and we have g(nτ) = 0
And we have 2n – 1
Lagrange points in the interval [(n – 1)τ, nτ]
The total number of such points in the interval [0, 18τ] is given by
= 1 + 3 + 5…….(18 terms)
= (18)² = 324.

7. Let f(x)=(x-frac{x^3}{3^2.2!}+frac{x^5}{5^2.4!}-frac{x^7}{7^2.6!}+…infty). Find a point nearest to c such that f'(c) = 1
a) 1
b) 0
c) 2.3445 * 10^-9
d) 458328.33 * 10^-3
Answer: d
Explanation: First find f'(x)
f'(x)=1-(frac{x^2}{3!}+frac{x^4}{5!}-frac{x^6}{7!}+…infty)
Multiplying and dividing by x We have the well known series expansion of (frac{sin(x)}{x})

We get
f'(x)=(frac{1}{x}times (x-frac{x^3}{3!}+frac{x^5}{5!}-frac{x^7}{7!}+…infty))
f'(x)=(frac{sin(x)}{x})
Equating this to 1 We have
(frac{sin(x)}{x}) = 1
We know the well known limit (lim_{x rightarrow 0} frac{sin(x)}{x}) = 1
Thus we have to choose a point nearer to 0 as our answer which is,
458328.33 * 10^-3.

8. A function f(x) with n roots should have n – 1 unique Lagrange points.
a) True
b) False
Answer: b
Explanation: If the roots of the polynomial are equal then we will have less than n – 1 Lagrange points that are unique. Hence, the right Option is False.

9. Let f(x)=(frac{x^{100}}{100}+frac{x^{101}}{101}+…infty). Find a point c ∈ (- ∞, ∞) such that f'(c) = 0
a) 1
b) 2
c) 0
d) -1
Answer: c
Explanation: f'(x) = x⁹⁹+x¹⁰⁰+x¹⁰¹+…∞
Using geometric series we have
f'(x)=(frac{x^{99}}{1-x})
Equating to 0 we have
x⁹⁹ = 0
Observe that x = 0 satisfies the equation. Neglect other roots as they are complex roots.