Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on Improper Integrals and Answers

Engineering Mathematics Multiple Choice Questions on “Improper Integrals – 2”.

1. Find the value of ∫tan-1⁡(x)dx.
a) sec-1 (x) – 1⁄2 ln⁡(1 + x2)
b) xtan-1 (x) – 1⁄2 ln⁡(1 + x2)
c) xsec-1 (x) – 1⁄2 ln⁡(1 + x2)
d) tan-1 (x) – 1⁄2 ln⁡(1 + x2)
Answer: b
Explanation: Add constant automatically
Given, ∫tan-1⁡(x)dx
Putting, x = tan(y),
We get, dy = sec2(y)dy,
∫ysec2(y)dy
By integration by parts,
ytan(y) – log⁡(sec⁡(y)) = xtan-1 (x) – 1⁄2 ln⁡(1 + x2).

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: Add constant automatically
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. Find the value of ∫x3 Sin(x)dx.
a) x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x)
b) – x3 Cos(x) + 3x2 Sin(x) – 6Sin(x)
c) – x3 Cos(x) – 3x2 Sin(x) + 6xCos(x) – 6Sin(x)
d) – x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x)
Answer: d
Explanation: Add constant automatically
Let f(x) = x3 Sin(x)
∫x3 Sin(x)dx = – x3 Cos(x) + 3∫x2 Cos(x)dx
∫x2 Cos(x)dx = x2 Sin(x) – 2∫xSin(x)dx
∫xSin(x)dx = – xCos(x) + ∫Cos(x)dx = – xCos(x) + Sin(x)
=> ∫x3 Sin(x)dx = – x3 Cos(x) + 3[x2 Sin(x) – 2[ – xCos(x) + Sin(x)]]
=> ∫x3 Sin(x)dx = – x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x).

4. Value of ∫uv dx,where u and v are function of x.
a) (sum_{i=1}^n(-1)^i u_i v^{i+1})
b) (sum_{i=0}^nu_i v^{i+1})
c) (sum_{i=0}^n(-1)^i u_i v^{i+1})
d) (sum_{i=0}^n(-1)^i u_i v^{n-i})
Answer: c
Explanation: Add constant automatically
Given, f(x)=(int uvdx=sum_{i=0}^n (-1)^i u_i v^{i+1})

5. Find the value of ∫x7 Cos(x) dx.
a) x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x)
b) x7 Sin(x) – 7x6 Cos(x) + 42x5 Sin(x) – 210x4 Cos(x) + 840x3 Sin(x) – 2520x2 Cos(x) + 5040xSin(x) – 5040Cos(x)
c) x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x)
d) x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 10080Cos(x)
Answer: a
Explanation: Add constant automatically
By, f(x)=(int uvdx=sum_{i=0}^n (-1)^i u_i v^{i+1})
Let, u = x7 and v = Cos(x),
∫x7 Cos(x) dx = x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x)

6. Find the value of ∫x3 ex e2x e3x….enx dx.
a) (frac{2}{n(n+1)} e^{frac{n(n+1)}{2}x} left [x^3+3x^2 [frac{2}{n(n+1)}]^1+6x[frac{2}{n(n+1)}]^2 +6[frac{2}{n(n+1)}]^3right ])
b) (frac{2}{n(n+1)} e^{frac{n(n+1)}{2}x} left [x^3+3x^2 [frac{2}{n(n+1)}]^1+6x[frac{2}{n(n+1)}]^2 +6[frac{2}{n(n+1)}]^3right ])
c)(frac{2}{n(n+1)} e^{frac{n(n+1)}{2}x} left [x^3+3x^2 [frac{2}{n(n+1)}]^1+6x[frac{2}{n(n+1)}]^2 +6[frac{2}{n(n+1)}]^3right ])
d)(frac{2}{n(n+1)} e^{frac{n(n+1)}{2}x} left [x^3+3x^2 [frac{2}{n(n+1)}]^1+6x[frac{2}{n(n+1)}]^2 +6[frac{2}{n(n+1)}]^3right ])
Answer: a
Explanation: Add constant automatically
By, f(x)=(int uvdx=sum_{i=0}^n (-1)^i u_i v^{i+1})
Let, u = x3 and v=ex e2x e3x…..enx=ex(1+2+3+…n)=(e^{frac{n(n+1)x}{2}}),
(int x^3 e^x e^2x e^3x……..e^nx dx)
(=x^3 frac{2}{n(n+1)} e^{frac{n(n+1)}{2}x}+3x^2 [frac{2}{n(n+1)}]^2 e^{frac{n(n+1)}{2}x})
(+6x[frac{2}{n(n+1)}]^3 e^{frac{n(n+1)}{2}x}+6[frac{2}{n(n+1)}]^4 e^{frac{n(n+1)}{2}x})
=(frac{2}{n(n+1)} e^{frac{n(n+1)}{2}x} left [x^3+3x^2 [frac{2}{n(n+1)}]^1+6x[frac{2}{n(n+1)}]^2+6[frac{2}{n(n+1)}]^3right])

7. Find the area of a function f(x) = x2 + xCos(x) from x = 0 to a, where, a>0.
a) a2⁄2 + aSin(a) + Cos(a) – 1
b) a3⁄3 + aSin(a) + Cos(a)
c) a3⁄3 + aSin(a) + Cos(a) – 1
d) a3⁄3 + Cos(a) + Sin(a) – 1
Answer: c
Explanation: Given, f(x) = x2 + xCos(x)
Hence, F(x) = ∫x2 + xCos(x) dx = x3⁄3 + xSin(x) + Cos(x)
Hence, area inside f(x) is,
F(a) – F(0) = a3⁄3 + aSin(a) + Cos(a) – 1.

8. Find the area ln(x)⁄x from x = x = aeb to a.
a) b2⁄2
b) b⁄2
c) b
d) 1
Answer: a
Explanation:
Let, F(x)=(int frac{ln⁡(x)}{x} dx)
Let, z=ln⁡(x)=>dz=dx/x
=F(x)=∫ zdz=(frac{z^2}{2}=frac{ln^2⁡(x)}{2})
Area inside curve from 4a to a is,
(F(ae^b)-F(a)=frac{ln^2⁡(ae^b )}{2}-frac{ln^2⁡(a)}{2}=frac{ln^2⁡(frac{ae^b}{a})}{2}=frac{ln^2⁡(e^b)}{2}=frac{b}{2})

9. Find the area inside a function f(t) = ( frac{t}{(t+3)(t+2)} dt) from t = -1 to 0.
a) 4 ln⁡(3) – 5ln⁡(2)
b) 3 ln⁡(3)
c)3 ln⁡(3) – 4ln⁡(2)
d) 3 ln⁡(3) – 5 ln⁡(2)
Answer: d
Explanation:
Now, F(t)=(int frac{t}{(t+3)(t+2)} dt)
F(t)=(int frac{t}{(t+3)(t+2)} dt)
=(int [frac{3}{t+3}-frac{2}{t+2}]dx)
=(int [frac{3}{t+3}]dx-int [frac{2}{t+2}]dx)
=3 ln⁡(t+3)-2ln⁡(t+2)
Now area inside a function is, F(0) – F(-1),
hence, F(0)-F(-1)=3 ln⁡(3)-2 ln⁡(2)-3 ln⁡(2)+2 ln⁡(1)=3 ln⁡(3)-5ln⁡(2)

10. Find the area inside integral f(x)=(frac{sec^4⁡(x)}{sqrt{tan⁡(x)}}) from x = 0 to π.
a) π
b) 0
c) 1
d) 2
Answer: b
Explanation:
Given,F(x)=(int frac{sec^4⁡ (x)}{sqrt{tan⁡(x)}} dx)
F(x)=(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})
F(x)=(frac{2}{5} sqrt{tan⁡(x)} [5+tan^2⁡(x)])
Now area inside a function f(x) from x=0 to π, is
F(π)-F(0)=0-0=0

11. Find the area inside function (frac{(2x^3+5x^2-4)}{x^2}) from x = 1 to a.
a) a2⁄2 + 5a – 4ln(a)
b) a2⁄2 + 5a – 4ln(a) – 11⁄2
c) a2⁄2 + 4ln(a) – 11⁄2
d) a2⁄2 + 5a – 11⁄2
Answer: b
Explanation: Add constant automatically
Given,
f(x) = (frac{(2x^3+5x^2-4)}{x^2}),
Integrating it we get, F(x) = x2⁄2 + 5x – 4ln⁡(x)
Hence, area under, x = 1 to a, is
F(a) – F(1)=a2⁄2 + 5a – 4ln(a) – 1/2 – 5=a2⁄2 + 5a – 4ln(a) – 11⁄2

12. Find the value of ∫(x4 – 5x2 – 6x)4 4x3 – 10x – 6 dx.
a) (frac{(x^4-5x^2-6x)^4}{4})
b) (frac{(x^4-5x^2-6x)^5}{5})
c) (frac{(4x^3-10x-6)^5}{5})
d) (frac{(4x^3-10x-6)^4}{4})
Answer: b
Explanation: Add constant automatically
Given, (int (x^4-5x^2-6x)^4 4x^3-10x-6 dx)
putting, (x^4-5x^2-6x=z), we get, (dz=4x^3-10x-6 dx)
(int z^4 dz=frac{z^5}{5}=frac{(x^4-5x^2-6x)^5}{5})

13. Temperature of a rod is increased by moving x distance from origin and is given by equation T(x) = x2 + 2x, where x is the distance and T(x) is change of temperature w.r.t distance. If, at x = 0, temperature is 40 C, find temperature at x=10.
a) 473 C
b) 472 C
c) 474 C
d) 475 C
Answer: a
Explanation: Temperature at distance x is,
T = ∫T(x) dx = ∫x2 + 2x dx = x3⁄3 + x2 + C
At x=0 given T = 40 C
C = T(x = 0) = 40 C
At x= 10,
T(x = 10) = 1000⁄3 + 100 + 43 = 473 C.

14. Find the value of (int frac{1}{16x^2+16x+10}dx).
a) 1⁄8 sin-1(x + 1⁄2)
b) 1⁄8 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}{16x^2+16x+10}dx=frac{1}{2}int frac{1}{4x^2+4x+5}dx)
=(int frac{1}{8(x^2+x+frac{5}{4}+frac{1}{4}+frac{1}{4})}dx=int frac{1}{8[(x+frac{1}{2})^2+1^2]}dx=frac{1}{8}tan^{-1}(x+frac{1}{2}))

250+ TOP MCQs on Formation of Ordinary Differential Equations by Elimination of Arbitrary Constants and Answers

Ordinary Differential Equations Questions and Answers for Entrance exams focuses on “Formation of Ordinary Differential Equations by Elimination of Arbitrary Constants”.

1. What is the slope of the equation, y= x2+8?
a) 2x
b) 0
c) 8
d) x
Answer: a
Explanation: The slope of the given equation, y= x2+8, is given by,
Slope= (frac{dy}{dx}=2x )

2. 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
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

3. 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
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.

4. What is the differential equation of a family of parabolas with the foci at the origin and axis along the X-axis?
a) 2xy’+ 4y(y’)2-y=0
b) xy’+ y(y’)2-y=0
c) 2xy’+ y(y’)2-y=0
d) 2xy’+2y(y’)2-y=0
Answer: c
Explanation: The equation is, y2=4ax+4a2……………………………………. (1)
Differentiating (1) with respect to x, we get,
2yy’=4a ………………………………………………………………………………………….. (2)
Therefore, substituting the value of 4a in (1), we get,
y2=2yy’x+(yy’)2
So, the required differential equation is given by,
2xy’+y(y’)2-y=0

5. What is the nature of the equation, (xy^3 (frac{dy}{dx})^2+yx^2+frac{dy}{dx}=0)?
a) Second order, third degree, linear differential equation
b) First order, third degree, non-linear differential equation
c) First order, third degree, linear differential equation
d) Second order, third degree, non-linear differential equation
Answer: b
Explanation: Since the equation has only first derivative, i.e. ((frac{dy}{dx}),) it is a first order equation.
Degree is defined as the highest power of the highest order derivative involved. Hence it is 2.
The equation has one/more terms having a variable of degree two/higher; hence it is non-linear.

6. 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
Answer: b
Explanation: There are 2 types of Iterative methods, (i) Interpolation methods (or Bracketing methods) and (ii) Extrapolation methods (or Open-end methods).

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
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. What is the solution of the given equation?
x6y6 dy + (x7y5 +1) dx = 0
a) (frac{(xy)^6}{6} + lnx = c)
b) (frac{(xy)^5}{6} + lny = c)
c) (frac{(xy)^5}{5} + lnx = c)
d) (frac{(xy)^6}{6} + lny = c)
Answer: a
Explanation: Given: (x6y6 + 1) dy + x7y5dx = 0, is an example of non-exact differential equation.
Dividing the equation by x we get,
x5y6 dy + x6y5dx + (frac{dx}{x} = 0)
x5y5 (ydy + xdx) + (frac{dx}{x} = 0 )
(xy)5(d(xy)) + (frac{dx}{x} = 0)
(frac{(xy)^6}{6} + lnx = c)

9. A rectangular frame is to be made of 240 cm long. Determine the value of the length of the rectangle required to maximize the area.
a) 24 cm
b) 60 cm
c) 240 cm
d) 120 cm
Answer: b
Explanation: Let us consider ‘x’ as length and ‘y’ as the breadth of the rectangle.
Given: Perimeter 2(x + y) = 240 cm
x + y = 120
y = 120 – x
Area of the rectangle, a = x*y = x(120-x) = 120x – x2
Finding the derivative, we get, (d(a))/dx = (d(120x – x2))/dx=120-2x
To find the value of x that maximizes the area, we substitute (d(a))/dx = 0.
Therefore, we get, 120 – 2x =0
2x = 120
x = 60 cm
To check if x = 60 cm is the value that maximizes the area, we find the second derivative of the area,
(d2 (a))/(dx2)= -2 < 0 …………………. (i)
We know that the condition for maxima is (d2 (f(x)))/(dx2)<0, which is satisfied by (i), therefore, x = 60 cm maximizes the area of the rectangle.

10. In the equation, y = x2+c,c is known as the parameter and x and y are known as the main variables.
a) True
b) False
Answer: a
Explanation: Given: y = x2+c, where c is known as an arbitrary constant. It is also referred to as the parameter to differentiate it from the main variables x and y.

Global Education & Learning Series – Ordinary Differential Equations.

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250+ TOP MCQs on Laplace Transform of Periodic Function and Answers

Ordinary Differential Equations Multiple Choice Questions on “Laplace Transform of Periodic Function”.

1. Find the laplace transform of f(t), where
f(t) = 1 for 0 < t < a
-1 for a < t < 2a
a) (frac{1}{s} coth⁡(frac{as}{2}))
b) (frac{1}{s} sinh⁡(frac{as}{2}))
c) (frac{1}{s} e^{-as})
d) (frac{1}{s} tanh⁡(frac{as}{2}))
Answer: d
Explanation: In the given question
f(t) is a periodic function having a period 2a
The formula for Laplace Transform is given by:
(L(f(t))=frac{1}{1-e^{-2as}} int_{0}^{2a}e^{-st} f(t)dt)
(L(f(t))=frac{1}{1-e^{-2as}} int_{0}^{a}e^{-st} (1)dt + frac{1}{1-e^{-2as}} int_{a}^{2a}e^{-st}(-1)dt)
=(begin{bmatrix}frac{1}{1-e^(-2as)}×frac{e^{-as}}{-s} – frac{1}{1-e^{-2as}} × frac{-1}{s}end{bmatrix} – begin{bmatrix}frac{1}{1-e^{-2as}} × frac{e^{-2as}}{-s} – frac{1}{1-e^{-2as}} × frac{e^{-as}}{-s}end{bmatrix})
= (frac{1}{1-e^{-2as}}×frac{1}{s}×(1-e^{-as})^2)
= (frac{1}{s}(frac{1+e^{-as}}{1-e^{-as}}))
Dividing both numerator and denominator by (e^{frac{-as}{2}})
= (frac{1}{s} tanh⁡(frac{as}{2}))
Thus, the correct answer is (frac{1}{s} tanh⁡(frac{as}{2})).

2. Find the laplace transform of f(t), where f(t) = |sin(pt)| and t>0.
a) (frac{p}{s^2+p^2}×cosh⁡(frac{spi}{2p}))
b) (frac{p}{s^2+p^2}×sinh⁡(frac{spi}{2p}))
c) (frac{p}{s^2+p^2}×coth⁡⁡(frac{spi}{2p}))
d) (frac{p}{s^2+p^2}×tanh⁡⁡(frac{spi}{2p}))
Answer: c
Explanation: From this question, we know –
Period of sin(t)=2π
Period of sin⁡(pt)=(frac{2pi}{p})
Period of |sin⁡(pt)|=(frac{pi}{p})
(L(f(t))=frac{1}{1-e^{frac{-pi}{ps}}} int_{0}^{frac{pi}{p}}e^{-st} f(t)dt)
Since |sin⁡(pt)| is positive in all quadrants
(L(f(t))=frac{1}{1-e^{frac{-pi}{ps}}} int_{0}^{frac{pi}{p}}e^{-st} sin⁡(pt)dt)
=(frac{1}{1-e^{frac{-pi}{ps}}}begin{bmatrix}frac{e^{frac{-sπ}{p}}}{s^2+p^2}×pend{bmatrix}-begin{bmatrix}frac{1}{s^2+p^2}×(-p)end{bmatrix})
=(frac{1}{1-e^{frac{-pi}{ps}}}×frac{p}{s^2+p^2}×(1+e^{frac{-π}{ps}}))
=(frac{p}{s^2+p^2}×coth⁡(frac{spi}{2p})), (Multiplying and dividing by (e^{frac{-sπ}{2p}}))
Thus, the answer is (frac{p}{s^2+p^2}×coth⁡⁡(frac{spi}{2p})).

Global Education & Learning Series – Ordinary Differential Equations.

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250+ TOP MCQs on Diagonalization Powers of a Matrix and Answers

Linear Algebra and Vector Calculus Multiple Choice Questions on “Diagonalization Powers of a Matrix”.

1. Which of the following is not a necessary condition for a matrix, say A, to be diagonalizable?
a) A must have n linearly independent eigen vectors
b) All the eigen values of A must be distinct
c) A can be an idempotent matrix
d) A must have n linearly dependent eigen vectors
Answer: d
Explanation: The theorem of diagonalization states that, ‘An n×n matrix A is diagonalizable, if and only if, A has n linearly independent eigenvectors.’ Therefore, if A has n distinct eigen values, say λ1, λ2, λ3…λn, then the corresponding eigen vectors are said to be linearly independent. Also, all idempotent matrices are said to be diagonalizable.

2. The geometric multiplicity of λ is its multiplicity as a root of the characteristic polynomial of A, where λ be the eigen value of A.
a) True
b) False
Answer: b
Explanation: The diagonalization theorem in terms of multiplicities of eigen values is defined as follows,
The algebraic multiplicity of λ is its multiplicity as a root of the characteristic polynomial of A.
The geometric multiplicity of λ is the dimension of the λ-eigenspace.

3. If A is diagonalizable then, ____________
a) An = (PDP-1)n = PDnPn
b) An = (PDP-1)n = PDnP1
c) An = (PDP-1)n = PDnP-1
d) An = (PDP-1)n = PDnP
Answer: c
Explanation: The definition of diagonalization states that, An n × n matrix A is diagonalizable if there exists an n × n invertible matrix P and an n × n diagonal matrix D such that,
P-1 AP = D
A = PDP-1
An = (PDP-1)n = PDnP-1

4. The computation of power of a matrix becomes faster if it is diagonalizable.
a) True
b) False
Answer: a
Explanation: Some of the applications of diagonalization of a matrix are:
The powers of a diagonalized matrix can be computed easily since the result is nothing but the powers of the diagonal elements obtained by diagonalization.
Reducing quadratic forms to canonical forms by orthogonal transformations.
In mechanics, it can be used to find the natural frequency of vibrations.

5. Find the invertible matrix P, by using diagonalization method for the following matrix.
A = (begin{bmatrix}
2 & 0 & 0 \
1 & 2 & 1\
-1 & 0 & 1
end{bmatrix} )
a) A = (begin{bmatrix}
-1 & -1 & 0 \
1 & 0 & -1\
-1 & 1 & 1
end{bmatrix} )
b) A = (begin{bmatrix}
0 & -1 & 0 \
1 & 0 & -1 \
0 & 1 & 1
end{bmatrix} )
c) A = (begin{bmatrix}
0 & 0 & 0\
1 & 1 & -1\
-1 & 0 & 1
end{bmatrix} )
d) A = (begin{bmatrix}
1 & 0 & 0\
1 & 0 & -1\
-1 & 0 & 1
end{bmatrix} )
Answer: b
Explanation: Procedure to find the invertible matrix is as follows,
Step 1: Find the eigen values of the given matrix.
A = (begin{bmatrix}
2 & 0 & 0 \
1 & 2 & 1\
-1 & 0 & 1
end{bmatrix} )
⎸A – λI ⎸ = 0
(begin{vmatrix}
2-λ & 0 & 0\
1 & 2-λ & 1\
-1 & 0 & 1-λ
end{vmatrix} ) = 0 ……………………… (i)
(2-λ) ((2- λ) (1- λ)) = 0
(2- λ)2 (1-λ) = 0
λ = 2, 2, 1
Step 2: Compute the eigen vectors
Consider λ = 2,
(A – λI) (vec{X} = vec{0} )
(begin{bmatrix}
0 & 0 & 0\
1 & 0 & 1 \
-1 & 0 & -1
end{bmatrix}
vec{X} = vec{0} quad ^{underrightarrow{Reducing ,further, ,we get,,}}
begin{bmatrix}0 & 0 & 0 \
1 & 0 & 1 \
0 & 0 & 0end{bmatrix}
vec{X}=vec{0}
)
x2 is the free variable, hence, x2 = s
Let x1 = -t, x3 = t, since x1+x3 = 0
(vec{X} = sbegin{bmatrix}
0 \
1 \
0
end{bmatrix} )
+ t(begin{bmatrix}
-1\
0\
1
end{bmatrix} )
(vec{X_1} = begin{bmatrix}
0\
1\
0
end{bmatrix} )
(vec{X_2} = begin{bmatrix}
-1\
0\
1end{bmatrix} )
Consider λ = 1
(A – λI)(vec{X} = vec{0} )
(begin{bmatrix}
1 & 0 & 0\
1 & 1 & 1 \
-1 & 0 & 0
end{bmatrix}
vec{X} = vec{0} quad ^{underrightarrow{Reducing ,further, ,we get,,}}
begin{bmatrix}1 & 0 & 0 \
0 & 1 & 1 \
0 & 0 & 0end{bmatrix}
vec{X}=vec{0}
)
x1 = 0, Let x2 = -s and x3 = s since x2+x3=0
(vec{X}= sbegin{bmatrix}
0\
-1\
1
end{bmatrix} )
(vec{X_3} = begin{bmatrix}
0\
-1\
1
end{bmatrix} )
Step 3: Formation of the invertible matrix.
(P = [vec{X_1} vec{X_2} vec{X_3}])
P = (begin{bmatrix}
0 & -1 & 0 \
1 & 0 & -1\
0 & 1 & 1
end{bmatrix} )

6. Determine the algebraic and geometric multiplicity of the following matrix.
(begin{bmatrix}
2 & 4 & -4 \
0 & 4 & 2\
-2 & 4 & 4
end{bmatrix} )
a) Algebraic multiplicity = 1, Geometric multiplicity = 2
b) Algebraic multiplicity = 1, Geometric multiplicity = 3
c) Algebraic multiplicity = 2, Geometric multiplicity = 2
d) Algebraic multiplicity = 2, Geometric multiplicity = 1
Answer: d
Explanation: The eigen values of the given matrix can be computed as,
⎸A – λI ⎸ = 0
(begin{vmatrix}
1-λ & 0 & 1\
3 & 3-λ & 0\
0 & 0 & 1-λ
end{vmatrix})= 0
(1-λ) ((3-λ) (1-λ)) = 0
(1-λ)2 (3-λ) = 0
λ = 1, 1, 3 are the eigen values of the matrix. So, the algebraic multiplicity of λ = 1 is two.
For λ = 3,
(A – λI) (vec{X} = vec{0} )
(begin{bmatrix}
-2 & 0 & 1 \
3 & 0 & 0 \
0 & 0 & -2
end{bmatrix}vec{X} = vec{0}quad ^{underrightarrow{Reducing ,further, ,we ,get,}}
begin{bmatrix}
1 & 0 & 0 \
0 & 0 & 1\
0 & 0 & 0
end{bmatrix}
vec{X} = vec{0} )
x1 = 0, x3 = 0
x2 is the free variable, therefore let x2 = s,
Hence, (vec{X_1}= sbegin{bmatrix}0 \ 1\ 0end{bmatrix} ≈ begin{bmatrix}0 \ 1 \0 end{bmatrix} )
For λ = 1,
(A – λI) (vec{X} = vec{0} )
(begin{bmatrix}
0 & 0 & 1 \
3 & 2 & 0 \
0 & 0 & 0
end{bmatrix} vec{X} = vec{0} quad ^{underrightarrow{Reducing ,further, ,we ,get,}}
begin{bmatrix}
1 & 2/3 & 0\
0 & 0 & 1\
0 & 0 & 0
end{bmatrix} vec{X} = vec{0} )
x1 + (frac{2}{3}) x2 = 0, x3 = 0
Let x1 = -2 and x3 = 3,
(vec{X_2} = begin{bmatrix}-2 \ 3\ 0 end{bmatrix})
Thus, there corresponds only one eigen vector for the repeated eigen value λ=1. Thus, the geometric multiplicity of λ = 2 is one.

7. Given P = ( begin{bmatrix}2 & -1 \ 5 & 1 end{bmatrix} , and, D = begin{bmatrix}6 & 0 \ 0 & -1end{bmatrix},) find A3.
a) (begin{bmatrix}
61 & 62 \ 156 & 154end{bmatrix})
b) (begin{bmatrix}
61 & 62 \ 155 & 154 end{bmatrix})
c) (begin{bmatrix}
61 & 60 \ 155 & 154 end{bmatrix})
d) (begin{bmatrix}
61 & 62\ 155 & 150end{bmatrix})
Answer: b
Explanation: From the theory of diagonalization, we know that,
A = PDP-1
An = PDnP-1
Given, P= (begin{bmatrix}
2 & -1\
5 & 1,end{bmatrix} hence P^{-1} = frac{1}{7}
begin{bmatrix}
1 & 1\
-5 & 2end{bmatrix})
Therefore, (A^3 = frac{1}{7} begin{bmatrix}
2 & -1\
5 & 1
end{bmatrix}
begin{bmatrix}
6 & 0\
0 & -1 end{bmatrix}^3
begin{bmatrix}
1 & 1\
-5 & 2 end{bmatrix})…………………………. since n=3
(A^3 = frac{1}{7}
begin{bmatrix}2 & -1 \ 5 & 1 end{bmatrix}
begin{bmatrix}216 & 0 \ 0 & -1 end{bmatrix}
begin{bmatrix}1 & 1 \ -5 & 2 end{bmatrix})
(A^3 = frac{1}{7}
begin{bmatrix} 2 & -1 \ 5 & 1 end{bmatrix}
begin{bmatrix} 216 & 216 \ 5 & -2 end{bmatrix})
(A^3 = frac{1}{7}
begin{bmatrix}427 & 434\ 1085 & 1078 end{bmatrix})
(A^3 = begin{bmatrix} 61 & 62\ 155 & 154 end{bmatrix})

8. Find the trace of the matrix (A = begin{bmatrix}1 & 0 & 6\
0 & 5 & 0\
0 & 4 & 4 end{bmatrix}.)
a) 0
b) 10
c) 4
d) 1
Answer: b
Explanation: The sum of the entries on the main diagonal is called the trace of matrix A.
Therefore, trace = 1+5+4 = 9.

9. The determinant of the matrix whose eigen values are 4, 2, 3 is given by, _______
a) 9
b) 24
c) 5
d) 3
Answer: b
Explanation: The product of the eigen values of a matrix gives the determinant of the matrix,
Therefore, ∆ = 24.

10. Which of the following relation is correct?
a) A = AT
b) A = -AT
c) A = A2T
d) A = AT/2
Answer: a
Explanation: To prove that A = AT, let us consider an example,
(A = begin{bmatrix} 1 & 0 \ 2 & 3 end{bmatrix})
⎸A – λI ⎸ = 0
( begin{vmatrix} 1-λ & 0 \ 2 & 3-λ end{vmatrix} = 0)
(1-λ) (3-λ) = 0
3 – λ – 3λ +λ2 = 0
λ2 – 4λ + 3 = 0
(λ – 3) (λ – 1) = 0
λ = 1, 3
Consider (A^T = begin{bmatrix}1 & 2\ 0 & 3 end{bmatrix})
⎸A – λI ⎸ = 0
(begin{bmatrix}1-λ & 2 \ 0 & 3-λ end{bmatrix} = 0)
(1-λ) (3-λ) = 0, which is similar to the result obtained for A, hence the eigen values are same.

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250+ TOP MCQs on Homogeneous Linear PDE with Constant Coefficient and Answers

Partial Differential Equations Assessment Questions and Answers focuses on “Homogeneous Linear PDE with Constant Coefficient”.

1. Homogeneous Equations are those in which the dependent variable (and its derivatives) appear in terms with degree exactly one.
a) True
b) False
View Answer

Answer: a
Explanation: Linear partial differential equations can further be classified as:

  • Homogeneous for which the dependent variable (and its derivatives) appear in terms with degree
    exactly one, and
  • Non-homogeneous which may contain terms which only depend on the independent variable

2. Which of the following is false regarding quasilinear equations?
a) All the terms with highest order derivatives of dependent variables occur linearly
b) The coefficients of terms with highest order derivatives of dependent variables are functions of only lower order derivatives of the dependent variables
c) Lower order derivatives can occur in any manner
d) All the terms with lower order derivatives of dependent variables occur linearly
View Answer

Answer: d
Explanation: A PDE is called as a quasi-linear if at the minimum one coefficient of the partial derivatives is a function of the dependent variable. For example, (frac{∂^2 u}{∂x^2}-u frac{∂^2 u}{∂y^2}=0. )

3. Which of the following is false regarding Bessel polynomials?
a) Krall and Fink (1949) defined the Bessel polynomials
b) The polynomials satisfy the recurrence formula
c) The solutions of homogeneous equations are closely related to Bessel polynomials
d) Bessel polynomials are not an orthogonal sequence of polynomials
View Answer

Answer: d
Explanation: The Bessel polynomials are an orthogonal sequence of polynomials. The most accepted definition for these series was put forth by Krall and Frink (1948) as,
(y_n (x)=∑_{k=0}^nfrac{(n+k)!}{(n-k)!k!} (frac{x}{2})^k.)

4. Which of the following is not a homogeneous equation?
a) (frac{∂^2 u}{∂t^2}-c^2 frac{∂^2 u}{∂x^2}=0)
b) (frac{∂^2 u}{∂x^2}+frac{∂^2 u}{∂y^2}=0)
c) (frac{∂^2 u}{∂x^2}+(frac{∂^2 u}{∂x∂y})^2+frac{∂^2 u}{∂y^2}=x^2+y^2)
d) (frac{∂u}{∂t} -Tfrac{∂^2 u}{∂x^2}=0)
View Answer

Answer: c
Explanation: As we know that homogeneous equations are those in which the dependent variable (and its derivatives) appear in terms with degree exactly one, hence the equation,
(frac{∂^2 u}{∂x^2}+(frac{∂^2 u}{∂x∂y})^2+frac{∂^2 u}{∂y^2}=x^2+y^2) is not a homogenous equation (since its degree is 2).

5. What is the general solution of the DE with n linearly independent solutions u1(t), …., un(t) of a nth order linear homogeneous DE?
a) (u(t)=u_1 (t)+⋯+c_{n+1} u_n (t)=∑_{k+1}^n=c_{k+1} u_k (t))
b) (u(t)=u_1 (t)+⋯+u_n (t)=∑_{k=1}^nu_k(t) )
c) (u(t)=c_1 u_1 (t)+⋯+c_n u_n (t)=∑_{k=1}^n c_k u_k (t) )
d) (u(t)=c_0 u_0 (t)+⋯+c_n u_n (t)=∑_{k=0}^∞c_k u_k (t) )
View Answer

Answer: c
Explanation: If we know n linearly independent solutions u1(t), …., un(t) of a nth order linear homogeneous DE, then the general solution of this DE has the form:
(u(t)=c_1 u_1 (t)+⋯+c_n u_n (t)=∑_{k=1}^nc_k u_k (t))

6. What is the degree of the homogeneous partial differential equation, (frac{∂^2 u}{∂t^2}-c^2 frac{∂^2 u}{∂x^2}=0)?
a) Second-degree
b) First-degree
c) Third-degree
d) Zero-degree
View Answer

Answer: b
Explanation: From the given equation, (frac{∂^2 u}{∂t^2}-c^2 frac{∂^2 u}{∂x^2}=0)we deduce that the power of the highest order term is 1, hence degree = 1.

7. The solution of an ODE contains arbitrary constants, the solution to a PDE contains arbitrary functions.
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.

8. Which of the following is not an example of 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.

9. A homogeneous linear differential equation has constant coefficients if it has the form
(a_0 y+a_1 y’+a_2 y”+⋯+a_n y^{(n)}=0,) where a1,…,an are (real or complex) numbers.
a) False
b) True
View Answer

Answer: b
Explanation: (a_0 y+a_1 y’+a_2 y”+⋯+a_n y^{(n)}=0,) is the form which represents a homogeneous linear differential equation which has constant coefficients. In other words, it has constant coefficients if it is defined by a linear operator with constant coefficients.

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. 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

12. 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.

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250+ TOP MCQs on Functions of a Complex Variable and Answers

Complex Analysis Multiple Choice Questions on “Functions of a Complex Variable”.

1. Find the domain of the function defined by f(z)=z/(z+z̅).
a) Im(z)≠0
b) Re(z)≠0
c) Im(z)=0
d) Re(z)=0
View Answer

Answer: b
Explanation: Write z=x+iy ⇒ f(x+iy)=(x+iy)/(x+iy+x-iy)=(x+iy)/2x
=1/2+iy/2x ⇒ x≠0 ⇒ Re(z)≠0 .

2. Let f(z)=z+1/z. What will be the definition of this function in polar form?
a) (r+1/r)cosθ+i(r-1/r)sinθ
b) (r-1/r)cosθ+i(r+1/r)sinθ
c) (r+1/r)sinθ+i(r-1/r)cosθ
d) (r+1/r)sinθ+i(r-1/r)cosθ
View Answer

Answer: a
Explanation: Write z=r(cosθ+isinθ), therefore, f(z)=z+1/z=r(cosθ+isinθ)+1/[r(cosθ+isinθ)]
=reiθ+(1/r)e-iθ=r(cosθ+isinθ)+1/r(cosθ-isinθ)=(r+1/r)cosθ+i(r-1/r)sinθ.

3. For the function f(z)=zi, what is the value of |f(ω)|+Arg f(ω), ω being the cube root of unity with Im(ω)>0?
a) e-2π/3
b) e2π/3
c) e-2π/3+2π/3
d) e-2π/3-2π/3
View Answer

Answer: a
Explanation: Let y=zi⇒ ln y=iln z=i(ln |z|+iarg z)=iln |z|-arg z
⇒ y=eiln |z|/earg z ⇒ |y|=earg z and Arg y=ln |z| ⇒ |f(ω)|+Arg f(ω)=e-2π/3+0=e-2π/3.

4. Let f(z)=(z2–z–1)7. If α2+α+1=0 and Im(α)>0, then find f(α).
a) 128α
b) -128α
c) 128α2
d) -128α2
View Answer

Answer: c
Explanation: Note that α=ω. Therefore, f(α)=f(ω)=(ω2–ω–1)7
=(ω2+ω2)7=(2ω2)7=27ω14=128ω2=128α2.

5. For all complex numbers z satisfying Im(z)≠0, if f(z)=z2+z+1 is a real valued function, then find its range.
a) (-∞, -1]
b) (-∞, 1/3)
c) (-∞, 1/2]
d) (-∞, 3/4)
View Answer

Answer: d
Explanation: Let y=f(z). then z2+z+1=y has imaginary roots (∵Im(z)≠0)
⇒ D<0 ⇒ 1–4(1–y)<0 ⇒ 4y<3 ⇒ y<3/4 . Also, putting Re z=-1/2 and Im z=∞, we get, f(z)=-∞.

6. Let x, y, z be integers, not all simultaneously equal. If ω is a cube root of unity with Im(ω)≠1, and if f(z)=az2+bz+c, then find the range of |f(ω)|.
a) (0, ∞)
b) [1, ∞)
c) (√3/2, ∞)
d) [1/2, ∞)
View Answer

Answer: b
Explanation: ω=-1/2+i√3/2. Therefore, |f(ω)|=|a+b(-1/2+i√3/2)+c(-1/2-i√3/2)|
=|(2a-b-c)/2+i(b√3-c√3)/2|=1/2[(2a-b-c)2+3(b-c)2]1/2={1/2[(a-b)2+(b-c)2+(c-a)2]}1/2. Putting b=c=0 and a=1 gives us the minimum value=1, while, a=∞ gives us the maximum value=∞.

7. Let f(z)=arg 1/(1 – z), then find the range of f(z) for |z|=1, z≠1.
a) (-∞, π/2)
b) (-π/2, π/2)
c) (-∞, ∞)
d) [0, π/2)
View Answer

Answer: b
Explanation: Let y=1/(1-z) ⇒ z=1-1/y
|z|=1 ⇒ |1-1/y|=1 ⇒ |y-1|=|y| ⇒ locus of y is the perpendicular bisector of line segment joining 0 and 1 ⇒ arg y ∈(-π/2, π/2).

8. Define f(z)=z2+bz−1=0 and g(z)=z2+z+b=0. If there exists α satisfying f(α)=g(α)=0, which of the following cannot be a value of b?
a) √3i
b) -√3i
c) 0
d) √3i/2
View Answer

Answer: d
Explanation: α2+bα−1=0 and α2+α+b=0 ⇒ (b−1)α−1−b=0 ⇒ α=(b+1)/(b-1)
⇒ (b+1)2/(b-1)2+(b+1)/(b-1)+b=0 ⇒ b=√3i, -√3i, 0.

9. Let f(z)=2(z+z̅)+3i(z-z̅) and g(z)=|z|. f(z)=2 divides the region g(z)≤6 into two parts. If Q={(2+3i/4), (5/2+3i/4), (1/4-i/4), (1/8+i/4)}, then find the number of elements of Q lying inside the smaller part.
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a) 1
b) 2
c) 3
d) 4
View Answer

Answer: b
Explanation: Writing z=x+iy, we get L as 2x–3y–1 and S as x2+y2–6, a point z1 lies in the smaller region if L1>0 and S1<0. ∴ (2+3i/4) and (1/4-i/4) lie in the smaller region.

10. Find the range of the function defined by f(z)=Re[2iz/(1-z2)].
a) (−∞, 0) ⋃ (0, ∞)
b) [2, ∞)
c) (−∞, −1] ⋃ [1, ∞)
d) (−∞, 0] ⋃ [2, ∞)
View Answer

Answer: c
Explanation: z=2i(x+iy)/(1-(x+iy)2)=2i(x+iy)/(1-(x2-y2+2ixy))
Using 1-x2=y2, z=(2ix-2y)/(2y2-2ixy)=-1/y
∵ –1≤y≤1 ⇒ –1/y≤-1 or -1/y≥1.

11. Let f(z)=|z|2+Re z(2(z+z̅)+3(z-z̅)/2i, the find the maximum value of |z|2/f(z).
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: b
Explanation: Write z=|z|(cosθ+isinθ) ⇒ |z|2/f(z)=1/(1+4cos2θ+3sinθcosθ)
=1/(1+4cos2θ+3/2sin2θ)=1/[2(1+cos2θ)+1+3/2sin2θ].
Now, 2(1+cos2θ)+1+3/2sin2θ=3+2cos2θ+3/2sin2θ≥3-(4+9/4)1/2=1/2.
Hence, maximum value is 2.

12. Consider a function f(z) of degree two, having real coefficients. If z1 and z2 satisfying f(z1)=f(z2)=0 are such that Re z1=Re z2=0 and if z3 satisfies f(f(z3))=0, then select the correct statement.
a) Re z3=0
b) Im z3=0
c) Re z3×Imz3≠0
d) Re z3=0 and Im z3=0
View Answer

Answer: c
Explanation: f(z)=az2+b, with a, b of same sign ⇒ f(f(z))=a(az2+b)2+b
If z∈R or iz∈R ⇒ z2∈R ⇒ f(z)∈R ⇒ f(f(z))≠0 ⇒ Hence real or purely imaginary number cannot satisfy f(f(z))=0.

13. Let f(z)=|1–z|, if zk=cos(2kπ/10)+isin(2kπ/10), then find the value of f(z1)×f(z2)×…×f(z9).
a) 10
b) 15
c) 20
d) 30
View Answer

Answer: a
Explanation: z10–1=(z-1)(z-z1)…(z-z9) ⇒ (z-z1)(z-z2) …(z-z9)=1+z+z2+…+z9.
Now, putting z=1, we get, (z-z1)(z-z2)…(z-z9)=f(z1)×f(z2)×…×f(z9)=10.

14. For a∈R, let f(z)=z5-5z+a. Select the correct statement for α satisfying f(α)=0.
a) α has exactly three possible real values for a>4
b) α has exactly one possible real value for a>4
c) α has exactly three possible real values for a<-4
d) α has exactly one possible real value for -4<a<4
View Answer

Answer: b
Explanation: z5-5z+a=0 ⇒ z5-5z=-a ⇒ z(z-51/4)(z+51/4)(z2+51/2)=-a f'(z)=5z4–5=0 ⇒ (z2+1)(z2-1)=0 ⇒ (z-1)(z+1)(z2+1)=0 ⇒ α has exactly one possible real value for a>4 and exactly three possible real values for -4<a<4.

15. Let f(z)=z4+a1z3+a2z2+a3z+a4=0; a1, a2, a3, a4 being real and non-zero. If f has a purely imaginary root, then what is the value of the expression a3/(a1a2)+ a1a4/(a2a3) ?
a) 0
b) 1
c) -2
d) 2
View Answer

Answer: b
Explanation: For real x(≠0), let ix be the root⇒x4-a1x3i- a2x2+a3xi+a4=0⇒x4-a2x2+a4=0 and a1x3-a3x=0
a1x3-a3x=0 ⇒ a1x2-a3=0 ⇒x2=a3/a1, putting this value in the equation, a3/(a1a2)+a1a4/(a2a3)=1.

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