300+ REAL TIME Engineering Mathematics Objective Questions & Answers 2023

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})).

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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…λ_n, 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 = A^T
b) A = -A^T
c) A = A^2T
d) A = A^T/2
Answer: a
Explanation: To prove that A = A^T, 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), …., u_n(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,…,a_n 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)]
=re^iθ+(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) e^2π/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=e^{iln |z|}/e^{arg z} ⇒ |y|=e^{arg 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 z_k=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-5^1/4)(z+5^1/4)(z²+5^1/2)=-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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250+ TOP MCQs on Leibniz Rule and Answers

Engineering Mathematics Interview Questions and Answers for freshers focuses on “Leibniz Rule – 3”.

1. Leibniz rule gives the
a) Nth derivative of addition of two function
b) Nth derivative of division of two functions
c) Nth derivative of multiplication of two functions
d) Nth derivative of subtraction of two function
Answer: c
Explanation: Leibniz rule is
(frac{d^n}{dx^n}(uv))
(= n_{C_0} u_n v+ n_{C_1} u_{(n-1)} v_1+ n_{C_2} u_{n-2} v_2+…+ n_{C_r} u_{n-r} v_r+…+ n_{C_n} u_0 v_n)
Hence Leibniz theorem gives nth derivative of multiplication of two functions u and v.
.

2. Leibniz theorem is applicable if n is a
a) Rational Number
b) Negative Integer
c) Positive Integer
d) Decimal Number
Answer: c
Explanation: Leibniz rule is
(frac{d^n}{dx^n}(uv))
(= n_{C_0} u_n v+ n_{C_1} u_{n-1} v_1+ n_{C_2} u_{n-2} v_2+…+ n_{C_r} u_{n-r} v_r+…+ n_{C_n} u_0 v_n)
For all n > 0, i.e n should be positive
Hence Leibniz theorem gives nth derivative of multiplication of two functions u and v if n is a positive integer.

3. If nth derivative of xy₃ + x²y₂ + x³y₀ = 0 then order of its nth differential equation is,
a) n
b) n+1
c) n+2
d) n+3
Answer: d
Explanation:
1. If we differentiate this equation n times then terms comes in nth order differential equation is y_n+3, y_n+2, y_n+1, y_n, y_n-1, y_n-2, y_n-3. Hence order of differential equation becomes n+3.
2. By Leibniz rule differentiating it n times, we get
Xy_n+3 + ny_n+2 + x²y_n+2 + 2nxy_n+1 + 2n(n-1)y_n + x³y_n + 3nx²y_n-1 + 6n(n-1)xy_n-2 + 6n(n-1)(n-2)y_n-3 = 0
Hence order of differential equation becomes n+3.

4. Find nth derivative of x²y₂ + xy₁ + y = 0
a) x²y_n+2 + (2n+1)xy_n+1 + (n²+1)y_n = 0
b) x²y_n+2 + nxy_n+1 + (n²+1)y_n = 0
c) x²y_n+2 + (2n+1)xy_n+1 + n²y_n = 0
d) x²y_n+2 + 2nxy_n+1 + n²y_n = 0
Answer: a
Explanation: x²y₂ + xy₁ + y = 0
By Leibniz theorem
x²y_n+2 + n(2x)(y_n+1) + n(n-1)(2)(y_n) + xy_n+1 + ny_n + y_n= 0

x²y_n+2 + xy_n+1(2n+1) + y_n(n²+1) = 0.

5. The nth derivative of x²y₂ + (1-x²)y₁ + xy = 0 is,
a) x²y_n+2 + y_n(2n²-2n-2nx+x) – y_n-1(2n²-3n)=0
b) x²y_n+2 + y_n+1(2nx-x²) + y_n(2n²-2n-2nx+x) – y_n-1(2n²-3n)=0
c) x²y_n+2 + y_n+1(2nx+1-x²) + y_n(2n²-2n-2nx+x)-y_{n – 1}(2n²-3n)=0
d) x²y_n+2 + y_n+1(2nx+1-x²) + xy_n2n²-y_{n – 1}(2n²-3n)=0
Answer: c
Explanation: x²y₂ + (1-x²)y₁ + xy = 0
Differentiating n times by Leibniz Rule
x²y_n+2 + 2nxy_n+1 + 2n(n-1)y_n + (1-x²)y_n+1 – 2nxy_n – 2n(n-1)y_n-1+ xy_n + ny_n-1 = 0
x²y_n+2 + y_n+1(2nx+1-x²) + y_n(2n²-2n-2nx+x) – y_n-1(2n²-3n)=0.

6. Find nth derivative of xⁿSin(nx)
a) (n!Sin(nx) + n_{C_1} n_{P_{n-1}}x^{n-1}nCos(nx) – n_{C_2}n_{P_{n-2}}x^{n-2}n^2Sin(nx))
(-n_{C_3}n_{P_{n-3} }x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+x^nn^nSin(nx+nπ/2))
b) (Sin(nx) + n_{C_1} n_{P_{n-1}}x^{n-1}nCos(nx) – n_{C_2}n_{P_{n-2}}x^{n-2}n^2Sin(nx))
(-n_{C_3}n_{P_{n-3}}x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+n!x^nn^nSin(nx+nπ/2))
c) (nSin(nx) + n_{C_1} n_{P_{n-1}}x^{n-1}nCos(nx) – n_{C_2}n_{P_{n-2}}x^{n-2}n^2Sin(nx))
(-n_{C_3}n_{P_{n-3}}x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+nx^nn^nSin(nx+nπ/2))
d) ((n-1)!Sin(nx) + n_{C_1} n_{P_{n-1}}x^{n-1}nCos(nx) – n_{C_2}n_{P_{n-2}}x^{n-2}n^2Sin(nx))
(-n_{C_3}n_{P_{n-3}}x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+x^nn^{n-1}Sin(nx+nπ/2))
Answer: a
Explanation: Y = xⁿSin(nx)
By Leibniz Rule , put u = xⁿ and v = Sin(nx), we get
(n!Sin(nx) + n_{C_1} n_(P_{n-1})x^{n-1}nCos(nx) – n_{C_2}n_(P_{n-2})x^{n-2}n^2Sin(nx))
(-n_{C_3}n_{P_{n-3}}x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+x^nn^nSin(nx+nπ/2))

7. If y(x) = tan^-1x then
a) (y_n+1)(0) = (n-1)(y_n-1)0
b) (y_n+1)(0) = n(n-1)(y_n-1)0
c) (y_n+1)(0) = -(n-1)(y_n-1)0
d) (y_n+1)(0) = -n(n-1)(y_n-1)0
Answer: d
Explanation: Y = tan^-1x
Y₁ = 1/(1+x²)
(1+x²)y₁ = 1
By Leibniz Rule,
(1+x²)y_n+1 + 2nxy_n + n(n-1)y_n-1 = 0
Put x=0, gives
→ (y_n+1)(0) = -n(n-1)(y_n-1)(0).

8. If y = sin^-1x, then
a) (1-x²)y_n+2 – xy_n+1(2n-1) = ny_n(2n-1)
b) x²y_n+2 – xy_n+1(2n-1) = ny_n(2n-1)
c) (1-x²)y_n+2 – 2nxy_n+1 = ny_n(2n-1)
d) (1-x²)y_n+2 – xy_n+1(2n-1) +ny_n(2n-1)=0
Answer: a
Explanation: Y = sin^-1x
Differentiating it
Y₁ = (frac{1}{sqrt{1-x^2}})
(1-x²)(y₁)²= 1
Again Differentiating we get
(1-x²)2y₁y₂ – 2x(y₁)² = 0
(1-x²)y₂ = xy₁
By Leibniz Rule, Diff it n times,
(1-x²)y_n+2 – 2xny_n+1 – 2n(n-1)y_n = xy_n+1 + ny_n
(1-x²) y_n+2 – xy_n+1(2n-1) = ny_n(2(n-1)+1)
(1-x²) y_n+2 – xy_n+1(2n-1) = ny_n(2n-1).

9. If y = cos^-1x, then
a) (y_n+2)(0) = -n(2n-1) y_n(0)
b) (y_n+2)(0) = n(2n-1) y_n(0)
c) (y_n+2)(0) = n(n-2) y_n(0)
d) (y_n+2)(0) = n(n-3) y_n(0)
Answer: b
Explanation: y = cos^-1x
Differentiating it
Y₁ = (frac{1}{sqrt{1-x^2}})
(1-x²)(y₁)²= 1
Again Differentiating we get
(1-x²)2y₁y₂ – 2x(y₁)² = 0
(1-x²)y₂ = xy₁
By Leibniz Rule, Diff it n times,
(1-x²)y_n+2 – 2xny_n+1 – 2n(n-1)y_n = xy_n+1 + ny_n
(1-x²) y_n+2 – xy_n+1(2n-1) = ny_n(2(n-1)+1)
(1-x²) y_n+2 – xy_n+1(2n-1) = ny_n(2n-1)
At x=0, we get
(y_n+2)(0) = n(2n-1) y_n(0).

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