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)]
=re+(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
=(ω22)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.

Global Education & Learning Series – Complex Analysis.

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

 

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 yn+3, yn+2, yn+1, yn, yn-1, yn-2, yn-3. Hence order of differential equation becomes n+3.
2. By Leibniz rule differentiating it n times, we get
Xyn+3 + nyn+2 + x2yn+2 + 2nxyn+1 + 2n(n-1)yn + x3yn + 3nx2yn-1 + 6n(n-1)xyn-2 + 6n(n-1)(n-2)yn-3 = 0
Hence order of differential equation becomes n+3.

4. Find nth derivative of x2y2 + xy1 + y = 0
a) x2yn+2 + (2n+1)xyn+1 + (n2+1)yn = 0
b) x2yn+2 + nxyn+1 + (n2+1)yn = 0
c) x2yn+2 + (2n+1)xyn+1 + n2yn = 0
d) x2yn+2 + 2nxyn+1 + n2yn = 0
Answer: a
Explanation: x2y2 + xy1 + y = 0
By Leibniz theorem
x2yn+2 + n(2x)(yn+1) + n(n-1)(2)(yn) + xyn+1 + nyn + yn= 0

x2yn+2 + xyn+1(2n+1) + yn(n2+1) = 0.

5. The nth derivative of x2y2 + (1-x2)y1 + xy = 0 is,
a) x2yn+2 + yn(2n2-2n-2nx+x) – yn-1(2n2-3n)=0
b) x2yn+2 + yn+1(2nx-x2) + yn(2n2-2n-2nx+x) – yn-1(2n2-3n)=0
c) x2yn+2 + yn+1(2nx+1-x2) + yn(2n2-2n-2nx+x)-yn – 1(2n2-3n)=0
d) x2yn+2 + yn+1(2nx+1-x2) + xyn2n2-yn – 1(2n2-3n)=0
Answer: c
Explanation: x2y2 + (1-x2)y1 + xy = 0
Differentiating n times by Leibniz Rule
x2yn+2 + 2nxyn+1 + 2n(n-1)yn + (1-x2)yn+1 – 2nxyn – 2n(n-1)yn-1 + xyn + nyn-1 = 0
x2yn+2 + yn+1(2nx+1-x2) + yn(2n2-2n-2nx+x) – yn-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) (yn+1)(0) = (n-1)(yn-1)0
b) (yn+1)(0) = n(n-1)(yn-1)0
c) (yn+1)(0) = -(n-1)(yn-1)0
d) (yn+1)(0) = -n(n-1)(yn-1)0
Answer: d
Explanation: Y = tan-1x
Y1 = 1/(1+x2)
(1+x2)y1 = 1
By Leibniz Rule,
(1+x2)yn+1 + 2nxyn + n(n-1)yn-1 = 0
Put x=0, gives
→ (yn+1)(0) = -n(n-1)(yn-1)(0).

8. If y = sin-1x, then
a) (1-x2)yn+2 – xyn+1(2n-1) = nyn(2n-1)
b) x2yn+2 – xyn+1(2n-1) = nyn(2n-1)
c) (1-x2)yn+2 – 2nxyn+1 = nyn(2n-1)
d) (1-x2)yn+2 – xyn+1(2n-1) +nyn(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)yn+2 – 2xnyn+1 – 2n(n-1)yn = xyn+1 + nyn
(1-x2) yn+2 – xyn+1(2n-1) = nyn(2(n-1)+1)
(1-x2) yn+2 – xyn+1(2n-1) = nyn(2n-1).

9. If y = cos-1x, then
a) (yn+2)(0) = -n(2n-1) yn(0)
b) (yn+2)(0) = n(2n-1) yn(0)
c) (yn+2)(0) = n(n-2) yn(0)
d) (yn+2)(0) = n(n-3) yn(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)yn+2 – 2xnyn+1 – 2n(n-1)yn = xyn+1 + nyn
(1-x2) yn+2 – xyn+1(2n-1) = nyn(2(n-1)+1)
(1-x2) yn+2 – xyn+1(2n-1) = nyn(2n-1)
At x=0, we get
(yn+2)(0) = n(2n-1) yn(0).

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250+ TOP MCQs on Derivative of Arc Length and Answers

Differential Calculus Question Bank focuses on “Derivative of Arc Length”.

1. For the cartesian curve y=f(x) with ‘s’ as arc length which of the following condition holds good?
a) (frac{ds}{dx} = sqrt{1+(frac{dy}{dx})^2})
b) (frac{d^2s}{dx^2} = sqrt{1-(frac{dy}{dx})^2})
c) (frac{dy}{dx} = sqrt{1+(frac{ds}{dx})^2})
d) (sqrt{(frac{ds}{dx})^2 + (frac{dy}{dx})^2} = 1)
View Answer

Answer: a

2. For the curve (y=a ,log ,sec(frac{x}{a})) what is the value of ( frac{ds}{dx})? (where φ is the angle made by tangent to the curve with x-axis)?
a) cos φ
b) sec φ
c) tan φ
d) cot φ
View Answer

Answer: b
Explanation: w.k.t (frac{ds}{dx} = sqrt{1+(frac{dy}{dx})^2})
(frac{dy}{dx} = a frac{tan frac{x}{a}.secfrac{x}{a}}{secfrac{x}{a}} frac{1}{a} = tan frac{x}{a} )
substituting we get
(frac{ds}{dx} = sqrt{1+(tan frac{x}{a})^2}=secfrac{x}{a}, ,but ,w.k.t, frac{dy}{dx} = tan φ = tanfrac{x}{a} ,thus, φ=frac{x}{a})
(frac{ds}{dx} = sec⁡φ.)

3. If the parametric equation of the curve is given by x=aet sin⁡t & y=aet cos⁡ t the value of (frac{ds}{dt}) is given by _____
a) aet
b) 2aet
c) √3 aet
d) √2 aet
View Answer

Answer: d
Explanation: w.k.t (frac{ds}{dt} = sqrt{(frac{dx}{dt})^2 + (frac{dy}{dt})^2}…(1))
(frac{dx}{dt} = ae^t (cos t + sin t), frac{dy}{dt} = ae^t (-sin⁡ t + cos⁡t))
substituting in (1) we get
(frac{ds}{dt} = ae^t sqrt{(cos t + sin t)^2+(-sin t+cos⁡t)^2})
(frac{ds}{dt} = ae^t sqrt{1+2 sin t cos + 1 – 2 sin⁡ t cos⁡t} = sqrt{2} ae^t…(cos^{2} t + sin{2} t = 1).)

4. For the curve in polar form (sqrt{frac{r}{a}} = sec⁡(frac{θ}{2}) ,the, ,value, ,of, frac{ds}{dθ}) is _____
a) r sec θ
b) r sec ((frac{θ}{2}))
c) r sec (2θ)
d) r cosec ((frac{θ}{2}))
View Answer

Answer: c
Explanation: Squaring the given curve on both side i.e (r=a sec ^2 (frac{θ}{2}))…(1)
(frac{dr}{dθ} = a.2 sec⁡(frac{θ}{2}) sec⁡(frac{θ}{2}) tan (frac{θ}{2}).1/2 = a sec^2 (frac{θ}{2}) tan(frac{θ}{2}) )
from (1)
(frac{dr}{dθ} = r tan (frac{θ}{2}) ,the, ,equation, ,for, frac{ds}{dθ} ,is, = sqrt{(r^2+(frac{dr}{dθ})^2)} = sqrt{r^2(1+ tan^2 (frac{θ}{2}))} )
(frac{ds}{dθ} = r sec (frac{θ}{2}).)

5. If r=b eθ cot⁡a, where a, b are constants then ‘s’ is represented by which one of the following equation?
a) s=r sec(a)+c
b) s=r cos(a)+c
c) s=r+c
d) s=r tan(a)+c
View Answer

Answer: a
Explanation: r=b eθ cot⁡a
(frac{dr}{dθ} = b e^{θ cot⁡a}(cot a)=r cot(a) )
w.k.t (frac{ds}{dr} = sqrt{1+r^2 (frac{dθ}{dr})^2} ,where, frac{dθ}{dr}=frac{1}{r} ,tan⁡a)
(frac{ds}{dr} = sqrt{1+tan^2 a} = sec ,a rightarrow s = int sec ,a ,dr , = r sec(a)+c,) where c is constant of integration.

250+ TOP MCQs on Errors and Approximations and Answers

Engineering Mathematics Multiple Choice Questions on “Errors and Approximations”.

1. Error is the Uncertainty in measurement.
a) True
b) False
Answer: a
Explanation: In the term of mathematics, “Error tells the person how much correct or certain its measurement is.”

2. Relative error in x is?
a) δx
b) δxx
c) δxx * 100
d) 0
Answer: b
Explanation: Option ‘δx’ is called absolute error.
Option ‘δxx’ is called relative error.
Option ‘δxx * 100’ is called Percentage error.

3. Find the percentage change power in the circuit if error in value of resistor is 1% and that of voltage source is .99%.
a) z should be homogeneous and of order n
b) z should not be homogeneous but of order n
c) z should be implicit
d) z should be the function of x and y only
Answer: a
Explanation: Power is given by P = V2R
Taking log on both sides,
log(P) = 2log(V) – log(r)
Differentiating it ,
δpp = 2δVVδrr
Multiplying by 100 we get,
%P = 2%V – %r = (2*.99) – 1 = 0.98%.

4. Magnitude of error can be negative or positive.
a) True
b) False
Answer: b
Explanation: Magnitude of error can not be negative.Negative or positive sign only shows the increase or decrease in the quatity.

5. Given the kinetic energy of body is T = 12 mv2. If the mass of body changes from 100 kg to 100 kg and 500 gm and velocity of a body changes from 1600 mt/sec to 1590 mt/sec. Then find the approximate change in T.
a) 960000 J decrease in value
b) 960000 J increase in value
c) 450000 J decrease in value
d) 450000 J increase in value
Answer: a
Explanation: Given T = 12 mv2
Now taking log and differentiating,
δT = 0.5[v2 δm + 2mvδv]
Now, v = 1600 mt/sec, m = 100kg, δv = -10, δm = 0.5
Then,
δT = -960000 J => decrese in value of T by 960000 J.

6. The speed of a boat is given by, v = k(1t – at), where k is the constant and l us the distance travel by boat in time t and a is the acceleration of water. If there is an change in ‘l’ from 2cm to 1cm in time 2sec to 1sec. If the acceleration of water changes from 0.95 mt/sec2 to 2 mt/sec2 find the motion of boat.
a) -2
b) 2
c) 0.5
d) -0.5
Answer: a
Explanation: Given, v = k(1t – at)
Differentiating it we get
(δv=kleft [frac{(tδl-lδt)}{t^2} – aδt – tδa right ])
Hence,
(frac{δv}{v}=frac{kleft [frac{(tδl-lδt)}{t^2} – aδt – tδa right ]}{v})
(frac{δv}{v}=frac{left [frac{(tδl-lδt)}{t^2} – aδt – tδa right ]}{(frac{l}{t}-at)})
Put, l = 2cm, t = 2sec, a = 0.95 mt/sec2
and δl = 1cm, δt = 1 sec⁡and δa = -1.05 mt/sec2
we get,
δvv = 2.

7. The relative error in the volume of figure having hemispherical ends and a body of right circular cylinder is, if error in radius(r) is 0 and in height(h) is 1.
a) 1/(h + 43 r)
b) 1/(h + 23 r)
c) h/(h + 43 r)
d) r/(h + 43 r)
Answer: a
Explanation: Given V = πr2 h + 43 πr3
Now since error in radius is zero , it should be treated as constant, Hence,
(δV=frac{πr^2 δh}{πr^2 (h+frac{4}{3} r)}=frac{1}{(h+frac{4}{3} r)})

8. If n resistors of unequal resistances are connected in parallel,and the percenrage error in all resistors are k then,total error in parallel combination is?
a) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
b) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=1-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
c) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=1+frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
d) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
Answer: b
Explanation: Given 1r = 1a + 1b + 1c +⋯.. + 1n
Differentiating all,
1r2 dr = – 1a2 da – 1b2 db – ….- 1n2 dn
Now,
1r2 dr = + 1a2 da + 1b2 db + ….+ 1n2 dn
Multiplying by 100 and putting da = db = ……. = dn =k.

9. The approximate value of function f(x + δx, y + δy) is?
a) f + ∂f∂x dx + ∂f∂y dy
b) ∂f∂x dx + ∂f∂y dy
c) f – ∂f∂x dx + ∂f∂y dy
d) ∂f∂x dx – ∂f∂y dy
Answer: a
Explanation: f(x + δx, y + δy) = f(x,y) + df = f + ∂f∂x dx + ∂f∂y dy.

10. At a distance of 120ft from the foot of a tower the elevation of its top is 60o. If the possible error in measuring the distance and elevation are 1 inch and 1 minute, find approximate error in calculated height.
a) 0.342
b) 0.284
c) 0.154
d) 0.986
Answer: b
Explanation: Tan(z) = hx
h = x Tan(z)
Taking log and then differentiate we get,
∂hh = ∂xx + 1Tan(z) Sec2 (z)δz
Now h = 120 tan(60o) = 120√3
Putting, δx = 112 ft, δz = π(60*180)
Putting the values we get,
δh = 0.284.

11. Find the approximate value of (1.04)3.01.
a) 1.14
b) 1.13
c) 1.11
d) 1.12
Answer: d
Explanation: Let, f(x,y) = xy
Now,
∂f∂x = yx(y-1) and ∂f∂y = xy log⁡(x)
Putting, x = 1, y = 3, δx = 0.04, δy = 0.01
Now, df = δx ∂f∂x + δy∂f∂y = 0.12
Hence, f(x + δx,y + δy) = (1.04)3.01 = 1.12.

12. Find the approximate value of [0.982+2.012+1.942 ](12).
a) 1.96
b) 2.96
c) 0.04
d) -0.04
Answer: b
Explanation: Let f(x,y,z) = (x2+y2+z2 )(12) ……………..(1)
Hence, x = 1, y = 2, z = 2 so that, dx = -0.02, dy = 0.01, dz = -0.06
From (1),
∂f∂x = xf
∂f∂y = yf
∂f∂z = zf
And df = ∂f∂x dx + ∂f∂y dy + ∂f∂z dz = ((xdx + ydy + zdz))/f = (-0.02 + 0.02 – 0.12)/3 = -0.04
Hence,
[0.982+2.012+1.942 ](12) = f(1, 2, 2) + df = 3-0.04 = 2.96.

13. Find the approximate value of log⁡(11.01-log⁡(10.1)), Given log(10) = 2.30 and and log(8.69) = 2.16, all the log are in base ‘e’.
a) 2.1654
b) 2.1632
c) 2.1645
d) 2.1623
Answer: d
Explanation: Let, f(x,y) = log(x-log(y))
Now by differentiating,
(frac{∂f}{∂x}=frac{1}{x-log⁡(y)}) and (frac{∂f}{∂y}=frac{1}{y(x-log⁡(y))})
Now, putting, x = 11, y = 10, δx=.01 and δy=.1
We get,
(frac{∂f}{∂x}∂f/∂x)=1/8.69 and (frac{∂f}{∂y}∂f/∂y)=1/86.9
Hence, df = 0.0023
Hence, f(x + δx, y + δy) = log⁡(11.01 – log⁡(10.1))= 2.16 + df = 2.1623.

14. Find approximate value of e10.19.09, given e90 = 1.22 * 1039.
a) 2.41 * 1039
b) 2.42 * 1039
c) 2.43 * 1039
d) 2.44 * 1039
Answer: b
Explanation: Let, f(x,y) = exy = exy
Now by differentiating,
∂f∂x = yexy and ∂f∂y = xexy
Now, putting, x = 10, y = 9, δx = .01 and δy = .09
We get,
∂f∂x = 1.09* 1040 and ∂f∂y = 1.22* 1040
Hence, df = 1.27* 1039
Hence, f(x + δx, y + δy) = e10.19.09 = 2.42 * 1039.