Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on First Order Nonlinear Differential Equations and Answers

Ordinary Differential Equations Multiple Choice Questions on “First Order Nonlinear Differential Equations”.

1. Find the solution for (lim_{xto 0}⁡ frac{ax}{ax+x}.)
a) 0
b) a
c) 1
d) (frac{a}{a+1})
Answer: d
Explanation: The given equation can be solved using L-Hospital’s Rule,
(frac{d(ax)}{dx}=a, frac{d(ax+x)}{dx}=a+1)
(lim_{xto 0}⁡frac{ax}{ax+x} → lim_{xto 0}⁡frac{a}{a+1} → frac{a}{a+1})

2. 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, (frac{d(a)}{dx}= frac{d(120x- x^2)}{dx}=120-2x )
To find the value of x that maximizes the area, we substitute (frac{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,
(frac{d^2 (a)}{dx^2}= -2) < 0 …………………. (i)
We know that the condition for maxima is (frac{d^2 (f(x))}{dx^2}<0,) which is satisfied by (i), therefore, x = 60 cm maximizes the area of the rectangle.

3. Find the solution of the system using Gauss Elimination method.

x – y + 2z = 8
y – z = 4
2x + 3z = 2

a) x = 18, y = -18, z = -22
b) x = -12, y = -18, z = 22
c) x = 34, y = -18, z = -22
d) x = -18, y = -12, z = 22
Answer: c
Explanation: Augmented Matrix of the given system is,
(
left[ begin{array}{ccc|c}
1 & -1 & 2 & 8 \
0 & 1 & -1 & 4 \
2 & 0 & 3 & 2
end{array} right])
Now, applying the steps for Gauss Elimination method (making the elements below the diagonal zero), we get,
(
left[ begin{array}{ccc|c}
1 & -1 & 2 & 8 \
0 & 1 & -1 & 4 \
2 & 0 & 3 & 2
end{array} right] quad ^{underrightarrow{R3 rightarrow R3 – 2R1}} )
(
left[ begin{array}{ccc|c}
1 & -1 & 2 & 8 \
0 & 1 & -1 & 4 \
0 & 2 & -1 & -14
end{array} right] quad ^{underrightarrow{R3 rightarrow R3 – 2R2}} )
(
left[ begin{array}{ccc|c}
1 & -1 & 2 & 8 \
0 & 1 & -1 & 4 \
0 & 0 & 1 & -22
end{array} right])
Now converting the augmented matrix back to set of equations, we get,
x – y + 2z = 8 …………………………. (i)
y – z = 4 …………………………………. (ii)
z = -22 …………………………………… (iii)
Substituting value of z in (ii), we get,
y + 22 = 4
y = -18
Substituting the value of y and z in (i), we get,
x + 18 + 2(-22) = 8
x – 26 = 8
x = 34
Therefore, the solution of the system is x = 34, y = -18, z = -22.

4. 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 )

5. Determine the current i(t) for the circuit shown, if the initial current is zero.

a) i(t) = 9 – 9e8t
b) i(t) = 9 – e-8t
c) i(t) = 9 – 9e-8t
d) i(t) = 8 – 9e-8t
Answer: c
Explanation: The given circuit is an application of First Order Differential equations (R-L Circuit).
Hence, we know that, (Lfrac{di(t)}{dt} + Ri(t) = E(t) )
(frac{1}{2} frac{di(t)}{dt} + 4i(t) = 10)
(frac{di(t)}{dt} + 8i(t) = 20) …………………………. (i)
Integrating Factor, (u(t) = e^{∫8 dt}= e^{8t}) …………………………… (ii)
Applying (ii) on both sides of (i) we get,
(e^{8t}.frac{di(t)}{dt} + e^{8t}. 8i(t) = 72 e^{8t} )
(frac{d(e^{8t}.i(t))}{dt} = 72 e^{8t} )
Integrating on both sides we get,
(∫frac{d(e^{8t}.i(t))}{dt} = ∫72 e^{8t}.dt)
(e^{8t} .i(t) = frac{72}{8} .e^{8t} + c)
i(t) = 9 + ce-8t
Given: i(0) = 0
i(0)= 9 + ce-8(0) = 0
c = -9
Therefore, i(t) = 9 – 9e-8t

6. (xy^3(frac{dy}{dx})^2+yx^2+frac{dy}{dx}=0) is a _____________
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.

7. Which of the following is one of the criterions for linearity of an equation?
a) The dependent variable and its derivatives should be of second order
b) The dependent variable and its derivatives should not be of same order
c) Each coefficient does not depend on the independent variable
d) Each coefficient depends only on the independent variable
Answer: d
Explanation: The two criterions for linearity of an equation are:
The dependent variable y and its derivatives are of first degree.
Each coefficient depends only on the independent variable

8. Beta function is not a symmetric function.
a) True
b) False
Answer: b
Explanation: Beta function is a symmetric function, i.e.,
β(x,y) = β(y,x), where x>0 and y>0.

9. Which of the following is the property of error function?
a) erf (0) = 1
b) erf (∞) = 1
c) erf (0) = ∞
d) erf (∞) = 0
Answer: b
Explanation: Error Function is given by, (erf(x) = frac{2}{sqrt π} ∫_0^xe^{-t^2}dt. )
Some of its properties are:
erf (0) = 0
erf (∞) = 1
erf (-x) = -erf(x)

10. The equation (2frac{dy}{dx} – xy = y^{-2},) is an example for Bernoulli’s equation.
a) False
b) True
Answer: b
Explanation: A first order, first degree differential equation of the form,
(frac{dy}{dx} + P(x). y = Q(x). y^a,) is known as Bernoulli’s equation.

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250+ TOP MCQs on Laplace Transform by Properties and Answers

Engineering Mathematics Questions and Answers for Campus interviews focuses on “Laplace Transform By Properties – 2”.

1. Transfer function may be defined as ____________
a) Ratio of out to input
b) Ratio of laplace transform of output to input
c) Ratio of laplace transform of output to input with zero initial conditions
d) None of the mentioned
Answer: c
Explanation: Transfer function may be defined as the ratio of laplace transform of output to input with zero initial conditions.

2. Poles of any transfer function is define as the roots of equation of denominator of transfer function.
a) True
b) False
Answer: a
Explanation: Let transfer function be defined as G(s)/H(s), then poles of transfer function may be defined as H(s)=0.

3. Zeros of any transfer function is define as the roots of equation of numerator of transfer function.
a) True
b) False
Answer: a
Explanation: Let transfer function be defined as G(s)/H(s), then zeros of transfer function may be defined as G(s)=0.

4. Find the poles of transfer function which is defined by input x(t)=5Sin(t)-u(t) and output y(t)=Cos(t)-u(t).
a) 4.79, 0.208
b) 5.73, 0.31
c) 5.89, 0.208
d) 5.49, 0.308
Answer: a
Explanation: Given ,y(t)=Cos(t) – u(t) and x(t) = 5Sin(t) – u(t),
Hence, transfer function H(s)=(frac{[frac{s}{s^2+1}-frac{1}{s}]}{[frac{5}{s^2+1}-frac{1}{s}]} =-frac{1}{frac{s(s^2+1)}{frac{(5s-s^2-1)}{s(s^2+1)}}}=frac{1}{S^2-5S+1})
Roots of equation s2 – 5s + 1 = 0 is s = 4.79, 0.208.

5. Find the equation of transfer function which is defined by y(t)-∫0t y(t)dt + d⁄dt x(t) – 5Sin(t) = 0.
a) (frac{s(e^{-as}-1)}{s-1})
b) (frac{(e^{-as}-s)}{s-1})
c) (frac{s(e^{-as}-s)}{s-1})
d) (frac{s(e^{-as}-s^2)}{s-1})
Answer: c
Explanation:
Given, (y(t)-∫_0^t y(t)dt+frac{d}{dt} x(t)-x(t-a)=0)
Taking Laplace, (Y(s)-frac{Y(s)}{s}+sX(s)-e^{-as} X(s)=0)
H(s)=Y(s)/X(s) =(frac{(e^{-as}-s)}{1-frac{1}{s}}=frac{s(e^{-as}-s)}{s-1})

6. Find the poles of transfer function given by system d2⁄dt2 y(t) – d⁄dt y(t) + y(t) – ∫0t x(t)dt = x(t).
a) 0, 0.7 ± 0.466
b) 0, 2.5 ± 0.866
c) 0, 0 .5 ± 0.866
d) 0, 1.5 ± 0.876
Answer: c
Explanation: We know that,
Given, (frac{d^2}{dt^2} y(t)-frac{d}{dt} y(t)+y(t)-int_0^t x(t)dt=x(t))
Now, Taking laplace, we get, ((s^2-s+1)Y(s)=(1+frac{1}{s})X(s))
H(s)=(frac{s+1}{s(s^2-s+1)})
Roots of s3-s2+s=0, are 0,.5±0.866

7. Find the transfer function of a system given by equation d2⁄dt2 y(t-a) + x(t) + 5 d⁄dt y(t) = x(t-a).
a) (e-as-s)/(1+e-as s2)
b) (e-as-5s)/(e-as s2)
c) (e-as-s)/(2+e-as s2)
d) (e-as-5s)/(1+e-as s2)
Answer: d
Explanation: Given, d2⁄dt2 y(t-a) + x(t) + 5 d⁄dt y(t) = x(t-a).

Taking laplace transform, s2 Y(s) e-sa + X(s) + 5sY(s) = e-as X(s)

Hence, H(s) = Y(s)⁄X(s) =(e-as-5s)/(1+e-as s2).

8. Any system is said to be stable if and only if ____________
a) It poles lies at the left of imaginary axis
b) It zeros lies at the left of imaginary axis
c) It poles lies at the right of imaginary axis
d) It zeros lies at the right of imaginary axis
Answer: a
Explanation: Any system is said to be stable if and only if it poles lies at the left of imaginary axis.

9. The system given by equation 5 d3⁄dt3 y(t) + 10 d⁄dt y(t) – 5y(t) = x(t) + ∫0t x(t)dt, is?
a) Stable
b) Unstable
c) Has poles 0, 0.455, -0.236±1.567
d) Has zeros 0, 0.455, -0.226±1.467
Answer: a
Explanation:
Given, 5 d3⁄dt3 y(t) + 10 d⁄dt y(t) – 5y(t) = x(t) + ∫0t x(t)dt,
By laplace transform, 5s3 Y(s)+10sY(s)-5Y(s)=X(s)+1/s X(s)
Hence,
(frac{Y(s)}{X(s)} =frac{[1+1/s]}{5s^3+10s-5}=frac{[s+1]}{s(5s^3+10s-5)})
Poles of (frac{Y(s)}{X(s)}) are, 0 ,0.455,-0.226±1.467
Hence system is Unstable.

10. Find the laplace transform of input x(t) if the system given by d3⁄dt3 y(t) – 2 d2⁄dt2 y(t) –d⁄dt y(t) + 2y(t) = x(t), is stable.
a) s + 1
b) s – 1
c) s + 2
d) s – 2
Answer: b
Explanation: d3⁄dt3 y(t) – 2 d2⁄dt2 y(t) – d⁄dt y(t) + 2y(t) = x(t),
Taking laplace transform,
(s3 – 2s2 – s + 2)Y(s) = X(s)
H(s) = Y(s)⁄X(s) = 1⁄(s-1)(s+1)(s+2)
For the system to be stable, X(s) = s – 1.

11. The system given by equation y(t – 2a) – 3y(t – a) + 2y(t) = x(t – a) is?
a) Stable
b) Unstable
c) Marginally stable
d) 0
Answer: a
Explanation: Given, y(t-2a)-3y(t-a)+2y(t)=x(t-a), then
Taking Laplace transform, e-2as Y(s)-3e-as Y(s)+2Y(s)=e-as X(s),
Hence, (H(s)=frac{e^{-as}}{(1-e^{-as})(2-e^{-as})})
To find the stability, we should have, (∫_{-∞}^∞ H(s)ds>0)
Hence, (int_{-infty}^infty frac{e^{-as}}{(1-e^{-as})(2-e^{-as})} ds)
Let, (e^{-as}=z=>-ae^{-as} ds=dz)
(frac{1}{a} int_0^∞ frac{1}{(1-z)(2-z)} dz=frac{1}{a} int_0^∞ left [frac{1}{(z-1) }-frac{1}{z-2} right ]dz=frac{1}{a} ln⁡left [frac{z-1}{z-2}right ])
putting, z=0, (frac{1}{a} ln⁡left [frac{z-1}{z-2}right ]=frac{1}{a} ln⁡(frac{1}{2}))
putting, z=∞, (frac{1}{a} ln⁡left [frac{z-1}{z-2}right ]=frac{1}{a} ln⁡[frac{1-1/z}{1-2/z}])=0
hence, (int_{-infty}^infty frac{e^{-as}}{(1-e^{-as})(2-e^{-as})} ds=frac{1}{a} ln⁡(frac{1}{2}))

It is stable.

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250+ TOP MCQs on Eigenvalues and Vectors of a Matrix and Answers

Linear Algebra Multiple Choice Questions on “Eigenvalues and Vectors of a Matrix”.

1. Find the Eigen values for the following 2×2 matrix.
A=(begin{bmatrix}1&8\2&1end{bmatrix}).
a) -3
b) 2
c) 6
d) 4
Answer: a
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
[A-λI]=(begin{bmatrix}1&8\2&1end{bmatrix} -lambda begin{bmatrix}1&0\0&1end{bmatrix})
[A-λI]=(begin{bmatrix}1-lambda &8\2&1-lambda end{bmatrix})
|A-λI|=(1-λ)(1-λ)-16=0
(1-λ)2=16
(1-λ)=±4
λ=-3 or λ=5.

2. Find the Eigenvalue for the given matrix.
A=(begin{bmatrix}4&1&3\1&3&1\2&0&5end{bmatrix}).
a) 13
b) -3
c) 7.1
d) 8.3
Answer: c
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
[A-λI]=(begin{bmatrix}4&1&3\1&3&1\2&0&5end{bmatrix}-lambda begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
The characteristic polynomial is given by-
λ3-(Sum of diagonal elements) λ2+(Sum of minor of diagonal element)λ-|A|=0
λ3-12λ2+40λ-39=0
λ=7.1 or λ=3.

3. Find the Eigen vector for value of λ=-2 for the given matrix.
A=(begin{bmatrix}3&5\3&1end{bmatrix}).
a) (begin{bmatrix}0\-1end{bmatrix})
b) (begin{bmatrix}1\-1end{bmatrix})
c) (begin{bmatrix}-1\-1end{bmatrix})
d) (begin{bmatrix}1\0end{bmatrix})
Answer: b
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
Given that, λ=-2
[A-λI]=(begin{bmatrix}3&5\3&1end{bmatrix}-(-2)begin{bmatrix}1&0\0&1end{bmatrix})
[A-λI]=(begin{bmatrix}3+2&5\3&1+2end{bmatrix})
[A-λI]=(begin{bmatrix}5&5\3&3end{bmatrix})
Since, [A-λI]X=0
(begin{bmatrix}5&5\3&3end{bmatrix}
begin{bmatrix}x\yend{bmatrix} = begin{bmatrix}0\0end{bmatrix})
Thus,
5x+5y=0 and 3x+3y=0
Let x=t,
Then, y=-t
X = (begin{bmatrix}t\-tend{bmatrix} = begin{bmatrix}1\-1end{bmatrix})

4. Find the Eigen vector for value of λ=3 for the given matrix.
A=(begin{bmatrix}3&10&5\-2&-3&-4\3&5&7end{bmatrix}).
a) (begin{bmatrix}-1\-1\2end{bmatrix})
b) (begin{bmatrix}-1\1\2end{bmatrix})
c) (begin{bmatrix}-1\-1\-2end{bmatrix})
d) (begin{bmatrix}-1\-2\2end{bmatrix})
Answer: a
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
Given that, λ=3
[A-λI]=(begin{bmatrix}3&10&5\-2&-3&-4\3&5&7end{bmatrix}-(3)begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
[A-λI]=(begin{bmatrix}3-3&10&5\-2&-3-3&-4\3&5&7-3end{bmatrix})
[A-λI]=(begin{bmatrix}0&10&5\-2&-6&-4\3&5&4end{bmatrix})
Since, [A-λI]X=0
(begin{bmatrix}0&10&5\-2&-6&-4\3&5&4end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})
Using Row transformation

(1) Interchanging R1 and R2/2
(begin{bmatrix}-1&-3&-2\0&10&5\3&5&4end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})
(2) R3=R3+3R1
(begin{bmatrix}0&10&5\0&10&5\0&-4&-2end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})
(3) R2=R2/5 and R3=R3+2R2
(begin{bmatrix}-1&-3&-2\0&2&1\0&0&0end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})
Thus,
-x-3y-2z=0 and 2y+z=0
Let z=t,then y=(frac{-t}{2}) and x=(frac{-t}{2})

X = (begin{bmatrix}-1\-1\2end{bmatrix})

5. Find the Eigen value and the Eigen Vector for the given matrix.
A=(begin{bmatrix}3&4&2\1&6&2\1&4&4end{bmatrix}).
a) 3, (begin{bmatrix}1\1\1end{bmatrix})
b) 9, (begin{bmatrix}1\1\1end{bmatrix})
c) 9, (begin{bmatrix}1\0\1end{bmatrix})
d) 2, (begin{bmatrix}1\0\1end{bmatrix})
Answer: b
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
[A-λI]=(begin{bmatrix}3&4&2\1&6&2\1&4&4end{bmatrix}-lambda begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
The characteristic polynomial is given by-
λ3-(Sum of diagonal elements) λ2+(Sum of minor of diagonal element)λ-|A|=0

λ3-13λ2+40λ-36=0
λ=9 or λ=2
For λ=9,
[A-λI]=(begin{bmatrix}3&4&2\1&6&2\1&4&4end{bmatrix}-9begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
[A-λI]=(begin{bmatrix}3-9&4&2\1&6-9&2\1&4&4-9end{bmatrix})
[A-λI]=(begin{bmatrix}-6&4&2\1&-3&2\1&4&-5end{bmatrix})
[A-λI]X=0
(begin{bmatrix}-6&4&2\1&-3&2\1&4&-5end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})

(1) Interchanging R1 and R2
(begin{bmatrix}1&-3&2\-6&4&2\1&4&-5end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})

(2) R2=R2+6R1 and R3=R3-R1
(begin{bmatrix}1&-3&2\0&-14&14\0&7&-7end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})

(3) R3=R3+R2/2
(begin{bmatrix}1&-3&2\0&-14&14\0&0&0end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})

-14y+14z=0
x-3y+2z=0
Let z=t
Then, y=t and x=t
X=(begin{bmatrix}1\1\1end{bmatrix}).

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250+ TOP MCQs on First Order Linear PDE and Answers

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

1. First order partial differential equations arise in the calculus of variations.
a) True
b) False
View Answer

Answer: a
Explanation: The calculus of variations is a type of analysis in the field of mathematics (branch of calculus) which is used to find maxima and minima of definite integrals.

2. 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.

3. What is the order of the equation, (xy^3(frac{∂y}{∂x})^2+yx^2+frac{∂y}{∂x}=0)?
a) Third Order
b) Second Order
c) First Order
d) Zero Order
View Answer

Answer: c
Explanation: The equation having only first derivative, i.e., (frac{∂y}{∂x}) are said to be first order differential equation. Since the given equation satisfies this condition, it is of first order.

4. 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
View Answer

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.

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

Answer: d
Explanation: The two criterions for linearity of an equation are:

  • The dependent variable y and its derivatives are of first degree.
  • Each coefficient depends only on the independent variable

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
View Answer

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

7. Which of the following is an example for first order linear partial differential equation?
a) Lagrange’s Partial Differential Equation
b) Clairaut’s Partial Differential Equation
c) One-dimensional Wave Equation
d) One-dimensional Heat Equation
View Answer

Answer: a
Explanation: Equations of the form Pp + Qq = R , where P, Q and R are functions of x, y, z, are known as Lagrange’s linear equation.

8. What is the nature of Lagrange’s linear partial differential equation?
a) First-order, Third-degree
b) Second-order, First-degree
c) First-order, Second-degree
d) First-order, First-degree
View Answer

Answer: d
Explanation: Lagrange’s linear equation contains only the first-order partial derivatives which appear only with first power; hence the equation is of first-order and first-degree.

9. Find the general solution of the linear partial differential equation, yzp+zxq=xy.
a) φ(x2-y2 – z2 )=0
b) φ(x2-y2, y2-z2 )=0
c) φ(x2-y2, y2-x2 )=0
d) φ(x2-z2, z2-x2 )=0
View Answer

Answer: b
Explanation: Given: yzp+zxq=xy
Here, the subsidiary equations are, (frac{dx}{yz}=frac{dy}{zx}=frac{dz}{xy} )
From the first two and last two terms, we get, respectively,
(frac{dx}{yz}=frac{dy}{zx’}) or xdx-ydy=0, and
(frac{dx}{zx}=frac{dy}{xy’}) or ydx-zdy=0,
Integrating these we get two solutions
x2-y2=a , y2-z2=b
Hence, the general solution of the given equation is,
φ(x2-y2, y2-z2 )=0.

10. The equation 2(frac{dy}{dx} – xy = y^{-2},) is an example for Bernoulli’s equation.
a) False
b) True
View Answer

Answer: b
Explanation: A first order, first degree differential equation of the form,
(frac{dy}{dx} + P(x). y = Q(x). y^a,) is known as Bernoulli’s equation.

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

Answer: b
Explanation: A particular solution for an equation is derived by substituting particular values to the arbitrary constants in the complete solution.

12. A partial differential equation is one in which a dependent variable (say ‘y’) depends on one or more independent variables (say ’x’, ’t’ etc.)
a) False
b) True
View Answer

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

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250+ TOP MCQs on Roots of Complex Numbers and Answers

Complex Analysis Multiple Choice Questions on “Roots of Complex Numbers.

1. The nth roots of any number are in ____________
a) arithmetic progression
b) geometric progression
c) harmonic progression
d) no specific pattern
View Answer

Answer: b
Explanation: We know that reiα=rei(α+2π). Hence, if reiα/n is a nth root of a number, then, re i(α/n+2π/n) is also a nth root of that number. Hence, the roots are in G.P. with common ratio= ei2π/n.

2. In the Argand Plane shown below, a,b,c,d are the 4-th roots of 16. Find the area of the closed Polygon having a,b,c,d as its vertices.

a) 2 sq. units
b) 4 sq. units
c) 8 sq. units
d) 16 sq. units
View Answer

Answer: c
Explanation: The complex numbers a,b,c,d are in G.P. with common ratio eiπ/2. Therefore, a square is formed having side of 81/2(since O-a length is 2, therefore, b-a length is 81/2) . Hence the required area=(81/2)2=8 sq. units.

3. For k=1,2,…9, if we define zk=cos(3kπ/10)+isin(2kπ/10), then is it true that for each zk, there exists zj satisfying zk× zj=1?
a) True
b) False
View Answer

Answer: a
Explanation: zk= ei(2kπ/10)⇒ zk× zj=ei(2π/10)(k+j), zk is 10th root of unity.
⇒(overline{z_k}) is also 10th root of unity. Taking zj as (overline{z_k}), we have zk×zj=1.

4. For k=1,2,…9, if we define zk=cos(3kπ/10)+isin(2kπ/10), then is it possible that z1×z=zk has no Solution z?
a) True
b) False
View Answer

Answer: b
Explanation: z=zk/z1=ei(2kπ/10-2π/10)=eiπ/5(k-1)
For k = 2; z=eiπ/5, therefore a solution always exists.

5. Find the value of the expression (-1/2+i√3/2)637+(-1/2-i√3/2)337.
a) -1
b) 0
c) 1
d) i
View Answer

Answer: a
Explanation: (-1/2+i√3/2)=ω and (-1/2-i√3/2)=ω2, ω being the complex cube root of unity.
Therefore, expression has the value ω637+ω337=ω+ω2=-1.

6. Let a and b be complex cube roots of unity. If x=7a+2b and y=2a+7b, then evaluate xy.
a) 9
b) 39
c) 45
d) 53
View Answer

Answer: d
Explanation: Write a=ω and b=ω2.
Now, xy=(7a+2b)(2a+7b)=(7ω+2ω2)(2ω+7ω2)=14ω2+14ω+53=39 (since, 1+ω+ω2=0).

7. For integral x,y,z, find the range of |x+yω+zω2| if it is not true that x=y=z.
a) [1, ∞)
b) [√3, ∞)
c) (0, √3)
d) (0, ∞)
View Answer

Answer: a
Explanation: Put a=∞ and b=c=0 to show that maximum value tends to infinity.
Now, z=|x+yω+zω2|, hence, z2=|x+yω+zω2|2=(x2+y2+z2-xy-yz-zx)=1/2{(x-y)2+(y-z)2+(z-x)2}
Now if x=y, then y≠z and x≠z (given that x,y,z are not all equal)⇒(y-z)2≥1, also (z-x)2≥1 and
(x-y)2=0. Therefore, z2≥½(0+1+1)=1, hence |z|≥1.

8. Find the value of (1+ω)(1+ω2)(1+ω4)(1+ω8)…to 2n factors.
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: a
Explanation: (1+ω)(1+ω2)(1+ω4)(1+ω8)…up to 2n factors
= (-ω2)(-ω)(1+ω)(1+ω2) …up to 2n factors
= 1×1×1×L up to n factors = 1.

9. If α,β,ȣ are the roots of equation x3–3x2+3x+7=0 and ω is cube root of unity, then evaluate
(α-1)/(β-1)+(β-1)/(ȣ-1)+(ȣ-1)/(α-1).
a) ω
b) ω2
c) 3ω
d) 3ω2
View Answer

Answer: d
Explanation: equation can be simplified to (x-1)3=-8⇒(x-1)/(-2)=(1)1/3⇒roots: 1,ω,ω2
⇒ α=-1, β=1-2ω, ȣ=1-2ω2. Therefore, required value=-2/(-2ω)+(-2ω)/(-2ω2)+(-2ω2)/(-2)
= 1/ω+1/ω+1/ω=3/ω=3ω2.

10. Find the possible value(s) of Re(i1/2)+|Im(i1/2)|.
a) -1, 1
b) 0, √2
c) 0, 1
d) 1, √2
View Answer

Answer: b
Explanation: We can write: i=eiπ/2⇒i1/2=eiπ/4=1/√2+i/√2, also i1/2=ei3π/4=-1/√2-i/√2
Hence, Re(i1/2)+|Im(i1/2)|=1/√2+1/√2=√2 OR -1/√2+1/√2=0.

11. Let ω and ω2 be the non-real cube roots of unity and 1/(a+ω)+1/(b+ω)+1/(c+ω)=2ω2 and 1/(a+ω2)+1/(b+ω2)+1/(c+ω2)=2ω, then calculate 1/(a+1)+1/(b+1)+1/(c+1).
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: b
Explanation: Given relations can be written as: 1/(a+ω)+1/(b+ω)+1/(c+ω)=2/ω, and
1/(a+ω2)+1/(b+ω2)+1/(c+ω2)=2/ω2⇒ω and ω2 are roots of 1/(a+x)+1/(b+x)+1/(c+x)=2/x.
⇒[3x2+2(a+b+c)x+bc+ca+ab]/[(a+x)(b+x)(c+x)]=2/x⇒x3+(bc+ca+ab)x-2abc=0.
Let 3rd root be α (apart from ω and ω2). Then, α+ ω+ ω2=coefficient of x2=0⇒α=1. Hence, 1/(a+1)+1/(b+1)+1/(c+1)=2/1=2.

12. Find the cube root of 8i lying in the first quadrant of the complex plane.
a) i-√3
b) 2i+√3
c) i+2√3
d) i+√3
View Answer

Answer: d
Explanation: Let z3=8i⇒z3=-8i3⇒[z/(-2i)]3=1.
⇒z/(-2i)=1 or ω or ω2⇒z=-2i or -2iω or -2iω2
⇒z=-2i or i+√3 or i-√3 out of which i+√3 lies in the first quadrant.

13. If ω is the complex cube root of unity, then which among the following is a factor of the polynomial x6+ 4x5+3x4+2x3+x+1?
a) x+ω
b) x+ω2
c) (x+ω)(x+ω2)
d) (x–ω)(x–ω2)
View Answer

Answer: d
Explanation: Let f(x)=x6+ 4x5+3x4+2x3+x+1. Therefore, f(ω)= ω6+4ω5+3ω4+2ω3+ω+1=0
Since, ω2 is the complex conjugate of ω, therefore ω2 is also a root (roots occur in conjugate pairs). Therefore (x–ω)(x–ω2) is a root. Also, f(-ω)=(-ω)6+4(-ω)5+3(-ω)4+2(-ω)3+(-ω)+1=1-4ω2+3ω-2-ω+1≠0⇒-ω and hence -ω2 are not the roots.

14. Find ∑r=1(ar+b) ωr-1 if ω is a complex nth root of unity.
a) n(n+1)a/2
b) nb/(1-n)
c) na/(ω-1)
d) n(n+1)a/(ω-1)
View Answer

Answer: c
Explanation: (a+b)+(2a+b)ω+(3a+b)ω2+…+(na+b)ωn-1=b(1+ω+ω2+….+ωn-1)+a(1+2ω+3ω2+…+nωn-1)
Let S=1+2ω+3ω2+…+nωn-1⇒Sω= ω+2ω2+…+(n-1)ωn-1+nωn⇒S(1-ω)=1+ω+ ω2+…+ωn-1–nωn=-n
⇒S=n/(ω-1)⇒E=na/(ω-1).

15. Which of the following is not equal to (-1)1/3?
a) -1
b) (-√3+i)/(2i)
c) (√3+i)/(2i)
d) (√3–i)/(2i)
View Answer

Answer: d
Explanation: Let x=(-1)1/3⇒x3=-1⇒(-x)3=1⇒-x=1, ω, ω2
Therefore, x=-1,- ω,- ω2=-1, (-√3+i)/(2i), (√3+i)/(2i).

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250+ TOP MCQs on Leibniz Rule and Answers

Engineering Mathematics Multiple Choice Questions on “Leibniz Rule – 1”.

1. Let f(x) = sin(x)/1+x2. Let y(n) denote the nth derivative of f(x) at x = 0 then the value of y(100) + 9900y(98) is
a) 0
b) -1
c) 100
d) 1729
Answer: a
Explanation:The key here is a simple manipulation and application of the Leibniz rule.
Rewriting the given function as
y(1 + x2) = sin(x)…….(1)
The Leibniz rule for two functions is given by
(uv)(n)=(c_{0}^{n}u(v)^{(n)}+c_{1}^{n}u^{(1)}(v)^{(n-1)}+….+c_{n}^{n}u^{(n)}v)
Differentiating expression (1) in accordance to Leibniz rule (upto the hundredth derivative) we have
(y(1+x2))(100) = (c_{0}^{100}y^{(100)}(1+x^2)+c_{1}^{100}y^(99)(2x)+c_{2}^{100}y^(98)(2)+0….+0)
(y(1+x2))=(sin(x))(100)=sin(x)
Now substituting gives us
y(100)+9900y(98)=sin(0)=0
Hence, Option 0 is the required answer.

2. Let f(x) = ln(x)/x+1 and let y(n) denote the nth derivative of f(x) at x = 1 then the value of 2y(100) + 100y(99)
a) (99)!
b) (-99)!
c) (100)!
d) (-98)!
Answer: b
Explanation:Assume f(x) = y
Rewrite the function as
y(x + 1 ) = ln(x)
Now differentiate both sides up to hundredth derivative in accordance to the Leibniz rule we have
(y(1+x))(100)=(c_0^{100}y^{(100)}(x+1)+c_1^{100}y^{(99)}+0+…..+0=(ln(x))^{(100)})
Using the nth derivative for ln(x+a) as (frac{d^n(ln(x+a))}{dx^n}=frac{(-1)^{n+1}times(n-1)!}{(x+a)^n})
we have the right hand side as
(y(1+x))(100) = (frac{(-99)!}{x^{100}})
Now substituting x = 1 yields
2y(100) + y(99) = (frac{(-99)!}{1})
= -(99)!

3. Let f(x) = (sqrt{1-x^2}) and let y(n) denote the nth derivative of f(x) at x = 0 then the value of 6y (1) y(2) + 2y(3) is
a) -998
b) 0
c) 998
d) -1
Answer: b
Explanation:Assume f(x) = y
Rewriting the function as
y2 = 1 – x2
Differentiating both sides of the equation up to the third derivative using leibniz rule we have
(y2)(3)=(c_0^3y^{(3)}y+c_1^3y^{(2)}y^{(1)}+c_2^3y^{(1)}y^{(2)}+c_3^3y^{(3)}y)
(y2)(3)=(2y^{(3)}y+6y^{(2)}y^{(1)})
(1-x2)(3)=0
Now substituting x = 0 in both the equations and equating them yields
2y (3) y + 6y (2) y(1) = 0.

4. Let f(x) = tan(x) and let y(n) denote the nth derivative of f(x) then the value of y(9998879879789776) is
a) 908090988
b) 0
c) 989
d) 1729
Answer: b
Explanation:Assume y = f(x) and we also know that tan(x)=(frac{sin(x)}{cos(x)})
Rewriting the function as y(cos(x))=sin(x)
Now differentiating on both sides upto nnt derivative we have
(y(cos(x))(n)=(c_0^ny^{(n)}cos(x)-c_1^ny^{(n-1)}sin(x)+….+(cos(x))^{(n)}y)
Now observe that y(0)=tan(0)=0….(1)
Now consider the second derivative at x=0 on both sides
(y(cos(x))(2)=(c_0^2y^{(2)}cos(0)-c_1^2y^{(1)}sin(0)-c_2^2ycos(0)=(sin(x))^{(2)}=0)
Using (1) and the above equation one can conclude that
y(2) = 0
This gives the value of second derivative to be zero
Similarly for any even value of n all the odd derivatives of y in the expression would have sin(x) as their coefficients and as the values of y(0) and y(2) are zero. Every even derivative of the tan(x) function has to be zero.
Thus, we have
y(9998879879789776) = 0.

5. If the first and second derivatives at x = 0 of the function f(x)=(frac{cos(x)}{x^2-x+1})
were 2 and 3 then the value of the third derivative is
a) -3
b) 3
c) 2
d) 1
Answer: b
Explanation: Assume
f(x) = y
Write the given function as
y(x2 – x + 1) = cos(x)
Now applying Leibniz rule up to the third derivative we get
((y(x^2-x+1))^{(3)}=c_0^3y^{(3)}(x^2-x+1)-c_1^3y^{(2)}(2x-1)-c_2^3y^{(1)}(2)(cos(x))^{(3)})=sin(x)
Equating both sides we have
(sin(x) = c_0^3y^{(3)}(x^2-x+1)-c_1^3y^{(2)}(2x-1)-c_2^3y^{(1)}(2))
Now in the question it is assumed that the y(1)=2 and y(2)=3
Substituting these values in (1) we have
(sin(x) = c_0^3y^{(3)}(x^2-x+1)-c_1^3(3)(2x-1)-c_2^3(2)(2))
Substituting x = 0 gives
sin(0) = y(3) + 9 -12
y(3) = 12 – 9 = 3.

6. For the given function f(x)=(sqrt{x^3+x^7}) the values of first and second derivative at x = 1 are assumed as 0 and 1 respectively. Then the value of the third derivative could be
a) 54√2
b) 2√2
c) √2
d) Indeterminate
Answer: a
Explanation:Rewriting the given function as
y2 = x3 + x7
Now applying the Leibniz rule up to the third derivative we have
(y2)(3)=(x3+x7)(3)
(c_0^3y^{(3)}y+c_1^3y^{(2)}y^{(1)}+c_2^3y^{(1)}y^{(2)}+c_3^3yy^{(3)}=3!+(7.6.5)x^4)….(1)
Equating both sides and substituting x = 1 we get
y(1) = 0
Now assumed in the question are the values y(1) = 0 and y(2) = 1
We also know y(1) = √2
Putting them in equation (1) we get
2√2 y(3) = 3! + 210 = 216
y(3) = 54√2.

7. Let f(x)=(frac{e^x times sin(x)}{x}) and let the nth derivative at x = 0 be given by y(n) Then the value of the expression for y(n) is given by
a) (frac{pi n}{4})
b) (sum_{i=0}^{i<=n}frac{(-1)^i c_{2i}^n}{2i+1})
c) πn
d) (frac{pi}{2n})
Answer: b
Explanation: Expanding sin(x)/x into Taylor series we have
(frac{sin(x)}{x}=frac{1}{1!}-frac{x^2}{3!}+frac{x^4}{5!}-…..infty)
Now Taking the nth derivative of function using Leibniz rule we have
((e^x(frac{sin(x)}{x}))^{(n)}=c_0^ne^x(frac{1}{1!}-frac{x^2}{3!}+frac{x^4}{5!}-…infty)+c_1^ne^x(frac{-2x}{3!}-frac{4x^3}{5!}-…infty)+….)
Now substituting x=0 we have
([(e^x(frac{sin(x)}{x}))^{(n)}]_{x=0} = c_0^n(frac{1}{1!})+c_2^n(frac{-2!}{3!})+c_4n(frac{4!}{5!})…infty)
([(e^x(frac{sin(x)}{x}))^{(n)}]_{x=0}=sum_{i=0}^{i<=n}frac{(-1)^i c_{2i}^n}{2i+1})
Hence, Option (sum_{i=0}^{i<=n}frac{(-1)^i c_{2i}^n}{2i+1}) is the right answer.

8. Let f(x) = ex sinh(x) / x, let y(n) denote the nth derivative of f(x) at x = 0 then the expression for y(n) is given by
a) (sum_{i=0}^n frac{c_{2i}^n}{2i+1})
b) (sum_{i=0}^n frac{1}{2i+1})
c) 1
d) Has no general form
Answer: a
Explanation: The key here is to find a separate Taylor expansion for sinh(x) / x which is
(frac{sin h(x)}{x}=frac{frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}…infty}{x})
(frac{sin h(x)}{x}=frac{x}{1!}+frac{x^2}{3!}+frac{x^4}{5!}…infty)
Now consider (((e^x)(frac{sin h(x)}{x}))) applying the Leibniz rule for nth derivative we have
((e^x(frac{sin(x)}{x}))^{(n)}=c_0^ne^x(frac{1}{1!}-frac{x^2}{3!}+frac{x^4}{5!}-…infty)+c_1^ne^x(frac{2x}{3!}+frac{4x^3}{5!}…infty))
(+c_2^n(frac{2}{3!} + frac{12x^2}{5!}….infty)+…)
Now substituting x=0 yields
((e^x(frac{sin(x)}{x}))^{(n)}=frac{c_0^n}{1} + frac{c_2^n}{3} + frac{c_4^n}{5})
(sum_{i=0}^n frac{c_{2i}^n}{2i+1})