300+ REAL TIME Engineering Mathematics Objective Questions & Answers 2023

250+ TOP MCQs on Geometrical Applications and Answers

Ordinary Differential Equations Multiple Choice Questions on “Geometrical Applications”.

1. Who invented the rectangular coordinate system?
a) Frans van Schooten
b) René Descartes
c) Isaac Newton
d) Gottfried Wilhelm Leibniz
Answer: b
Explanation: The rectangular coordinate system was invented by French mathematician and philosopher René Descartes, who published this idea in 1637.

2. What is the formula for the intercept of tangent line on x-axis?
a) (x-frac{y}{y’})
b) (y-frac{x}{y’})
c) y – xy’
d) y – x’y
Answer: a
Explanation: Intercept of a tangent line on the:
x-axis: (x-frac{y}{y’})
y-axis: y – xy’

3. The intercept of a normal line on y axis is given by, (y+frac{x}{y’}.)
a) True
b) False
Answer: a
Explanation: Intercept of a normal line on the:
x-axis: x + yy’
y-axis: (y+frac{x}{y’})

4. What is the time taken by the particle to cover a distance of 98 meters, if the particle is moving with a velocity given by, (frac{dx}{dt}= frac{1}{x} ) (x is the distance)?
a) 48 s
b) 50 s
c) 98 s
d) 49 s
Answer: d
Explanation: Given: (frac{dx}{dt}= frac{1}{x} )
dt=xdx
Integrating, (∫dt=∫_0^{98}xdx)
(t= (frac{x^2}{2})_0^{98}=frac{98}{2}=49 s)

5. Which of the following relations hold true?
a) i × i = j × j = k × k = 1
b) i × j = -k, j × i = k
c) i × i = j × j = k × k = 0
d) k × i = -j, i × k = j
Answer: c
Explanation: One of the properties of vector or cross product is, for the orthogonal vectors, i, j, and k, we have the relation,
i × i = j × j = k × k = 0,
i × j = k, j × i = -k,
j × k = i, k × j = -i,
k × i = j, i × k = -j

6. In recurrence relation, each further term of a sequence or array is defined as a function of its succeeding terms.
a) True
b) False
Answer: b
Explanation: An equation that gives a sequence such that each next term of the sequence or array is defined as a function of its preceding terms, is called a recurrence relation.

7. What is the maximum area of the rectangle with perimeter 650 mm?
a) 26,406.25 mm²
b) 26,406 mm²
c) 24,000 mm²
d) 24,075 mm²
Answer: a
Explanation: Let x be the length of the rectangle and y be the width of the rectangle. Then, Area A is,
A=x*y …………………………………………………. (1)
Given: Perimeter of the rectangle is 620 mm. Therefore,
P=2(x+y)
650=2(x+y)
x+y=325
y=325-x
We can now substitute the value of y in (1)
A=x*(325-x)
A=325x-x²
To find maximum value we need derivative of A,
(frac{dA}{dx}=325-2x)
To find maximum value, (frac{dA}{dx}=0)
325-2x=0
2x=325
x=162.5 mm
Therefore, when the value of x=162.5 mm and the value of y=325-162.5=162.5 mm, the area of the rectangle is maximum, i.e., A=162.5*162.5=26,406.25 mm²

8. What is the derivative of z=3x*logx+5x⁶ with respect to x?
a) 3+30x⁵ e^x
b) 3+5x⁶ e^x+30x⁵ e^x
c) 3+5x⁶ e^x
d) 3+3logx+30x⁵ e^x
Answer: d
Explanation: Given: z = 3x*logx+5x⁶ e^x
(frac{dz}{dx}=3x(frac{1}{x}))+3logx+30x⁵ e^x
(frac{dz}{dx}) = 3+3logx+30x⁵ e^x

9. A sphere with the dimensions is shown in the figure. What is the error that can be incorporated in the radius such that the volume will not change more than 10%?

a) 0.3183%
b) 0.03183%
c) 31.83%
d)3.183%
Answer: b
Explanation: We know that volume of the sphere is,
(V = frac{4}{3} πR^3 )
Differentiating the above equation with respect to R we get,
(frac{dV}{dR}= frac{4}{3} π×3R^2=4πR^2 )
Since the volume of the sphere should not exceed more than 10%,
(dR=frac{dV}{4πR^2}=frac{0.1}{4π(5)^2}=0.0003183)
Error in radius = 0.03183%

10. What is the degree of the differential equation, x³-6x³y³+2xy=0?
a) 3
b) 5
c) 6
d) 8
Answer: c
Explanation: The degree of an equation that has not more than one variable in each term is the exponent of the highest power to which that variable is raised in the equation. But when more than one variable appears in a term, it is necessary to add the exponents of the variables within a term to get the degree of the equation. Hence, the degree of the equation, x³-6x³y³+2xy=0, is 3+3 = 6.

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250+ TOP MCQs on General Properties of Inverse Laplace Transform and Answers

Ordinary Differential Equations Multiple Choice Questions on “General Properties of Inverse Laplace Transform”.

1. Find the (L^{-1} (frac{s+3}{4s^2+9})).
a) (frac{1}{4} cos⁡(frac{3t}{2})+frac{1}{2} cos⁡(frac{3t}{2}))
b) (frac{1}{4} cos⁡(frac{3t}{4})+frac{1}{2} sin⁡(frac{3t}{2}))
c) (frac{1}{2} cos⁡(frac{3t}{2})+frac{1}{2} sin⁡(frac{3t}{2}))
d) (frac{1}{4} cos⁡(frac{3t}{2})+frac{1}{2} sin⁡(frac{3t}{2}))
Answer: d
Explanation: In the given question
=(frac{1}{4} L^{-1}left (frac{s+3}{s^2+frac{9}{4}}right ))
=(frac{1}{4} Big{L^{-1}left (frac{s}{s^2+frac{9}{4}}right)+L^{-1}left (frac{3}{s^2+frac{9}{4}}right)Big})
=(frac{1}{4} Big{cos⁡(frac{3t}{2})+2 sin⁡(frac{3t}{2})Big})
=(frac{1}{4} cos⁡(frac{3t}{2})+frac{1}{2} sin⁡(frac{3t}{2})).

2. Find the (L^{-1} (frac{1}{(s+2)^4})).
a) (e^{-2t}×3)
b) ⁡(e^{-2t}×frac{t^3}{3})
c) (e^{-2t}×frac{t^3}{6})
d) (e^{-2t}×frac{t^2}{6})
Answer: c
Explanation: In the given question,
(L^{-1} (frac{1}{(s+2)^4})=e^{-2t} L^{-1} frac{1}{s^4}) —————– By the first shifting property
=(e^{-2t}×frac{t^3}{3!})
=(e^{-2t}×frac{t^3}{6}).

3. Find the (L^{-1} (frac{s}{(s-1)^7})).
a) (e^{-t} left (frac{t^6}{5!}+frac{t^5}{6!}right ))
b) (e^t left (frac{t^6}{5!}+frac{t^5}{6!}right ))
c) (e^t left (frac{t^6}{6!}+frac{t^5}{5!}right ))
d) (e^{-t} left (frac{t^6}{6!}+frac{t^5}{5!}right ))
Answer: c
Explanation: In the given question,
=(L^{-1} left (frac{s-1+1}{(s-1)^7}right))
=(e^t L^{-1} left (frac{s+1}{s^7}right))
=(e^t L^{-1} left (frac{1}{s^7}+frac{1}{s^6}right))
=(e^t left (frac{t^6}{6!}+frac{t^5}{5!}right))

4. Find the (L^{-1} (frac{s}{2s+9+s^2})).
a) (e^{-t} {cos⁡(2sqrt{2t})-sin⁡(sqrt{2t})})
b) (e^{-t} {cos⁡(2sqrt{2t})-sin⁡(2sqrt{2t})})
c) (e^{-t} {cos⁡(2sqrt{2t})-cos(sqrt{2t})})
d) (e^{-2t} {cos⁡(2sqrt{2t})-sin⁡(2sqrt{2t})})
Answer: b
Explanation: In the given question,
(L^{-1} left (frac{s}{2s+9+s^2}right )=L^{-1} left (frac{s}{(s+1)^2}+8)right ))
=(L^{-1} left (frac{(s+1)-1}{(s+1)^2+8}right ))
=(e^{-t} L^{-1} left (frac{(s-1)}{s^2+8}right )) ———————–By First Shifting Property
=(e^{-t} L^{-1} left (frac{s}{s^2+8}right )-e^{-t} L^{-1} left (frac{1}{s^2+8}right ))
=(e^{-t} {cos⁡(2sqrt{2t})-sin⁡(2sqrt{2t})}).

5. Find the (L^{-1} left (frac{(s+1)}{(s+2)(s+3)}right )).
a) 2e^-3t-e^-2t
b) 3e^-3t-e^-2t
c) 2e^-3t-3e^-2t
d) 2e^-2t-e^-t
Answer: a
Explanation: In the given question,
(L^{-1} left (frac{(s+1)}{(s+2)(s+3)}right )=L^{-1} left (frac{2(s+2)-(s+3)}{(s+2)(s+3)}right ))
=(L^{-1} left (frac{2}{(s+3)}right )+L^{-1} left (frac{1}{(s+2)}right ))
=2e^-3t-e^-2t.

6. Find the (L^{-1} left (frac{(3s+9)}{(s+1)(s-1)(s-2)}right )).
a) e^-t+6e^t+5e^2t
b) e^-t-e^t+5e^2t
c) e^-3t-6e^t+5e^2t
d) e^-t-6e^t+5e^2t
Answer: d
Explanation: In the given question,
(L^{-1} left (frac{(3s+9)}{(s+1)(s-1)(s-2)}right ))
=(L^{-1} left (frac{1}{(s+1)}right )-6L^{-1} left (frac{-6}{(s-1)}right )+5L^{-1} left (frac{-6}{(s-2)}right ) )————-Using properties of Partial Fractions
=e^-t-6e^t+5e^2t.

7. Find the (L^{-1} (frac{1}{(s^2+4)(s^2+9)})).
a) (frac{1}{5} left (frac{sin⁡(2t)}{2}-frac{sin⁡(t)}{3}right ))
b) (frac{1}{5} left (frac{sin⁡(2t)}{2}+frac{sin⁡(3t)}{3}right ))
c) (frac{1}{5} left (frac{sin⁡(t)}{2}-frac{sin⁡(3t)}{3}right ))
d) (frac{1}{5} left (frac{sin⁡(2t)}{2}-frac{sin⁡(3t)}{3}right ))
Answer: d
Explanation: In the given question,
(L^{-1} left (frac{1}{(s^2+4)(s^2+9)}right))
=(frac{1}{5} L^{-1} left (frac{5}{(s^2+4)(s^2+9)}right))
=(frac{1}{5} L^{-1} left (frac{(s^2+9)-(s^2+4)}{(s^2+4)(s^2+9)}right))
=(frac{1}{5} L^{-1} left (frac{1}{(s^2+4)}right )-frac{1}{5} L^{-1} left (frac{1}{(s^2+9)}right))
=(frac{1}{5} left (frac{sin⁡(2t)}{2}-frac{sin⁡(3t)}{3}right)).

8. Find the (L^{-1} left (frac{s}{(s^2+1)(s^2+2)(s^2+3)}right )).
a) (frac{1}{2} cos⁡(t)-cos⁡(sqrt3t)-frac{1}{2} cos⁡(sqrt3t))
b) (frac{1}{2} cos⁡(t)+cos⁡(sqrt2t)-frac{1}{2} cos⁡(sqrt3t))
c) (frac{1}{2} cos⁡(t)-cos⁡(sqrt2t)-frac{1}{2} cos⁡(sqrt3t))
d) (frac{1}{2} cos⁡(t)+cos⁡(sqrt2t)+frac{1}{2} cos⁡(sqrt3t))
Answer: c
Explanation: In the given question,
(L^{-1} left (frac{s}{(s^2+1)(s^2+2)(s^2+3)}right ))
=(L^{-1} left (frac{frac{1}{2}}{(s^2+1)}+frac{(-1)}{(s^2+2)}+frac{frac{(-1)}{2}}{(s^2+3)}right )) ——————-By method of Partial fractions
=(frac{1}{2} cos⁡(t)-cos⁡(sqrt2t)-frac{1}{2} cos⁡(sqrt3t)).

9. Find the (L^{-1} left (frac{s+1}{(s-1)(s+2)^2}right )).
a) (frac{2}{7} e^t-frac{2}{9} e^{-2t}+frac{1}{3} e^{-2t}×t)
b) (frac{2}{9} e^t-frac{2}{9} e^{-2t}+frac{1}{3} e^{-2t}×t)
c) (frac{2}{9} e^t-frac{2}{9} e^{-3t}+frac{1}{3} e^{-2t}×t)
d) (frac{2}{9} e^t-frac{2}{9} e^{-2t}+frac{1}{3} e^{-2t})
Answer: b
Explanation: In the given question,
(L^{-1} left (frac{s+1}{(s-1)(s+2)^2}right ))
Using properties of partial fractions-
s+1=A(s+2)²+B(s-1)(s+2)+C(s-1)
At s=1, A=(frac{2}{9})
At s=2, C=(frac{1}{3})
At s=0, B=(frac{-2}{9})
Re substituting all these values in the original fraction,
=(L^{-1} left (frac{2}{9(s-1)} + frac{-2}{9(s+2)} + frac{1}{3(s+2)^2}right))
=(frac{2}{9} e^t-frac{2}{9} e^{-2t}+frac{1}{3} e^{-2t}×t).

10. The (L^{-1} left (frac{3s+8}{s^2+4s+25}right )) is (e^{-st} (3cos⁡(sqrt{21}t+frac{2sin⁡(sqrt{21}t)}{sqrt{21}})). What is the value of s?
a) 0
b) 1
c) 2
d) 3
Answer: c
Explanation: In the given question,
(L^{-1} left (frac{3s+8}{s^2+4s+25}right )=L^{-1} left (frac{3(s+2)+2}{(s+2)^2+21}right ))
By the first shifting property
=(e^{-2t} L^{-1} left (frac{3s+2}{s^2+21}right ))
=(e^{-2t} L^{-1} left (frac{3s}{s^2+21}right )+e^{-2t} L^{-1} left (frac{2}{s^2+21}right ))
=(e^{-2t} (3cos⁡(sqrt{21}t+frac{2sin⁡(sqrt{21}t)}{sqrt{21}})).

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250+ TOP MCQs on Real Matrices: Symmetric, Skew-symmetric, Orthogonal Quadratic Form and Answers

Linear Algebra and Vector Calculus Multiple Choice Questions on “Real Matrices: Symmetric, Skew-symmetric, Orthogonal Quadratic Form”.

1. Who introduced the term matrix?
a) James Sylvester
b) Arthur Cayley
c) Girolamo Cardano
d) Paul Erdos
Answer: a
Explanation: The term ‘matrix’ was introduced by James Sylvester during the 19^th century. Later in 1850, Arthur Cayley developed the algebraic aspect of matrices in two of his papers.

2. Which among the following is listed under the law of transposes?
a) (A^T)^T = A^I
b) (A*)* = A*
c) (cA)^T = cA^T
d) (AB)^T = A^TB^T
Answer: c
Explanation: The following are listed under the law of transposes:
(A*)* = A
(A^T)^T = A
(AB)^T = B^T A^T
(cA)^T = cA^T
(A ± B)^T = A^T ± B^T (and for ∗)
If A is symmetric, A = A^T

3. The matrix which remains unchanged under transposition is known as skew symmetric matrix.
a) False
b) True
Answer: a
Explanation: The matrix which remains unchanged under transposition is known as symmetric matrix.
For example, if we consider a symmetric matrix A = (begin{bmatrix} 2 & 3 & 4 \
3 & 5 & 6\
4 & 6 & 7 end{bmatrix} ) and take the transpose of it, we get,
A^T = (begin{bmatrix}
2 & 3 & 4 \
3 & 5 & 6 \
4 & 6 & 7 end{bmatrix} )

4. Find the values of x and y in the matrix below if the matrix is a skew symmetric matrix.
P = (begin{bmatrix}
0 & y & -4\
-5 & 0 & 8 \
x+y&-8 & 0
end{bmatrix} )
a) x = -1, y = 5
b) x = -9, y = -5
c) x = 9, y = 5
d) x = 1, y = -5
Answer: a
Explanation: The general form of a skew symmetric matrix is given by,
(begin{bmatrix}
0 & w1 & -w2\
-w1 & 0 & w3 \
w2&-w3 & 0
end{bmatrix} )
Therefore, from the given matrix,
y = 5,
(x+y = 4 rightarrow x+5=4 rightarrow x = -1)

5. Which of the following is known as the reversal rule?
a) (AB)^-1 = B^-1A^-1
b) (AB)^-1 = A^-1 B^-1
c) (BA)^-1 = B^-1 A^-1
d) (BA)^-1 = B^-1 A
Answer: a
Explanation: The reversal rule of matrix multiplication states that, ‘the inverse of the product of two matrices is equal to the product of their individual inverses, taken in the reverse order’.

6. Every Identity matrix is an orthogonal matrix.
a) True
b) False
Answer: b
Explanation: An orthogonal matrix can be defined as ‘a matrix having its entries as orthogonal unit vectors’ or it can also be defined as ‘a matrix whose transpose is equal to its inverse’. Since this property is satisfied by an identity matrix, every identity matrix is an orthogonal matrix.

7. Which of the following matrix is orthogonal?
a) (begin{bmatrix}
0.33 & 0.67 & -0.67 \
0.67 & 0.67 & 0.33 \
0.67 & 0.33 & 0.67
end{bmatrix} )
b) (begin{bmatrix}
0.33 & 0.67 & -0.67\
-0.67 & 0.67 & 0.33 \
0.67 & 0.33 & 0.67
end{bmatrix} )
c) (begin{bmatrix}
0.33 & 0.67 & -0.67\
-0.67 & 0.67 & 0.33 \
0.67 & 0.33 & 0.67
end{bmatrix} )
d) (begin{bmatrix}
0.33 & 0.67 & -0.67\
-0.67 & 0.67 & 0.33 \
0.67 & 0.33 & 0.67
end{bmatrix} )
Answer: b
Explanation: Out of the given options,(begin{bmatrix}
0.33 & 0.67 & -0.67\
-0.67 & 0.67 & 0.33 \
0.67 & 0.33 & 0.67
end{bmatrix} ) satisfies the condition for orthogonality, i.e. AA^T = I
(begin{bmatrix}
0.33 & 0.67 & -0.67\
-0.67 & 0.67 & 0.33 \
0.67 & 0.33 & 0.67
end{bmatrix} )
(begin{bmatrix}
0.33 & -0.67 & 0.67 \
0.67 & 0.67 & 0.33\
-0.67 & 0.33 & 0.67
end{bmatrix} ) = (begin{bmatrix}
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 1
end{bmatrix} )

8. Find the symmetric matrix for the given quadratic form, (Q = 2x_1^2 + 2x_2^2 + 5x_3^2 – 3x_1 x_2+7x_3 x_1.)

a) (begin{bmatrix} 1 & frac{-7}{2} & frac{3}{2} \ frac{-7}{2} & 2 & 0 \ frac{3}{2} & 0 & frac{5}{2}end{bmatrix} )
b) (begin{bmatrix} 1 & frac{-3}{2} & frac{7}{2} \ frac{-3}{2} & 2 & 0 \ frac{7}{2} & 0 & frac{5}{2}end{bmatrix} )
c) (begin{bmatrix} 1 & frac{3}{2} & frac{-7}{2}\ frac{3}{2} & 2 & 0 \ -frac{7}{2} & 0 & frac{5}{2}end{bmatrix} )
d) (begin{bmatrix} 1 & frac{-3}{2} & frac{7}{2} \ frac{-3}{2} & 2 & 0 \ frac{7}{2} & 0 & 5end{bmatrix} )
Answer: b
Explanation: The following steps need to be followed to obtain the symmetric matrix:
Step 1: There are three variables in the given quadratic equation. Hence, the symmetric matrix to be formed should be of dimension 3×3 and the general form can be written as,
Q = (begin{bmatrix}c_{11} & c_{12} & c_{13} \ c_{21} & c_{22} & c_{23} \ c_{31} & c_{32} & c_{33} end{bmatrix} )
Step 2: Place the square term coefficients of the quadratic equation (2, 2, 5) on the diagonal of the matrix.
Q = (begin{bmatrix}2 & c_{12} & c_{13} \ c_{21} & 2 & c_{23} \ c_{31} & c_{32} & 5end{bmatrix} )
Step 3: Place the remaining coefficients of (x_i x_j ,at, c_{ij},) i.e. coefficient of x₁ x₂ (-3) at c₁₂ and so on.
Q = (begin{bmatrix}2&-3 & 0 \ 0 & 2 & 0 \ 7 & 0 & 5end{bmatrix} )
Step 4: For a symmetric matrix, S = (frac{1}{2} (Q+ Q^T))
S = (frac{1}{2} begin{bmatrix}2&-3 & 0 \ 0 & 2 & 0 \ 7 & 0 & 5end{bmatrix} + begin{bmatrix}2 & 0 & 7 \ -3 & 2 & 0 \ 0 & 0 & 5end{bmatrix} )
(S = frac{1}{2} begin{bmatrix} 2 & -3 & 7 \ -3 & 4 & 0 \ 7 & 0 & 10end{bmatrix} )
S = (begin{bmatrix}1 & frac{-3}{2} & frac{7}{2} \ frac{-3}{2} & 2 & 0 \ frac{7}{2} & 0 & frac{5}{2}end{bmatrix} )

9. Which of the following is known as Hadamard matrix?
a) (begin{bmatrix}1 & 0 & 1 & 0\ 0 & 1 & 0 & 1\ 1 & 0 & 1 & 0\ 0 & 1 & 0 & 1end{bmatrix} )
b) (begin{bmatrix}1 & -1 & 1 & -1\ 1 & -1 & 1 & -1\ 1 & -1 & 1 & -1\ 1 & -1 & 1 & -1 end{bmatrix} )
c) (begin{bmatrix}1 & 1 & 1 & 1\ 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1\ 1 & 1 & 1 & 1 end{bmatrix} )
d) (begin{bmatrix}0 & 0 & 0 & 0\ 0 & 1 & 1 & 0 \ 0 & 1 & 1 & 0\ 0 & 0 & 0 & 0 end{bmatrix} )
Answer: b
Explanation: Hadamard matrix is named after a famous French mathematician, Jacques Hadamard. It is defined as ‘a square matrix whose entries are only 1 or -1 and whose column (or row) vectors orthogonal’.

10. The sum of two skew-symmetric matrices is also a skew-symmetric matrix.
a) False
b) True
Answer: b
Explanation: To prove the above statement, let us consider an example,
A = (begin{bmatrix} 0 & -4 & 1 \ 4 & 0 & -3 \-1 & 3 & 0end{bmatrix} )

Therefore, A + A = (begin{bmatrix} 0 & -4 & 1 \ 4 & 0 & -3 \ -1 & 3 & 0end{bmatrix} ) + (begin{bmatrix}0&-4 & 1 \ 4 & 0&-3 \-1 & 3 & 0end{bmatrix} ) = (begin{bmatrix} 0 & -8 & 2 \ 8 & 0 & -9\ -2 & 6 & 0end{bmatrix} ) which is also a skew-symmetric matrix.

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

Partial Differential Equations Questions and Answers for Experienced people focuses on “Non-Homogeneous Linear PDE with Constant Coefficient”.

1. Non-homogeneous which may contain terms which only depend on the independent variable.
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 a non-homogeneous equation?
a) (frac{∂^2 u}{∂t^2}-c^2frac{∂^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}-T frac{∂^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).

3. What is the general form of the general solution of a non-homogeneous DE (u_h(t)= general solution of the homogeneous equation, u_p(t)= any particular solution of the non-homogeneous equation)?
a) u(t)=u_h (t)/u_p (t)
b) u(t)=u_h (t)*u_p (t)
c) u(t)=u_h (t)+u_p (t)
d) u(t)=u_h (t)-u_p (t)
View Answer

Answer: c
Explanation: The general solution of an inhomogeneous ODE has the general form: u(t)=u_h (t)+u_p (t), where u_h (t) is the general solution of the homogeneous equation, u_p (t) is any particular solution of the non-homogeneous equation.

4. While an ODE of order m has m linearly independent solutions, a PDE has infinitely many.
a) False
b) True
View Answer

Answer: b
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.

5. Which of the following methods is not used in solving non-homogeneous equations?
a) Exponential Response Formula
b) Method of Undetermined Coefficients
c) Orthogonal Method
d) Variation of Constants
View Answer

Answer: c
Explanation: There are several methods in solving a non-homogeneous equation. Some of them are:

Exponential Response Formula
Method of Undetermined Coefficients
Variation of Constants
annihilator method

6. What is the order of the non-homogeneous partial differential equation,
(frac{∂^2 u}{∂x^2}+(frac{∂^2 u}{∂x∂y})^2+frac{∂^2 u}{∂y^2}=x^2+y^2)?
a) Order-3
b) Order-2
c) Order-0
d) Order-1
View Answer

Answer: b
Explanation: The order of an equation is defined as the highest order derivative present in the equation and hence from the equation, (frac{∂^2 u}{∂x^2}+(frac{∂^2 u}{∂x∂y})^2+frac{∂^2 u}{∂y^2}=x^2+y^2), it is clear that it of 2^nd order.

7. What is the degree of the non-homogeneous partial differential equation,
((frac{∂^2 u}{∂x∂y})^5+frac{∂^2 u}{∂y^2}+frac{∂u}{∂x}=x^2-y^3)?
a) Degree-2
b) Degree-1
c) Degree-0
d) Degree-5
View Answer

Answer: d
Explanation: Degree of an equation is defined as the power of the highest derivative present in the equation. Hence from the equation,((frac{∂^2 u}{∂x∂y})^5+frac{∂^2 u}{∂y^2}+frac{∂u}{∂x}=x^2-y^3), the degree is 5.

8. The Integrating factor of a differential equation is also called the primitive.
a) True
b) False
View Answer

Answer: b
Explanation: The general solution of a differential equation is also called the primitive. The solution of a partial differential equation obtained by eliminating the arbitrary constants is called a general solution.

9. A particular solution for an equation is derived by substituting particular values to the arbitrary constants in the complete solution.
a) True
b) False
View Answer

Answer: a
Explanation: A solution which does not contain any arbitrary constants is called a general solution whereas a particular solution is derived by substituting particular values to the arbitrary constants in this solution.

10. What is the complete solution of the equation, (q= e^frac{-p}{α})?
a) (z=ae^frac{-a}{α} y)
b) (z=x+e^frac{-a}{α} y)
c) (z=ax+e^frac{-a}{α} y+c)
d) (z=e^frac{-a}{α} y)
View Answer

Answer: c
Explanation: Given: (q= e^frac{-p}{α})
The given equation does not contain x, y and z explicitly.
Setting p = a and q = b in the equation, we get (b= e^frac{-p}{α}.)
Hence, a complete solution of the given equation is,
z=ax+by+c, with (b= e^frac{-a}{α})
(z=ax+e^frac{-a}{α} y+c.)

11. In recurrence relation, each further term of a sequence or array is defined as a function of its succeeding terms.
a) True
b) False
View Answer

Answer: b
Explanation: An equation that gives a sequence such that each next term of the sequence or array is defined as a function of its preceding terms, is called a recurrence relation.

12. What is the degree of the differential equation, x³-6x³ y³+2xy=0?
a) 3
b) 5
c) 6
d) 8
View Answer

Answer: c
Explanation: The degree of an equation that has not more than one variable in each term is the exponent of the highest power to which that variable is raised in the equation. But when more than one variable appears in a term, it is necessary to add the exponents of the variables within a term to get the degree of the equation. Hence, the degree of the equation, x³-6x³ y³+2xy=0, is 3+3 = 6.

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250+ TOP MCQs on Continuity and Answers

Complex Analysis Multiple Choice Questions on “Continuity”.

1. The function f(x, y)=(frac{x^2(2xy-3x)+9y}{3x-2y}) is which of the following?
a) Discontinuous at origin
b) Discontinuous at (1, 1)
c) Continuous at origin
d) Discontinuous at (2, 2)
Answer: a
Explanation: Origin implies the coordinates (0, 0)
Case 1: (lim_{x to 0 \ y to 0}frac{x^2(2xy-3x)+9y}{3x-2y})
(lim_{y to 0}frac{9y}{-2y})
=(frac{-9}{2})
Case 2: (lim_{y to 0 \ x to 0}frac{x^2(2xy-3x)+9y}{3x-2y})
(lim_{x to 0}frac{-3x^3}{3x})
=0
Since, both the limits are not equal, the function is not continuous at origin.

2. The function f(x, y)=(frac{2xy-3x}{3x-2y}) is continuous at (1, 1).
a) True
b) False
Answer: b
Explanation: Case 1: (lim_{x to 1 \ y to 1}frac{2xy-3x}{3x-2y})
(lim_{y to 1}frac{2y-3}{3-2y})=(frac{-1}{1})=-1
Case 2: (lim_{y to 1 \ x to 1}frac{2xy-3x}{3x-2y})
(lim_{x to 1}frac{2x-3x}{3x-2})=(frac{-1}{1})=-1
Case 3: (lim_{y to mx \ x to 0}frac{2xy-3x}{3x-2y})
(lim_{x to 0}frac{2mx^2-3x}{3x-2mx})=(frac{-3}{3-2m})
Since the cases are not same, f(x, y) is discontinuous at x=1.

3. f(x, y)=(frac{3xy+3y+3x}{5x+5y}) continuous at origin.
a) True
b) False
Answer: b
Explanation: Case 1: (lim_{x to 0 \ y to 0}frac{3xy+3y+3x}{5x+5y})
(lim_{y to 0}frac{3y}{5y})=(frac{3}{5})
Case 2: (lim_{y to 0 \ x to 0}frac{3xy+3y+3x}{5x+5y})
(lim_{x to 0}frac{3x}{5x})=(frac{3}{5})
Case 3: Along a linear path, (lim_{y to mx \ x to 0}frac{3xy+3y+3x}{5x+5y})
(lim_{x to 0}frac{3mx^2+3mx+3x}{5x})=(frac{3(m+1)}{5})
Since, the cases are not equal, f(x, y) is discontinuous at x=0.

4. If f(x)=(frac{x+3}{x^2-5x+6}) and g(x)=(frac{x-3}{x^2-6x+8}), then which of the following is correct?
a) f(x) is continuous at x=2 and g(x) is continuous at x=2
b) f(x) is discontinuous at x=3 and g(x) is discontinuous at x=2
c) f(x) is continuous at x=4 and g(x) is discontinuous at x=4
d) f(x) is discontinuous at x=3 and g(x) is continuous at x=2
Answer: b
Explanation: The denominator of f(x) is (x²-5x=6). When we factorize this, we get,
(x²-5x+6)=x²-2x-3x+6
(x²-5x+6)=(x-2)(x-3)
Since the denominator becomes 0 at x=2 and x=3, f(x) is discontinuous at x=2 and x=3.
The denominator of g(x) is (x²-6x+8). When we factorize this, we get,
(x²-6x+8)=x²-2x-4x+8
(x²-6x+8)=(x-2)(x-4)
Since the denominator becomes 0 at x=2 and x=4, f(x) is discontinuous at x=2 and x=4.

5. If f(x)=(frac{x+3}{x^2-5x})and g(x)=(frac{x-3}{x^2-6x}), then which of the following is correct?
a) f(x) is continuous at x=1 and g(x) is continuous at x=0
b) f(x) is discontinuous at x=6 and g(x) is discontinuous at x=5
c) f(x) is continuous at x=5 and g(x) is discontinuous at x=0
d) f(x) is discontinuous at x=5 and g(x) is continuous at x=5
Answer: b
Explanation: The denominator of f(x) is (x²-5x). When we factorize this, we get,
(x²-5x)=x(x-5)
Since the denominator becomes 0 at x=0 and x=5, f(x) is discontinuous at x=0 and x=5.
The denominator of g(x) is (x²-6x). When we factorize this, we get,
(x²-6x+8)=x(x-6)
Since the denominator becomes 0 at x=0 and x=6, f(x) is discontinuous at x=0 and x=6.

6. The function f(x)=(frac{1}{x^3-3x^2-10x+24})is continuous at which of the following?
a) 1
b) 2
c) 3
d) 4
Answer: a
Explanation: When we substitute x=2, x=3 or x=4 in the function, we get the denominator as 0. There by, making the function discontinuous at these points.
At x = 1, we get f(1)=(frac{1}{12}) which is continuous.

7. The function f(x)=(frac{e^{-x}}{x(x-5)(x+2)}) is continuous at which of the following?
a) 0
b) 2
c) 5
d) The function is continuous at all points
Answer: b
Explanation: When we substitute x=0, x=5 or x=-2 in the f(x), we get the denominator as 0. This is because these are the points of discontinuity.
At x=2, we get f(2)=(frac{-1}{24}) which is continuous.

8. Which of the following is true about f(x)=tanx?
a) Continuous at x=(frac{pi}{2})
b) Continuous at x=(-frac{3pi}{2})
c) Continuous at x=0
d) Continuous at x=(frac{3pi}{2})
Answer: c
Explanation: tanx=(frac{sinx}{cosx})
cosx=0 when x=(frac{pm (n+1)pi}{2}) which are points of discontinuity for the function, f(x)=tanx.

9. Which of the following is true about f(x)=sinx+cosx?
a) Continuous everywhere
b) Continuous at x=(frac{pm npi}{2}), where n is any integer
c) Continuous at x=(frac{pm (n+1)pi}{2}), where n is any integer
d) Continuous at x=0
Answer: a
Explanation: The values of sinx and cosx never become infinity or non-determinable for any value of x. Therefore, the function f(x)=sinx+cosx is continuous at all points of x.

10. Which of the following is false if both f(x) and g(x) are continuous?
a) f(x)+g(x) is continuous
b) f(x)-g(x)is continuous
c) f(x)*g(x)is continuous
d) (frac{f(x)}{g(x)}) is continuous
Answer: d
Explanation: For (frac{f(x)}{g(x)}) to be continuous, the denominator needs to be greater than 0. If not, the function becomes infinity and cannot be determined, thereby becoming discontinuous.

11. Which of the following is discontinuous at x=9?
a) (frac{x^2-81}{x-9})
b) (frac{x^3-20x^2+117x-162}{x-9})
c) (frac{x^2-11x+18}{x(x-9)})
d) (frac{x^2-13x-90}{x-9})
Answer: d
Explanation: (frac{x^2-81}{x-9})=(frac{(x-9)(x+9)}{x-9})=(x-9) which is continuous at x = 9.
(frac{x^3-20x^2+117x-162}{x-9})=(frac{(x-9)^2(x-2)}{(x-9)})=(x-9)(x-2) which is continuous at x = 9.
(frac{x^2-11x+18}{x(x-9)})=(frac{(x-2)(x-9)}{x(x-9)})=(frac{(x-2)}{x}) which is continuous at x = 9.
(frac{x^2-13x-90}{x-9})=(frac{(x-18)(x+5)}{x-9}) which is discontinuous at x = 9, since the denominator becomes 0.

12. Which of the following functions are continuous at x=11?
a) (frac{1}{x^2+x-132})
b) (frac{1}{x^3-7x^2-40x-44})
c) (frac{1}{x^2-2x-143})
d) (frac{x-11}{x^3-23x^2+143x-121})
Answer: c
Explanation: (frac{1}{x^2+x-132})=(frac{1}{(x-11)(x+12)}) is discontinuous at x=11 since the denominator becomes 0.
(frac{1}{x^3-7x^2-40x-44})=(frac{1}{(x-11)(x+2)^2})is discontinuous at x=11 since the denominator becomes 0.
(frac{x-11}{x^3-23x^2+143x-121})=(frac{(x-11)}{(x-11)^2(x-1)})=(frac{1}{(x-11)(x-1)})is discontinuous at x=11 since the denominator becomes 0.
(frac{1}{x^2-2x-143})is continuous at x=11.

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250+ TOP MCQs on Rolle’s Theorem and Answers

Engineering Mathematics Multiple Choice Questions on “Rolle’s Theorem – 1”.

1. For y = -x² + 2x there exist a c in the interval [- 19765, 19767] Such that f'(c) = 0.
a) True
b) False
Answer: a
Explanation: The key here is to rewrite the function as y = -(x – 1)² + 1
Observe here that on substituting – 19765 and 19767 in the equation we get
(-19766)² + 1 and (-19766)² + 1 respectively.
As we are dealing with their squared values they have to be equal
We have
f(-19765) = f(19767)
Polynomial functions are continuous and differentiable over the whole domain and hence by Rolles Theorem we must have a c such that f'(c) = 0 in the interval [-19765, 19767]
Hence, the claim is true.

2. For the function f(x) = (frac{sin(x)}{x^2}) How many points exist in the interval [0, 7π] Such that f'(c) = 0.
a) 8
b) 0
c) 7
d) 6
Answer: d
Explanation: We know that sine is a periodic function and it is divided by x².
Observe that the sine takes the value of zero at integral arguments of π hence at every interval of the form [nπ, (n + 1)π]
We have f(nπ) = f((n + 1) π)
The sine and the polynomial combination is continuous and differentiable at every point except x = 0
Every such interval has a point such that f'(c) = 0
Hence, by Rolles theorem, in every interval of the form [nπ, (n + 1)π] we must have a point such that f'(c) = 0
Leaving the interval [0, π] we are left with six such intervals from 0 to 7π.

3. f(x) = (frac{sin(x)}{x}), How many points exist such that f'(c) = 0 in the interval [0, 18π].
a) 18
b) 17
c) 8
d) 9
Answer: a
Explanation: We have the sine function that takes the value of zero at Integral multiples of π
But for (frac{sin(x)}{x}) we have the exceptional value of (lim_{x rightarrow 0}frac{sin(x)}{x}) reaching one.
So leaving the first interval [0, π], for every other interval of the form [nπ (n + 1)π] we must have f(nπ) = f((n + 1)π)
By Rolles theorem we have
f’ (c) = 0 For every interval of the form [nπ (n + 1)π]
There are 17 such intervals.

4. Let f(x) = x + sin(x) Every point on the graph is rotated by 45 degree with respect to the origin along the radius equal to the radius vector at that point. How many c that belong to [0, 11π] exist Such that f'(c) = 0.
a) 10
b) 11
c) 110
d) 9
Answer: b
Explanation: The function has the beautiful property that the sine function is built over the line y = x
In other words the axis for the sine graph is no more on the x axis but is the line y = x.
When this graph is rotated by 45 degrees which Is equal to the slope of the line y = x we must get the original graph sin(x) again. The number of c in the interval [0, 11π] is then obviously equal to eleven.

5. A Function f(x) has the property f(a) = f(b) for ∀a,b…∊….I and a + b = 20 then which of the following even degree polynomials could be f(x)?
a) ^x⁴⁄₄ – 10x³ + 150x² – 1000x + 10131729
b) x² + 5x + 6
c) x² + x + 1
d) Polynomial functions are inadequate representations
Answer: a
Explanation: Given that f(a) = f(b) for all integral values of a, b, one additional constraint is that (frac{a+b}{2}) = 10
This means that the graph is symmetric about the line x = 10
This means that the even degree polynomial has a Rolle point(the only Rolle point) at x = 10
We need to differentiate the functions given in the options and observe which of these has a single(OR repeated root), which is equal to 10.
The first Option when differentiated yields
f'(x) = x³ – 30x² + 300x – 1000
Equating to zero, we see that x = 10
satisfies the equation
(10)³ – 30(10)² + 300(10) – 1000 = 0.

6. For some function f(x) we have f(a) = f(b) for ∀a,b…∊….I and a + b = 2 then which of the following even degree polynomials could f(x) be
a) x² + 3x +1
b) ^5x²⁄₂ – 5x + 101
c) x² + 2x + 1
d) Even degree polynomials of such kind cannot exist
Answer: b
Explanation: Given that f(a) = f(b) for all integral values of a, b with the condition (frac{a+b}{2}) = 1
We know that the function is symmetric about the line x = 1
Thus we must have a Rolle point at 1 for any function that satisfies these conditions.
Differentiating the given functions in the options we have for a
2x + 3 = 0
x = ^-3⁄₂ ≠ 1
Doing the same for (b) we get
5x – 5 = 0
x = 1.

7. For all second degree polynomials with y = ax² + bx + k, it is seen that the Rolles’ point is at c = 0. Also the value of k is zero. Then what is the value of b?
a) 0
b) 1
c) -1
d) 56
Answer: a
Explanation: Given that k = 0 we must rewrite the function to get
y = ax² + bx
Now differentiating the function yields
y’ = 2ax + b = 0
Equating it to zero we get the Rolle point which is also zero
2a(0) + b = 0
b = 0.

8. For second degree polynomial it is seen that the roots are equal. Then what is the relation between the Rolles point c and the root x?
a) c = x
b) c = x²
c) They are independent
d) c = sin(x)
Answer: a
Explanation: For a polynomial with equal roots we can write the polynomial as f(x) = (x – a)²
Where a is the repeated root.
Differentiating it and equating the function to zero we get the Rolles point,
f'(x) = 2(x – a) = 0
x = a = c.

9. For any second degree polynomial with two real unequal roots. The relation between Rolles point r₁ and the two roots r₂ is
a) They are independent
b) c = r₁ – r₂
c) c = r₁ * ¹⁄_r₂
d) c = (frac{r_1 + r_2}{2})
Answer: d
Explanation: Given the polynomial has two real and unequal roots, we can write the polynomial in the following form
y = (x – a)(x – b)
Where a ≠ b are the two, unequal, real roots.
Now differentiation and equating to zero yields
y = (x – b) + (x -a ) = 0
Put x = c (because is the Rolles point here)
(c – a) + (c – b) = 0
2c – (a + b) = 0
c = a + b / 2 = (frac{r_1 + r_2}{2}).

10. If the domain of a function can be broken into infinite number of disjoint subsets such that every subset has a Rolles point then the function cannot be in a polynomial structure.
a) True
b) False
Answer: b
Explanation: The function could be an infinite degree polynomial (similar to Taylor series), such that after differentiating once we get another polynomial with infinite degree.
Hence, we may conclude that it has infinite distinct roots (with proper choice of polynomial) and hence, the domain can be broken into epsilon (very small) intervals around the roots of the differentiated polynomial to get infinite intervals containing the Rolles point.