300+ REAL TIME Engineering Mathematics Objective Questions & Answers 2023

250+ TOP MCQs on Method of Undetermined Coefficients and Answers

Ordinary Differential Equations Interview Questions and Answers for freshers focuses on “Method of Undetermined Coefficients”.

1. Solution of the D.E y’’ – 4y’ + 4y = e^x when solved using method of undetermined coefficients is _____
a) y = (c₁ + c₂)e^2x + 2e^x – 1
b) y = (c₁ + c₂ x)e^2x + 4e^x – 4
c) y = (c₁ + c₂ x)e^2x + e^x
d) y = (c₁ + c₂ x)e^x + 4e^x
Answer: c
Explanation: We have (D² – 4D + 4)y = e^x
A.E is m² – 4m + 4 = 0 –> (m – 2)² = 0 –> m = 2,2
thus y_c = (c₁ + c₂ x) e^2x, ∅(x) = e^x and 1 is not a root of the A.E
we assume P.I in the form y_p = ae^x…(1) to find a such that y_p’’ – 4y_p’ + 4y_p = e^x….(2)
y_p’ = ae^x and y_p’ = ae^x now (2) becomes ae^x – 4ae^x + 4ae^x = e^x –> a = 1
substituting the value of a in (1) we get y_p = e^x
thus the solution is y = y_c + y_p –> y = (c₁ + c₂ x) e^2x + e^x.

2. Solution of the D.E y’’ + 3y’ + 2y = 12x² when solved using the method of undetermined coefficients is ________
a) y = c₁ e^x + c₂ e^2x + 2 – 11x + x²
b) y = c₁ e^{– x} + c₂ e^{– 2x} + 18 + 21x + 3x²
c) y = c₁ e^x + c₂ e^{– 2x} + 11 + 18x + 2x²
d) y = c₁ e^{– x} + c₂ e^{– 2x} + 21 – 18x + 6x²
Answer: d
Explanation: We have (D² + 3D + 2)y = 12x²
A.E is m² + 3m + 2 = 0 –> (m + 1)(m + 2) = 0 –> m = – 1, – 2
y_c = c₁ e^{– x} + c₂ e^{– 2x} and ∅(x) = 12x² and 0 is not a root of the A.E,
we assume P.I in the form y_p = a + bx + cx²….(1)to find a,b & c such that
y_p’’ + 3y_p’ + 2y_p = 12x²….(2), y_p‘ = b + 2cx, y_p” = 2c now (2) becomes
2c + 3(b + 2cx) + 2(a + bx + cx²) = 12x²
(2a + 3b + 2c) + (2b + 6c)x + (2c)x² = 12x²
2a + 3b + 2c = 0, 2b + 6c = 0, 2c = 12 –> c = 6, b = – 18, a = 21 hence (1) becomes
y_p = 21 – 18x + 6x² thus complete solution is
y = y_c + y_p –> c₁ e^{– x} + c₂ e^{– 2x} + 21 – 18x + 6x².

3. Find the Particular integral solution of the D.E (D² – 4D + 3)y = 20 cos x by the method of undetermined coefficients.
a) y_p = 4 cos⁡x – 3 sin⁡x
b) y_p = 2 sin⁡x – 4 cos⁡x
c) y_p = – 3 cos⁡x + 4 sin⁡x
d) y_p = 2 cos⁡x – 4 sin⁡x
Answer: d
Explanation: ∅(x) = 20 cos x,we assume P.I in the form y_p = a cos⁡x + b sin⁡x ….(1)
and since A.E has m = 1,3 as roots,∓i are not roots of A.E.we have to find a and b
such that y_p” – 4y_p‘ + 3y_p = 20 cos⁡x……(2)
from (1) we get y_p‘ = – a sin⁡x + b cos⁡x, y_p” = – a cos⁡x – b sin⁡x, (2) becomes
– a cos⁡x – b sin⁡x – 4( – a sin⁡x + b cos⁡x) + 3(a cos⁡x + b sin⁡x) = 20 cos x
(2a – b)cos x + (4a + 2b)sin x = 20 cos x –> 2a – b = 20 and 4a + 2b = 0, by solving we get
a = 2, b = – 4 from (1) y_p = 2 cos⁡x – 4 sin⁡x is the particular integral solution.

4. Using the method of undetermined coefficients find the P.I for the D.E x’’’(t) – x’’(t) = 3e^t + sin⁡t.
a) x_p = 3e^t + (frac{t}{2}) (cos⁡t – 2 sin⁡t )
b) x_p = 3te^t + (frac{1}{2}) (cos⁡t + sin⁡t )
c) x_p = 3te^t + (frac{t}{3}) (4cos⁡t + 2sin⁡t )
d) x_p = 3e^t + (frac{1}{2}) (cos⁡t – sin⁡t )
Answer: b
Explanation: We have (D³ – D²)x(t) = 3e^t + sin⁡t, where D = d/dt, A.E is m³ – m² = 0
m² (m – 1) = 0 –> m = 0, 0, 1 –> x_c (t) = (c₁ + c₂ t) + c₃ e^t
∅(t) = 3e^t + sin⁡t we assume for P.I in the form x_p = ate^t + b cos⁡t + c sin⁡t …(1)
since 1 is a root and ∓i are not a roots of the A.E. To find a, b& c such that
x_p’’’(t) – x_p’’(t) = 3e^t + sin⁡t……(2)
from (1) we have x_p‘ = a(te^t + e^t) – b sin⁡t + c cos⁡t
x_p” = a(te^t + 2e^t) – b cos⁡t – c sin⁡t
x_p”’ = a(te^t + 3e^t) + b sin⁡t – c cos⁡t,now (2) becomes
ate^t + 3ae^t + b sin⁡t – c cos⁡t, – ate^t – 2ae^t + b cos⁡t + c sin⁡t = 3e^t + sin⁡t
ae^t + (b + c) sin⁡t + (b – c) cos⁡t = 3e^t + sin⁡t
– – > a = 3 and b + c = 1, b – c = 0 –> a = 3 and b = 1/2, c = 1/2
hence from (1) x_p = 3te^t + 1/2 (cos⁡t + sin⁡t) is the particular integral solution.

5. What is the solution of D.E (D² – 2D)y = e^x sin⁡x when solved using the method of undetermined coefficients?
a) ( y = c_1 + c_2 ,e^{2x} – frac{e^x (xsin x + cos⁡x)}{2})
b) (y = c_1 + c_2 ,e^{2x} – frac{e^x sin x}{2})
c) (y = c_1 + c_2 ,e^{2x} – frac{e^x cos x}{2})
d) (y = c_1 + c_2 ,e^{2x} – frac{e^x (sin x + x cos⁡x)}{4})
Answer: b
Explanation: A.E is m² – 2m = 0 or m(m – 2) = 0 –> m = 0,2
y_c = c₁ + c₂ e^2x and ∅(x) = e^x sin⁡x. we assume PI in the form
y_p = e^x (a cos x + b sin x)….(1) since 1±i are not roots of the A.E.
we have to find a, b such that y_p” – 2y_p‘ = e^x sin⁡x…..(2)
from (1) y_p‘ = e^x (- a sin x + b cos x) + e^x (a cos x + b sin x)
y_p” = e^x (- a sin x + b cos x) + e^x (a cos x + b sin x) + e^x (- a cos x – b sin x) + e^x (- a sin x + b cos x) = 2e^x (- a sin⁡x + b cos⁡x) hence (2) becomes
2e^x (- a sin⁡x + b cos⁡x) – 2e^x (- a sin⁡x + b cos⁡x)
– 2e^x (a cos⁡x + b sin⁡x) = e^x sin⁡x i.e – 2ae^x cos⁡x – 2be^x sin⁡x = e^x sin⁡x
–> – 2a = 0, – 2b = 1 –> a = 0, b = – 1/2 hence (1) becomes (y_p = frac{-e^x sin x}{2})
y = y_c + y_p = c₁ + c₂ e^2x – (frac{e^x sin x}{2})
.

Global Education & Learning Series – Ordinary Differential Equations.

To practice all areas of Ordinary Differential Equations for Interviews,

250+ TOP MCQs on Finding Inverse and Rank of Matrix and Answers

Linear Algebra Interview Questions and Answers focuses on “Finding Inverse and Rank of Matrix”.

1. If A (α, β) = (begin{bmatrix}
cosalpha & sinalpha & 0\
-sinalpha & cosalpha & 0\
0 & 0 & e^{beta}\
end{bmatrix}) then A(α, β)^-1 is equal to?
a) A (-α, β)
b) A (-α, -β)
c) A (α, -β)
d) A (α, β)
Answer: b
Explanation: Here (|A| = cosα(e^β cosα) – sinα(e^β.-sinα) = e^β )
(And, adjA = begin{bmatrix}
e^{beta} cosalpha & e^{beta} sinalpha & 0\
-e^{beta} sinalpha & e^{beta} cosalpha & 0\
0 & 0 & 1\
end{bmatrix})
( ∴ A(α,β)^{-1} = frac{1}{e^β} begin{bmatrix}
e^{beta} cosalpha & e^{beta} sinalpha & 0\
-e^{beta} sinalpha & e^{beta} cosalpha & 0\
0 & 0 & 1\
end{bmatrix})
( = A (-α,-β).)

2. If (A = begin{bmatrix}
a+ib & c+id\
-c+id & a-ib\
end{bmatrix} and , a^2+b^2+c^2+d^2=1) then find A^-1.
a) (begin{bmatrix}
a+ib & -c+id\
-a+id & a-ib\
end{bmatrix} )
b) (begin{bmatrix}
a-ib & c-id\
-c-id & a+ib\
end{bmatrix} )
c) (begin{bmatrix}
a-ib & -c-id\
c-id & a+ib\
end{bmatrix} )
d) (begin{bmatrix}
a+ib & c+id\
-c+id & a-ib\
end{bmatrix} )
Answer: c
Explanation: We have
(|A| = (a+ib) (a-ib) – (-c+id) (c+id) )
( = a^2+b^2+c^2+d^2 )
(s=1 )
Also (adj (A) = begin{bmatrix}
a-ib & -c-id\
c-id & a+ib\
end{bmatrix} )
∴ A^-1 = (frac{1}{|A|} .adj (A) = frac{1}{1} begin{bmatrix}
a-ib & -c-id\
c-id & a+ib\
end{bmatrix} )
(begin{bmatrix}
a-ib & -c-id\
c-id & a+ib\
end{bmatrix} ).

3. The inverse of a symmetric matrix (if it exists) is?
a) A symmetric matrix
b) A skew symmetric matrix
c) A diagonal matrix
d) A triangular matrix
Answer: a
Explanation: Let A be an invertible matrix.
We have AA^-1 = A^-1A = I_N
(AA^-1)’ = (A^-1A)’ = (I_N)’
(A^-1)’ A’ = A’ (A^-1)’ = I_N
(A^-1)’ A = A (A^-1)’ = I_N
(A^-1)’ = A^-1 (inverse of a matrix is unique).

4. Find the rank of the matrix (begin{bmatrix}
4 & 2 & -1 & 2\
1 & -1 & 2 & 1\
2 & 2 & -2 & 0\
end{bmatrix} )?
a) 0
b) 1
c) 2
d) 3
Answer: d
Explanation: Say,
A = (begin{bmatrix}
4 & 2 & -1 & 2\
1 & -1 & 2 & 1\
2 & 2 & -2 & 0\
end{bmatrix} )
( ~ begin{bmatrix}
1 & -1 & 2 & 1\
4 & 2 & -1 & 2\
2 & 2 & -2 & 0\
end{bmatrix} )R₂↔R₁
( ~ begin{bmatrix}
1 & -1 & 2 & 1\
0 & 6 & -9 & -2\
0 & 4 & -6 & -2\
end{bmatrix}) R₂-4R₁, R₃-2R₁
( ~ begin{bmatrix}
1 & -1 & 2 & 1\
0 & 6 & -9 & -2\
0 & 0 & 0 & -4\
end{bmatrix}) 6R₃-4R₂
(∴ ρ(A)= 3.)

5. Rank of the following matrix is A = (begin{bmatrix}
3 & 2 & -1\
4 & 2 & 6\
7 & 4 & 5\
end{bmatrix} ) is?
a) 1
b) 2
c) 3
d) 0
Answer: b
Explanation: (|A| = 3(10-24) + 2(42-20) – (16-14) = 10)
(∵ |A| = 0)
( ~ begin{bmatrix}
3 & 2\
4 & 2\
end{bmatrix} ≠0 = -2 )
(∴ ρ (A) = 2.)

6. If every minor of order ‘r’ of a matrix is zero then ρ (A) =?
a) >r
b) =r
c) ≤r
d) Answer: d
Explanation: By the definition of ‘Rank of a matrix’
A matrix is said to have rank ‘r’ if
(i) At least one minor of order r is non-zero
(ii) All minors of order r+1 is zero
∴ The given matrix (ii) condition seems to be applied
Hence, rank of matrix ρ (A) = < r.

7. Find the inverse of the matrix by using Cayley Hamilton Theorem.
A = ( ~ begin{bmatrix}
1 & 0 & 0\
0 & 1 & 1\
0 & -2 & 4\
end{bmatrix} )
a) (frac{1}{6} begin{bmatrix}
1 & 0 & 0\
0 & -1 & -1\
0 & 2 & -4\
end{bmatrix} )
b) (frac{1}{6} begin{bmatrix}
1 & 0 & 0\
0 & 1 & 1\
0 & 2 & 4\
end{bmatrix} )
c) (frac{1}{6} begin{bmatrix}
1 & 0 & 0\
0 & -1 & -1\
0 & -2 & -4\
end{bmatrix} )
d) (frac{1}{6} begin{bmatrix}
1 & 0 & 0\
0 & 1 & 1\
0 & -2 & 4\
end{bmatrix} )
Answer: a
Explanation: By Cayley Hamliton Theorem
We have λ³ – S₁ λ² + S₂ λ – |A| = 0 characteristics equation of matrix A
Also A satisfies above equation according to theorem.
∴ A³– S₁A²+ S₂A – 6 = 0……… (i) Where S₁=6, S₂=6 & |A|=6
∴ A³ – 6A² + 6A – 6 = 0
Multiplying equation (i) by A^-1
A^-1 (A³) – 6 A^-1 (A²) + A^-1 (A) – 6 A^-1 = 0
=> A² – 6A + 6I – 6 A^-1 = (frac{1}{6}) A^-1
=> A^-1 = (frac{1}{6} Bigg{begin{bmatrix}
1 & 0 & 0\
0 & -1 & 5\
0 & -10 & 14\
end{bmatrix} – begin{bmatrix}
6 & 0 & 0\
0 & 6 & 6\
0 & -12 & -24\
end{bmatrix} + begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 1\
end{bmatrix} Bigg} )
(⇨ A^{-1} = frac{1}{6} begin{bmatrix}
1 & 0 & 0\
0 & -1 & -1\
0 & 2 & -4\
end{bmatrix} ).

Global Education & Learning Series – Linear Algebra.

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250+ TOP MCQs on Divergence and Curl of a Vector Field and Answers

Vector Calculus Multiple Choice Questions on “Divergence and Curl of a Vector Field”.

1. What is the divergence of the vector field ( vec{f} = 3x^2 hat{i}+5xy^2hat{j}+xyz^3hat{k} ) at the point (1, 2, 3).
a) 89
b) 80
c) 124
d) 100
View Answer

Answer: b
Explanation: ( vec{f} = 3x^2 hat{i}+5xy^2hat{j}+xyz^3hat{k} )
∴ div (vec{f}= ∇.vec{f} = (frac{∂}{∂x}hat{i}+ frac{∂}{∂y}hat{j} +frac{∂}{∂z}hat{k}).(3x^2hat{i} + 5xy^2hat{j} + xyz^3hat{k}) )
(= 6x +10xy+ 3xyz^2 )
At the point (1, 2, 3)
div (vec{f} = 6(1)+10(1)(2)+3(1)(2)(3)^2 = 80. )

2. Divergence of ( vec{f}(x,y,z) = frac{(xhat{i}+yhat{j}+zhat{k})}{(x^2+y^2+z^2)^{3/2}}, (x, y, z) ≠ (0, 0, 0).)
a) 0
b) 1
c) 2
d) 3
View Answer

Answer: a
Explanation: Here (vec{f}(x,y,z) = frac{x}{r^3}hat{i} + frac{y}{r^3}hat{j} + frac{z}{r^3}hat{k},) where (r=(x^2+y^2+z^2) )
Also here we can see that
( frac{∂r}{∂x} = frac{x}{r} , frac{∂r}{∂y} = frac{y}{r} , frac{∂r}{∂z} = frac{z}{r} )
∴ div (vec{f} = frac{∂}{∂x} (frac{x}{r}) + frac{∂}{∂y} (frac{y}{r}) + frac{∂}{∂z} (frac{z}{r}) )
(frac{∂}{∂x} (frac{x}{r^3}) = frac{r^3-x 3r^2 (x/r)}{(r^3)^2} = frac{1}{r^3} -frac{3x^2}{r^5} )
(frac{∂}{∂y} (frac{y}{r^3}) = frac{1}{r^3} – frac{3y^2}{r^5} , frac{∂}{∂y} (frac{y}{r}) = frac{1}{r^3} – frac{3z^2}{r^5} )
∴ we get div (vec{f} = frac{3}{r^3} – frac{3x^2+3y^2+3z^2}{r^5} = 0. )

3. Divergence of (vec{f} (x, y, z) = e^{xy} hat{i} -cos⁡y hat{j}+(sinz)^2 hat{k}.)
a) ye^xy+ cos⁡y + 2 sinz.cosz
b) ye^xy– sin⁡y + 2 sinz.cosz
c) 0
d) ye^xy+ sin⁡y + 2 sinz.cosz
View Answer

Answer: d
Explanation: (div vec{f} = (frac{∂}{∂x}hat{i} + frac{∂}{∂y}hat{j}+ frac{∂}{∂z}hat{k}) .( e^{xy} hat{i} – cos⁡y hat{j} + (sinz)^2 hat{k}) )
( = frac{∂}{∂x}(e^{xy}) + frac{∂}{∂y}(-cos⁡y) + frac{∂}{∂z}((sinz)^2) )
( = ye^{xy} + sin⁡y + 2 sinz.cosz .)

4. Curl of (vec{f} (x, y, z) = 2xy hat{i}+ (x^2+z^2)hat{j} + 2zyhat{k} ) is ________
a) (xy^2hat{i} – 2xyzhat{k}) & irrotational
b) 0 & irrotational
c) (xy^2hat{i} – 2xyz hat{k} ) & rotational
d) 0 & rotational
View Answer

Answer: b
Explanation: Curl (vec{f} = ∇ ⤫ vec{f} = begin{vmatrix}
i & j & k\
frac{∂}{∂x} & frac{∂}{∂y} & {∂}{∂z}\
2xy & x^2 + z^2 & 2zy\
end{vmatrix} )
( = (2z – 2z) hat{i} – (0-0)hat{j} + (2x-2x)hat{k} )
(= 0)
Hence F is irrotational field as Curl (vec{f} = 0.)

5. Chose the curl of (vec{f} (x ,y ,z) = x^2 hat{i} + xyz hat{j} – z hat{k} ) at the point (2, 1, -2).
a) (2hat{i} + 2hat{k} )
b) (-2hat{i} – 2hat{j} )
c) (4hat{i} – 4hat{j} + 2hat{k} )
d) (-2hat{i} – 2hat{k} )
View Answer

Answer: d
Explanation:(vec{f} (x ,y ,z) = x^2 hat{i} + xyz hat{j} – z hat{k} )
⇨ Curl (vec{f} = ∇ ⤫ vec{f} = begin{vmatrix}
i & j & k\
frac{∂}{∂x} & frac{∂}{∂y} & frac{∂}{∂z}\
x^2 & xyz & -z\
end{vmatrix} )
( = bigg{frac{∂}{∂y} (-z)-frac{∂}{∂z}(xyz)) hat{i} + (frac{∂}{∂z} (x^2 )- frac{∂}{∂x}(-z)) hat{j} + (frac{∂}{∂x} (xyz)- frac{∂}{∂y} (x^2 ) hat{k}bigg} )
( = (0-xy) hat{i} + (0-0) hat{j} + (yz-0) hat{k} )
( = – xy hat{i} + yz hat{k} )
(∇ ⤫ vec{f} |_{(1, -1, 2)} = -2(1) hat{i} + (1) (-2) hat{k} = -2hat{i} – 2hat{k}. )

6. A vector field which has a vanishing divergence is called as ____________
a) Solenoidal field
b) Rotational field
c) Hemispheroidal field
d) Irrotational field
View Answer

Answer: a
Explanation: By the definition: A vector field whose divergence comes out to be zero or
Vanishes is called as a Solenoidal Vector Field.
i.e.
If (∇. vec{f} = 0 ↔ vec{f} ) is a Solenoidal Vector field.

7. Divergence and Curl of a vector field are ___________
a) Scalar & Scalar
b) Scalar & Vector
c) Vector & Vector
d) Vector & Scalar
View Answer

Answer: b
Explanation: Let (vec{f} = a_1hat{i} + a_2hat{j} + a_3hat{k} )
div(vec{f} = (frac{∂}{∂x} hat{i} + frac{∂}{∂y} hat{j}+ frac{∂}{∂z} hat{k}).(a_1hat{i} + a_2hat{j} + a_3hat{k}) )
( = frac{∂a_1}{∂x} + frac{∂a_2}{∂y} + frac{∂a_3}{∂z} ) which is a scalar quantity.
Also curl (vec{f} = begin{vmatrix}
i & j & k\
frac{∂}{∂x} & frac{∂}{∂y} & frac{∂}{∂z}\
a1 & a2 & a3\
end{vmatrix} ) = (b_1hat{i} + b_2hat{j} + b_3hat{k} )
Which is going to be a vector quantity as cross product of two vectors is again a vector, where dot product gives a scalar outcome.

8. A vector field with a vanishing curl is called as __________
a) Irrotational
b) Solenoidal
c) Rotational
d) Cycloidal
View Answer

Answer: a
Explanation: By the definition: Mathematically,
If (∇ ⤫ vec{f} = 0 ↔ vec{f} ) is an Irrotational Vector field.

9. The curl of vector field (vec{f} (x,y,z) = x^2hat{i} + 2z hat{j} – y hat{k} ) is _________
a) (-3hat{i} )
b) (-3hat{j} )
c) (-3hat{k} )
d) 0
View Answer

Answer: a
Explanation:Curl,
(vec{f} = ∇ ⤫ vec{f} = begin{vmatrix}
i & j & k\
frac{∂}{∂x} & frac{∂}{∂y} & frac{∂}{∂z}\
x^2 & 2z & -y\
end{vmatrix} )
( = bigg{frac{∂}{∂y} (-y)-frac{∂}{∂z}(2z)) hat{i} + (frac{∂}{∂z} (x^2 )-frac{∂}{∂x}(-y)) hat{j} + (frac{∂}{∂x} (2z)-frac{∂}{∂y} (x^2 ) hat{k}bigg} )
( = (-1-2) hat{i} + (0-0) hat{j} + (0-0) hat{k} = -3hat{i}. )

Global Education & Learning Series – Vector Calculus.

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250+ TOP MCQs on Derivation and Solution of Two-dimensional Heat Equation and Answers

Fourier Analysis and Partial Differential Equations Multiple Choice Questions on “Derivation and Solution of Two-dimensional Heat Equation”.

1. Who was the first person to develop the heat equation?
a) Joseph Fourier
b) Galileo Galilei
c) Daniel Gabriel Fahrenheit
d) Robert Boyle
View Answer

Answer: a
Explanation: The heat equation was first put forth by Jean-Baptiste Joseph Fourier (21 March 1768 – 16 May 1830) in 1822. He was a French mathematician and physicist born in Auxerre. It was developed to describe heat flow.

2. Which of the following is not a field in which heat equation is used?
a) Probability theory
b) Histology
c) Financial Mathematics
d) Quantum Mechanics
View Answer

Answer: b
Explanation: The heat equation is used is several fields, some of which include,

Probability theory (to study Brownian motion)
Financial Mathematics (for solving Black- Scholes PDE)
Quantum Mechanics (to determine spread of wave function)

3. Under ideal assumptions, what is the two-dimensional heat equation?
a) u_t = c∇² u = c(u_xx + u_yy)
b) u_t = c² u_xx
c) u_t = c² ∇² u = c² (u_xx + u_yy)
d) u_t = ∇² u = (u_xx + u_yy)
View Answer

Answer: c
Explanation: Consider a rectangular plate (thermally conductive material), with dimensions a × b. The plate is heated and then insulated.
We let u (x, y, t) = temperature of plate at position (x, y) and time t.
For a fixed t, the height of the surface z = u (x, y, t) gives the temperature of the plate at time t and position (x, y).
Under ideal assumptions (e.g. uniform density, uniform specific heat, perfect insulation, no internal heat sources etc.) one can show that u satisfies the two-dimensional heat equation,
u_t = c² ∇² u = c² (u_xx + u_yy) for 0 < x < a, 0 < y < b.

4. In mathematics, an initial condition (also called a seed value), is a value of an evolving variable at some point in time designated as the initial time (t=0).
a) False
b) True
View Answer

Answer: b
Explanation: Another definition of initial condition may be stated as, ‘any of a set of starting-point values belonging to or imposed upon the variables in an equation that has one or more arbitrary constants’

5. What is another name for heat equation?
a) Induction equation
b) Condenser equation
c) Diffusion equation
d) Solar equation
View Answer

Answer: c
Explanation: The heat equation is also known as the diffusion equation and it describes a time-varying evolution of a function u(x, t) given its initial distribution u(x, 0).

6. Heat Equation is an example of elliptical partial differential equation.
a) True
b) False
View Answer

Answer: b
Explanation: Jean-Baptiste Joseph Fourier (21 March 1768 – 16 May 1830) was a French mathematician and physicist born in Auxerre who was the first person to develop heat equation. The heat equation is the prototypical example of a parabolic partial differential equation.

7. What is the half-interval method in numerical analysis is also known as?
a) Newton-Raphson method
b) Regula Falsi method
c) Taylor’s method
d) Bisection method
View Answer

Answer: d
Explanation: The Bisection method, also known as binary chopping or half-interval method, is a starting method which is used, where applicable, for few iterations, to obtain a good initial value.

8. Which of the following represents the canonical form of a second order parabolic PDE?
a) (frac{∂^2 z}{∂η^2}+⋯=0 )
b) (frac{∂^2 z}{∂ζ∂η}+⋯=0)
c) (frac{∂^2 z}{∂α^2}+frac{∂^2 z}{∂β^2}…=0)
d) (frac{∂^2 z}{∂ζ^2}+⋯=0)
View Answer

Answer: a
Explanation: A second order linear partial differential equation can be reduced to so-called canonical form by an appropriate change of variables ξ = ξ(x, y), η = η(x, y).

9. Which of the following is the condition for a second order partial differential equation to be hyperbolic?
a) b²-ac<0
b) b²-ac=0
c) b²-ac>0
d) b²-ac=<0
View Answer

Answer: c
Explanation: For a second order partial differential equation to be hyperbolic, the equation should satisfy the condition, b²-ac>0.

10. What is the order of the partial differential equation, (frac{∂^2 z}{∂x^2}-(frac{∂z}{∂y})^5+frac{∂^2 z}{∂x∂y}=0)?
a) Order-5
b) Order-1
c) Order-4
d) Order-2
View Answer

Answer: d
Explanation: The order of an equation is defined as the highest derivative present in the equation. Hence, in the given equation, (frac{∂^2 z}{∂x^2}-(frac{∂z}{∂y})^5+frac{∂^2 z}{∂x∂y}=0), the order is 2.

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250+ TOP MCQs on Generalized Mean Value Theorem and Answers

Differential and Integral Calculus Multiple Choice Questions on “Differential Calculus Questions and Answers – Generalized Mean Value Theorem”.

1. Taylor’s theorem was stated by the mathematician _____________
a) Brook Taylor
b) Eva Germaine Rimington Taylor
c) Sir Geoffrey Ingram Taylor
d) Michael Eugene Taylor
View Answer

Answer: a
Explanation:

Brook Taylor (18 August 1685 – 29 December 1731) was an English mathematician, best known for Taylor’s theorem and the Taylor series.
Eva Germaine Rimington Taylor (1879–1966) was an English geographer and historian of science.
Sir Geoffrey Ingram Taylor (7 March 1886 – 27 June 1975) was a British physicist and mathematician, and a major figure in fluid dynamics and wave theory.
Michael Eugene Taylor (born 1946) is an American mathematician who is working in partial differential equations.

2. Lagrange’s Remainder for Maclaurin’s Theorem is given by _____________
a) (frac{x^n}{(n-1)!}f^{(n)}(θx) )
b) (frac{x^n}{n!} f^{(n)}(θx))
c) (frac{x^{n-1}}{n!} f^{(n)}(θx))
d) (frac{x^n}{n!}f^{(n-1)}(θx))
View Answer

Answer: b
Explanation: Maclaurin’s Theorem is a special case of Taylor’s Theorem; hence Schlomilch’s Remainder for Maclaurin’s Theorem is given by, (frac{x^n(1-θ)^{n-p}}{(n-1)!p} f^{(n)}(θx).) To obtain Lagrange’s Remainder for Maclaurin’s Theorem, we put p=n, which gives us, (frac{x^n}{n!} f^{(n)}(θx).)

3. Cauchy’s Remainder for Maclaurin’s Theorem is given by (frac{x^n(1-θ)^{n-1}}{(n-1)!}f^{(n)}(θx)).
a) True
b) False
View Answer

Answer: a
Explanation: Schlomilch’s Remainder for Maclaurin’s Theorem is given by, (frac{x^n(1-θ)^{n-p}}{(n-1)!p} f^{(n)}(θx).)
To obtain Cauchy’s Remainder for Maclaurin’s Theorem, we put p=1, which gives us,
(frac{x^n(1-θ)^{n-1}}{(n-1)!}f^{(n)}(θx).)

4. What is the Taylor series expansion of f(x)= x²-x+1 about the point x=-1?
a) f(x) = -3-3(x+1)+(x+1)²
b) f(x) = -3-(x+1)+(x+1)²
c) f(x) = -3-3(x+1)+2(x+1)²
d) f(x) = -1-3(x+1)+2(x+1)²
View Answer

Answer: a
Explanation: Given, f(x)= x²-x+1 , f(-1)=1+1+1=3
f'(x) = 2x-1,f'(-1)=2(-1)-1= -3
f”(x) = 2, f”(-1)=2
fⁿ (x)=0, for n > 2
The Taylor series expansion of f(x) about x=a is,
f(x) = f(a)+(x-a) f'(a)+(x-a)² (frac{f”}{2!}) (a)+⋯
Here a=-1,
f(x) = f(-1)+(x+1) f'(-1)+(x+1)² (frac{f”}{2!}) (-1) since,for n > 2, fⁿ (x) = 0.
f(x) = -3-3(x+1)+(x+1)²

5. What is the first term in the Taylor series expansion of f(x) = 8x⁵-3x²-5x about x=2?
a) 232
b) 244
c) 234
d) 222
View Answer

Answer: c
Explanation: Given, f(x) = 8x⁵-3x²-5x
First term in the Taylor series for f(x) is f(a). Here, a=2. Therefore,
f(2) = 8(2)⁵-3(2)²-5(2)=8(32)-3(4)-10=234.

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
View Answer

Answer: b
Explanation: A recurrence relation is an equation that recursively defines a sequence or multidimensional array of values, once one or more initial terms are given; each further term of the sequence or array is defined as a function of its preceding terms.

7. The initial condition for the recurrence relation of a factorial is ___________
a) 0!=0
b) 0!=1
c) 1!=1
d) 1!=0
View Answer

Answer: b
Explanation: Factorial of a number is defined by the recurrence relation as, n!=n(n-1)! for n>0 and the initial condition is 0!=1.

8. For the power series of the form, (∑_{i=0}^∞ a_i z^i,) which one of the following may not be true?
a) The series converges only for z=0
b) The series converges absolutely for all z
c) The series converges absolutely for all z in some finite open interval (-R, R) and diverges if z<-R or z>R
d) At the points z=R and z=-R, the series will diverge
View Answer

Answer: d
Explanation: A power series in a variable z is in the form of an infinite sum as given, i.e., (∑_{i=0}^∞ a_i z^i,), where ai are integers, real numbers, complex numbers, or any other quantities of a given type.
For any power series, one of the following is true:
i. The series converges only for z=0.
ii. The series converges absolutely for all z.
iii. The series converges absolutely for all z in some finite open interval (-R, R) and diverges if z<-R or z>R. At the points z=R and z=-R, the series may converge absolutely, converge conditionally, or diverge.

9. Maclaurin Series is named after ______________
a) Colin Maclaurin
b) Normand Maclaurin
c) Ian Maclaurin
d) Richard Cockburn Maclaurin
View Answer

Answer: a
Explanation:

Colin Maclaurin (February 1698 – 14 June 1746), was a Scottish Mathematician. The Maclaurin Series was named after him.
Normand Maclaurin (10 December 1835 – 24 August 1914) was a Scottish-born physician, company director, Australian politician and university administrator.
Ian Maclaurin, born 30 March 1937, is a British businessman, who has been chairman of Vodafone and chairman and chief executive of Tesco.
Richard Cockburn Maclaurin (June 5, 1870 – January 15, 1920) was a Scottish-born U.S. educator and mathematical physicist.

10. Maclaurin’s Theorem is a special type of Taylor’s Theorem.
a) False
b) True
View Answer

Answer: b
Explanation: The Taylor series expansion of f(x) about x=a is,
f(x)= f(a)+(x-a) f'(a)+(x-a)²(frac{f”}{2!}) (a)+⋯
If we put x=0, we get the Maclaurin series which is given by, f(x)= f(0)+xf'(0)+x² (frac{f”}{2!}) (0)+⋯

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250+ TOP MCQs on Limits and Derivatives of Several Variables and Answers

Engineering Mathematics Multiple Choice Questions on “Limits and Derivatives of Several Variables – 2”.

1. Two men on a surface want to meet each other. They have taken the point (0, 0) as meeting point. The surface is 3-D and its equation is f(x,y) = (frac{x^{frac{-23}{4}}y^9}{x+(y)^{frac{4}{3}}}). Given that they both play this game infinite number of times with their starting point as (908, 908) and (90, 180)
(choosing a different path every time they play the game). Will they always meet?
a) They will not meet every time
b) They will meet every time
c) Insufficient information
d) They meet with probability ¹⁄₂
Answer: a
Explanation: The question is asking us to simply find the limit of the given function exists as the pair (x, y) tends to (0, 0) (The two men meet along different paths taken or not)
Thus, put x = t : y = a(t)^³⁄₄
(=lt_{(x,y)rightarrow(0,0)}=lt_{trightarrow 0}frac{a^9.t^{frac{27}{4}}.t^{frac{-23}{4}}}{t+a^{frac{4}{3}}.t^{frac{1}{1}}})
(=lt_{trightarrow 0}frac{t}{t} times frac{a^9}{1+a^{frac{4}{3}}})
(=lt_{trightarrow 0} frac{a^9}{1+a^{frac{4}{3}}})
By putting different values of a we get different limits
Thus, there are many paths that do not go to the same place.
Hence, They will not meet every time is the right answer.

2. Find (lt_{(x,y,z)rightarrow(0,0,0)}frac{y^2.z^2}{x^3+x^2.(y)^{frac{4}{3}}+x^2.(z)^{frac{4}{3}}})
a) 1
b) 0
c) ∞
d) Does Not Exist
Answer: d
Explanation: Put x = t : y = a₁ * t^³⁄₄ : z = a₂ * t^³⁄₄
(lt_{(x,y,z)rightarrow(0,0,0)}frac{(a_1)^2.t^{frac{3}{2}}.(a_2)^2.t^{frac{3}{2}}}{t^3+t^2.t.(a_1)^{frac{4}{3}}+t^2.t.(a_2)^{frac{4}{3}}})
(lt_{(x,y,z)rightarrow(0,0,0)}frac{t^3}{t^3}times frac{(a_1)^2.(a_2)^2}{1+(a_1)^{frac{4}{3}}+(a_2)^{frac{4}{3}}})
(lt_{(x,y,z)rightarrow(0,0,0)} frac{(a_1)^2.(a_2)^2}{1+(a_1)^{frac{4}{3}}+(a_2)^{frac{4}{3}}})
By varying a₁ : a₂ one can get different limit values.

3. Find (lt_{(x,y,z)rightarrow(0,0,0)}frac{sin(x).sin(y)}{x.z})
a) ∞
b) ¹⁄₃
c) 1
d) Does Not Exist
Answer: d
Explanation: Put x = t : y = at : z = t
=(lt_{trightarrow 0}frac{sin(t).sin(at)}{t^2})
=(lt_{trightarrow 0}frac{sin(t)}{t}times (a) times lt_{trightarrow 0}frac{sin(at)}{at})
= (1) * (a) * (1) = a

4. Find (lt_{(x,y,z)rightarrow(2,2,4)}frac{x^2+y^2-z^2+2xy}{x+y-z})
a) ∞
b) 123
c) 9098
d) 8
Answer: d
Explanation: Simplifying the expression yields
(lt_{(x,y,z)rightarrow(0,0,0)}frac{(x+y)^2-z^2}{(x+y)-z})
(lt_{(x,y,z)rightarrow(0,0,0)}frac{(x+y+z).(x+y-z)}{(x+y-z)})
(lt_{(x,y,z)rightarrow(0,0,0)}(x+y+z)=2+2+4)
=8

5. Find (lt_{(x,y,z,w)rightarrow(0,0,0,0)}frac{x^{-6}.y^2.(z.w)^3}{x+y^2+z-w})
a) 1990
b) ∞
c) Does Not Exist
d) 0
Answer: c
Explanation: Put x = t : y = a₁.t^¹⁄₂ : z = a₂.t : w = a₃.t
(lt_{(x,y,z,w)rightarrow(0,0,0,0)}frac{t^{-6}.t.(a_1)^2.t^6.(a_2)^3.(a_3)^3}{t+t.(a_1)^2+a_2.t-a_3.t})
(lt_{(x,y,z,w)rightarrow(0,0,0,0)}frac{t}{t}times frac{(a_1)^2.(a_2)^3.(a_3)^3}{1+(a_1)^2+a_2-a_3})
(lt_{(x,y,z,w)rightarrow(0,0,0,0)} frac{(a_1)^2.(a_2)^3.(a_3)^3}{1+(a_1)^2+a_2-a_3})
By changing the values of a₁ : a₂ : a₃ we get different values of limit.
Hence, Does Not Exist is the right answer.

6. Find (lt_{(x,y,z,w)rightarrow(3,1,1,11)}frac{x^4+y^2+z^2+2x^2y+2yz+2x^2z-(w)^2}{x^2+y+z-w})
a) 700
b) 701
c) 699
d) 22
Answer: d
Explanation: Simplifying the expression we have
(lt_{(x,y,z,w)rightarrow(3,1,1,11)}frac{(x^2+y+z)^2-(w)^2}{x^2+y+z-w})
(lt_{(x,y,z,w)rightarrow(3,1,1,11)}frac{(x^2+y+z+w).(x^2+y+z-w)}{x^2+y+z-w})
(lt_{(x,y,z,w)rightarrow(3,1,1,11)}(x^2+y+z+w))=(3²+1+1+11)
=9+1+1+11=22

7. Given that limit exists find (lt_{(x,y,z)rightarrow(-2,-2,-2)}frac{sin((x+2)(y+5)(z+1))}{(x+2)(y+7)})
a) 1
b) ³⁄₅
c) ¹⁄₂
d) 0
Answer: b
Explanation: Given that limit exists we can parameterize the curve
Put x = t : y = t : z = t
(lt_{trightarrow -2}frac{sin((t+2)(t+5)(t+1))}{(t+2)(t+7)})
(lt_{trightarrow -2}frac{sin((t+2)(t+5)(t+1))}{(t+2)(t+5)(t+1)}times lt_{trightarrow -2}frac{(t+5)(t+1)}{(t+7)})
((1)times frac{(-2+5)(-2+1)}{(-2+7)})
=(frac{(3).(1)}{(5)}=frac{3}{5})

8. Given that limit exist find (lt_{(x,y,z)rightarrow(-9,-9,-9)}frac{tan((x+9)(y+11)(z+7))}{(x+9)(y+10)})
a) 2
b) 1
c) 4
d) 3
Answer: c
Explanation: We can parameterize the curve by
x = y = z = t
(lt_{trightarrow -9}frac{tan((t+9)(t+11)(t+7))}{(t+9)(t+10)})
(lt_{trightarrow -9}frac{tan((t+9)(t+11)(t+7))}{(t+9)(t+11)(t+7)}times lt_{trightarrow -9}frac{tan((t+11)(t+7))}{t+10})
(=frac{(-9+11)(-9+7)}{(-9+10)}=frac{(2)(2)}{(1)})
=4

9. Given that limit exists find (lt_{(x,y,z)rightarrow(-1,-1,-1)}frac{tan((x-1)(y-2)(z-3))}{(x-1)(y-6)(z+7)})
a) 1
b) ¹⁄₂
c) ¹⁄₇
d) ²⁄₇
Answer: d
Explanation: We can parameterize the curve by
x=y=z=t
(lt_{trightarrow -1}frac{tan((t-1)(t-2)(t-3))}{(t-1)(t-6)(t+7)})
(lt_{trightarrow -1}frac{tan((t-1)(t-2)(t-3))}{(t-1)(t-2)(t-3)}times lt_{trightarrow -1}frac{(t-2)(t-3)}{(t-6)(t+7)} )
=(frac{(-1-2)(-1-3)}{(-1-6)(-1+7)}=frac{(3)(4)}{(7)(6)})
=(frac{12}{42}=frac{2}{7})

10. Given that limit exists find (lt_{(x,y,z)rightarrow(2,2,2)}left (frac{ln(1+frac{xy-2x-y+z}{xz-2x-6z+12}+frac{xz-5x-2z+10}{xy-7y-2x+14}}{(x-2)(y-2)(z-2)}right ))
a) ∞
b) 1
c) 0
d) ln(⁴⁄₅)
Answer: a
Explanation: We can parameterize the curve by
x = y = z = t
(lt_{trightarrow 2}left (frac{ln(1+frac{xy-2x-y+z}{xz-2x-6z+12}+frac{xz-5x-2z+10}{xy-7y-2x+14}}{(x-2)(y-2)(z-2)}right ))
(=lt_{trightarrow 2}left (frac{ln(1+frac{t}{(t-6)}+frac{(t-5)}{(t-7)}}{(t-2)^3}right ))
(=lt_{trightarrow 2}left (frac{ln(1+frac{2}{(t-6)}+frac{(2-5)}{(2-7)})}{(2-2)^3}right )=frac{ln(frac{4}{5})}{0}rightarrowinfty)

11. Given that limit exists (lt_{(x,y,z)rightarrow(0,0,0)}left (frac{cos(frac{pi}{2}-x).tan(y).cot(frac{pi}{2}-z)}{sin(x).sin(y).sin(z)}right ))
a) 99
b) 0
c) 1
d) 100
Answer: c
Explanation: Put x = y = z = t
(lt_{trightarrow 0}left (frac{cos(frac{pi}{2}-t).tan(y).cot(frac{pi}{2}-t)}{sin(x).sin(y).sin(z)}right))
(lt_{trightarrow 0}frac{(sin(t))(tan^2(t))}{sin^3(t)})
(=lt_{trightarrow 0}frac{tan^2(t)}{sin(t)} = lt_{trightarrow 0}frac{1}{cos^2(t)})
(=frac{1}{cos^2(0)}=frac{1}{1}=1)

12. Two men on a 3-D surface want to meet each other. The surface is given by (f(x,y)=frac{x^{-6}.y^7}{x+y}). They make their move horizontally or vertically with the X-Y plane as their reference. It was observed that one man was initially at (200, 400) and the other at (100, 100). Their meet point is decided as (0, 0). Given that they travel in straight lines, will they meet?
a) They will meet
b) They Will not meet
c) They meet with probability ¹⁄₂
d) Insufficient information
Answer: b
Explanation: The problem asks us to find the limit of the function f(x, y) along two lines y = x and y = 2x
For the first line (first person)
x = t : y = 2t
(=lt_{trightarrow 0}frac{x^{-6}.2^7.t^7}{t+2t}=lt_{trightarrow 0}frac{2^7t}{3t})
=(frac{2^7}{3})
For the second line (Second Person)
x = t = y
=(lt_{trightarrow 0}frac{t^{-6}.t^7}{t+t}=lt_{trightarrow 0}frac{t}{2t})
=1/2
The limits are different and they will not meet.

13. Two men on a 3-D surface want to meet each other. The surface is given by (f(x,y)=frac{x^6.y^7}{x^{13}+y^{13}}). They make their move horizontally or vertically with the X-Y plane as their reference. It was observed that one man was initially at (400, 1600) and the other at (897, 897). Their meet point is decided as (0, 0). Given that they travel in straight lines, will they meet?
a) They will meet
b) They will not meet
c) They meet with probability ¹⁄₂
d) Insufficient information
Answer: b
Explanation: The problem asks us to find the limit of the function f(x, y) along two lines y = x and y = 4x
For the first line (first person)
x = t : y = 4t
=(lt_{trightarrow 0}frac{t^6.4^7.t^7}{t^{13}+4^{13}t^{13}}=lt_{trightarrow 0}frac{4^7t}{t(1+4^{13})})
=(frac{4^7}{1+4^{13}})
For the second line (Second Person)
x = t = y
=(lt_{trightarrow 0}frac{t^6.t^7}{t^{13}+t^{13}}=lt_{trightarrow 0}frac{t^{13}}{2t^{13}})
= ¹⁄₂
The limits are different and the will not meet.

14. Observe the figure. It is given that the function has no limit as (x, y) → (0 ,0) along the paths given in the figure. Then which of the following could be f(x, y)
a) (f(x,y) = frac{x^7.y^8}{(x+y)})
b) f(x,y) = x²y⁷
c) (f(x,y) = frac{xy^2}{(x^2+y^2)})
d) (f(x,y) = frac{x^6.y^2}{(y^5+x^{10})})
Answer: d
Explanation: The curves in the given graph are parabolic and thus they can be parameterized by
x = t : y = at²
Substituting in Option (f(x,y) = frac{x^6.y^2}{(y^5+x^{10})}) we get
=(lt_{trightarrow 0}frac{t^6.a^2.t^4}{a^5.t^{10}+t^{10}})
(lt_{trightarrow 0}frac{t^{10}}{t^{10}}times frac{a^2}{a^5+1})
(lt_{trightarrow 0} frac{a^2}{a^5+1})
By varying a we get different limits