Engineering Mathematics Interview Questions and Answers

This set of Engineering Mathematics Interview Questions and Answers focuses on “The nth Derivative of Some Elementary Functions – 2”.

1. nth derivative of Sinh(x) is
a) 0.5(e^x – e^-x)
b) 0.5(e^-x – e^x)
c) 0.5(e^x – (-1)ⁿ e^-x)
d) 0.5((-1)^-n e^-x -e^x)
View Answer

Answer: c
Explanation: Y = Sinh(x)
Y = 0.5[e^x – e^-x].
y₁ = 0.5 [e^x – (-1)e^-x].
y₂ = 0.5 [e^x – (-1)² e^-x].
Similarly,
y_n = 0.5 [e^x – (-1)ⁿ e^-x].

2. If y=log⁡(^x⁄_{(x² – 1)}), then nth derivative of y is ?
a) (-1)^(n-1) (n-1)!(x^(-n) + (x-1)^(-n) + (x+1)^(-n))
b) (-1)ⁿ (n)! (x^(-n-1) + (x-1)^(-n-1) + (x+1)^(-n-1))
c) (-1)⁽ⁿ⁺¹⁾ (n+1)!(x^(-n) + (x-1)^(-n) + (x+1)^(-n))
d) (-1)ⁿ(n)! (x^(-n-1) + (x-1)^(-n+1) + (x+1)^(-n+1))
View Answer

Answer: a
Explanation: Y=log(x) – log(x² – 1)
y₁ = x^(-1)-2x/(x²-1)
y₁ = x^(-1)-(x-1)^(-1) + (x+1)^(-1)
y_n = (-1)^(n-1) (n-1)!(x^(-n)-(x-1)^(-n) + (x+1)^(-n)).

3. If x = a(Cos(t) + t²) and y = a(Sin(t) + t² + t³) then dy/dx equals to
a) (Cos(t) + 3t² + 2t) / (-Sin(t) + 2t)
b) (Sin(t) + 3t² + 2t) / (-Cos(t) + 2t)
c) (Sin(t) + 3t² + 2t) / (Cos(t) + 2t)
d) (Cos(t) + 3t² + 2t) / (Sin(t) + 2t)
View Answer

Answer: a
Explanation: dx/dt = a(-Sin(t) + 2t)
dy/dt = a(Cos(t) + 2t + 3t²)
Then,
dy/dx = (Cos(t) + 3t²+2t)/(-Sin(t) + 2t).

4. If y=tan^(-1)⁡(x) , then which one is correct ?
a) y₃ + y₁² + 4xy₂ y₁=0
b) y₃ + y₁² + xy₂ y₁=0
c) y₃ + 2y₁² + xy₂ y₁=0
d) y₂ + 2y₁² + 4xy₂ y₁=0
View Answer

Answer: d
Explanation: y=tan^-1⁡(x)
\(y_1 = \frac{1}{1+x^2}\)
\(y_2 = \frac{-2x}{(1+x^2)^2}\)
\(y_3 = -2[\frac{(1+x^2)^2-4x^2 (1+x^2)}{(1+x^2)^4}]\)
\(y_3 = -2[\frac{1}{1+(x^2)^2}-\frac{(4x^2)}{(1+x^2)^3}]\)
\(y_3+2y_1^2 + 4xy_2 y_1\)=0

5. What is the value of \(\frac{d^n (x^m)}{dx^n}\) for m<n, m=n, m>n?
a) 0, n!, m_{P_n} x^(m-n)
b) m_{P_n} x^(m-n), n!, 0
c) 0, n!, m_{C_n} x^(m-n)
d) m_{C_n} x^(m-n), n!, 0
View Answer

Answer: a
Explanation: For, m > n
\(\frac{d^n (x^m)}{dx^n} = m \frac{d^{n-1} (x^{m-1})}{dx^{n-1}}=m(m-1)\frac{d^{n-2} (x^{m-2})}{dx^{n-2}}\)=……..
Since m>n, m-n=0 hence this cycle will moves upto (m-n) times and at last
\(\frac{d^n (x^m)}{dx^n}=m(m-1)(m-2)….(m-(n-1)) x^{m-n}\)
Hence,
\(\frac{d^n (x^m)}{dx^n}=m_{P_n} x^{(m-n)}\) ………. (1)
For m=n, from equation 1,
\(\frac{d^n (x^n)}{dx^n}=n_{P_n} x^{(n-n)}=n!\)
From m<n, from equation 1,
\(\frac{d^n (x^m)}{dx^n}=m_{P_n} x^{(m-n)}=0\) (Because, \(m_{P_n}=0\) for m<n)

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6. Which of the following is true
a) Value of \(\frac{d^m (Sin⁡(nx))}{dx^m}\) is always positive for m=0, 1, 4, 5, 8, 9… for 0 < nx < ^π⁄₂ and n<0
b) Value of \(\frac{d^m (Sin⁡(nx))}{dx^m}\) is always positive for m=2, 3, 6, 7, 10, 11… for 0 < nx < ^π⁄₂ and n>0
c) Value of \(\frac{d^m (Sin⁡(nx))}{dx^m}\) is always positive for m=0, 1, 4, 5, 8, 9… for 0 < nx < ^π⁄₂ and n>0
d) Value of \(\frac{d^m (Sin⁡(nx))}{dx^m}\) is always positive for m=2, 3, 6, 7, 10, 11… for 0 < nx < ^π⁄₂ and n<0
View Answer

Answer: c
Explanation: Here,
\(\frac{d(Sin⁡(nx))}{dx} = n Cos(nx)\) …………….(m=1)
\(\frac{d^2 (Sin⁡(nx))}{dx^2} = -n^2 Sin(nx)\) …..(m=2)
\(\frac{d^3 (Sin⁡(nx))}{dx^3} = -n^3 Cos(nx)\) …..(m=3)
\(\frac{d^4 (Sin⁡(nx))}{dx^4} = n^4 Sin(nx)\) ……(m=4)
So the value of \(\frac{d^m (Sin⁡(nx))}{dx^m} = \begin{cases}n^m Cos(nx) \,\,\, m=1,5,9,….\\-n^m Sin(nx) \,\,\, m=2,6,10…\\-n^m Cos(nx)\,\,\,m=3,7,11…..\\n^m Sin(nx)\,\,\,m=4,8,12….\end{cases} \)
Hence, for n>0 and 0<nx<(π)/2 only \(\frac{d^m (Sin⁡(nx))}{dx^m}\) is positive only when m = 1,4,5,8,9,….. otherwise negative.

7. If nth derivative of e^ax sin⁡(bx+c) cos⁡(bx+c) is, e^ax rⁿ sin⁡(bx+c+^nα⁄₂) cos⁡(bx+c+^nα⁄₂) then,
a) r = \(\sqrt{a^2+b^2}, \alpha=tan^{-1}\frac{⁡b}{a}\)
b) r = \(\sqrt{a^2+4b^2}, \alpha=tan^{-1}⁡\frac{2b}{a}\)
c) r = \(\sqrt{a^2+8b^2}, \alpha=tan^{-1}⁡\frac{4b}{a}\)
d) r = \(\sqrt{a^2+16b^2}, \alpha=tan^{-1}\frac{⁡4b}{a}\)
View Answer

Answer: b
Explanation: y = e^ax sin⁡(bx+c) cos⁡(bx+c)
y = e^ax sin⁡2(bx+c)/2
y_n = e^ax rⁿ sin⁡(2(bx+c+nα/2))/2
y_n = e^ax rⁿ sin⁡(bx+c+nα/2) cos⁡(bx+c+nα/2)
where
r = \(\sqrt{a^2+4b^2}\), α = tan^-1⁡2b/a.

8. If y=^x⁴⁄_x²-1, then?
a) 0.5*(-1)ⁿ (n-1)! [(x-1)^-n-1 + (x+1)^-n-1]
b) 0.5*(-1)ⁿ (n-1)! [x^{– n-1} + (x-1)^-n-1 + (x+1)^-n-1]
c) 0.5*(-1)ⁿ (n-1)! [(x-1)^-n + (x+1)^-n)]
d) 0.5*(-1)ⁿ (n-1)! [x^-n + (x-1)^-n + (x+1)^-n]
View Answer

Answer: a
Explanation: \(y = x^2+\frac{x^2}{x^2-1}\)
\(y = x^2+\frac{1}{x^2-1}\)+1
\(y = x^2+1+\frac{1}{(x-1)(x+1)}\)
\(y = x^2+1+0.5[\frac{1}{(x-1)}-\frac{1}{(x+1)}]\)
\(y_n=0.5*(-1)^n (n-1)![(x-1)^{-n-1}+(x+1)^{-n-1}]\)

9. If y=sin^(-1)⁡(x) then select the true statement.
a) y₂ = xy₁³
b) y₃ = xy₂³
c) y₂ = xy₁²
d) y₃ = xy₁²
View Answer

Answer: a
Explanation: y=sin^-1⁡(x)
\(y_1=\frac{1}{\sqrt{1-x^2}}\)
\(y_2=\frac{x}{(1-x^2)^{3/2}}\)
\(y_2=\frac{x}{(1-x^2)^{3/2}} = xy_1^3\)

10. nth derivative of y = sin²x cos³x is
a) ¹⁄₈ cos⁡(x + ^nπ⁄₂) –¹⁄₁₆ 5ⁿ cos⁡(x + ^nπ⁄₂) – ¹⁄₁₆ 3ⁿ cos⁡(3x + ^nπ⁄₂)
b) ¹⁄₈ sin⁡(x+^nπ⁄₂) –¹⁄₁₆ 5ⁿ cos⁡(x + ^nπ⁄₂) – ¹⁄₁₆ 3ⁿ cos⁡(3x + ^nπ⁄₂)
c) ¹⁄₈ cos⁡(x+^nπ⁄₂) –¹⁄₁₆ 5ⁿ sin⁡(x + ^nπ⁄₂) – ¹⁄₁₆ 3ⁿ sin⁡(3x + ^nπ⁄₂)
d) ¹⁄₈ sin⁡(x + ^nπ⁄₂) –¹⁄₁₆ 5ⁿ sin⁡(x + ^nπ⁄₂) – ¹⁄₁₆ 3ⁿ sin⁡(3x + ^nπ⁄₂)
View Answer

Answer: a
Explanation: y = sin²x cos²x cos(x)
y = ¹⁄₄ sin²2x cos²x cos(x)
y = ¹⁄₈ (2sin²2x) cos(x)
y = ¹⁄₈ (1 – cos4x) cos(x)
y = ¹⁄₈ (1 – cos4x) cos(x)
y = ¹⁄₈ cos(x) – ¹⁄₈ cos4x cos(x)
y = ¹⁄₈ cos(x) – ¹⁄₁₆ (cos5x + cos(3x))
Now, nth derivative is
y_n = ¹⁄₈ cos⁡(x + ^nπ⁄₂) – ¹⁄₁₆ 5ⁿ cos⁡(x + ^nπ⁄₂) – ¹⁄₁₆ 3ⁿ cos⁡(3x + ^nπ⁄₂).

11. If In=e^nxTan(x), and \(\frac{I_{n+2}-2nI_{n+1}+n^2 I_n}{nI_{n-I}}\) = c (1+x²)\(\frac{d^2}{dx^2} (\frac{1}{1+x^2})\), Then value of ‘c’ equals to
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: a
Explanation: In=enxTan(x)
⇨I_n+1=ne^xnTan(x) + e^xn/(1+x2)
⇨I_n+1=nI_n + e^xn/(1+x2)
⇨I_n+2=nI_n+1 + \(\frac{ne^nx}{1+x^2} – \frac{2xe^nx}{(1+x^2)^2}\)
⇨I_n+2=nI_n+1 + n[I_n+1-nI_n] – \(\frac{2xe^nx}{(1+x^2)^2}\)
⇨\(\frac{I_{n+2}-2nI_{n+1}+n^2 I_n}{nI_{n-I}}=(1+x^2)\frac{d^2}{dx^2} (\frac{1}{1+x^2})\), Hence c=1.

12. Nth derivative of \(\frac{1}{(1+x^2)}\)?
a) (-1)ⁿn! r^-n-1 Sin(n+1)ϑ
b) (-1)ⁿ(n)! r^-n-1 Sin(-n-1)ϑ
c) (-1)ⁿ⁺¹(n+1)! r^-n-1 Sin(n+1)ϑ
d) (-1)ⁿ⁺¹(n+1)! r^-n-1 Sin(-n-1)ϑ
View Answer

Answer: a
Explanation: Here,
Y = \(\frac{1}{(x+i)(x-i)}\)
Y = \(\frac{1}{2i(x-i)} – \frac{1}{2i(x+i)}\)
\(y_n = \frac{-1^n n!}{2i}[(x-i)^{-n-1}-(x+i)^{-n-1}]\)
put x=rcos ϑ and y=rsinϑ, we get
\(y_n=\frac{(-1)^n n!}{2i}[(re^{-iθ})^{-n-1}-(re^{iθ})^{-n-1}]\) (By Eulers Identity)
\(y_n=\frac{(-1)^n n!r^{-n-1}}{2i}[e^{iθ(n+1)}-e^{-iθ(n+1)}]\)
\(y_n=(-1)^n n!r^{-n-1} Sin((n+1)ϑ)\)

13. Find nth derivative of y = Sin(x) Cos³(x)
a) (1/4) 2ⁿSin(2x+nπ/2) + (1/8) 4ⁿSin(4x+nπ/2)
b) (1/4) 2ⁿCos(2x+nπ/2) + (1/8) 4ⁿSin(4x+nπ/2)
c) (1/4) 2ⁿSin(2x+nπ/2) + (1/8) 4ⁿCos(4x+nπ/2)
d) (1/4) 2ⁿCos(2x+nπ/2) + (1/8) 4ⁿCos(4x+nπ/2)
View Answer

Answer: a
Explanation: y = Sin(x) Cos³(x)
y = (1/2)[2Sin(x) Cos(x)] Cos²(x)
y = (1/4)Sin(2x)(Cos(2x)+1)
y = (1/4)[Sin(2x)Cos(2x)+Sin(2x)] y = (1/8)Sin(4x) + (1/4)Sin(2x)
Hence nth derivative of y is
Y_n = (1/4) 2ⁿSin(2x+nπ/2) + (1/8) 4ⁿSin(4x+nπ/2)

14. If nth derivative of y = \(\frac{x}{(x+1)(x+2)}\) is y_n = a(-1)^n+! n! (x+1)^-n-1 + b(-1)ⁿn!(x+2)^-n-1 then find the value of a and b.
a) -1, -2
b) 2, 1
c) 1, 2
d) -2, -1
View Answer

Answer: c
Explanation: y = \(\frac{x}{(x+1)(x+2)}\)
y = \(\frac{(-1)}{(x+1)} + \frac{2}{(x+2)}\)
y = (-1)(-1)ⁿn!(x+1)^-n-1 + 2(-1)ⁿn!(x+2)^-n-1
y = (-1)ⁿ⁺¹n!(x+1)^-n-1 + 2(-1)ⁿn!(x+2)^-n-1
Hece, a=1 and b=2.

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