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(ex – e-x)
b) 0.5(e-x – ex)
c) 0.5(ex – (-1)n e-x)
d) 0.5((-1)-n e-x -ex)
Answer: c
Explanation: Y = Sinh(x)
Y = 0.5[ex – e-x].
y1 = 0.5 [ex – (-1)e-x].
y2 = 0.5 [ex – (-1)2 e-x].
Similarly,
yn = 0.5 [ex – (-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))
Answer: a
Explanation: Y=log(x) – log(x2 – 1)
y1 = x(-1)-2x/(x2-1)
y1 = x(-1)-(x-1)(-1) + (x+1)(-1)
yn = (-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)
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
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
a) 0, n!, mPn x(m-n)
b) mPn x(m-n), n!, 0
c) 0, n!, mCn x(m-n)
d) mCn x(m-n), n!, 0
Answer: a 6. Which of the following is true 7. If nth derivative of eax sin(bx+c) cos(bx+c) is, eax rn sin(bx+c+nα⁄2) cos(bx+c+nα⁄2) then, 8. If y=x4⁄x2-1, then?
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
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
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
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})
Answer: b
Explanation: y = eax sin(bx+c) cos(bx+c)
y = eax sin2(bx+c)/2
yn = eax rn sin(2(bx+c+nα/2))/2
yn = eax rn sin(bx+c+nα/2) cos(bx+c+nα/2)
where
r = (sqrt{a^2+4b^2}), α = tan-12b/a.
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]