Determinants Matrices Question 62
Question: If $ f(x)= \begin{vmatrix} {x^{n-1}} & \cos x & \frac{1}{x+3} \\ 0 & \cos \frac{n\pi }{2} & \frac{{{(-1)}^{n}}n!}{{3^{n+1}}} \\ \alpha & {{\alpha }^{3}} & {{\alpha }^{5}} \\ \end{vmatrix} $ then $ \frac{d^{n}}{dx^{n}}{{[f(x)]} _{x=0}}= $
Options:
A) $ 1 $
B) $ -1 $
C) $ 0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] We have, $ \frac{d^{n}}{dx^{n}}[f(x)]= \begin{vmatrix} 0 & \cos ( x+\frac{n\pi }{2} ) & \frac{{{(-1)}^{n}}n!}{{{(x+3)}^{n+1}}} \\ 0 & \cos \frac{n\pi }{2} & \frac{{{(-1)}^{n}}n!}{{3^{n+1}}} \\ \alpha & {{\alpha }^{3}} & {{\alpha }^{5}} \\ \end{vmatrix} $
$ \therefore \frac{d^{n}}{dx^{n}}{{[f(x)]} _{x=0}}= \begin{vmatrix} 0 & \cos \frac{n\pi }{2} & \frac{{{(-1)}^{n}}n!}{{3^{n+1}}} \\ 0 & \cos \frac{n\pi }{2} & \frac{{{(-1)}^{n}}n!}{{3^{n+1}}} \\ \alpha & {{\alpha }^{3}} & {{\alpha }^{5}} \\ \end{vmatrix}=0 $
$ (\because R_1 \text{ and } R_2 \text{ are identical}) $
BETA
