Differentiation Question 399
Question: Let $ f(x)= \begin{vmatrix} x^{3} & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3} \\ \end{vmatrix} $ , where p is a constant. Then $ \frac{d^{3}}{dx^{3}}{ f(x) } $ at $ x=0 $ is
[IIT 1997 Cancelled]
Options:
A) p
B) $ p+p^{2} $
C) $ p+p^{3} $
D) Independent of p
Show Answer
Answer:
Correct Answer: D
Solution:
$ {f}’’’(x)= \begin{vmatrix} \frac{d^{3}}{dx^{3}}x^{3} & \frac{d^{3}}{dx^{3}}\sin x & \frac{d^{3}}{dx^{3}}\cos x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3} \\ \end{vmatrix} = \begin{vmatrix} 6 & -\cos x & \sin x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3} \\ \end{vmatrix} $
$ \therefore {f}’’’(0)= \begin{vmatrix} 6 & -1 & 0 \\ 6 & -1 & 0 \\ p & p^{2} & p^{3} \\ \end{vmatrix} =0 $ , which is independent of p.
BETA
