Question

Let $f(x) = \left| \begin{matrix} x^{3}\sin x\cos x \\ 6 - 10 \\ 1p^{2}p^{3} \end{matrix} \right|$ where p is a constant. Then

$\frac{d^{3}}{dx^{3}}\lbrack f(x)\rbrack$ at $x = 0$ is

• A. p
• B. p + p²
• C. p + p³
• D. Independent of p

Answer 1

Answer: (D)

Independent of p

Answer 2

Given $f(x) = \left| \begin{matrix} x^{3}\sin x\cos x \\ 6 - 10 \\ 1p^{2}p^{3} \end{matrix} \right|$, 2^nd and 3^rd rows are constant, so only 1^st row will take part in differentiation

$\therefore$ $\frac{d^{3}}{dx^{3}}f(x) = \left| \begin{matrix} \frac{d^{3}}{dx^{3}}x^{3}\frac{d^{3}}{dx^{3}}\sin x\frac{d^{3}}{dx^{3}}\cos x \\ \begin{aligned} & \\ & 6 - 10 \end{aligned} \\ \begin{aligned} & \\ & 1p^{2}p^{3} \end{aligned} \end{matrix} \right|$

We know that $\frac{d^{n}}{dx^{n}}x^{n} = n!,\frac{d^{n}}{dx^{n}}\sin x = \sin(x + \frac{n\pi}{2})$

and $\frac{d^{n}}{dx^{n}}\cos x = \cos(x + \frac{n\pi}{2})$

Using these results,

$\frac { d ^ { 3 } } { d x ^ { 3 } } f ( x ) = \left| \begin{array} { c c c } 3 ! & \sin \left( x + \frac { 3 \pi } { 2 } \right) & \cos \left( x + \frac { 3 \pi } { 2 } \right) \\ 6 & - 1 & 0 \\ 1 & p ^ { 2 } & p ^ { 3 } \end{array} \right|$ $\left. \ \frac{d^{3}}{dx^{3}}f(x) \right|_{\text{at }x = 0} = \left| \begin{matrix} 6 - 10 \\ 6 - 10 \\ 1p^{2}p^{3} \end{matrix} \right|$ = 0 i.e., independent of p.