Question

Let $f(x)=\left|\begin{array}{ccc}{x}^3& \sin x& \cos x\\ 6& -1& 0\\ p& p^2& p^3\end{array}\right|$, where $p$ is constant.
Then, $\frac{d^3}{d{x}^3}f(x)$ at $x=0$ is

• A. $p$
• B. $p+p^2$
• C. $p+p^3$
• D. independent of $p$

Answer 1

Answer: (D)

Given, $f(x)=\left|\begin{array}{ccc}{x}^3& \sin x& \cos x\\ 6& -1& 0\\ p& p^2& p^3\end{array}\right|$
On differentiating w.r.t. $x$, we get
$f^{\prime }(x)=\left|\begin{array}{ccc}3x^2& \cos x& -\sin x\\ 6& -1& 0\\ p& p^2& p^3\end{array}\right|+\left|\begin{array}{ccc}{x}^3& \sin x& \cos x\\ 0& 0& 0\\ p& p^2& p^3\end{array}\right|+\left|\begin{array}{ccc}{x}^3& \sin x& \cos x\\ 6& -1& 0\\ 0& 0& 0\end{array}\right|$
$\Rightarrow f{}^{\prime }(x)=\left|\begin{array}{ccc}{x}^3& \cos x& -\sin x\\ 6& -1& 0\\ p& p^2& p^3\end{array}\right|$
$\Rightarrow f^{\prime \prime }(x)=\left|\begin{array}{ccc}6x& -\sin x& -\cos x\\ 6& -1& 0\\ p& p^2& p^3\end{array}\right|+0+0$
and $f^{\prime \prime \prime }(x)=\left|\begin{array}{ccc}6& -\cos x& \sin x\\ 6& -1& 0\\ p& p^2& p^3\end{array}\right|+0+0$
$\therefore f{"}^{\prime }(0)=\left|\begin{array}{ccc}6& -1& 0\\ 6& -1& 0\\ p& p^2& p^3\end{array}\right|=0=$ independent of $p$