Limit Continuity And Differentiability Ques 11
- 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
(1997, 2M)
(a) $p$
(b) $p+p^2$
(c) $p+p^3$
(d) independent of $p$
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Answer:
Correct Answer: 11.(d)
Solution: (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
$ \begin{aligned} f^{\prime}(x)=\left|\begin{array}{ccc} 3 x^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| \end{aligned} $
$\Rightarrow \quad f^{\prime}(x)=\left|\begin{array}{ccc}3 x^2 & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|$
$\Rightarrow \quad f^{\prime \prime}(x)=\left|\begin{array}{ccc}6 x & -\sin x & -\cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|+0+0$
and $f^{\prime \prime \prime}(x)=\left|\begin{array}{ccc}6 & -\cos x & \sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|+0+0$
$\therefore \quad f^{\prime \prime}(0)=\left|\begin{array}{ccc}6 & -1 & 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|=0=$ independent of $p$
BETA
