Limit Continuity and Differentiability 7 Question 19
19. The left hand derivative of $f(x)=[x] \sin (\pi x)$ at $x=k, k$ is an integer, is
(a) $(-1)^{k}(k-1) \pi$
(b) $(-1)^{k-1}(k-1) \pi$
(c) $(-1)^{k} k \pi$
(d) $(-1)^{k-1} k \pi$
(2001, 2M)
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Answer:
Correct Answer: 19. (b)
Solution:
- 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
$$ f^{\prime}(x)=\left|\begin{array}{ccc} 3 x^{2} & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3} \end{array}\right|+\left|\begin{array}{ccc} x^{3} & \sin x & \cos x \\ 0 & 0 & 0 \\ p & p^{2} & p^{3} \end{array}\right| $$
$$ +\left|\begin{array}{ccc} x^{3} & \sin x & \cos x \\ 6 & -1 & 0 \\ 0 & 0 & 0 \end{array}\right| $$
$$ \begin{aligned} \Rightarrow & f^{\prime}(x) & =\left|\begin{array}{ccc} 3 x^{2} & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3} \end{array}\right| \\ \Rightarrow & f^{\prime \prime}(x) & =\left|\begin{array}{ccc} 6 x & -\sin x & -\cos x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3} \end{array}\right|+0+0 \end{aligned} $$
$$ \begin{aligned} & \text { and } f^{\prime \prime \prime}(x)=\left|\begin{array}{ccc} 6 & -\cos x & \sin x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3} \end{array}\right|+0+0 \\ & \therefore \quad f^{\prime \prime \prime}(0)=\left|\begin{array}{ccc} 6 & -1 & 0 \\ 6 & -1 & 0 \\ p & p^{2} & p^{3} \end{array}\right|=0=\text { independent of } p \end{aligned} $$
