Explanation

(a)
Here, $f(x) = \left\{ \begin{array}{r}
\sin{x,\ x > 0} \\
0,\ \ x = 0 \\
- \sin{x,\ \ \ x < 0}
\end{array} \right.\$
RHD$= \lim_{h \rightarrow 0}\ \frac{\sin{|0 + h| - \sin{(0)}}}{h}$
$= \lim_{h \rightarrow 0}\ \ \frac{\sin h}{h} = 1$
LHD$= \lim_{h \rightarrow 0}\frac{\sin{\left| (0 - h) \right| - \sin{(0)}}}{- h}$
$= \frac{- \sin h}{h} = - 1$
$\therefore$ LHD$\neq$RHD at $x = 0$
$\therefore\ \ f(x)$ is not derivable at $x = 0$
Alternate
[image:5_image2.emf]
It is clear from the graph that $f(x)$ is not differentiable at $x = 0$