Question

$ f(x)=\sin |x| $ . Then $ f(x) $ is not differentiable at

• A. $ x=0 $ only
• B. all $ x $
• C. multiples of $ \pi $
• D. multiples of $ \frac{\pi }{2} $

Answer 1

Answer: (A)

Any function $f(x)$ is derivable at $x=a$, if $\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h}=\lim _{h \rightarrow 0^{-}} \frac{f(a-h)-f(a)}{-h}$ Given,
$
\begin{array}{l}
f(x)=\sin |x| \Rightarrow f(x)=\left\{\begin{array}{cl}
\sin x & x > 0 \\
0 & x=0 RHD =\lim _{h \rightarrow 0} \frac{\sin |(0+h)|-\sin (0)}{-} \\
-\sin x & x < 0
\end{array}\right. \\
R H D=\lim _{h \rightarrow 0} \frac{\sin ((0+h) \mid-\sin (0)}{h} \\
=\lim _{h \rightarrow 0} \frac{\sinh }{h}=1 \\
L H D=\lim _{h \rightarrow 0} \frac{\sin |(0-h)|-\sin (0)}{-h} \\
=\frac{-\sin h}{h}=-1 \\
\therefore L H D \neq R H D \text { at } x=0 \\
\therefore f(x) \text { is not derivable at } x=0 .
\end{array}
$