Question

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

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

Answer 1

Answer: (C)

The function has smooth curve at all the points of multiples of $\pi $ except at $0$ .
Hence, the function is continuous and differentiable at all the points of multiples of $\pi $ except at $0$ .

At $x=0$ for $f(x)$ to be continuous
$\displaystyle\lim_{x\rightarrow 0} f(0)^- =\displaystyle\lim_{x\rightarrow 0} f(0)^+$
$f(x)=0$ at $x=0$
R. H. L $=\displaystyle\lim _{x \rightarrow 0^{+}} \sin |x|=\displaystyle\lim _{h \rightarrow 0} \sin |0+h|=\displaystyle\lim _{h \rightarrow 0} \sin h =0$
L.H. L $=\displaystyle\lim _{x \rightarrow 0^{-}} \sin |x|=\displaystyle\lim _{h \rightarrow 0} \sin |0-h|=\displaystyle\lim _{h \rightarrow 0} \sin h =0$
Hence, $f(x)$ is continuous at $x=0$.
Now, checking differentiability at $x=0 f(x)$ is said to be differentiable at $x=0$ if $R . H . D= L.H. D=$ finite i.e. $f^{\prime}\left(a^{+}\right)=f^{\prime}\left(a^{-}\right)=$finite
$f^{\prime}\left(0^{+}\right)=\displaystyle\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{\sin |h|-\sin 0}{h}=\displaystyle\lim _{h \rightarrow 0} \frac{\sinh }{h}=1$
$f^{\prime}\left(0^{-}\right)=\displaystyle\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\displaystyle\lim _{h \rightarrow 0} \frac{\sin |-h|-\sin 0}{-h}=\displaystyle\lim _{h \rightarrow 0} \frac{\sinh }{-h}=-1$
Clearly, R.H.D $ \neq$ L.H.D.
Hence, $f(x)$ is not differentiable at $x=0$.