Question

$f\left(x\right)=sin|x|$ . $f\left(x\right)$ 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
$\underset{x\to 0}{\lim }f(0{)}^{-}=\underset{x\to 0}{\lim }f(0{)}^{+}$
$f(x)=0$ at $x=0$
R. H. L $=\underset{x\to 0^{+}}{\lim }\sin |x|=\underset{h\to 0}{\lim }\sin |0+h|=\underset{h\to 0}{\lim }\sin h=0$
L.H. L $=\underset{x\to 0^{-}}{\lim }\sin |x|=\underset{h\to 0}{\lim }\sin |0-h|=\underset{h\to 0}{\lim }\sin h=0$
Hence, $f(x)$ is continuous at $x=0$.
Now, checking differentiability at $x=0f(x)$ 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
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{\lim }\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\lim }\frac{\sin |h|-\sin 0}{h}=\underset{h\to 0}{\lim }\frac{\sinh }{h}=1$
$f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{\lim }\frac{f(0-h)-f(0)}{-h}=\underset{h\to 0}{\lim }\frac{\sin |-h|-\sin 0}{-h}=\underset{h\to 0}{\lim }\frac{\sinh }{-h}=-1$
Clearly, R.H.D $\ne$ L.H.D.
Hence, $f(x)$ is not differentiable at $x=0$.