Question

Which of the following functions is not differentiable at $x=0 ?$
(i) $f(x)=\min \{x, \sin x\}$
(ii) $f(x)=\begin{cases} 0, & x \geq 0 \\ x^{2}, & x<0 \end{cases}$
(iii) $f(x)=x^{2} {\text{sgn}} (x)$

• A. Only (i)
• B. Only (ii)
• C. Only (iii)
• D. All of these

Answer 1

Answer: (D)

(i) Graph of $f(x)=\min \{x, \sin x\}$ is as follows :
From the graph, $f(x)=\begin{cases}x, & x<0 \\ \sin x, & x \geq 0\end{cases}$

$\therefore f'(x)=\begin{cases}1, & x<0 \\ \cos x, & x>0\end{cases}$
$f'\left(0^{+}\right)=f'\left(0^{-}\right)=1 .$
Hence, $f(x)$ is differentiable at $x=0 .$
(ii) $ f(x)=\begin{cases} 0, & x \geq 0 \\ x^{2}, & x<0 \end{cases}$
Here, $f(x)$ is continuous at $x=0$. Now,
$f'(x)=\begin{cases}0, & x>0 \\2 x, & x<0\end{cases}$
$f'\left(0^{+}\right)=0$ and $f'\left(0^{-}\right)=0$
Hence, $f(x)$ is differentiable at $x=0$.
(iii) $f(x)=x^{2} {\text{sgn}}(x)=\begin{cases}x^{2} ; & x \geq 0 \\ -x^{2} ; & x<0\end{cases}$, whichis
continuous as well as differentiable at $x=0$.