Question

Which of the following functions is not differentiable at $x=0?$
(i) $f(x)=\min \{x,\sin x\}$
(ii) $f(x)=\{\begin{array}{ll}0,& x\ge 0\\ x^2,& x<0\end{array}$
(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, $f(x)=\{\begin{array}{ll}x,& x<0\\ \sin x,& x\ge 0\end{array}$

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