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If $f(x)={{x}^{3}}sgn \ (x)$ then.
A. f is differentiable at $x=0$
B. f is continuous but not differentiable at $x=0$
C. $f'\left( {{o}^{-}} \right)=1$
D. none of these

Answer Verified Verified
Hint: The sgn function written in question is called sign function or signum function. It is defined as
$ sgn \left( x \right)=\left\{ \begin{matrix}
   -1 & \text{at} & x<0 \\
   0 & \text{at} & x=0 \\
   1 & \text{at} & x>0 \\
\end{matrix} \right.$

Complete step-by-step answer:
 we have a function
$f\left( x \right)={{x}^{3}}sgn \ (x)$
$f\left( x \right)$ can be defined similarly as signum function
\[\Rightarrow f\left( x \right)={{x}^{3}}sgn \left( x \right)=\left\{ \begin{matrix}
   -{{x}^{3}} & \text{at} & x<0 \\
   0 & \text{at} & x=0 \\
   {{x}^{3}} & \text{at} & x>0 \\
\end{matrix} \right.\]
Now the above function may be discontinuous at $x=0$
So let us check its continuity at $x=0$
$\underset{x\to 0}{\mathop{\lim }}\,\ \ f\left( x \right)=$ $\underset{\text{n}\to 0}{\mathop{\lim }}\,f\left( 0-\text{h} \right)=$ $\underset{\text{n}\to 0}{\mathop{\lim }}\,f\left( -\text{h} \right)$
$\Rightarrow \underset{\text{n}\to 0}{\mathop{\lim }}\,f\left( -\text{n} \right)=-{{\text{n}}^{3}}=0$
Similarly we can prove that
$\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0$
$\therefore \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\ \ f\left( x \right)=\ \ \ \ \ \ \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\ \ f\left( x \right)=\ \ \ \ \ \ f\left( 0 \right)=0$
$\therefore f\left( 0 \right)=0$
Therefore $f\left( x \right)$ is continuous at $x=0$
Now if we differentiate the above function we will get
$f'\left( x \right)=\left\{ \begin{matrix}
   -3{{x}^{2}} & \text{at} & x<0 \\
   0 & \text{at} & x=0 \\
   3{{x}^{2}} & \text{at} & x>0 \\
\end{matrix} \right.$
Now according to the option we need to check its differentiable at $x=0$. For the function to be differentiable at $x=0$ it should satisfy
From above relation we can say that
$f'\left( {{o}^{-}} \right)=f'\left( {{o}^{+}} \right)=o$
$\therefore f\left( x \right)$ is differentiable at $x=0$
Since the function is continuous at $x=0$ and it is also differentiable at $x=0$
$\therefore $ Option B is eliminated similarly $f'\left( {{o}^{-}} \right)=o$, therefore, option C is also eliminated.
$\therefore $ correct option is A

Note: Sign um function re $ sgn \left( x \right)$ is alone not continuous at $x=0$ but when multiplied by ${{x}^{3}}$ the function $f\left( x \right)={{x}^{3}}sgn \left( x \right)$ is continuous at $x=0$ therefore, be careful while checking the continuity of a function when sign um function is involved.
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