Answer

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**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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