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If $\mathrm{f(x)=x^3 \operatorname{sgn} x}$ , then
Option: 1
f is derivable but not continuous at x=0.

Option: 2
f is continuous but not derivable at x=0.

Option: 3
LHD at x=0 is 1.

Option: 4
RHD at x=0 is 0.

Answers (1)

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$\mathrm{\text { We have, } f(x)=\left\{\begin{aligned} x^3, & x>0 \\ 0, & x=0 \\ -x^3, & x<0 \end{aligned}\right.}$

Clearly f(x) is continuous at x=0.
$\mathrm{( LHD ~at ~x=0)=\left[\frac{d}{d x}\left(-x^3\right)\right]_{x=0}=\left[-3 x^2\right]_{x=0}=0}$
Similarly, ( RHD at x=0)=0.
So, f(x) is differentiable at x=0.

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