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Option (a) f (x) and g (x) both are continuous at x = 0

Given: $f(x)=|x|, g(x)=\left|x^{3}\right|$

We know $|x|$ is continuous at $\mathrm{x}=0$ but not differentiable at $\mathrm{x}=0$ as $(\mathrm{LHD}$ at $\mathrm{x}=0) \neq(\mathrm{RHD}$ at $\mathrm{x}=0)$.

Now, for the function $g(x)=\left|x^{3}\right|= \begin{cases}x^{3}, & x \geq 0 \\ -x^{3}, & x<0\end{cases}$

Continuity at $x=0$ :

$(\mathrm{LHL}$ at $\mathrm{x}=0)=\lim _{x \rightarrow 0^{-}} g(x)=\lim _{h \rightarrow 0} g(0-h)=\lim _{h \rightarrow 0}-\left(-h^{3}\right)=\lim _{h \rightarrow 0} h^{3}=0$

$(\mathrm{RHL}$ at $\mathrm{x}=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} h^{3}=0 .$

and $g(0)=0$

Thus, $\lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{+}} g(x)=g(0)$.

Hence, $g(x)$ is continuous at $\mathrm{x}=0$.

Differentiability at $x=0$ :

$(\mathrm{LHD}$ at $\mathrm{x}=0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0}=\lim _{h \rightarrow 0} \frac{h^{3}-0}{-h}=0$

(RHD at $\mathrm{x}=0$ ) $=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h-0}=\lim _{h \rightarrow 0} \frac{h^{3}-0}{h}=\lim _{h \rightarrow 0} \frac{h^{3}}{h}=0$

Thus, $(\mathrm{LHD}$ at $\mathrm{x}=0)=(\mathrm{RHD}$ at $\mathrm{x}=0)$.

Hence, the function $g(x)$ is differentiable at $\mathrm{x}=0$.