Let f (x) = |x| and g (x) = |x³|, then
(a) f (x) and g (x) both are continuous at x = 0
(b) f (x) and g (x) both are differentiable at x = 0
(c) f (x) is differentiable but g (x) is not differentiable at x = 0
(d) f (x) and g (x) both are not differentiable at x = 0

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

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

We know $\left|x\right|$ is continuous at x=0 but not differentiable at x = 0 as (LHD at x = 0) ≠ (RHD at x = 0).
Now, for the function $g\left(x\right)=\left|x^3\right|=\left\{\begin{array}{l}{x}^3,x\ge 0\\ -x^3,x<0\end{array}\right.$
Continuity at x = 0:

(LHL at x = 0) = $\underset{x\to 0^{-}}{\lim }g\left(x\right)=\underset{h\to 0}{\lim }g\left(0-h\right)=\underset{h\to 0}{\lim }-\left(-h^3\right)=\underset{h\to 0}{\lim }{h}^3=0.$

(RHL at x = 0) = $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0+h\right)=\underset{h\to 0}{\lim }{h}^3=0.$

and $g\left(0\right)=0.$
Thus, $\underset{x\to 0^{-}}{\lim }g\left(x\right)=\underset{x\to 0^{+}}{\lim }g\left(x\right)=g\left(0\right)$.
Hence, $g\left(x\right)$ is continuous at x = 0.

Differentiability at x = 0:

(LHD at x = 0) = $\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{h\to 0}{\lim }\frac{f\left(0-h\right)-f\left(0\right)}{0-h-0}=\underset{h\to 0}{\lim }\frac{h^3-0}{-h}=0.$

(RHD at x = 0) = $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{0+h-0}=\underset{h\to 0}{\lim }\frac{h^3-0}{h}=\underset{h\to 0}{\lim }\frac{h^3}{h}=0$
Thus, (LHD at x = 0) = (RHD at x = 0).
Hence, the function $g\left(x\right)$ is differentiable at x = 0.