$\text{ If }f\left(x\right)=x^2-2\left|x\right|\text{ and }$$g\left(x\right)=\left\{f\right(t):-2\le t\le x,-2\le x\le 0\}, f\left(t\right):0\le t\le x, 0\le x\le 3\}$. Discuss continuity of $f\left(x\right), g\left(x\right).$

 (i) Graph of $f\left(x\right)=\left\{\begin{array}{l}{x}^2-2x, x\ge 0\\ x^2+2x, x<0\end{array}\right.$ is shown as

Above graph shows $f\left(x\right)$ is continuous for all $x\in R$ and differentiable for all $x\in R-\left\{0\right\}$.

(ii) We know that,

If $f\left(x\right)$ is an increasing function on $[a,b]$, then:

$\max \left\{f\right(t);a\le t\le x,a\le x\le b\}=f\left(x\right)$

$\min \left\{f\right(t);a\le t\le x,a\le x\le b\}=f\left(a\right)$

If $f\left(x\right)$ is decreasing function on $[a,b]$ then

$\max \left\{f\right(t);a\le t\le x,a\le x\le b\}=f\left(a\right)$

$\min \left\{f\right(t);a\le t\le x,a\le x\le b\}=f\left(x\right)$ 

From graph of $f\left(x\right)$,

We get:

$g\left(x\right)$ is:

$\Rightarrow x^2+2x,$ for $-2\le x<-1$

$\Rightarrow -1,$for $-1\le x<0$

$\Rightarrow 0,$for $0\le x<1$

$\Rightarrow x^2-2x,$ for $x\ge 1.$

Thus, graph of $g\left(x\right)$ is:

From above figure, it is clear that $g\left(x\right)$ is not continuous or differentiable at $x=0,1$. 

Therefore, correct option is $\left(1\right).$