If $f\left(x\right)=\left\{\begin{array}{l} \\ \end{array}\right.\begin{array}{ll}{x}^2(\mathrm{sgn}[x]+\{x\left\}\right),& 0\le x<2\\ \sin x+|x-3|,& 2\le x<4\end{array}$, where [] and {} represent the greatest integer and the fractional part function, respectivley.

For continuity at x = 1

$\Rightarrow \underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(x^2\mathrm{sgn}[x]+\{x\left\}\right)=1+0=1$

$\Rightarrow \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(x^2\mathrm{sgn}[x]+\{x\left\}\right)=1\mathrm{sgn}\left(0\right)+1=1$

Also, f(1) = 1 $\therefore$L.H.L = R.H.L = f(1). 

Now for differentiability, $f^{\mathrm{\prime }}\left(1^{+}\right)=\underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{h}$

$=\underset{h\to 0}{lim}\frac{(1+h{)}^2\mathrm{sgn}[1+h]+\{1+h\}-1}{h}$

$=\underset{h\to 0}{lim}\frac{(1+h{)}^2+h-1}{h}=\underset{h\to 0}{lim}\frac{h^2+3h}{h}\text{and}$

$f^{\mathrm{\prime }}\left(1^{-}\right)=\underset{h\to 0}{lim}\frac{(1-h{)}^2\mathrm{sgn}[1-h]+\{1-h\}-1}{-h}$

$=\underset{h\to 0}{lim}\frac{(1-h{)}^2+1-h-1}{-h}=\underset{h\to 0}{lim}\frac{h^2-3h}{-h}=3$

$f^{\mathrm{\prime }}\left(1^{+}\right)=f^{\mathrm{\prime }}\left(1^{-}\right)$

Hence, $f\left(x\right)$ is differentiable at $x =1$. Now at $x =2$,

$\Rightarrow \underset{x\to 2}{lim}f\left(x\right)=\underset{x\to 2}{lim}(x^2\mathrm{sgn}[x]+\{x\left\}\right)=4\times 1+1=5$

$\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}(\sin x+|x-3\left|\right)=\sin 2+1$

Hence $\mathrm{LHL}\ne \mathrm{RHL}$

Hence, $f\left(x\right)$ is discontinuous at $x =2$ and then $f\left(x\right)$is also non differentiable at $x = 2$.