Consider the following functions which are defined to be 0 at x=0,  $f_1\left(x\right)=x^2$ sgn(x),  $f_2\left(x\right)=x^{-1/3}\left|\sin x\right|$  and  $f_3\left(x\right)=x^3[-x]$ , where for  $t\in R,\left[t\right]$ denotes the greatest integer less than or equal to t. $f_4\left(x\right)={\int }_x^{2x}\frac{dt}{t},$  then match the elements of List I with the elements of List II.

	LIST-I		LIST-II
I	The function  $f_1\left(x\right)$ is	P	Discontinuous at x=0
II	The function $f_2\left(x\right)$  is	Q	Continuous but not differentiable at x=0
III	The function $f_3\left(x\right)$  is	R	First derivative exists at x=0 but second derivative does not exist
IV	The function $f_4\left(x\right)$   is	S	Second derivative exist at x=0

The correct option is 

We have, 
 $$\begin{gathered} f_1\left(x\right)=x^2\mathrm{sgn}\left(x\right)=\{\begin{array}{l}{x}^2,x\ge 0\\ -x^2,x<0\end{array} \\ \Rightarrow f_1\text{'}\left(x\right)=\{\begin{array}{l}2x,x>0\\ -2x,x<0\end{array} \\ \Rightarrow f_1\text{'}\left(0^{+}\right)=0=f{\text{'}}_1\left(0^{-}\right) \\ \Rightarrow f_1\text{'}\left(0\right)=0 \end{gathered}$$
NOW, $f_1\text{'}\text{'}\left(x\right)=\{\begin{array}{l}2,x>0\\ -2,x<0\end{array}$
 $\Rightarrow f_1\text{'}\text{'}\left(0^{+}\right)=2$
AND  $f_1\text{'}\text{'}\left(0^{-}\right)=-2$
So, first derivative exists at x=0 but second derivative does not exist there.
II. We have  $f_2\left(x\right)=\{\begin{array}{l}{x}^{-1/3}\left|\sin x\right|,x\ne 0\\ 0,x=0\end{array}$
Clearly  $\underset{x\to 0^{-}}{\lim }{f}_2\left(x\right)=\underset{x\to \infty }{\lim }{x}^{-1/3}(-\sin x)$
 $=-\underset{x\to 0^{-}}{\lim }\frac{\sin x}{x}.\underset{x\to 0^{-}}{\lim }{x}^{2/3}=0$
Similarly,  $=\underset{x\to 0^{+}}{\lim }{f}_2\left(x\right)=0$
$\therefore f_2\left(x\right)$  is continuous at x=0.
 $f{\text{'}}_2\left(0^{+}\right)=\underset{x\to 0^{+}}{\lim }\frac{x^{-1/3}\sin x}{x}$
Which does not exist.
So, $f_2\left(x\right)$  is continuous at x=0 but differentiable there.
III. we have,  $f_3\left(x\right)=x^3[-x]=\{\begin{array}{l}0,-1<x\le 0\\ -x^3,0<x<1\end{array}$