Let f 1 :   R → R ,   f 2 - π 2 , π 2   → R ,   f 3 : - 1 ,  …

(i) $f_1\left(x\right)=\sin \sqrt{1-e^{-x^2}}$. Clearly $f_1\left(x\right)$ is continuous at $x=0$ and $f\left(0\right)=0$
$f_1^{\prime }\left(0\right)=\underset{x\to 0}{\lim }\frac{\sin \sqrt{1-e^{-x^2}}}{\sqrt{1-e^{-x^2}}}·\sqrt{\frac{1-e^{-x^2}}{x^2}}·\frac{\left|x\right|}{x}$

$=1·1·\underset{x\to 0}{\lim }\frac{\left|x\right|}{x}$, which does not exist. So it is not differentiable at $x=0$.
So, $P\to 2$
(ii) $f_2\left(x\right)=\left\{\begin{array}{cc}\frac{\left|\sin x\right|}{{\tan }^{-1}x},& x\ne 0\\ 0& x=0\end{array}\right.$
$\underset{x\to 0^{+}}{\lim }\frac{\sin x}{x}\frac{x}{{\tan }^{-1}x}=1$ and $\underset{x\to 0^{-}}{\lim }-\frac{\sin x}{x}\times \frac{x}{{\tan }^{-1}x}=-1$
$\Rightarrow$ $f_2\left(x\right)$ is not continuous at $x=0$
So $Q\to 1$
(iii) In the close neighborhood of $x=0$ given function $f_3\left(x\right)=0$ hence ${f^{\prime }}_3\left(x\right)=0$ also ${f^{\prime }}_3\left(x\right)$ is continuous at $x=0$
So, $R\to 4$
(iv) $f_4\left(x\right)=\left\{\begin{array}{cc}{x}^2\sin \frac{1}{x},& x\ne 0\\ 0,& x=0\end{array}\right.$

Although $f_4\text{'}\left(0\right)=0$ due to sandwich theorem, but $\underset{x\to 0}{\lim }{f}_4\text{'}\left(x\right)$ does not exist because $\cos \left(\frac{1}{x}\right)$ oscillates infinitely many times near $x=0$.
So $S\to 3$