$( S 1): A \cap B =(1, \infty)-N$ and
$( S 2): A \cup B=(1, \infty)$
If $x \in I \lceil x \rceil=[ x ]$ (greatest integer function)
If $x \notin I \lceil x \rceil=[ x ]+1$
$\Rightarrow f(x)=\left\{\begin{array}{l}\frac{1}{\sqrt{[x]-x}}, x \in I \frac{1}{\sqrt{[x]+1-x}}, x \notin I\end{array}\right.$
$\begin{aligned} & \Rightarrow f(x)=\left\{\begin{array}{l}\frac{1}{\sqrt{-\{x\}}}, x \in I, \text { (does not exist) } \\ \frac{1}{\sqrt{1-\{x\}}}, x \notin I\end{array}\right. \\ & \Rightarrow \text { domain of } f(x)=R-I\end{aligned}$
$\text { Now, } f(x)=\frac{1}{\sqrt{1-\{x\}}}, x \notin I$
$\Rightarrow 0 < \{x\} < 1$
$\Rightarrow 0 < \sqrt{1-\{x\}} < 1$
$\Rightarrow \frac{1}{\sqrt{1-\{x\}}} > 1$
$\Rightarrow \text { Range }(1, \infty)$
$\Rightarrow A=R-I$
$B=(1, \infty)$
$\text { So, } A \cap B=(1, \infty)-N$
$A \cup B \neq(1, \infty)$
$\Rightarrow S 1 \text { is only correct }$
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$(i)$ $f _1(x)=\sin \left(\sqrt{1- e ^{-x^2}}\right)$
$(ii)$ $f_2(x)=\left\{\begin{array}{ll}\frac{|\sin x|}{\tan ^{-1} x} & \text { if } x \neq 0 \\ 1 & \text { if } x=0\end{array}\right.$, where the inverse trigonometric function of $\tan ^{-1} x$
assume values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,
$(iii)$ $f_3(x)=\left[\sin \left(\log _c(x+2)\right)\right]$, where, for $t \in R ,[t]$ denotes the greatest integer less than or equal to $t$,
(iv) $f_4(x)=\left\{\begin{array}{ll}x^2 \sin \left(\frac{1}{x}\right) & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{array}\right.$
| $LIST I$ | $LIST II$ |
| $P$ The function $f _1$ is | $1$ $NOT$ continuous at $x=0$ |
| $Q$ The function $f _2$ is | $2$ continuous at $x =0$ and $NOT$ differentiable at $x =0$ |
| $R$ The function $f_3$ is | $3$ differentiable at $x=0$ and its derivative is $NO$T continuous at $x =0$ |
| $S$ The function $f _4$ is | $4$ differentiable at $x =0$ and its derivative is continuous at $x =0$ |
The correct option is: