Sol. $f(x)=\frac{1}{\sqrt{[x]-x}}$ If $\mathrm{x} \in \mathrm{I}\lceil\mathrm{x}\rceil=[\mathrm{x}]$ (greatest integer function) If $x \notin I\lceil x\rceil=[x]+1$ $$ \begin{aligned} & \Rightarrow f(x)=\left\{\begin{array}{c} \frac{4}{\sqrt{x\rceil-x}}, x \in I \\ \frac{4}{\sqrt{x\rceil+1-x}}, x \notin I \end{array}\right. \\ & \Rightarrow f(x)=\left\{\begin{array}{c} \frac{1}{\sqrt{-\{x\}}}, x \in I, \text { (does not exist) } \\ \frac{1}{\sqrt{1-\{x\}}}, x \notin I \end{array}\right. \\ & \Rightarrow \text { domain of } f(x)=R-I \\ & \text { Now, } f(x)=\frac{1}{\sqrt{1-\{x\}}}, x \notin I \\ & \Rightarrow 0<\{x\}<1 \end{aligned} $$ $$ \begin{aligned} & \Rightarrow 0<\sqrt{1-\{x\}}<1 \\ & \Rightarrow \text { Range }(1, \infty) \\ & \Rightarrow A=R-1 \\ & B=(1, \infty) \end{aligned} $$ So, $A \cap B=(1, \infty)-N$ $A \cup B(1, \infty)$ $\Rightarrow \mathrm{S} 1$ is only correct