Let the sets A and B denote the domain and range respectively of the function $f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}$, where $\lceil x\rceil$ denotes the smallest integer greater than or equal to $x$. Then among the statements

(S1) : $A \cap B=(1, \infty)-\mathbb{N}$ and

(S2) : $A \cup B=(1, \infty)$

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