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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
$( S 1): A \cap B =(1, \infty)-N$ and
$( S 2): A \cup B=(1, \infty)$
only $(S1)$ is true
both $(S1)$ and $(S2)$ are true
neither $(S1)$ nor $(S2)$ is true
only $(S2)$ is true
Solution
$f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}$
If $x \in I \lceil x \rceil=[ x ]$ (greatest integer function)
If $x \notin I \lceil x \rceil=[ x ]+1$
$\Rightarrow f(x)=\left\{\begin{array}{l}\frac{1}{\sqrt{[x]-x}}, x \in I \frac{1}{\sqrt{[x]+1-x}}, x \notin I\end{array}\right.$
$\begin{aligned} & \Rightarrow f(x)=\left\{\begin{array}{l}\frac{1}{\sqrt{-\{x\}}}, x \in I, \text { (does not exist) } \\ \frac{1}{\sqrt{1-\{x\}}}, x \notin I\end{array}\right. \\ & \Rightarrow \text { domain of } f(x)=R-I\end{aligned}$
$\text { Now, } f(x)=\frac{1}{\sqrt{1-\{x\}}}, x \notin I$
$\Rightarrow 0 < \{x\} < 1$
$\Rightarrow 0 < \sqrt{1-\{x\}} < 1$
$\Rightarrow \frac{1}{\sqrt{1-\{x\}}} > 1$
$\Rightarrow \text { Range }(1, \infty)$
$\Rightarrow A=R-I$
$B=(1, \infty)$
$\text { So, } A \cap B=(1, \infty)-N$
$A \cup B \neq(1, \infty)$
$\Rightarrow S 1 \text { is only correct }$
