Let the sets A and B denote the domain and range respectively of the function [f(x)=1 x-x]…

Question

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

Accepted Answer

$$
f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}
$$

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 \mathrm{f}(\mathrm{x})=\left\{\[\begin{gathered} \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{gathered}\end{aligned}$$\right. \\
\& \Rightarrow \mathrm{f}(\mathrm{x})=\left\{$$\begin{gathered} \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{gathered}$$\right. \\
\& \Rightarrow \text { domain of } \mathrm{f}(\mathrm{x})=\mathrm{R}-\mathrm{I}

\]

$$
\begin{aligned}
& \text { Now, } \mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{1-\{\mathrm{x}\}}}, \mathrm{x} \notin \mathrm{I} \\
& \Rightarrow 01 \\
& \Rightarrow \text { Range }=(1, \infty) \\
& \Rightarrow \mathrm{A}=\mathrm{R}-\mathrm{I}
\end{aligned}
$$

$$
\begin{aligned}
& B=(1, \infty) \\
& \text { So, } A \cap B=(1, \infty)-N \\
& A \cup B \neq(1, \infty) \\
& \Rightarrow S 1 \text { is only correct }
\end{aligned}
$$

Solutions · 1