Let the sets A and B denote the domain and range respectively of the function <img alt="\mathrm{f(x)=\frac{1}{\sqrt{[x]-x}}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%

Let the sets A and B denote the domain and range respectively of the function $\mathrm{f(x)=\frac{1}{\sqrt{[x]-x}}}$ , where $\mathrm{[x]}$ denotes the smallest integer greater than or equal to $\mathrm{\mathrm{x}}$ . Then among the statements :
$(\mathrm{S} 1): \mathrm{A} \cap \mathrm{B}=(1, \infty)-\mathrm{N}\: \: and$

$(\mathrm{S} 2): A \cup B=(1, \infty)$

Option: 1
only (S1) is true

Option: 2
neither (S1) nor (S2) is true

Option: 3
only (S2) is true

Option: 4
both (S1) and (S2) are true

Answers (1)

$\mathrm{f(x)=\frac{1}{\sqrt{[x]-x}}}$

If $\mathrm{x} \in \mathrm{I}[\mathrm{x}]=[\mathrm{x}]$ (greatest integer function)
If $\mathrm{x} \notin \mathrm{I}[\mathrm{x}]=[\mathrm{x}]+1$

$\begin{aligned} & \Rightarrow f(x)=\left\{\begin{array}{l} \frac{1}{\sqrt{[x]-x}}, x \in I \\ \frac{1}{\sqrt{[x]+1-x}}, x \notin I \end{array}\right. \end{}$
$\begin{aligned} & \Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \frac{1}{\sqrt{-\{x\}}}, \mathrm{x} \in \mathrm{I}, \text { (doesnot exist) } \\ \frac{1}{\sqrt{1-\{x\}}}, x \notin \mathrm{I} \end{array}\right. \end{}$
$\Rightarrow \text { domain of } \mathrm{f}(\mathrm{x})=\mathrm{R}-\mathrm{I} \\$
$\text { Now, } \mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{1-\{\mathrm{x}\}}}, \mathrm{x} \notin \mathrm{I} \\$
$\Rightarrow \mathrm{x}<\{\mathrm{x}\}<1 \\$
$\Rightarrow 0<1 \sqrt{1-\{\mathrm{x}\}}<1 \\$
$\Rightarrow \frac{1}{\sqrt{1-\{\mathrm{x}\}}}>1 \\$
$\Rightarrow \text { Range }(1, \infty) \\$
$\Rightarrow \mathrm{A}=\mathrm{R}-\mathrm{I} \\$
$\mathrm{ B=(1 \infty) }$
$\mathrm{ \text { So, } A \cap B=(1, \infty)-N }$
$\mathrm{ A \cup B \neq(1, \infty) }$