JEE MAIN - Mathematics (2023 - 6th April Evening Shift - No. 13)
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) : $$A \cup B=(1, \infty)$$
only $$(\mathrm{S} 2)$$ is true
only (S1) is true
neither (S1) nor (S2) is true
both (S1) and (S2) are true
Explanation
$$
f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}
$$
If $\mathrm{x} \in \mathrm{I},\lceil\mathrm{x}\rceil=[\mathrm{x}]$ (greatest integer function)
If $x \notin I,\lceil x\rceil=[x]+1$
$$ \begin{aligned} & \Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \frac{1}{\sqrt{[\mathrm{x}]-\mathrm{x}}}, \mathrm{x} \in \mathrm{I} \\\\ \frac{1}{\sqrt{[\mathrm{x}]+1-\mathrm{x}}}, \mathrm{x} \notin \mathrm{I} \end{array}\right. \\\\ & \Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \frac{1}{\sqrt{-\{\mathrm{x}\}}}, \mathrm{x} \in \mathrm{I}, \text { (does not exist) } \\\\ \frac{1}{\sqrt{1-\{\mathrm{x}\}}}, \mathrm{x} \notin \mathrm{I} \end{array}\right. \\\\ & \Rightarrow \text { domain of } \mathrm{f}(\mathrm{x})=\mathrm{R}-\mathrm{I} \end{aligned} $$
$$ \begin{aligned} & \text { Now, } \mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{1-\{\mathrm{x}\}}}, \mathrm{x} \notin \mathrm{I} \\\\ & \Rightarrow 0<\{\mathrm{x}\}<1 \\\\ & \Rightarrow 0<\sqrt{1-\{\mathrm{x}\}}<1 \\\\ & \Rightarrow \frac{1}{\sqrt{1-\{\mathrm{x}\}}}>1 \\\\ & \Rightarrow \text { Range }=(1, \infty) \\\\ & \Rightarrow \mathrm{A}=\mathrm{R}-\mathrm{I} \end{aligned} $$
$$ \begin{aligned} & B=(1, \infty) \\\\ & \text { So, } A \cap B=(1, \infty)-N \\\\ & A \cup B \neq(1, \infty) \\\\ & \Rightarrow S 1 \text { is only correct } \end{aligned} $$
If $\mathrm{x} \in \mathrm{I},\lceil\mathrm{x}\rceil=[\mathrm{x}]$ (greatest integer function)
If $x \notin I,\lceil x\rceil=[x]+1$
$$ \begin{aligned} & \Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \frac{1}{\sqrt{[\mathrm{x}]-\mathrm{x}}}, \mathrm{x} \in \mathrm{I} \\\\ \frac{1}{\sqrt{[\mathrm{x}]+1-\mathrm{x}}}, \mathrm{x} \notin \mathrm{I} \end{array}\right. \\\\ & \Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \frac{1}{\sqrt{-\{\mathrm{x}\}}}, \mathrm{x} \in \mathrm{I}, \text { (does not exist) } \\\\ \frac{1}{\sqrt{1-\{\mathrm{x}\}}}, \mathrm{x} \notin \mathrm{I} \end{array}\right. \\\\ & \Rightarrow \text { domain of } \mathrm{f}(\mathrm{x})=\mathrm{R}-\mathrm{I} \end{aligned} $$
$$ \begin{aligned} & \text { Now, } \mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{1-\{\mathrm{x}\}}}, \mathrm{x} \notin \mathrm{I} \\\\ & \Rightarrow 0<\{\mathrm{x}\}<1 \\\\ & \Rightarrow 0<\sqrt{1-\{\mathrm{x}\}}<1 \\\\ & \Rightarrow \frac{1}{\sqrt{1-\{\mathrm{x}\}}}>1 \\\\ & \Rightarrow \text { Range }=(1, \infty) \\\\ & \Rightarrow \mathrm{A}=\mathrm{R}-\mathrm{I} \end{aligned} $$
$$ \begin{aligned} & B=(1, \infty) \\\\ & \text { So, } A \cap B=(1, \infty)-N \\\\ & A \cup B \neq(1, \infty) \\\\ & \Rightarrow S 1 \text { is only correct } \end{aligned} $$
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