Verified Solution • 2025–26 Session moderate

QJEEM23Macb74: JEE Main 2023 Mathematics - Class 11 JEEM23Macb74 Solution: Let the sets A and B denote the domain and range respectively…

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

(S1) : A∩B=(1,∞)−NA \cap B=(1, \infty)-\mathbb{N} and

(S2) : A∪B=(1,∞)A \cup B=(1, \infty)

Multiple Choice Options

  • A only (S2)(\mathrm{S} 2) is true
  • B only (S1) is true
  • C neither (S1) nor (S2) is true
  • D both (S1) and (S2) are true

Step-by-Step Verified Solution

f(x)=1⌈x⌉−xf(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}

If x∈I,⌈x⌉=[x]\mathrm{x} \in \mathrm{I},\lceil\mathrm{x}\rceil=[\mathrm{x}] (greatest integer function)

If x∉I,⌈x⌉=[x]+1x \notin I,\lceil x\rceil=[x]+1

amp;⇒f(x)={1[x]−x,x∈I1[x]+1−x,x∉Iamp;⇒f(x)={1−{x},x∈I, (does not exist) 11−{x},x∉Iamp;⇒ domain of f(x)=R−I\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}

amp; Now, f(x)=11−{x},x∉Iamp;⇒0lt;{x}lt;1amp;⇒0lt;1−{x}lt;1amp;⇒11−{x}gt;1amp;⇒ Range =(1,∞)amp;⇒A=R−I\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}

amp;B=(1,∞)amp; So, A∩B=(1,∞)−Namp;A∪B≠(1,∞)amp;⇒S1 is only correct \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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