JEE Mains · Maths · STD 12 - 1. relation and function
Let \([x]\) denote the greatest integer \(\leq x\), where \(x \in R\). If the domain of the real valued function \(\mathrm{f}(\mathrm{x})=\sqrt{\frac{[\mathrm{x}] \mid-2}{\sqrt{[\mathrm{x}] \mid-3}}}\) is \((-\infty, \mathrm{a}) \cup[\mathrm{b}, \mathrm{c}) \cup[4, \infty), \mathrm{a}\,<\,\mathrm{b}\,<\,\mathrm{c}\), then the value of \(\mathrm{a}+\mathrm{b}+\mathrm{c}\) is:
- A \(-3\)
- B \(1\)
- C \(-2\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(-2\)
Step-by-step Solution
Detailed explanation
For domain, \(\frac{|[x]|-2}{|[x]|-3} \geq 0\) Case \(I:\) When \(|[x]|-2 \geq 0\) and \(|[x]|-3\,>\,0\) \(\therefore x \in(-\infty,-3) \cup[4, \infty] \ldots . .(1)\) Case \(II:\) When \(|[x]|-2 \leq 0\) and \(|[x]|-3\,<\,0\) \(\therefore \mathrm{x} \in[-2,3) \quad \ldots(2)\)…
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