Let [x] denote greatest integer ≤ x , where x ∈ R . If domain of real valued function mathrm{f}(

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1.Relation and Function

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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

$8$

(JEE MAIN-2021)

Solution

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 $|[x]|-3\,<\,0$

$\therefore \mathrm{x} \in[-2,3) \quad \ldots(2)$

So, from $(1)$ and $(2)$

We get

Domain of function

$=(-\infty,-3) \cup[-2,3) \cup[4, \infty)$

$\therefore(a+b+c)=-3+(-2)+3=-2(a\,<\,b\,<\,c)$

Standard 12

Mathematics

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