Let $[\mathrm{x}]$ denote the greatest integer $\leq \mathrm{x}$, where $x \in \mathbf{R}$. If the domain of the real valued function $f(x)=\sqrt{\frac{\mid x] \mid-2}{|[x]|-3}}$
is $(-\infty, a) \cup[b, c) \cup[4, \infty), a
Correct Option: , 3
For domain,
$\frac{|[x]|-2}{|[x]|-3} \geq 0$
Case I : When $|[x]|-2 \geq 0$
and $|[x]|-3>0$
$\therefore x \in(-\infty,-3) \cup[4, \infty)$.....(1)
Case II : When $|[\mathrm{x}]|-2 \leq 0$
and $|[x]|-3<0$
$\therefore x \in[-2,3)$..(2).
So, from (1) and (2)
we get
Domain of function
$=(-\infty,-3) \cup[-2,3) \cup[4, \infty)$
$\therefore(\mathrm{a}+\mathrm{b}+\mathrm{c})=-3+(-2)+3=-2(\mathrm{a}<\mathrm{b}<\mathrm{c})$
$\Rightarrow$ Option (3) is correct.
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