Let (x) denote the


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

  1. 8

  2. 1

  3. -2

  4. -3

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

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


$\Rightarrow$ Option (3) is correct.

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