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Q. Let $[x]$ denote the greatest integer $\leq x$, where $x \in R$. If the domain of the real valued function $f(x)=\sqrt{\frac{|[x]|-2}{|[x]|-3}}$ is $(-\infty, a) \cup[b, c) \cup[4, \infty), a < b < c$, then the value of $a+b+c$ is:

JEE MainJEE Main 2021Relations and Functions - Part 2

Solution:

For domain,
$\frac{|[x]|-2}{|[x]|-3} \geq 0$
Case 1 : When $|[x]|-2 \geq 0$
and $|[x]|-3 > 0$
$\therefore x \in(-\infty,-3) \cup[4, \infty) \ldots \ldots . .(1)$
Case II : When $|[x]|-2 \geq 0$
and $|[x]|-3 < 0$
$\therefore x \in[-2,3) \ldots \ldots .(2)$
So, from (1) and (2)
we get
Domain of function
$=(-\infty,-3) \cup[-2,3) \cup[4, \infty)$
$\therefore (a+b+c)=-3+(-2)+3$
$=-2(a < b < c)$