Question

Let $[x]$ denote the greatest integer $\le x$, 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:

• A. 8
• B. 1
• C. -2
• D. -3

Answer 1

Answer: (C)

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