Question

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

• A. 8
• B. 1
• C. \(-2\)
• D. \(-3\)

Hint:

Solve \(\frac{|[x]|-2}{|[x]|-3}\ge0\) for the integer \([x]\) and convert back to intervals of \(x\).

Answer 1

The Correct Option is C

Let \(t=[x]\), an integer. We need \(\frac{|t|-2}{|t|-3}\ge0\), i.e. \(|t|\le2\) or \(|t|\gt3\).
\(|t|\le2\): \(t\in\{-2,-1,0,1,2\}\), i.e. \(x\in[-2,3)\).
\(|t|\gt3\) means \(t\ge4\) or \(t\le-4\): \(x\ge4\) or \(x\lt-3\).
Domain \(=(-\infty,-3)\cup[-2,3)\cup[4,\infty)\), so \(a=-3,\ b=-2,\ c=3\).
\(a+b+c=-3-2+3=-2\).
Answer: (3) \(-2\)