Let \([x]\) denote the greatest integer \(\le x\), where \(x\in R\). If the domain of the real valued function \(f(x)=\sqrt{\dfrac{|[x]|-2}{|[x]|-3}}\) is \((-\infty,a]\cup[b,c]\cup[4,\infty)\) where \(a<b<c\), then the value of \(a+b+c\) is:
-2
For \(f(x)\) to be real, we need \(\dfrac{|[x]|-2}{|[x]|-3}\ge 0\) with \(|[x]|\neq 3\). Let \(t=|[x]|\ge 0\).
\(\dfrac{t-2}{t-3}\ge 0\) holds when both factors have the same sign (or numerator is zero): this gives \(t\le 2\) or \(t>3\).
Case \(t=|[x]|\le 2\): this means \([x]\in\{-2,-1,0,1,2\}\), i.e. \(x\in[-2,3)\).
Case \(t=|[x]|>3\): since \([x]\) is an integer, this means \([x]\ge 4\) or \([x]\le -4\), i.e. \(x\ge 4\) or \(x<-3\).
Combining, the domain is \((-\infty,-3)\cup[-2,3)\cup[4,\infty)\), which matches the given form \((-\infty,a]\cup[b,c]\cup[4,\infty)\) with boundary values \(a=-3\), \(b=-2\), \(c=3\).
Therefore \(a+b+c=-3+(-2)+3=-2\).
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