The domain of the function $f(x)=\sqrt{\frac{4-x^2}{[x]+2}}$, where $[x]$ denotes the greatest integer not…
The domain of the function $f(x)=\sqrt{\frac{4-x^2}{[x]+2}}$, where $[x]$ denotes the greatest integer not more than $x$, is
$(-\infty,-2) \cup(1,2)$
$(-\infty,-2) \cup(-1,2)$
$(-\infty,-2), \cup[-1,2]$
$(-\infty,-1) \cup(1,2)$
Solution
Given function $f(x)=\sqrt{\frac{4-x^2}{[x]+2}}$, is define if
$
\frac{4-x^2}{[x]+2} \geq 0 \Rightarrow \frac{x^2-4}{[x]+2} \leq 0
$
So, either $x^2-4 \leq 0$
From intervals Eqs. (iii) and (iv),
$
x \in(-\infty,-2) \cup[-1,2]
$
