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

• A

  $(-\infty,-2) \cup(1,2)$
• B

  $(-\infty,-2) \cup(-1,2)$
• Correct answer
  C

  $(-\infty,-2), \cup[-1,2]$
• D

  $(-\infty,-1) \cup(1,2)$

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), $$ x \in(-\infty,-2) \cup[-1,2] $$

Hence, the correct answer is option: 'C'.