Question

The domain of $f\left(x\right)=\sqrt{\frac{4-x^2}{\left[x\right]+2}}$ is
(where [.] represents the greatest integer function)

• A. $(-\infty ,1)$
• B. $(-\infty ,-2)\cup [-1,2]$
• C. $(-\infty ,-1)\cup [2,\infty )$
• D. $(-\infty ,1)\cup [2,\infty )$

Answer 1

The correct option is B $(-\infty ,-2)\cup [-1,2]$

Given : $f\left(x\right)=\sqrt{\frac{4-x^2}{\left[x\right]+2}}$
For $f\left(x\right)$ to be defined, we get
$\frac{4-x^2}{\left[x\right]+2}\ge 0\text{and}\left[x\right]+2\ne 0\Rightarrow \left[x\right]\ne -2\therefore x\notin [-2,-1)$

Case $1:4-x^2\ge 0,\left[x\right]+2>0$
$\Rightarrow x\in [-2,2],x\in [-1,\infty )\therefore x\in [-1,2]\cdots \left(1\right)$

Case $2:4-x^2<0,\left[x\right]+2<0$
$\Rightarrow x\in (-\infty ,-2)\cup (2,\infty ),x\in (-\infty ,-2)\therefore x\in (-\infty ,-2)\cdots \left(2\right)$

So, from $\left(1\right)$ and $\left(2\right)$, we get
$x\in (-\infty ,-2)\cup [-1,2]$