Question

Domain of $f\left(x\right)=\frac{\sqrt{9-x^2}}{\sqrt{\left[x\right]+3}}$ is

• A. $(\infty ,2]$
• B. $(-\infty ,-2)\cup [-1,2]$
• C. $(-\infty ,-1)\cup [2,\infty )$
• D. $(-\infty ,-3)\cup [-2,3]$

Answer 1

The correct option is D $(-\infty ,-3)\cup [-2,3]$

$f\left(x\right)=\frac{\sqrt{9-x^2}}{\sqrt{\left[x\right]+3}}=\sqrt{\frac{9-x^2}{\left[x\right]+3}}$
$\Rightarrow \frac{9-x^2}{\left[x\right]+3}\ge 0,\left[x\right]+3\ne 0$
$\Rightarrow x\notin [-3,-2)$
Case : $1$
$9-x^2\ge 0,\left[x\right]+3>0$
$x\in [-3,3],x\in [-2,\infty )$
$\therefore x\in [-2,3]....\left(1\right)$
Case : $2$
$9-x^2\le 0,\left[x\right]+3<0$
$x\in (-\infty ,-3]\cup [3,\infty ),x\in (-\infty ,-3)$
$\Rightarrow x\in (-\infty ,-3)....\left(2\right)$
from $\left(1\right)$ and $\left(2\right)$
$x\in (-\infty ,-3)\cup [-2,3]$