Question

The domain of $f\left(x\right)=\frac{1}{\sqrt{9-x^2}}+\sqrt{x^2-4}$ is

• A. $(-4,-2)\cup (2,4)$
• B. $(-3,-2]\cup [2,3)$
• C. $(-\infty ,-3)\cup (2,\infty )$
• D. $(-\infty ,\infty )$

Answer 1

Answer: (B)

$(-3,-2]\cup [2,3)$

$f$ is defined if both terms under square root exist.

$\Rightarrow 9-x^2>0$ and $x^2-4\ge 0$

$\Rightarrow (3-x)(3+x)>0$ and $(x-2)(x+2)\ge 0$

$\Rightarrow (x-3)(3+x)<0$ and $(x-2)(x+2)\ge 0$

$\Rightarrow -3<x<3$ and $x<-2$ and $x>2$

$x\in (-3,3)$ and $x\in [-\infty ,-2]\cup [2,\infty ]$

Intersection of both gives the following solution

$\Rightarrow x\in (-3,-2]\cup [2,3)$