Question

The domain of the function $f(x)=\frac{1}{\sqrt{9-x^2}}$ is

• A. $-3\le x\le 3$
• B. $-3<x<3$
• C. $-9\le x\le 9$
• D. $-9<x<9$
• E. $-\infty <x<\infty$

Answer 1

Answer: (B)

$f(x)=\frac{1}{\sqrt{9-x^2}}$
For existence of $f(x),$ $9-x^2>0$ $\Rightarrow$ $x^2-9<0$ $\Rightarrow$ $(x-3)(x+3)<0$
$\Rightarrow$ $x\in (-3,3)$ or $-3<x<3$ Which is the required domain of $f(x)$ .