Question

The domain of $f\left(x\right)={\sin }^{-1}\left(\frac{1}{x^2+3}\right)$ is

• A. $[-\frac{\pi }{2},\frac{\pi }{2}]$
• B. $(0,\frac{1}{3}]$
• C. $R$
• D. $[-1,1]$

Answer 1

The correct option is C $R$

Given, $f\left(x\right)={\sin }^{-1}\left(\frac{1}{x^2+3}\right)$
We know that, $0\le x^2<\infty$
$\Rightarrow 3\le x^2+3<\infty$
$\Rightarrow 0<\frac{1}{x^2+3}\le \frac{1}{3}$

$\Rightarrow (0,\frac{1}{3}]$ is the subset of $[-1,1]$

$\therefore$ Domain $=R$