Question

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

• A. None of these
• B. $(0,1]$
• C. $(-1,1)-\left\{0\right\}$
• D. $[-1,1]-\left\{0\right\}$

Answer 1

The correct option is D $[-1,1]-\left\{0\right\}$

$f\left(x\right)=\frac{{\sin }^{-1}x}{x^2}$
For the function $f\left(x\right)$ to be defined $x\ne 0$
And domain of ${\sin }^{-1}x$ is $[-1,1]$
$\therefore$ Domain of $f\left(x\right)=[-1,1]-\left\{0\right\}$