Question

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

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

Answer 1

The correct option is D None of these

$f\left(x\right)=\frac{{\sin }^{-1}x}{x}=\frac{g\left(x\right)}{t\left(x\right)},t\left(x\right)\ne 0$

$t\left(x\right)\ne 0$

$\Rightarrow x\ne 0$

Domain of $t\left(x\right)$ is $R-\left\{0\right\}$

Again domain of $g\left(x\right)$ is $[-1,1]$

Now using concept domain of $f\left(x\right)=\frac{g\left(x\right)}{t\left(x\right)}$

domain of $f\left(x\right)=$ Domain of $g\left(x\right)$ - set of those points for which $t\left(x\right)=0$ if $g\left(x\right),t\left(x\right)$ is defined $\forall xϵR$

$\therefore$ Domain of $f\left(x\right)=[-1,1]-\left\{0\right\}$