Question

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

• A. $[-1,1]$
• B. $\varphi$
• C. $\{-1,1\}$
• D. $\left\{1\right\}$

Answer 1

Answer: (C)

$\{-1,1\}$

$f\left(x\right)={\sin }^{-1}\left(\frac{1+x^2}{2x}\right)={\sin }^{-1}\left[\frac{1}{2}(x+\frac{1}{x})\right]$

we know $|x+\frac{1}{x}|\ge 2$ and ${\sin }^{-1}x$ only exist if $-1\le x\le 1$

$\Rightarrow$ This in only possible if $x=-1\text{or}1$

Hence domain of $f\left(x\right)$ is $\{-1,1\}$