Question

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

• A. $\left\{1\right\}$
• B. $(-1,1)$
• C. $\{1,-1\}$
• D. none of these

Answer 1

Answer: (C)

$\{1,-1\}$

Let $g\left(x\right)={\sin }^{-1}\left(\frac{1+x^2}{2x}\right)$
and $h\left(x\right)=\sqrt{1-x^2}$.

Hence $f\left(x\right)=g\left(x\right)+h\left(x\right)$ and the domain set of $f\left(x\right)$ is the intersection of the domain sets of

$g\left(x\right)$ and $h\left(x\right)$.

Now, the domain of $h\left(x\right)$ is $\left[-1,1\right]$.

Since, $1+x^2\ge 2x$ for all real values of $x$,

we have $\frac{1+x^2}{2x}\ge 1$ for $x>0$ and $\frac{1+x^2}{2x}\le -1$ for $x<0$,

Also the domain of $si{n}^{-1}\left(x\right)$ is $\left[-1,1\right]$, only possible values for which ${\sin }^{-1}\left(\frac{1+x^2}{2x}\right)$ is defined is $\left\{-1,1\right\}$.

That is the domain of $g\left(x\right)$ is $\left\{-1,1\right\}$.

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