Question

Find the domain of function ${\sin }^{-1}\left[\frac{1+x^2}{2x}\right].$

• A. $[-1,1]$
• B. $(-1,1)$
• C. $\{-1,1\}$
• D. $\left\{0\right\}$

Answer 1

The correct option is C $\{-1,1\}$

${\sin }^{-1}\left(\frac{1+x^2}{2x}\right)$
$-1\le \frac{1+x^2}{2x}\le 1$
$=-2x^2\le x(x^2+1)\le 2x^2$
$-2x^2\le x(x^2+1)$
$\Rightarrow x(x^2+1+2x)\ge 0$
$\Rightarrow x{(x+1)}^2\ge 0$
$whichistrueforx\ge 0$
$x=-1$
$x(x^2+1)\le 2x^2$
$\Rightarrow x(x^2+1-2x)\le 0$
$\Rightarrow x{(x-1)}^2\le 0$
$whichistrueforx\le 0,x=1$
$Hencedomain:\{-1,1\}$