Question

Find the domain of the function $si{n}^{-1}\left(\frac{1+x^2}{2x}\right)$

Answer 1

$si{n}^{-1}\left(\frac{1+x^2}{2x}\right)$
$-1\le \left(\frac{1+x^2}{2x}\right)\le 1$
$\Rightarrow \left|\frac{1+x^2}{2x}\right|\le 1$
$\frac{1+x^2}{2x}\le \pm 1$
$1+x^2\le \pm 2x$
$1+x^2\le 2x$ $1+x^2\le 2x$
$1+x^2-2x\le 0$ $1+2x+x^2\le 0$
$(1-x{)}^2\le 0$ $(1+x{)}^2\le 0$
$1-x\le 0$ $1+x\le 0$
$1\le x$ $x\le -1$
$\Rightarrow xϵ[1,\infty )$ $\Rightarrow xϵ(-\infty ,-1]$
$\therefore Domain=(-\infty ,-1]\cup [1,\infty )$