Question

Find the domain $\frac{x-1}{\sqrt{x^2-1}}$

Answer 1

$y=\frac{x-1}{\sqrt{x^2-1}}$

For existance of $y$,

$\Rightarrow x^2-1>0$

$\Rightarrow (x-1)(x+1)>0$

$\Rightarrow x\in (-\infty ,-1)\cup (1,\infty )$

Range: $(-\infty ,-1)\cup (1,\infty )$