Question

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

• A. $(\sqrt{2},+\infty )$
• B. $(0,+\infty )$
• C. $(1,+\infty )$
• D. none of these

Answer 1

Answer: (A)

$(\sqrt{2},+\infty )$

${\log }_{x^2-1}\left(x\right)\ge 0$

Hence,
$x\ge 1\dots \left(1\right)$

Now, $x^2-1>1$

$x^2>2$

$x<-\sqrt{2}$ and $x>\sqrt{2}\dots \left(2\right)$

Hence from $\left(1\right)$ and $\left(2\right)$,

$x\in (\sqrt{2},\infty )$