Question

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

• A. $(1,\infty )$
• B. $(\sqrt{2},\infty )$
• C. $(0,\infty )$
• D. $(0,\sqrt{2})$

Answer 1

The correct option is B $(\sqrt{2},\infty )$

For the function $f\left(x\right)=\sqrt{{\log }_{x^2-1}x}$ to be defined,
$\left(i\right)x>0\text{and}{x}^2-1>0\text{and}{x}^2-1\ne 1$
$\Rightarrow x>0\text{and}x\in (-\infty ,-1)\cup (1,\infty )\text{and}x\ne \pm \sqrt{2}$
$\Rightarrow x\in (1,\sqrt{2})\cup (\sqrt{2},\infty )\dots \left(1\right)$

$\left(ii\right){\log }_{(x^2-1)}x\ge 0$
Let $x\in (1,\sqrt{2})\Rightarrow x^2-1\in (0,1)$
So, ${\log }_{(x^2-1)}x\ge 0\Rightarrow x\le 1$
$\Rightarrow x\in \varphi \dots \left(2\right)$

Let $x\in (\sqrt{2},\infty )\Rightarrow x^2-1\in (1,\infty )$
So, ${\log }_{(x^2-1)}x\ge 0\Rightarrow x\ge 1$
$\therefore x\in (\sqrt{2},\infty )\dots \left(3\right)$
$\therefore$Domain of the given function is $x\in (\sqrt{2},\infty )$