Question
The domain of $$f\left( x \right) = \sqrt {{{\log }_{{x^2} - 1}}\left( x \right)} $$ is :
A.
$$\left( {\sqrt 2 ,\, + \infty } \right)$$
B.
$$\left( {0,\, + \infty } \right)$$
C.
$$\left( {1,\, + \infty } \right)$$
D.
none of these
Answer :
$$\left( {\sqrt 2 ,\, + \infty } \right)$$
Solution :
$${x^2} - 1 > 0,\,{x^2} - 1 \ne 1,\,x > 0\,{\text{ and }}{\log _{{x^2} - 1}}\left( x \right) \geqslant 0$$
The first three imply $$x > 1{\text{ but }}x \ne \sqrt 2 $$
$$\eqalign{
& {\log _{{x^2} - 1}}\left( x \right) \geqslant 0 \Rightarrow x \geqslant 1{\text{ if }}{x^2} - 1 > 1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \leqslant 1{\text{ if }}{x^2} - 1 < 1 \cr} $$
Now, $${x^2} - 1 > 1$$ or $$x > \sqrt 2 \,\,\left( {\because x > 1} \right)$$ then $$x \geqslant 1$$ which is true for all $$x > \sqrt 2 $$
$${x^2} - 1 < 1$$ or $$1 < x < \sqrt 2 \,\,\left( {\because x > 1} \right)$$ then $$x \leqslant 1$$ which is not true for any $$x$$ in $$\left( {1,\,\sqrt 2 } \right)$$
This gives no value of $$x.$$ Thus $$x > \sqrt 2 $$