Question

The domain of the function $f\left(x\right)={\sin }^{-1}\frac{1}{|x^2-1|}+\frac{1}{\sqrt{{\sin }^{2}x+\sin x+1}}$ is

• A. $(-\infty ,\infty )$
• B. $(-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty )$
• C. $(-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty )\cup \left\{0\right\}$
• D. none of these

Answer 1

The correct option is C $(-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty )\cup \left\{0\right\}$

As, ${\sin }^{2}x+\sin x+1>0,\forall x\in R$
$\therefore \frac{1}{\sqrt{{\sin }^{2}x+\sin x+1}}$ is always exists.
$\therefore$ For ${\sin }^{-1}\left(\frac{1}{|x^2-1|}\right)$ to exists,
$0<\frac{1}{|x^2-1|}\le 1\Rightarrow |x^2-1|\ge 1$
$\Rightarrow x^2-1\le -1$ or $x^2-1\ge 1$
$\Rightarrow x^2\le 0$ or $x^2\ge 2$
$\Rightarrow x=0$ or $(x\le -\sqrt{2}$ or $x\ge \sqrt{2})$
$\therefore x\in (-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty )\cup \left\{0\right\}$
Hence, (c) is the correct answer.