Question

The domain of the function $f(x)={\sin }^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is $(-\infty ,-a]\cup [a,\infty ).$ Then a is equal to :

• A. $\frac{1+\sqrt{17}}{2}$
• B. $\frac{\sqrt{17}-1}{2}$
• C. $\frac{\sqrt{17}}{2}+1$
• D. $\frac{\sqrt{17}}{2}$

Answer 1

Answer: (A)

$f(x)=\sin \left(\frac{|x|+5}{x^2+1}\right)$
For domain :
$-1\le \frac{|x|+5}{x^2+1}\le 1$
Since $|x|+5\&x^2+1$ is always positive
So $\frac{|x|+5}{x^2+1}\ge 0\forall x\in R$
So for domain:
$\frac{|x|+5}{x^2+1}\le 1$
$\Rightarrow |x|+5\le x^2+1$
$\Rightarrow 0\le x^2-|x|-4$
$\Rightarrow 0\le \left(|x|-\frac{1+\sqrt{17}}{2}\right)\left(|x|-\frac{1-\sqrt{17}}{2}\right)$
$\Rightarrow |x|\ge \frac{1+\sqrt{17}}{2}$ or $|x|\le \frac{1-\sqrt{17}}{2}$ (Rejected)
$\Rightarrow x\in \left(-\infty ,-\frac{1+\sqrt{17}}{2}\right]\cup \left[\frac{1+\sqrt{17}}{2},\infty \right)$
So, $a=\frac{1+\sqrt{17}}{2}$