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 :
Correct Option: 1
$f(x)=\sin \left(\frac{|x|+5}{x^{2}+1}\right)$
For domain :
$-1 \leq \frac{|x|+5}{x^{2}+1} \leq 1$
So for domain :
$\frac{|x|+5}{x^{2}+1} \leq 1$
$\Rightarrow|x|+5 \leq x^{2}+1$
$\Rightarrow 0 \leq x^{2}-|x|-4$
$\Rightarrow 0 \leq\left(|x|-\frac{1+\sqrt{17}}{2}\right)\left(|x|-\frac{1-\sqrt{17}}{2}\right)$
$\Rightarrow|x| \geq \frac{1+\sqrt{17}}{2}$ or $|x| \leq \frac{1-\sqrt{17}}{2}$ (Rejected)
$\Rightarrow x \in\left(-\infty,-\frac{1+\sqrt{17}}{2}\right) \cup\left[\frac{1+\sqrt{17}}{2}, \infty\right)$
So, $a=\frac{1+\sqrt{17}}{2}$
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