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

\(f(x)=\sin ^{-1}\left(\frac{|x|+5}{x^2+1}\right)\)
\(\mathrm{f}(x)\) is defined, if
\(\begin{aligned}
& -1 \leq \frac{|x|+5}{x^2+1} \leq 1 \\
& \Rightarrow 0 \leq \frac{|x|+5}{x^2+1} \leq 1 \\
& \Rightarrow x^2-|x|-4 \geq 0 \\
& \Rightarrow\left(|x|-\frac{1-\sqrt{17}}{2}\right)\left(|x|-\frac{1+\sqrt{17}}{2}\right) \geq 0 \\
& \Rightarrow|x| \geq \frac{1+\sqrt{17}}{2} \\
& \text { or }|x| \leq \frac{1-\sqrt{17}}{2}, \text { which is not possible } \\
& \Rightarrow x \in\left(-\infty,-\frac{1+\sqrt{17}}{2}\right] \cup\left[\frac{1+\sqrt{17}}{2}, \infty\right) \\
& \Rightarrow \mathrm{a}=\frac{1+\sqrt{17}}{2}
\end{aligned}\)