The domain of the function

$f(x)=\sin \left(\frac{|x|+5}{x^{2}+1}\right)$

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, $a=\frac{1+\sqrt{17}}{2}$