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