The domain of the function f(x)= sin^-1( |x|+5 x^2+1) is (-∞ ,-a]∪ [a,∞ ), then a is equal to

$$f(x)={\mathrm{sin}}^{-1}(\frac{|x|+5}{{x}^{2}+1})$$f(x)=sin−1(x2+1∣x∣+5​)

$\Rightarrow \begin{matrix}-1\leq \frac{|x|+5}{{x}^{2}+1}\leq 1\end{matrix},\because x\in [-1,1]\mathrm{for}{\mathrm{sin}}^{-1}x$⇒−1≤x2+1∣x∣+5​≤1​,∵x∈[−1,1]forsin−1x to exist.

Case $1$1

$$-{x}^{2}-1\leq |x|+5$$−x2−1≤∣x∣+5

$\Rightarrow {x}^{2}+1+|x|+5\geq 0$⇒x2+1+∣x∣+5≥0, true for all $x\in R$x∈R.

Case $2$2

$$|x|+5\leq {x}^{2}+1$$∣x∣+5≤x2+1

$$\Rightarrow {x}^{2}-|x|-4\geq 0$$⇒x2−∣x∣−4≥0

Let $|x|=t\Rightarrow {t}^{2}-t-4\geq 0$∣x∣=t⇒t2−t−4≥0

$$\Rightarrow (|x|+\frac{\sqrt{17}-1}{2})(|x|-\frac{\sqrt{17}+1}{2})\geq 0$$⇒(∣x∣+217​−1​)(∣x∣−217​+1​)≥0

$$\Rightarrow x\in (-\infty ,-\frac{\sqrt{17}+1}{2}]\cup [\frac{\sqrt{17}+1}{2},\infty )$$⇒x∈(−∞,−217​+1​]∪[217​+1​,∞)

$$\therefore a=\frac{1+\sqrt{17}}{2}$$∴a=21+17​​