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

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Aicharya · Accepted Answer

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

The domain of the function $\mathrm{f}(x)=\sin ^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is $(-\infty,-a]…

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