The domain of the function f(x) = sin^{-1}lef | vedclass

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(C) $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is defined if $-1 \leq \frac{|x|+5}{x^2+1} \leq 1$. Since $|x|+5 > 0$ and $x^2+1 > 0$,the left

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$\frac{1 + \sqrt{17}}{2}$

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(C) $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is defined if $-1 \leq \frac{|x|+5}{x^2+1} \leq 1$. Since $|x|+5 > 0$ and $x^2+1 > 0$,the left inequality $\frac{|x|+5}{x^2+1} \geq -1$ is always true. We only need to solve $\frac{|x|+5}{x^2+1} \leq 1$. $|x|+5 \leq x^2+1$ $x^2 - |x| - 4 \geq 0$. Let $t = |x|$,where $t \geq 0$. Then $t^2 - t - 4 \geq 0$. The roots of $t^2 - t - 4 = 0$ are $t = \frac{1 \pm \sqrt{1 - 4(1)(-4)}}{2} = \frac{1 \pm \sqrt{17}}{2}$. Since $t \geq 0$,we have $t \geq \frac{1 + \sqrt{17}}{2}$. Thus,$|x| \geq \frac{1 + \sqrt{17}}{2}$,which implies $x \in \left(-\infty, -\frac{1 + \sqrt{17}}{2}\right] \cup \left[\frac{1 + \sqrt{17}}{2}, \infty\right)$. Comparing this with $(-\infty, -a] \cup [a, \infty)$,we get $a = \frac{1 + \sqrt{17}}{2}$.

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

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