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Step-by-Step Solution
Step 1: Understand the Function
We have the function
$f(x) = \sin^{-1}\Bigl(\frac{|x|+5}{x^2 + 1}\Bigr)$.
To find its domain, we need the expression inside the $\sin^{-1}$ (arcsine) function to lie within the interval $[-1,1]$. That is:
$-1 \;\le\; \frac{|x| + 5}{x^2 + 1} \;\le\; 1$.
Step 2: Analyze the Lower Bound
The numerator $|x| + 5$ is always nonnegative, and the denominator $x^2 + 1$ is always positive for all real $x$. Therefore,
$\frac{|x| + 5}{x^2 + 1}\ge 0$.
Since $0 \ge -1$, the lower bound
$\frac{|x| + 5}{x^2+1} \ge -1$
is automatically satisfied for all real $x$.
Hence, we focus on the upper bound condition.
Step 3: Apply the Upper Bound Condition
We need
$\frac{|x| + 5}{x^2 + 1} \le 1.$
Multiplying both sides by $(x^2 + 1) > 0$ gives:
$|x| + 5 \;\le\; x^2 + 1.$
Rearranging,
$x^2 - |x| - 4 \;\ge\; 0.$
Step 4: Solve the Inequality for |x|
Consider the expression in terms of $|x|$:
\[
x^2 - |x| - 4 \ge 0.
\]
We treat $|x|$ as a single variable (say $t = |x| \ge 0$). Then the inequality becomes
\[
t^2 - t - 4 \ge 0 \quad \text{where} \; t \ge 0.
\]
Solve the quadratic $t^2 - t - 4 = 0$:
$t \;=\; \frac{1 \pm \sqrt{1 + 16}}{2} \;=\; \frac{1 \pm \sqrt{17}}{2}.
The two roots are $\frac{1 - \sqrt{17}}{2}$ and $\frac{1 + \sqrt{17}}{2}$. Because $\sqrt{17}$ is a little more than 4, the value $\frac{1 - \sqrt{17}}{2}$ is negative. Since $t = |x| \ge 0$, the negative root is not valid as a boundary where $|x|$ can lie. Consequently, the valid solution is:
$|x| \;\ge\; \frac{1 + \sqrt{17}}{2}.
Step 5: Interpret the Result for x
If $|x| \ge \frac{1 + \sqrt{17}}{2}$, then $x$ lies in either
$x \leq -\,\frac{1 + \sqrt{17}}{2}$ or $x \geq \frac{1 + \sqrt{17}}{2}$. Hence, the domain of $f(x)$ is:
\[
(-\infty, \,-a] \;\cup\; [\,a, \,\infty),
\]
where
\[
a = \frac{1 + \sqrt{17}}{2}.
\]
Step 6: Final Answer
The value of $a$ is
$\displaystyle \frac{1 + \sqrt{17}}{2}.$
