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

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Question

(A) For the function $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ to be defined,the argument must satisfy $-1 \leq \frac{|x|+5}{x^2+1} \leq 1$.S

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For the function $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ to be defined,the argument must satisfy $-1 \leq \frac{|x|+5}{x^2+1} \leq 1$.Since $|x|+5 > 0$ and $x^2+1 > 0$ for all $x \in \mathbb{R}$,the condition $\frac{|x|+5}{x^2+1} \geq -1$ is always satisfied.Thus,we only need to solve $\frac{|x|+5}{x^2+1} \leq 1$.Multiplying by $x^2+1$ (which is positive),we get $|x|+5 \leq x^2+1$.Rearranging the terms,we have $x^2 - |x| - 4 \geq 0$.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 $t = |x| \geq 0$,we must have $t \geq \frac{1+\sqrt{17}}{2}$.Therefore,$|x| \geq \frac{1+\sqrt{17}}{2}$,which implies $x \in \left(-\infty, -\frac{1+\sqrt{17}}{2}\right] \cup \left[\frac{1+\sqrt{17}}{2}, \infty\right)$.Comparing this with the given domain $(-\infty, -a] \cup [a, \infty)$,we find $a = \frac{1+\sqrt{17}}{2}$.

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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