Question

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

• A. $\frac{\sqrt{17-1}}{2}$
• B. $\frac{\sqrt{17}}{2}$
• C. $\frac{1+\sqrt{17}}{2}$
• D. $\frac{\sqrt{17}}{2}+1$

Answer 1

The correct option is C

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

Explanation for correct answer Option(s)

Find the domain of the function $f\left(x\right)$

Consider the given function as,

$f\left(x\right)={\sin }^{-1}\left[\frac{\left|x\right|+5}{x^2+1}\right]$

Defined the above function as,

$$\begin{gathered} \Rightarrow \frac{\left|x\right|+5}{x^2+1}\le 1 \\ \Rightarrow \left|x\right|+5\le x^2+1 \\ \Rightarrow x^2-\left|x\right|-4\ge 0 \end{gathered}$$

Let $\left|x\right|=t$

$\Rightarrow t^2-t-4\ge 0$

We know that

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Then,

$$\begin{gathered} t=\frac{-1\pm \sqrt{-1^2-4\left(1\right)\left(-4\right)}}{2\left(-1\right)} \\ t=\frac{-1\pm \sqrt{1+16}}{-2} \\ t=\frac{1\pm \sqrt{17}}{2} \end{gathered}$$

Thus, $t=\frac{1+\sqrt{17}}{2}\text{or}t=\frac{1-\sqrt{17}}{2}$

Hence, the domain of the given function is $\frac{1+\sqrt{17}}{2}$

Therefore, the correct answer is Option C.