Question

The domain of the function $f\left(x\right)={\text{sin}}^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ 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 $\frac{1+\sqrt{17}}{2}$

$-1\le \frac{|x|+5}{x^2+1}\le 1$
$\Rightarrow -x^2-1\le |x|+5\le x^2+1$

Case $-I:-x^{2-}1\le |x|+5$
This inequality is always right $\forall x\in R$

Case $-II:|x|+5\le x^2+1$

$0\le |x{|}^2-|x|-4$

$\Rightarrow (|x|-\left(\frac{1+\sqrt{17}}{2}\right))(|x|-\left(\frac{1+\sqrt{17}}{2}\right))\le 0$
$\Rightarrow |x|\ge \frac{1+\sqrt{17}}{2}$
$\because |x|\le \frac{1-\sqrt{17}}{2}\text{Not possible}$

$\therefore x\in (-\infty ,\frac{-1-\sqrt{17}}{2}]\cup [\frac{1+\sqrt{17}}{2},\infty )$

Hence $a=\frac{1+\sqrt{17}}{2}$