Question:
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 :
Correct Option: , 3
Solution:
$\because f(x)=\sin ^{-1}\left(\frac{|x|+5}{x^{2}+1}\right)$
$\therefore-1 \leq \frac{|x|+5}{x^{2}+1} \leq 1$
$\Rightarrow|x|+5 \leq x^{2}+1$ $\left[\because x^{2}+1 \neq 0\right]$
$\Rightarrow x^{2}-|x|-4 \geq 0$
$\Rightarrow\left(|x|-\frac{1-\sqrt{17}}{2}\right)\left(|x|-\frac{1+\sqrt{17}}{2}\right) \geq 0$
$\Rightarrow x \in\left(-\infty,-\frac{1+\sqrt{17}}{2}\right) \cup\left[\frac{1+\sqrt{17}}{2}, \infty\right)$
$\therefore a=\frac{1+\sqrt{17}}{2}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.