The domain of the function


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 :

  1. (1) $\frac{\sqrt{17}}{2}$

  2. (2) $\frac{\sqrt{17}-1}{2}$

  3. (3) $\frac{1+\sqrt{17}}{2}$

  4. (4) $\frac{\sqrt{17}}{2}+1$

Correct Option: , 3


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

$\therefore-1 \leq \frac{|x|+5}{x^{2}+1} \leq 1$

$\Rightarrow|x|+5 \leq x^{2}+1$  $\left[\because x^{2}+1 \neq 0\right]$

$\Rightarrow x^{2}-|x|-4 \geq 0$

$\Rightarrow\left(|x|-\frac{1-\sqrt{17}}{2}\right)\left(|x|-\frac{1+\sqrt{17}}{2}\right) \geq 0$

$\Rightarrow x \in\left(-\infty,-\frac{1+\sqrt{17}}{2}\right) \cup\left[\frac{1+\sqrt{17}}{2}, \infty\right)$

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


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