Q64 / 150 · MHT CET 2023 (10 May Shift 2) · Mathematics

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

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

Solution

$f(x)=\sin ^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ $\mathrm{f}(x)$ is defined, if $\begin{aligned} & -1 \leq \frac{|x|+5}{x^2+1} \leq 1 \\ & \Rightarrow 0 \leq \frac{|x|+5}{x^2+1} \leq 1 \\ & \Rightarrow x^2-|x|-4 \geq 0 \\ & \Rightarrow\left(|x|-\frac{1-\sqrt{17}}{2}\right)\left(|x|-\frac{1+\sqrt{17}}{2}\right) \geq 0 \\ & \Rightarrow|x| \geq \frac{1+\sqrt{17}}{2} \\ & \text { or }|x| \leq \frac{1-\sqrt{17}}{2}, \text { which is not possible } \\ & \Rightarrow x \in\left(-\infty,-\frac{1+\sqrt{17}}{2}\right] \cup\left[\frac{1+\sqrt{17}}{2}, \infty\right) \\ & \Rightarrow \mathrm{a}=\frac{1+\sqrt{17}}{2} \end{aligned}$

Hence, the correct answer is option: 'C'.

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