Find the domain of $\sin^{-1}(x^2 - 4)$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $[-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$ ##
$-1 \le (x^2 - 4) \le 1$
$\Rightarrow 3 \le x^2 \le 5$
$\Rightarrow \sqrt{3} \le |x| \le \sqrt{5}$
$\Rightarrow x \in [-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$
So, required domain is $[-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$