The domain of the function \(f(x) = \cos^{-1}\left(x^{2} - 3\right)\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\left[-2, -\sqrt{2}\right] \cup \left[\sqrt{2}, 2\right]\)
\(\cos^{-1} t\) is defined only for \(-1 \le t \le 1\). So the argument \(x^{2} - 3\) must lie in \([-1, 1]\):
\(-1 \le x^{2} - 3 \le 1\)
Adding 3 throughout:
\(2 \le x^{2} \le 4\)
\(x^{2} \le 4\) gives \(-2 \le x \le 2\), and \(x^{2} \ge 2\) gives \(x \le -\sqrt{2}\) or \(x \ge \sqrt{2}\). Both must hold:
\(-2 \le x \le -\sqrt{2}\) or \(\sqrt{2} \le x \le 2\)
Hence the domain is \(\left[-2, -\sqrt{2}\right] \cup \left[\sqrt{2}, 2\right]\).
Why the other options are incorrect
Option (2) \(\left[\sqrt{2}, 2\right]\) — Incorrect. It leaves out the negative values; for example \(x = -2\) gives \(x^{2} - 3 = 1\), which is allowed.
Option (3) \([-2, 2]\) — Incorrect. It includes \(x = 0\), where \(x^{2} - 3 = -3\) is less than \(-1\).
Option (4) \([2, 4]\) — Incorrect. This is the set of values of \(x^{2}\), not of x; at \(x = 4\), \(x^{2} - 3 = 13\).