The domain of \(y =\cos^{-1}(x^2-4)\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \([-\sqrt{5},-\sqrt{3}] \cup [\sqrt{3},\sqrt{5}]\)
The domain of \(\cos^{-1}\) is [−1, 1], so the expression inside must satisfy
\(-1 \le x^2 - 4 \le 1\)
\(3 \le x^2 \le 5\)
\( \displaystyle \sqrt{3} \le |x| \le \sqrt{5} \)
So x lies in \([-\sqrt{5}, -\sqrt{3}]\) or in \([\sqrt{3}, \sqrt{5}]\).
Why the other options are incorrect
Option (1) \([0, \pi]\) — Incorrect. This is the range of \(\cos^{-1}\), not a set of allowed x. For example, x = 0 gives \(\cos^{-1}(-4)\), which is not defined.
Option (3) \([-\sqrt{5},-\sqrt{3}]\cap[-\sqrt{5},\sqrt{3}]\) — Incorrect. This intersection is only \([-\sqrt{5}, -\sqrt{3}]\). It leaves out \([\sqrt{3}, \sqrt{5}]\), although x = 2 gives \(\cos^{-1}0\), which is defined.
Option (4) \([-1, 1]\) — Incorrect. This is the domain of \(\cos^{-1}t\) itself, the condition on \(x^2 - 4\), not on x. For x = 0, \(x^2 - 4 = -4\).
NCERT Exemplar:
Inverse Trigonometric Functions
