The domain of the function \(f(x) = \tan^{-1}\left(\displaystyle\frac{1}{\sqrt{4 - x^{2}}}\right)\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \((-2, 2)\)
\(\tan^{-1} t\) is defined for every real number t, so the only conditions come from the inside expression \(\frac{1}{\sqrt{4 - x^{2}}}\).
The square root must be real and, being in the denominator, must not be 0:
\(4 - x^{2} \gt 0\)
\(x^{2} \lt 4\)
\(-2 \lt x \lt 2\)
Hence the domain is \((-2, 2)\).
Why the other options are incorrect
Option (1) \([-2, 2]\) — Incorrect. At \(x = \pm 2\), \(\sqrt{4 - x^{2}} = 0\) is in the denominator.
Option (2) the set of all real numbers — Incorrect. \(\tan^{-1}\) accepts all reals, but the inside expression does not: at \(x = 3\), \(4 - 9 = -5\) has no real square root.
Option (4) \((-\infty, -2) \cup (2, \infty)\) — Incorrect. These are exactly the x with \(4 - x^{2} \lt 0\), where the square root is not real.