The domain of the function \(f(x) = \sqrt{\sin^{-1} x}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \([0, 1]\)
Two conditions are needed: \(\sin^{-1} x\) must be defined, so \(-1 \le x \le 1\), and the quantity under the square root must not be negative:
\(\sin^{-1} x \ge 0\)
The graph of \(y = \sin^{-1} x\) is increasing and passes through the origin, since \(\sin^{-1} 0 = 0\). So \(\sin^{-1} x\) is negative for \(-1 \le x \lt 0\) and not negative for \(0 \le x \le 1\):
\(0 \le x \le 1\)
Hence the domain is \([0, 1]\).
Why the other options are incorrect
Option (1) \([-1, 1]\) — Incorrect. At \(x = -1\), \(\sin^{-1}(-1) = -\frac{\pi}{2}\) is negative, so the square root is not real.
Option (3) \(\left[0, \frac{\pi}{2}\right]\) — Incorrect. These are the values of \(\sin^{-1} x\), not of x; for \(x = \frac{\pi}{2} \approx 1.57\), \(\sin^{-1} x\) is not defined.
Option (4) \([-1, 0]\) — Incorrect. Here \(\sin^{-1} x \le 0\); for example \(\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}\).