The domain of the function \(y = \sin^{-1}(x-1) + \cos^{-1}\sqrt{x-1}\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \([1, 2]\)
The function is defined only where both terms are defined.
For \(\sin^{-1}(x - 1)\):
\( -1 \le x - 1 \le 1 \)
\( 0 \le x \le 2 \)
For \(\cos^{-1}\sqrt{x - 1}\), the root must be real and lie in \([-1, 1]\):
\( \displaystyle 0 \le \sqrt{x - 1} \le 1 \)
\( 0 \le x - 1 \le 1 \)
\( 1 \le x \le 2 \)
The common part of \([0, 2]\) and \([1, 2]\) is \([1, 2]\).
Why the other options are incorrect
Option (1) \([-1, 1]\) — Incorrect. At x = 0, \(\sqrt{x - 1} = \sqrt{-1}\) is not real.
Option (2) \([0, 1]\) — Incorrect. For \(x \lt 1\), \(\sqrt{x - 1}\) is not real; only x = 1 works.
Option (4) \([0, 2]\) — Incorrect. This is the domain of \(\sin^{-1}(x - 1)\) alone. It ignores the square root, which needs \(x \ge 1\).