The domain of the function \(f(x) = \sin^{-1}(2x) + \cos^{-1}(x - 1)\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\left[0, \displaystyle\frac{1}{2}\right]\)
A sum is defined only where both terms are defined, so the domain is the common part of the two domains.
For \(\sin^{-1}(2x)\):
\(-1 \le 2x \le 1\)
\( \displaystyle -\frac{1}{2} \le x \le \frac{1}{2} \)
For \(\cos^{-1}(x - 1)\):
\(-1 \le x - 1 \le 1\)
\(0 \le x \le 2\)
The common part of \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) and \([0, 2]\) is \(\left[0, \frac{1}{2}\right]\).
Why the other options are incorrect
Option (2) \(\left[-\frac{1}{2}, 2\right]\) — Incorrect. This joins the two domains instead of taking their common part; at \(x = 2\), \(\sin^{-1}(4)\) is not defined.
Option (3) \([0, 2]\) — Incorrect. This is the domain of the second term only; at \(x = 1\), \(\sin^{-1}(2)\) is not defined.
Option (4) \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) — Incorrect. This is the domain of the first term only; at \(x = -\frac{1}{2}\), \(\cos^{-1}\left(-\frac{3}{2}\right)\) is not defined.