The range of the function \(f(x) = \sin^{-1} x + \tan^{-1} x\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\left[-\displaystyle\frac{3\pi}{4}, \displaystyle\frac{3\pi}{4}\right]\)
\(\sin^{-1} x\) is defined only for \(-1 \le x \le 1\), so the domain of f is \([-1, 1]\).
On this interval both \(\sin^{-1} x\) and \(\tan^{-1} x\) are increasing, so their sum f is increasing. Its least value is at \(x = -1\) and its greatest value at \(x = 1\):
\( \displaystyle f(-1) = -\frac{\pi}{2} - \frac{\pi}{4} = -\frac{3\pi}{4} \)
\( \displaystyle f(1) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4} \)
f is continuous, so it takes every value in between. Hence the range is \(\left[-\frac{3\pi}{4}, \frac{3\pi}{4}\right]\).
Why the other options are incorrect
Option (1) \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) — Incorrect. This is the range of \(\sin^{-1} x\) alone; \(f(1) = \frac{3\pi}{4}\) is outside it.
Option (2) \((-\pi, \pi)\) — Incorrect. For \(-1 \le x \le 1\), \(\tan^{-1} x\) stays in \(\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]\), so f never goes beyond \(\frac{3\pi}{4}\); for example \(\frac{9\pi}{10}\) is not reached.
Option (3) \(\left[-\frac{\pi}{4}, \frac{3\pi}{4}\right]\) — Incorrect. The least value is \(f(-1) = -\frac{3\pi}{4}\), not \(-\frac{\pi}{4}\).