If \(-\displaystyle\frac{1}{2} \le x \le \displaystyle\frac{\sqrt{3}}{2}\), then the set of values of \(\cos^{-1} x\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\left[\displaystyle\frac{\pi}{6}, \displaystyle\frac{2\pi}{3}\right]\)
The graph of \(y = \cos^{-1} x\) falls steadily from \(\pi\) at \(x = -1\) to 0 at \(x = 1\), so \(\cos^{-1}\) is a decreasing function on \([-1, 1]\). On an interval its largest value is at the left end and its smallest value at the right end.
At the left end:
\( \displaystyle \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \)
At the right end:
\( \displaystyle \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \)
The function is continuous, so it takes every value between these two. Hence \(\cos^{-1} x\) takes all values in \(\left[\frac{\pi}{6}, \frac{2\pi}{3}\right]\).
Why the other options are incorrect
Option (1) \(\left[-\frac{\pi}{6}, \frac{\pi}{3}\right]\) — Incorrect. These are the values of \(\sin^{-1}\) at the two ends; \(\cos^{-1}\) never takes negative values.
Option (2) \([0, \pi]\) — Incorrect. This is the full range, reached only when x runs over all of \([-1, 1]\).
Option (3) \(\left[0, \frac{2\pi}{3}\right]\) — Incorrect. \(\cos^{-1} x = 0\) needs \(x = 1\), but here x stops at \(\frac{\sqrt{3}}{2}\).