The set of all real values of x for which \(\cos^{-1} x \lt \displaystyle\frac{2\pi}{3}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\left(-\displaystyle\frac{1}{2}, 1\right]\)
\(\cos^{-1} x\) is defined only for \(-1 \le x \le 1\), and on this interval it is a decreasing function: a larger x gives a smaller angle.
The boundary value is
\( \displaystyle \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \)
So \(\cos^{-1} x \lt \cos^{-1}\left(-\frac{1}{2}\right)\) holds exactly when x is to the right of \(-\frac{1}{2}\):
\( \displaystyle x \gt -\frac{1}{2} \)
Together with \(x \le 1\), the solution set is \(\left(-\frac{1}{2}, 1\right]\).
Why the other options are incorrect
Option (2) \(\left[-1, -\frac{1}{2}\right)\) — Incorrect. This treats \(\cos^{-1}\) as increasing; for example \(\cos^{-1}(-1) = \pi\), which is more than \(\frac{2\pi}{3}\).
Option (3) \(\left(\frac{1}{2}, 1\right]\) — Incorrect. It misses values such as \(x = 0\), where \(\cos^{-1} 0 = \frac{\pi}{2} \lt \frac{2\pi}{3}\).
Option (4) \(\left(-\frac{1}{2}, \infty\right)\) — Incorrect. For \(x \gt 1\), \(\cos^{-1} x\) is not defined.