The function \(f(x) = 2\sin x - x\), \(x \in (0, \pi)\), is strictly decreasing in the interval
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\left(\displaystyle\frac{\pi}{3}, \pi\right)\)
Differentiating:
\( f'(x) = 2\cos x - 1 \)
f is strictly decreasing where \(f'(x) \lt 0\):
\( 2\cos x - 1 \lt 0 \)
\( \displaystyle \cos x \lt \frac{1}{2} \)
On \((0, \pi)\), cos x decreases from 1 to −1 and equals \(\frac{1}{2}\) at \(x = \frac{\pi}{3}\). So \(\cos x \lt \frac{1}{2}\) exactly for \(\frac{\pi}{3} \lt x \lt \pi\).
Why the other options are incorrect
Option (1) \(\left(0, \frac{\pi}{3}\right)\) — Incorrect. Here \(\cos x \gt \frac{1}{2}\), so f′(x) > 0 and f is increasing.
Option (3) \(\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)\) — Incorrect. It contains \(\left(\frac{\pi}{6}, \frac{\pi}{3}\right)\), where f is increasing; \(\frac{\pi}{6}\) is where sin x, not cos x, equals \(\frac{1}{2}\).
Option (4) \(\left(0, \frac{2\pi}{3}\right)\) — Incorrect. f is increasing on \(\left(0, \frac{\pi}{3}\right)\), which lies inside this interval.