The function \(f(x) = \sin x + \sqrt{3}\cos x\), \(x \in [0, 2\pi]\), is strictly decreasing in the interval
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\left(\displaystyle\frac{\pi}{6}, \frac{7\pi}{6}\right)\)
Differentiating:
\( f'(x) = \cos x - \sqrt{3}\sin x \)
Writing it as a single cosine, with \(\cos\frac{\pi}{3} = \frac{1}{2}\) and \(\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\):
\( \displaystyle f'(x) = 2\left(\frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x\right) \)
\( \displaystyle = 2\cos\left(x + \frac{\pi}{3}\right) \)
f is strictly decreasing where \(f'(x) \lt 0\), that is, where \(\cos\left(x + \frac{\pi}{3}\right) \lt 0\).
For \(0 \le x \le 2\pi\), the angle \(x + \frac{\pi}{3}\) runs from \(\frac{\pi}{3}\) to \(\frac{7\pi}{3}\). Its cosine is negative exactly when
\( \displaystyle \frac{\pi}{2} \lt x + \frac{\pi}{3} \lt \frac{3\pi}{2} \)
\( \displaystyle \frac{\pi}{6} \lt x \lt \frac{7\pi}{6} \)
So f is strictly decreasing in \(\left(\frac{\pi}{6}, \frac{7\pi}{6}\right)\) and strictly increasing in \(\left(0, \frac{\pi}{6}\right)\) and \(\left(\frac{7\pi}{6}, 2\pi\right)\).
Why the other options are incorrect
Option (1) \(\left(\frac{\pi}{3}, \frac{4\pi}{3}\right)\) — Incorrect. It contains \(\left(\frac{7\pi}{6}, \frac{4\pi}{3}\right)\), where f is increasing; for example \(f'\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}} \gt 0\).
Option (2) \(\left(0, \frac{\pi}{6}\right)\) — Incorrect. Here \(x + \frac{\pi}{3} \lt \frac{\pi}{2}\), so f′(x) > 0 and f is increasing; for example f′(0) = 1.
Option (3) \(\left(\frac{5\pi}{6}, \frac{11\pi}{6}\right)\) — Incorrect. It contains \(\left(\frac{7\pi}{6}, \frac{11\pi}{6}\right)\), where f is increasing; for example \(f'\left(\frac{3\pi}{2}\right) = 0 + \sqrt{3} \gt 0\).