The maximum value of \(f(x) = \sin x\,(1 + \cos x)\) on the interval \([0, \pi]\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{3\sqrt{3}}{4}\)
Writing \(f(x) = \sin x + \frac{1}{2}\sin 2x\) and differentiating:
\( f'(x) = \cos x + \cos 2x \)
\( = 2\cos^{2} x + \cos x - 1 \)
\( = (2\cos x - 1)(\cos x + 1) \)
In \([0, \pi]\), f′(x) = 0 at \(x = \frac{\pi}{3}\) (\(\cos x = \frac{1}{2}\)) and at \(x = \pi\). Comparing with the end points:
\( f(0) = 0,\ f(\pi) = 0 \)
\( \displaystyle f\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \times \frac{3}{2} = \frac{3\sqrt{3}}{4} \)
So the maximum value is \(\frac{3\sqrt{3}}{4}\).
Why the other options are incorrect
Option (2) 2 — Incorrect. This takes \(\sin x = 1\) and \(1 + \cos x = 2\) together, but they never occur at the same x.
Option (3) 1 — Incorrect. This is \(f\left(\frac{\pi}{2}\right)\); x = π/2 is not a critical point.
Option (4) \(\displaystyle\frac{\sqrt{3}}{4}\) — Incorrect. This is \(\frac{1}{2}\sin\frac{2\pi}{3}\), only the second term at \(x = \frac{\pi}{3}\).