The maximum value of the function \( f(x)=\sin x(1+\cos x) \) in the interval \( [0,\pi] \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{3\sqrt{3}}{4} \)
By the product rule,
\( \displaystyle f'(x)=\cos x(1+\cos x)+\sin x(-\sin x) \)
\( \displaystyle =\cos x+\cos^{2}x-\sin^{2}x \)
With \( \sin^{2}x=1-\cos^{2}x \):
\( \displaystyle f'(x)=2\cos^{2}x+\cos x-1 \)
\( \displaystyle =(2\cos x-1)(\cos x+1) \)
f′(x) = 0 gives \( \cos x=\frac{1}{2} \) or \( \cos x=-1 \), i.e. \( x=\frac{\pi}{3} \) or \( x=\pi \) in \( [0,\pi] \).
On a closed interval, the maximum is the largest of the values at the critical points and the end points:
\( \displaystyle f(0)=0\cdot 2=0 \)
\( \displaystyle f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\left(1+\frac{1}{2}\right) \)
\( \displaystyle =\frac{3\sqrt{3}}{4} \)
\( \displaystyle f(\pi)=0\cdot 0=0 \)
The maximum value is \( \frac{3\sqrt{3}}{4} \).