Question

The function $f(x) = \sin x(1 + \cos x)$, $0 \le x \le \dfrac{\pi}{2}$, has a maximum value when $x$ equals

• A. $\dfrac{\pi}{6}$
• B. $\dfrac{\pi}{4}$
• C. $\dfrac{\pi}{3}$
• D. $\dfrac{\pi}{2}$

Hint:

Use $\cos 2x = 2\cos^2 x - 1$ to convert $f'(x)$ into a quadratic in $\cos x$, then solve by factoring.

Answer 1

The Correct Option is C

Step 1: Conceptual Understanding:
Differentiate $f(x)$ and set $f'(x) = 0$ to find critical points.
Step 2: Explanation in Detail:
$f'(x) = \cos x(1+\cos x) - \sin^2 x = \cos x + \cos^2 x - \sin^2 x = \cos x + \cos 2x$.
$= \cos x + 2\cos^2 x - 1 = 0 \Rightarrow (2\cos x - 1)(\cos x + 1) = 0$.
In $[0,\pi/2]$: $\cos x = 1/2 \Rightarrow x = \pi/3$.
Step 3: Therefore, Stating the Final Answer
The maximum occurs at $x = \dfrac{\pi}{3}$.