Key Concepts and Formulas

• Finding Absolute Extrema: To find the absolute maximum or minimum of a continuous function $f(x)$ on a closed interval $[a, b]$, evaluate $f(x)$ at all critical points in $(a, b)$ and at the endpoints $a$ and $b$. The largest value is the absolute maximum, and the smallest is the absolute minimum.
• Trigonometric Identities:
  • $\sin 2x = 2\sin x \cos x$
  • $\cos 2x = 2\cos^2 x - 1$
  • $\cos 3x = 4\cos^3 x - 3\cos x$
• Derivatives:
  • $\frac{d}{dx} \sin x = \cos x$
  • $\frac{d}{dx} \cos x = -\sin x$

Step-by-Step Solution

Step 1: Simplify the Function

The given function is $f(x) = x - 2\sin x \cos x + \frac{1}{3} \sin 3x$. We simplify it using the identity $\sin 2x = 2\sin x \cos x$: $f(x) = x - \sin 2x + \frac{1}{3} \sin 3x$ This simplification makes differentiation easier.

Step 2: Find the First Derivative

To find the critical points, we need to find the first derivative $f'(x)$: $f'(x) = \frac{d}{dx} \left( x - \sin 2x + \frac{1}{3} \sin 3x \right)$ Applying the derivative rules: $f'(x) = 1 - 2\cos 2x + \cos 3x$

Step 3: Find Critical Points by Setting $f'(x) = 0$

We set $f'(x) = 0$ to find the critical points: $1 - 2\cos 2x + \cos 3x = 0$ We use the trigonometric identities $\cos 2x = 2\cos^2 x - 1$ and $\cos 3x = 4\cos^3 x - 3\cos x$ to express the equation in terms of $\cos x$. $1 - 2(2\cos^2 x - 1) + (4\cos^3 x - 3\cos x) = 0$ $1 - 4\cos^2 x + 2 + 4\cos^3 x - 3\cos x = 0$ $4\cos^3 x - 4\cos^2 x - 3\cos x + 3 = 0$ Let $c = \cos x$. Then we have: $4c^3 - 4c^2 - 3c + 3 = 0$ We factor by grouping: $4c^2(c - 1) - 3(c - 1) = 0$ $(4c^2 - 3)(c - 1) = 0$ So, $c = 1$ or $4c^2 = 3 \implies c^2 = \frac{3}{4} \implies c = \pm \frac{\sqrt{3}}{2}$. Thus, $\cos x = 1, \frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}$.

Now, we find the values of $x$ in the interval $0 \leq x \leq \pi$:

• $\cos x = 1 \implies x = 0$
• $\cos x = \frac{\sqrt{3}}{2} \implies x = \frac{\pi}{6}$
• $\cos x = -\frac{\sqrt{3}}{2} \implies x = \frac{5\pi}{6}$

The critical points are $x = 0, \frac{\pi}{6}, \frac{5\pi}{6}$.

Step 4: Evaluate the Function at Critical Points and Endpoints

We evaluate $f(x)$ at the critical points and endpoints $x = 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi$:

• $f(0) = 0 - \sin(0) + \frac{1}{3}\sin(0) = 0$
• $f\left(\frac{\pi}{6}\right) = \frac{\pi}{6} - \sin\left(\frac{\pi}{3}\right) + \frac{1}{3}\sin\left(\frac{\pi}{2}\right) = \frac{\pi}{6} - \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{\pi + 2 - 3\sqrt{3}}{6}$
• $f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} - \sin\left(\frac{5\pi}{3}\right) + \frac{1}{3}\sin\left(\frac{5\pi}{2}\right) = \frac{5\pi}{6} - \left(-\frac{\sqrt{3}}{2}\right) + \frac{1}{3}(1) = \frac{5\pi}{6} + \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{5\pi + 2 + 3\sqrt{3}}{6}$
• $f(\pi) = \pi - \sin(2\pi) + \frac{1}{3}\sin(3\pi) = \pi - 0 + 0 = \pi$

We want to find the maximum of $0$, $\frac{\pi + 2 - 3\sqrt{3}}{6}$, $\frac{5\pi + 2 + 3\sqrt{3}}{6}$, and $\pi$. Since $\pi \approx 3.14$, $3\sqrt{3} \approx 3(1.73) = 5.19$, $\frac{\pi + 2 - 3\sqrt{3}}{6} \approx \frac{3.14 + 2 - 5.19}{6} = \frac{0.05}{6} > 0$. Also $\frac{5\pi + 2 + 3\sqrt{3}}{6} \approx \frac{5(3.14) + 2 + 5.19}{6} = \frac{15.7 + 2 + 5.19}{6} = \frac{22.89}{6} \approx 3.815$. Since $\pi \approx 3.14$, we compare $\pi$ and $\frac{5\pi + 2 + 3\sqrt{3}}{6}$. We also check if $\frac{5\pi + 2 + 3\sqrt{3}}{6} > \pi$. $5\pi + 2 + 3\sqrt{3} > 6\pi$ $2 + 3\sqrt{3} > \pi$ Since $\pi \approx 3.14$ and $2 + 3\sqrt{3} \approx 2 + 3(1.732) = 2 + 5.196 = 7.196$, $2 + 3\sqrt{3} > \pi$. Therefore, $\frac{5\pi + 2 + 3\sqrt{3}}{6}$ is the maximum value.

Common Mistakes & Tips

• Trigonometric Identities: Ensure you use the correct trigonometric identities when simplifying the function and its derivatives.
• Factoring: Factoring the cubic equation can be tricky. Grouping terms is a useful technique.
• Approximations: Be careful when using approximations to compare values. It's better to compare them analytically if possible.

Summary

We found the maximum value of the function $f(x) = x - 2\sin x \cos x + \frac{1}{3}\sin 3x$ on the interval $[0, \pi]$ by finding the critical points, simplifying the function using trigonometric identities, and evaluating the function at the critical points and endpoints. The maximum value occurs at $x=\frac{5\pi}{6}$ and is equal to $\frac{5\pi + 2 + 3\sqrt{3}}{6}$.

Final Answer

The final answer is \boxed{\frac{5 \pi+2+3 \sqrt{3}}{6}}, which corresponds to option (A).