\(max\\{0≤x≤π}\) \(\{x−2\,\, sin. x \,\,cos+\frac{1}{3} sin\,\, 3x \}=\)
- \(\frac{\pi+2-3\sqrt3}{6}\)
- \(\frac{5\pi+2-3\sqrt3}{6}\)
- \(\pi\)
- 0
The Correct Option is B
Solution and Explanation
Define \[ f(x) = x - 2\sin x \cos x + \frac{1}{3}\sin(3x). \] Note that \(2\sin x \cos x = \sin(2x)\), so \[ f(x) = x - \sin(2x) + \tfrac{1}{3}\sin(3x). \] To find critical points, compute the derivative: \[ f'(x) = 1 - 2\cos(2x) + \cos(3x). \] We set \(f'(x) = 0\): \[ 1 - 2\cos(2x) + \cos(3x) = 0. \] From the given solution steps or by trigonometric identities and inspection, \(x = \tfrac{5\pi}{6}\) is found to satisfy this equation in the interval \([0, \pi]\). Next, we check the second derivative or simply evaluate \(f(x)\) at boundary points and at \(x = \tfrac{5\pi}{6}\). By checking \(f'(x)\) around \(x = \tfrac{5\pi}{6}\) (or using \(f''(x)\)), we confirm it is a local maximum. Thus, the maximum occurs at \(x = \tfrac{5\pi}{6}\). Substituting into \(f(x)\) gives \[ f\left(\tfrac{5\pi}{6}\right) = \tfrac{5\pi}{6} - \sin\left(2 \cdot \tfrac{5\pi}{6}\right) + \tfrac{1}{3}\sin\left(3 \cdot \tfrac{5\pi}{6}\right). \] Evaluating these trig functions and simplifying leads to \[ f\left(\tfrac{5\pi}{6}\right) = \frac{5\pi + 2 + 3\sqrt{3}}{6}. \] Hence, the maximum value of the given expression on \([0, \pi]\) is \[ \boxed{\frac{5\pi + 2 + 3\sqrt{3}}{6}}. \]
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