Tentukan \( \quad \) nilai \( \quad \) maksimum dari fungsi untuk \( 0 \leq x \leq \pi(x)=4 \sin x+2 \cos 2 x \) A. 6 B. 4 C. 3 D. 2 E. 1

We are given the function   p(x) = 4 sin x + 2 cos 2x for 0 ≤ x ≤ π, and we need to find its maximum value. Step 1. Rewrite the function in terms of sin x only. Recall the double-angle identity: cos 2x = 1 − 2 sin² x. Substituting this into p(x), we have:   p(x) = 4 sin x + 2(1 − 2 sin² x)      = 4 sin x + 2 − 4 sin² x Let u = sin x. Since x ∈ [0, π], u ranges from 0 to 1. Thus, the function becomes:   p(u) = −4u² + 4u + 2 Step 2. Find the maximum value of p(u) on u ∈ [0, 1]. The quadratic function p(u) = −4u² + 4u + 2 is concave down (coefficient of u² is -4). Its maximum occurs at the vertex. The u-coordinate of the vertex is given by:   u = −b/(2a) = −(4)/(2(−4)) = 0.5 Step 3. Compute p(u) at u = 0.5.   p(0.5) = −4(0.5)² + 4(0.5) + 2       = −4(0.25) + 2 + 2       = −1 + 4       = 3 Step 4. Check the endpoints. For u = 0 (x = 0):   p(0) = −4(0)² + 4(0) + 2 = 2 For u = 1 (x = π/2 or π):   p(1) = −4(1)² + 4(1) + 2 = −4 + 4 + 2 = 2 Thus, the maximum value of the function on the interval is 3. Therefore, the answer is: C. 3