The absolute maximum value of f(x) = x + sin 2x on the interval [0, 2…

The correct answer is Option (2) → \(2\pi\)

Setting the derivative equal to zero:

    \(f'(x) = 1 + 2\cos 2x = 0\)

    \( \displaystyle \cos 2x = -\frac{1}{2} \)

With 2x in \([0, 4\pi]\), this gives

    \( \displaystyle x = \frac{\pi}{3}, \ \frac{2\pi}{3}, \ \frac{4\pi}{3}, \ \frac{5\pi}{3} \)

The absolute maximum is the largest of the values at these points and at the end points:

    \( \displaystyle f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \approx 1.91 \)

    \( \displaystyle f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \approx 1.23 \)

    \( \displaystyle f\left(\frac{4\pi}{3}\right) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2} \approx 5.05 \)

    \( \displaystyle f\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2} \approx 4.37 \)

    \( f(0) = 0 \), \( f(2\pi) = 2\pi \approx 6.28 \)

The largest is \(f(2\pi) = 2\pi\).

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{\pi}{3} + \frac{\sqrt{3}}{2}\) — Incorrect. This is the local maximum value at \(x = \frac{\pi}{3}\), about 1.91, far below 2π.

• Option (3) \(\displaystyle\frac{4\pi}{3} + \frac{\sqrt{3}}{2}\) — Incorrect. This is the larger local maximum value, about 5.05, but the end value f(2π) ≈ 6.28 is bigger.

• Option (4) \(\displaystyle\frac{5\pi}{3} - \frac{\sqrt{3}}{2}\) — Incorrect. This is a local minimum value, about 4.37.