To find the absolute maximum value of the function $f(x) = 2\sin x + 2\sqrt{3} \cos x - 1$ on the interval $[\pi, 2\pi]$, let's follow these steps:

Step 1: Simplify the Function

The function $f(x) = 2\sin x + 2\sqrt{3} \cos x - 1$ can be rewritten in terms of a single trigonometric function by combining $\sin x$ and $\cos x$ using an amplitude-phase form.

We know that an expression of the form $a \sin x + b \cos x$ can be rewritten as: $R \sin(x + \phi)$ where $R = \sqrt{a^2 + b^2}$ and $\tan \phi = \frac{b}{a}$.

In this case:

• $a = 2$
• $b = 2\sqrt{3}$

Thus, $R = \sqrt{(2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$

And $\tan \phi = \frac{2\sqrt{3}}{2} = \sqrt{3} \Rightarrow \phi = \frac{\pi}{3}$

Therefore, $f(x) = 4 \sin\left(x + \frac{\pi}{3}\right) - 1$

Step 2: Analyze $f(x) = 4 \sin\left(x + \frac{\pi}{3}\right) - 1$

The function $f(x) = 4 \sin\left(x + \frac{\pi}{3}\right) - 1$ oscillates with an amplitude of 4, so the range of $4 \sin\left(x + \frac{\pi}{3}\right)$ is from -4 to 4.

Thus, $f(x) \text{ ranges from } -4 - 1 = -5 \text{ to } 4 - 1 = 3$

Step 3: Check the Interval $[\pi, 2\pi]$

Since $f(x)$ ranges from -5 to 3, the maximum value of $f(x)$ over this interval will be at its highest point, which is 3.

Conclusion

The absolute maximum value of $f(x)$ on the interval $[\pi, 2\pi]$ is: $\boxed{3}$

None of the given options directly matches this value, suggesting a possible error in the choices provided. If you need further details or explanations, feel free to ask.

Would you like further explanation on any part? Here are some related questions that might interest you:

1. How can we determine maximum and minimum values of trigonometric functions over specific intervals?
2. What is the amplitude-phase form of a trigonometric function, and why is it useful?
3. How does the phase shift affect the graph of a sine function?
4. How can we verify maximum values using derivatives in trigonometric functions?
5. What techniques are available to evaluate absolute extrema for periodic functions on a closed interval?

Tip: When transforming a trigonometric function into amplitude-phase form, always verify your values for $R$ and $\phi$ carefully, as this affects the range of possible function values.