If \( m \) and \( M \) are the absolute minimum and absolute maximum values of the function \( f(x) = 2\sqrt{2 \sin x - \tan x} \) in the interval \( \left[0, \frac{\pi}{3} \right] \), then \( m + M = \)
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- \( -1 \)
- \( 0 \)
- \( 1 \)
- \( 2 \)
The Correct Option is C
Solution and Explanation
Step 1: Analyze the function on the given interval.
We are given the function: \[ f(x) = 2\sqrt{2} \sin x - \tan x \] We examine it in the interval \( \left[0, \frac{\pi}{3} \right] \).
Step 2: Take the derivative to find critical points. \[ f'(x) = 2\sqrt{2} \cos x - \sec^2 x \] Set \( f'(x) = 0 \): \[ 2\sqrt{2} \cos x = \sec^2 x \Rightarrow 2\sqrt{2} \cos x = \frac{1}{\cos^2 x} \Rightarrow 2\sqrt{2} \cos^3 x = 1 \Rightarrow \cos x = \sqrt[3]{\frac{1}{2\sqrt{2}}} \] Step 3: Evaluate \( f(x) \) at endpoints and critical points.
Compute values at: \( x = 0 \): \( f(0) = 0 \)
\( x = \frac{\pi}{3} \): \( f\left( \frac{\pi}{3} \right) = 2\sqrt{2} \cdot \frac{\sqrt{3}}{2} - \sqrt{3} = \sqrt{6} - \sqrt{3} \approx 0.717 \)
At critical point: use calculator to get local min/max.
Using all, we find \( m + M = 1 \).
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