Question

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 \( m + M = \)

• A. \( -1 \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( 2 \)

Hint:

For absolute extrema on closed intervals, always evaluate the function at endpoints and critical points within the interval.

Answer 1

The Correct Option is C

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 \( m + M = 1 \).