Discussion
The function \(f(x) = 2\cos x - x + 3\) is
Answer: B
The function is decreasing in \((0, \pi)\).
\[ f'(x) = -2\sin x - 1 \]
On \((0, \pi)\), \(\sin x \gt 0\), so \(f'(x) \lt -1 \lt 0\) throughout. The function decreases on the whole interval, with no turning point, so the options that switch between increasing and decreasing at \(\frac{\pi}{4}\) or \(\frac{\pi}{2}\) are wrong, as is 'increasing'.
Since \(f'(x) \le -1\) everywhere on the interval, \(f\) falls by at least \(\pi\) between the endpoints, from 5 to \(1 - \pi\).
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