Note $ s = \frac{1}{2} \sin 2x $, and since $ 0 < x < \frac{\pi}{2} $, $ 0 < 2x < \pi $, so $ 0 < \sin 2x \leq 1 $, hence $ 0 < s \leq \frac{1}{2} $.

Fri, 17 Apr 2026 13:10:32 +0000

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