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已知函数$f(x)=2x-x^2$,函数$g(x)=sin(\dfrac{πx}{2})$,证明: ⅰ.当$x∈[0,2]$时,$f(x)\geqslant g(x)$ ⅱ. 当$x∈[0,2]$时,$f(x)\leqslant\sqrt{g(x)}$ ⅲ. 当$x∈[0,2]$时,$\sqrt{f(x)}\leqslant g(\sqrt{x})$ ... 青青子衿 发表于 2013-10-2 10:20
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2025-6-4 21:24 GMT+8
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