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[主观题]
设f(x)在[0,a]上可导,且f(0)=0,0<f'(x)≤1,试证
设f(x)在[0,a]上可导,且f(0)=0,0<f'(x)≤1,试证
提问人:网友anonymity
发布时间:2022-12-27
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[证明] 对
,先证F'(u)>0,u∈(0,a).
由于![](https://img2.soutiyun.com/latex/latex.action)
又f(u)>0(因f'(x)>0,0=f(0)<f(x)),只要证明![](https://img2.soutiyun.com/latex/latex.action)
因为G'(u)=2f(u)[1-f'(u)]>0,G(u)单调增加,G(u)>G(0)=0.于是F'(u)>0,F(u)单调增加,F(u)>F(0)=0,u∈(0,a].取u=a,得F(a)>0.即
作
,要证F(a)≥0,只要证F(u)≥0即可
由于
又f(u)>0(因f'(x)>0,0=f(0)<f(x)),只要证明
因为G'(u)=2f(u)[1-f'(u)]>0,G(u)单调增加,G(u)>G(0)=0.于是F'(u)>0,F(u)单调增加,F(u)>F(0)=0,u∈(0,a].取u=a,得F(a)>0.即
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