设f(x)在[0,t](t>0)上有n阶导数且非负,已知f(0)=f’+(0)=f”+(0)=…=f+(n-2)(0)=0,f(n)(x)>0. (I)求F(t)=∫0tsf(x)dx-t∫0tf(x)dx(n为大于1的正整数)的n阶导数; (Ⅱ)证明:(

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问题 设f(x)在[0,t](t>0)上有n阶导数且非负,已知f(0)=f’+(0)=f”+(0)=…=f+(n-2)(0)=0,f(n)(x)>0.
(I)求F(t)=∫0tsf(x)dx-t∫0tf(x)dx(n为大于1的正整数)的n阶导数;
(Ⅱ)证明:(Ⅰ)中的F(t)>0.

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答案(Ⅰ)F(t)=∫0txf(x)dxt-[*]t∫0tf(x)dx变形为 (n+1)F(t)=(n+1)∫0txf(x)dx-nt∫0tf(x)dx, 则 [(n+1)F(t)]’=(n+1)tf(t)-n[∫0tf(x)dx+tf(t)] =tf(t)-n∫0tf(x)dx, [(n+1)F(t)]”=f(t)+tf’(t)-nf(t)=(1-n)f(t)+tf’(f), [(n+1)F(t)]’”=(1-n)f’(t)+f’(t)+f”(t)=(2-n)f’(t)+tf”(t), 依此类推,得 [(n+1)F(t)](n))=(n-1-n)f(n-2)(t)+tf(n-1)(t), 故[F(t)](n)=[*] (Ⅱ)由f0(n-2)(0)=0,应用拉格朗日中值定理,有 [F(t)](n) [*] 由f(n))>0,知f(n-1)(x)单调增加,故[F(t)](n)>0,所以[F(t)](n-1)单调增加.又[F(0)](n-1)=0,知[F(t)](n-1)>[F(0)](n-1)=0.依此类推,可得F(t)>F(0)=0.

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