设函数f(x)在[0,+∞)内二阶可导,并当x>0时满足 xf"(x)+3x[f’(x)]2≤1一e—x. (Ⅰ)求证:当x>0时f"(x)<1. (Ⅱ)又设f(0)=f’(0)=0,求证:当x>0时f(x)<x2.

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问题 设函数f(x)在[0,+∞)内二阶可导,并当x>0时满足
    xf"(x)+3x[f’(x)]2≤1一e—x
(Ⅰ)求证:当x>0时f"(x)<1.
(Ⅱ)又设f(0)=f’(0)=0,求证:当x>0时f(x)<x2

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答案 (Ⅰ)由假设条件有 [*] 令F(x)=x一(1一e—x)=x+e—x一1, → F(0)=0, F’(x)=1—e—x>0(x>0) → F(x)在[0,+∞)单调增加,F(x)>F(0)=0(x>0), [*]

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