2. 考研
  3. 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．

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．
选项
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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