专题一第6讲母题突破1　导数与不等式的证明_2.2025数学总复习_赠品通用版（老高考）复习资料_二轮复习_2023年高考数学二轮复习讲义+课件（全国版文科）_学生版_学生用书Word版文档_520

第 6 讲 导数的综合应用
[考情分析] 1.利用导数研究函数的单调性与极值(最值)是高考的常见题型，而导数与函数、
不等式、方程、数列等的交汇命题是高考的热点和难点.2.多以解答题的形式压轴出现，难度
较大．
母题突破 1 导数与不等式的证明
母题 (2022·安康模拟)已知函数f(x)＝ln x－x－1，g(x)＝xex－2x－m，若m≤1，证明：当
x>0时，f(x)<g(x)．
思路分析
❶m≤1
↓
❷xex－2x－m≥xex－2x－1
↓
❸xex－2x－1≥xx＋1－2x－1
↓
❹构造hx＝x2－ln x.求最值
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[子题1] 已知函数f(x)＝ex－ax－a，当a＝1时，令g(x)＝.求证：当x>0时，g(x)<1.
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[子题2] (2022·九江模拟)已知函数f(x)＝ln x.
求证：e2f(x)<ex.
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________________________________________________________________________规律方法 利用导数证明不等式问题的方法
(1)直接构造函数法：证明不等式 f(x)>g(x)(或 f(x)<g(x))转化为证明 f(x)－g(x)>0(或 f(x)－
g(x)<0)，进而构造辅助函数h(x)＝f(x)－g(x)．
(2)适当放缩构造法：一是根据已知条件适当放缩；二是利用常见放缩结论．
(3)构造“形似”函数，稍作变形再构造，对原不等式同结构变形，根据相似结构构造辅助函
数．
1．已知函数f(x)＝.
(1)求f(x)在R上的极值；
(2)求证：<x2－2ln x.
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2．已知函数f(x)＝ex－a－ln(x＋a)．当a≤1时，证明：f(x)>0.
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