所属成套资源:2024年高考数学复习二轮讲义(考前回顾+思想方法+六专题)
专题一 第5讲 母题突破1 导数与不等式的证明--2024年高考数学复习二轮讲义
展开这是一份专题一 第5讲 母题突破1 导数与不等式的证明--2024年高考数学复习二轮讲义,共3页。
母题突破1 导数与不等式的证明
母题 (2023·十堰调研)已知函数f(x)=(2-x)ex-ax-2.
(1)若f(x)在R上是减函数,求a的取值范围;
(2)当0≤a<1时,求证:f(x)在(0,+∞)上只有一个零点x0,且x0
❶f′x≤0恒成立
❷f′xmax≤0求解
❸0
❻2-x0≤e
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[子题1] (2023·哈师大附中模拟)已知函数f(x)=ex+exln x(其中e是自然对数的底数).
求证:f(x)≥ex2.
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[子题2] 已知函数f(x)=ln x,g(x)=ex.证明:f(x)+eq \f(2,ex)>g(-x).
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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)
(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.(2023·桂林模拟)已知函数f(x)=x2-cs x,求证:f(x)+2-eq \f(x,ex-1)>0.
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2.(2023·南昌模拟)已知函数f(x)=a(x2-1)-ln x(x>0).
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