2024年数学高考大一轮复习第三章 §3.5　利用导数研究恒(能)成立问题

§3.5　利用导数研究恒(能)成立问题

考试要求　恒(能)成立问题是高考的常考考点，其中不等式的恒(能)成立问题经常与导数及其几何意义、函数、方程等相交汇，综合考查学生分析问题、解决问题的能力，一般以压轴题的形式出现，试题难度略大．

题型一　分离参数求参数范围

例1 (2020·全国Ⅰ)已知函数f(x)＝ex＋ax2－x.

(1)当a＝1时，讨论f(x)的单调性；

(2)当x≥0时，f(x)≥x3＋1，求a的取值范围．

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思维升华 分离参数法解决恒(能)成立问题的策略

(1)分离变量，构造函数，直接把问题转化为函数的最值问题．

(2)a≥f(x)恒成立⇔a≥f(x)max；

a≤f(x)恒成立⇔a≤f(x)min；

a≥f(x)能成立⇔a≥f(x)min；

a≤f(x)能成立⇔a≤f(x)max.

跟踪训练1 (2023·苏州质检)已知函数f(x)＝ax－ex(a∈R)，g(x)＝.

(1)当a＝1时，求函数f(x)的极值；

(2)若存在x∈(0，＋∞)，使不等式f(x)≤g(x)－ex成立，求实数a的取值范围．

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题型二　等价转化求参数范围

例2 (2023·柳州模拟)已知函数f(x)＝ax－ln x.

(1)讨论函数f(x)的单调性；

(2)若x＝1为函数f(x)的极值点，当x∈[e，＋∞)时，不等式x[f(x)－x＋1]≤m(e－x)恒成立，求实数m的取值范围．

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思维升华 根据不等式恒成立构造函数转化成求函数的最值问题，一般需讨论参数范围，借助函数单调性求解．

跟踪训练2 已知函数f(x)＝(x＋a－1)ex，g(x)＝x2＋ax，其中a为常数．若对任意的x∈[0，＋∞)，不等式f(x)≥g(x)恒成立，求实数a的取值范围．

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题型三　双变量的恒(能)成立问题

例3 (2023·石家庄质检)已知函数f(x)＝ax2ln x与g(x)＝x2－bx.

(1)若f(x)与g(x)在x＝1处有相同的切线，求a，b，并证明f(x)≥g(x)；

(2)若对∀x∈[1，e]，都∃b∈使f(x)≥g(x)恒成立，求a的取值范围．

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思维升华 “双变量”的恒(能)成立问题一定要正确理解其实质，深刻挖掘内含条件，进行等价变换，常见的等价转换有

(1)∀x1，x2∈D，f(x1)>g(x2)⇔f(x)min>g(x)max.

(2)∀x1∈D1，∃x2∈D2，f(x1)>g(x2)⇔f(x)min>g(x)min.

(3)∃x1∈D1，∀x2∈D2，f(x1)>g(x2)⇔f(x)max>g(x)max.

跟踪训练3 已知函数f(x)＝ln x－mx，g(x)＝x－(a>0)．

(1)求函数f(x)的单调区间；

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(2)若m＝，对∀x1，x2∈[2,2e2]都有g(x1)≥f(x2)成立，求实数a的取值范围．

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