专题一　第5讲　母题突破2　恒成立问题与能成立问题--2024年高考数学复习二轮讲义

这是一份专题一　第5讲　母题突破2　恒成立问题与能成立问题--2024年高考数学复习二轮讲义，共3页。

思路分析一
❶fx≥0恒成立
❷fxmin≥0
❸分类讨论求fxmin
思路分析二
❶fx≥0恒成立
❷求证x－ln x>0
❸分离参数构造新函数
❹求新函数最值
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[子题1] (2023·青岛模拟)已知函数f(x)＝ex－a－ln x．若存在x0∈[e，＋∞)，使f(x0)<0，求a的取值范围．
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[子题2] (2023·全国乙卷改编)已知函数f(x)＝eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(1,x)＋a))ln(1＋x)．若f′(x)≥0在区间(0，＋∞)上恒成立，求a的取值范围．
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规律方法 (1)由不等式恒成立求参数的取值范围问题的策略
①求最值法：将恒成立问题转化为利用导数求函数的最值问题．
②分离参数法：将参数分离出来，进而转化为a>f(x)max或a(2)不等式有解问题可类比恒成立问题进行转化，要理解清楚两类问题的差别．
1．已知函数f(x)＝(x－4)ex－x2＋6x，g(x)＝ln x－(a＋1)x，a>－1.若存在x1∈[1,3]，对任意的x2∈[e2，e3]，使得不等式g(x2)>f(x1)成立，求实数a的取值范围．(e3≈20.09)
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2．(2023·全国甲卷)已知函数f(x)＝ax－eq \f(sin x,cs3x)，x∈eq \b\lc\(\rc\)(\a\vs4\al\c1(0，\f(π,2))).
(1)当a＝8时，讨论f(x)的单调性；
(2)若f(x)________________________________________________________________________
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