专题一　培优点4　极值点偏移问题--2024年高考数学复习二轮讲义

这是一份专题一　培优点4　极值点偏移问题--2024年高考数学复习二轮讲义，共3页。

考点一 对称化构造函数
例1 (2023·唐山模拟)已知函数f(x)＝xe2－x.
(1)求f(x)的极值；
(2)若a>1，b>1，a≠b，f(a)＋f(b)＝4，证明：a＋b<4.
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规律方法 对称化构造函数法构造辅助函数
(1)对结论x1＋x2>2x0型，构造函数F(x)＝f(x)－f(2x0－x)．
(2)对结论x1x2>xeq \\al(2,0)型，方法一是构造函数F(x)＝f(x)－f eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(x\\al(2,0),x)))，通过研究F(x)的单调性获得不等式；方法二是两边取对数，转化成ln x1＋ln x2>2ln x0，再把ln x1，ln x2看成两变量即可．
跟踪演练1 (2022·全国甲卷)已知函数f(x)＝eq \f(ex,x)－ln x＋x－a.
(1)若f(x)≥0，求a的取值范围；
(2)证明：若f(x)有两个零点x1，x2，则x1x2<1.
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考点二 比值代换
例2 (2023·沧州模拟)已知函数f(x)＝ln x－ax－1(a∈R)．若方程f(x)＋2＝0有两个实根x1，x2，且x2>2x1，求证：x1xeq \\al(2,2)>eq \f(32,e3).(参考数据：ln 2≈0.693，ln 3≈1.099)
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规律方法 比值代换法是指通过代数变形将所证的双变量不等式通过代换t＝eq \f(x1,x2)化为单变量的函数不等式，利用函数单调性证明．
跟踪演练2 (2023·淮北模拟)已知a是实数，函数f(x)＝aln x－x.
(1)讨论f(x)的单调性；
(2)若f(x)有两个相异的零点x1，x2且x1>x2>0，求证：x1x2>e2.
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