专题一　第5讲　母题突破3　零点问题--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3　零点问题--2024年高考数学复习二轮讲义，共4页。

思路分析
❶等价转换f(x)＝0
❷判断g(x)＝ex·sin x－x＋1的零点
❸讨论g(x)在eq \b\lc\(\rc\](\a\vs4\al\c1(0，\f(π,2)))上的零点个数
❹讨论g(x)在eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)，π))上的零点个数
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[子题1] (2023·安庆模拟)已知函数f(x)＝eln x＋bx2e1－x.若f(x)的导函数f′(x)恰有两个零点，求b的取值范围．
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[子题2] 设函数f(x)＝aln(x＋1)＋x2(a∈R)，函数g(x)＝ax－1.证明：当a≤2时，函数H(x)＝f(x)－g(x)至多有一个零点．
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规律方法 (1)求解函数零点(方程根)个数问题的步骤
①将问题转化为函数的零点问题，进而转化为函数的图象与x轴(或直线y＝k)在该区间上的交点问题．
②利用导数研究该函数在该区间上的单调性、极值(最值)、端点值等性质．
③结合图象求解．
(2)已知零点求参数的取值范围
①结合图象与单调性，分析函数的极值点．
②依据零点确定极值的范围．
③对于参数选择恰当的分类标准进行讨论．
1．(2023·郑州模拟)已知函数f(x)＝xln x＋a－ax(a∈R)．若函数f(x)在区间[1，e]上有且只有一个零点，求实数a的取值范围．
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2．(2023·商洛模拟)已知函数f(x)＝(x－2)ex，其中e为自然对数的底数．函数g(x)＝f(x)－ln x，证明：g(x)有且仅有两个零点．
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