2024学生版大二轮数学新高考提高版（京津琼鲁辽粤冀鄂湘渝闽苏浙黑吉晋皖云豫新甘贵赣桂）专题一　第5讲　母题突破3　零点问题13

母题突破3零点问题母题已知函数f(x)＝sinx－eq\f(x－1,ex)，判断f(x)在(0，π)上零点的个数，并说明理由．思路分析❶等价转换f(x)＝0❷判断g(x)＝ex·sinx－x＋1的零点❸讨论g(x)在eq\b\lc\(\rc\](\a\vs4\al\co1(0，\f(π,2)))上的零点个数❹讨论g(x)在eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)，π))上的零点个数________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子题1](2023·安庆模拟)已知函数f(x)＝elnx＋bx2e1－x.若f(x)的导函数f′(x)恰有两个零点，求b的取值范围．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[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．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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)＝xlnx＋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．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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)－lnx，证明：g(x)有且仅有两个零点．____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