专题一　培优点2　隐零点问题--2024年高考数学复习二轮讲义

这是一份专题一　培优点2　隐零点问题--2024年高考数学复习二轮讲义，共3页。
考点一 不含参函数的隐零点问题
例1 (2023·咸阳模拟)已知f(x)＝(x－1)2ex－eq \f(a,3)x3＋ax(x>0)(a∈R)．
(1)讨论函数f(x)的单调性；
(2)当a＝0时，判定函数g(x)＝f(x)＋ln x－eq \f(1,2)x2零点的个数，并说明理由．
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规律方法 已知不含参函数f(x)，导函数方程f′(x)＝0的根存在，却无法求出，利用零点存在定理，判断零点存在，设方程f′(x)＝0的根为x0，则①有关系式f′(x0)＝0成立，②注意确定x0的合适范围．
跟踪演练1 (2023·天津模拟)已知函数f(x)＝ln x－ax＋1，g(x)＝x(ex－x)．
(1)若直线y＝2x与函数f(x)的图象相切，求实数a的值；
(2)当a＝－1时，求证：f(x)≤g(x)＋x2.
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考点二 含参函数的隐零点问题
例2 (2023·包头模拟)已知函数f(x)＝aex－ln(x＋1)－1.
(1)当a＝e时，求曲线y＝f(x)在点(0，f(0))处的切线与两坐标轴所围成的三角形的面积；
(2)证明：当a>1时，f(x)没有零点．
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规律方法 已知含参函数f(x，a)，其中a为参数，导函数方程f′(x，a)＝0的根存在，却无法求出，设方程f′(x)＝0的根为x0，则①有关系式f′(x0)＝0成立，该关系式给出了x0，a的关系；②注意确定x0的合适范围，往往和a的范围有关．
跟踪演练2 (2023·石家庄模拟)已知函数f(x)＝x－ln x－2.
(1)讨论函数f(x)的单调性；
(2)若对任意的x∈(1，＋∞)，都有xln x＋x>k(x－1)成立，求整数k的最大值．
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