2023版考前三个月冲刺专题练　第9练　导数与不等式证明【无答案版】

这是一份2023版考前三个月冲刺专题练　第9练　导数与不等式证明【无答案版】，共2页。
第9练　导数与不等式证明[考情分析]　导数与不等式证明是高考考查的重点内容，在解答题中一般会考查函数的单调性、极值和最值的综合运用，试题难度较大，多以压轴题出现． 一、单变量函数不等式的证明例1　(2022·新高考全国Ⅱ)已知函数f(x)＝xeax－ex.(1)当a＝1时，讨论f(x)的单调性；(2)当x>0时，f(x)<－1，求a的取值范围；(3)设n∈N*，证明：＋＋…＋>ln(n＋1)．______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________规律方法　用导数证明不等式一般有以下方法(1)构造函数法．(2)由结论出发，通过对函数变形，证明不等式．(3)分成两个函数进行研究．(4)利用图象的特点证明不等式．(5)利用放缩法证明不等式．跟踪训练1　(2022·宜宾第四中学模拟)已知函数f(x)＝－ax2＋xln x＋2.(1)若f(x)有两个极值点，求实数a的取值范围；(2)当a＝0时，证明：f(x)>x－.______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________二、双变量函数不等式的证明例2　(2022·全国甲卷)已知函数f(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.______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________规律方法　破解含双参不等式的证明的关键：一是转化，即由已知条件入手，寻找双参所满足的关系式，并把含双参的不等式转化为含单参的不等式；二是构造函数，借助导数，判断函数的单调性，从而求其值；三是回归含双参的不等式的证明，把所求的最值应用到含参的不等式中，即可证得结果．跟踪训练2　(2022·绍兴模拟)已知函数f(x)＝ex－a－x2(a∈R)有两个极值点x1，x2(x1<x2)，其中e＝2.718 28…为自然对数的底数．(1)记f′(x)为f(x)的导函数，证明：f′(1)<0；(2)证明：<.____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________