《2025年高考帮数学》

训练1 已知函数$f(x)=(x - 2)e^{x}+a(x - 1)^{2}$有两个零点,$a＞0$,设$x_{1},x_{2}$是$f(x)$的两个零点,证明:$x_{1}+x_{2}＜2$.

答案： 因为$f'(x)=(x - 1)(e^{x}+2a)$,且$a\gt0$,所以当$x\lt1$时,$f'(x)\lt0$,当$x\gt1$时,$f'(x)\gt0$,所以$f(x)$的极小值点为$x = 1$.
$f(x_{1})=f(x_{2})=0$,不妨设$x_{1}\lt1\lt x_{2}$,要证$x_{1}+x_{2}\lt2$,即证$x_{2}\lt2 - x_{1}$.
构造函数$F(x)=f(2 - x)-f(x),x\lt1$,代入整理得$F(x)=-xe^{-x + 2}-(x - 2)e^{x}$.
求导得$F'(x)=(1 - x)(e^{x}-e^{-x + 2})$.
当$x\lt1$时,$F'(x)\lt0$,则$F(x)$在$(-\infty,1)$上单调递减,于是$F(x)\gt f(2 - 1)-f(1)=0$,
则$f(2 - x)-f(x)\gt0$,即$f(2 - x)\gt f(x)(x\lt1)$.
将$x_{1}$代入,则$f(x_{1})\lt f(2 - x_{1})$,
又$f(x_{1})=f(x_{2})$,所以$f(x_{2})\lt f(2 - x_{1})$.
又函数$f(x)$在$(1,+\infty)$上单调递增,且$x_{2},2 - x_{1}\in(1,+\infty)$,所以$x_{2}\lt2 - x_{1}$,即$x_{1}+x_{2}\lt2$得证.

例2 已知函数$f(x)=2ax-\ln x$,其中$a\in R$.
(1)讨论函数$f(x)$的单调性.
(2)记函数$f(x)$的导函数为$f'(x)$.当$a＞0$时,若$x_{1},x_{2}(0＜x_{1}＜x_{2})$满足$f(x_{1})=f(x_{2})$,证明:$f'(x_{1})+f'(x_{2})＜0$.

答案： 函数$f(x)=2ax-\ln x$的定义域为$(0,+\infty)$,
且$f'(x)=2a-\frac{1}{x}=\frac{2ax - 1}{x},x\gt0$.
①当$a\leq0$时,$f'(x)\lt0$在$(0,+\infty)$上恒成立,
所以$f(x)$在$(0,+\infty)$上单调递减.
②当$a\gt0$时,令$f'(x)=0$,可得$x=\frac{1}{2a}$,
当$x\in(0,\frac{1}{2a})$时,$f'(x)\lt0$;当$x\in(\frac{1}{2a},+\infty)$时,$f'(x)\gt0$.
所以$f(x)$在$(0,\frac{1}{2a})$上单调递减,在$(\frac{1}{2a},+\infty)$上单调递增.
综上所述,当$a\leq0$时,$f(x)$在$(0,+\infty)$上单调递减;当$a\gt0$时,$f(x)$在$(0,\frac{1}{2a})$上单调递减,在$(\frac{1}{2a},+\infty)$上单调递增.
(2)由$x_{1},x_{2}(0\lt x_{1}\lt x_{2})$满足$f(x_{1})=f(x_{2})$,可得$2ax_{1}-\ln x_{1}=2ax_{2}-\ln x_{2}$,即$\frac{\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}=2a$.
因为$f'(x_{1})+f'(x_{2})=2a-\frac{1}{x_{1}}+2a-\frac{1}{x_{2}}=4a-\frac{x_{1}+x_{2}}{x_{1}x_{2}}$,
所以欲证$f'(x_{1})+f'(x_{2})\lt0$,即证$\frac{x_{1}+x_{2}}{x_{1}x_{2}}\gt4a$,
即证$\frac{x_{1}+x_{2}}{x_{1}x_{2}}\gt\frac{2(\ln x_{1}-\ln x_{2})}{x_{1}-x_{2}}$,
即证$2\ln\frac{x_{1}}{x_{2}}\gt\frac{x_{1}^{2}-x_{2}^{2}}{x_{1}x_{2}}=\frac{x_{1}}{x_{2}}-\frac{x_{2}}{x_{1}}$.
设$\frac{x_{1}}{x_{2}}=t(0\lt t\lt1)$,即证$2\ln t+\frac{1}{t}-t\gt0$.
设$h(t)=2\ln t+\frac{1}{t}-t(0\lt t\lt1)$,则$h'(t)=\frac{2}{t}-\frac{1}{t^{2}}-1=\frac{-(t - 1)^{2}}{t^{2}}\lt0$在$(0,1)$上恒成立,
所以$h(t)$在$(0,1)$上单调递减,所以$h(t)\gt0$,
所以$2\ln t+\frac{1}{t}-t\gt0$,即$f'(x_{1})+f'(x_{2})\lt0$.

训练2 已知函数$f(x)=\frac{\ln x + 1}{ax}$.
(1)讨论$f(x)$的单调性;
(2)若$(ex_{1})^{x_{2}}=(ex_{2})^{x_{1}}$(e是自然对数的底数),且$x_{1}＞0,x_{2}＞0,x_{1}\neq x_{2}$,证明:$x_{1}^{2}+x_{2}^{2}＞2$.

答案： 由$f(x)=\frac{\ln x + 1}{ax}$得$f'(x)=\frac{-\ln x}{ax^{2}}$,
令$f'(x)=0$,得$x = 1$.
当$a\gt0$时,若$x\in(0,1)$,则$f'(x)\gt0$;
若$x\in(1,+\infty)$,则$f'(x)\lt0$.
所以$f(x)$在$(0,1)$上单调递增,在$(1,+\infty)$上单调递减.
当$a\lt0$时,若$x\in(0,1)$,则$f'(x)\lt0$;
若$x\in(1,+\infty)$,则$f'(x)\gt0$.
所以$f(x)$在$(0,1)$上单调递减,在$(1,+\infty)$上单调递增.
综上,当$a\gt0$时,$f(x)$在$(0,1)$上单调递增,在$(1,+\infty)$上单调递减;当$a\lt0$时,$f(x)$在$(0,1)$上单调递减,在$(1,+\infty)$上单调递增.
(2)由$(ex_{1})^{x_{2}}=(ex_{2})^{x_{1}}$,两边取对数,得$x_{2}\ln(ex_{1})=x_{1}\ln(ex_{2})$,
即$x_{2}(\ln x_{1}+1)=x_{1}(\ln x_{2}+1)$,所以$\frac{\ln x_{1}+1}{x_{1}}=\frac{\ln x_{2}+1}{x_{2}}$,
即当$a = 1$时,存在$x_{1}\gt0,x_{2}\gt0,x_{1}\neq x_{2}$,满足$f(x_{1})=f(x_{2})$.
由
(1)可知,当$a = 1$时,$f(x)$在$(0,1)$上单调递增,在$(1,+\infty)$上单调递减,
所以$f(x)\leq f(1)=1$.
当$x\in(1,+\infty)$时,$f(x)=\frac{\ln x + 1}{x}\gt0$,
由于$f(\frac{1}{e})=0$,故$f(x)$在$(\frac{1}{e},1)$上恒有$f(x)\gt0$.
不妨令$x_{1}\lt x_{2}$,记$f(x_{1})=f(x_{2})=m$,则$m\in(0,1)$,且$\ln x_{1}+1=mx_{1}$ ①,$\ln x_{2}+1=mx_{2}$ ②,
①+②得$\ln(x_{1}x_{2})=m(x_{1}+x_{2})-2$ ③,
①-②得$\ln\frac{x_{1}}{x_{2}}=m(x_{1}-x_{2})$,则$m=\frac{1}{x_{1}-x_{2}}\ln\frac{x_{1}}{x_{2}}$,代入③得$\ln(x_{1}x_{2})=\frac{(x_{1}+x_{2})}{x_{1}-x_{2}}\ln\frac{x_{1}}{x_{2}}-2$.
记$t=\frac{x_{1}}{x_{2}},0\lt t\lt1$,则$\ln(x_{1}x_{2})=\frac{t + 1}{t - 1}\ln t-2=\frac{(t + 1)\ln t-2(t - 1)}{t - 1}$.
设$g(t)=(t + 1)\ln t-2(t - 1)(0\lt t\lt1)$,则$g'(t)=\ln t+\frac{1}{t}-1$.
设$h(t)=\ln t+\frac{1}{t}-1(0\lt t\lt1)$,则$h'(t)=\frac{1}{t}-\frac{1}{t^{2}}=\frac{t - 1}{t^{2}}\lt0$,所以$h(t)$在$(0,1)$上单调递减.又因为$h(1)=0$,所以当$t\in(0,1)$时,$h(t)=g'(t)\gt0$,即$g(t)$在$(0,1)$上单调递增,
又$g(1)=0$,所以当$t\in(0,1)$时,$g(t)\lt0$,
所以当$t\in(0,1)$时,$\ln(x_{1}x_{2})=\frac{(t + 1)\ln t-2(t - 1)}{t - 1}\gt0$,
所以$x_{1}x_{2}\gt1$,所以$x_{1}^{2}+x_{2}^{2}\gt2x_{1}x_{2}\gt2$.