《2025年世纪金榜高中全程复习方略高中数学A版》

视角一 利用双变量的关系化为单变量
[例1](2023·长春模拟)已知$f(x)=\frac{1}{2}x^{2}-2x + 2a\ln x$有两个极值点$x_{1},x_{2}$，证明：$f(x_{1})+f(x_{2})＞-3$.

答案： [证明]$f'(x)=x - 2+\frac{2a}{x}=\frac{x^{2}-2x + 2a}{x}(x＞0)$。因为函数$f(x)$有两个极值点$x_{1}$，$x_{2}$，所以方程$x^{2}-2x + 2a = 0$有两个正根$x_{1}$，$x_{2}$，所以$\begin{cases}x_{1}+x_{2}=2,\\x_{1}\cdot x_{2}=2a＞0,\end{cases}$且$\Delta = 4 - 8a＞0$，解得$0＜a＜\frac{1}{2}$。由题意得$f(x_{1})+f(x_{2})=\frac{1}{2}x_{1}^{2}-2x_{1}+2a\ln x_{1}+\frac{1}{2}x_{2}^{2}-2x_{2}+2a\ln x_{2}=\frac{1}{2}(x_{1}^{2}+x_{2}^{2})-2(x_{1}+x_{2})+2a\ln(x_{1}\cdot x_{2})=\frac{1}{2}(x_{1}+x_{2})^{2}-x_{1}\cdot x_{2}-2(x_{1}+x_{2})+2a\ln(x_{1}\cdot x_{2})=2a\ln(2a)-2a - 2$，令$h(a)=2a\ln(2a)-2a - 2(0＜a＜\frac{1}{2})$，则$h'(a)=2\ln(2a)＜0$，所以$y = h(a)$在$(0,\frac{1}{2})$上单调递减，所以$h(a)＞h(\frac{1}{2})=-3$，所以$f(x_{1})+f(x_{2})＞-3$。

视角二 作比整体代换法
[例2]已知函数$f(x)=x\ln x-\frac{1}{2}ax^{2}-x,a\in\mathbf{R}$.若$f(x)$的两个极值点为$x_{1},x_{2}$，且$x_{1}＜x_{2}$，求证$x_{1}x_{2}＞e^{2}$.

答案： [证明]要证$x_{1}x_{2}＞e^{2}$，只需证$\ln x_{1}+\ln x_{2}＞2$。$f(x)$的两个极值点$x_{1}$，$x_{2}$，即函数$f'(x)$的两个零点。因为$f(x)=\ln x - ax$，所以$x_{1}$，$x_{2}$是方程$f'(x)=0$的两个不同实数根。于是$\begin{cases}\ln x_{1}-ax_{1}=0,\\\ln x_{2}-ax_{2}=0.\end{cases}$两式相加，整理得$a=\frac{\ln x_{1}+\ln x_{2}}{x_{1}+x_{2}}$，两式相减，整理得$a=\frac{\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}$，因此$\frac{\ln x_{1}+\ln x_{2}}{x_{1}+x_{2}}=\frac{\ln x_{2}-\ln x_{1}}{x_{2}-x_{1}}$，于是$\ln x_{1}+\ln x_{2}=\frac{(\ln x_{2}-\ln x_{1})(x_{2}+x_{1})}{x_{2}-x_{1}}=\frac{(1+\frac{x_{2}}{x_{1}})\ln\frac{x_{2}}{x_{1}}}{\frac{x_{2}}{x_{1}}-1}$，又$0＜x_{1}＜x_{2}$，可设$t=\frac{x_{2}}{x_{1}}$，则$t＞1$。因此$\ln x_{1}+\ln x_{2}=\frac{(1 + t)\ln t}{t - 1}(t＞1)$。要证$\ln x_{1}+\ln x_{2}＞2$，即证$\frac{(1 + t)\ln t}{t - 1}＞2(t＞1)$。设$g(t)=\ln t-\frac{2(t - 1)}{t + 1}(t＞1)$，$g'(t)=\frac{1}{t}-\frac{2(t + 1)-2(t - 1)}{(t + 1)^{2}}=\frac{(t - 1)^{2}}{t(t + 1)^{2}}＞0$，所以$g(t)$在$(1,+\infty)$上单调递增，$g(t)＞g(1)=0$，于是，当$t＞1$时，有$\ln t＞\frac{2(t - 1)}{t + 1}$，所以$\ln x_{1}+\ln x_{2}＞2$成立，即$x_{1}x_{2}＞e^{2}$。

[例3]已知函数$f(x)=(a + 1)\ln x+ax^{2}+1$，若$a＜-1$，对$\forall x_{1},x_{2}\in(0,+\infty),|f(x_{1})-f(x_{2})|\geq4|x_{1}-x_{2}|$，求$a$的取值范围.

答案： [解析]$f(x)$的定义域为$(0,+\infty)$，$f'(x)=\frac{a + 1}{x}+2ax$。当$a＜-1$时，$f'(x)＜0$，所以函数$f(x)$在$(0,+\infty)$上单调递减。不妨设$x_{1}\geq x_{2}$，从而对$\forall x_{1}$，$x_{2}\in(0,+\infty)$，$|f(x_{1})-f(x_{2})|\geq4|x_{1}-x_{2}|$等价于$f(x_{2})+4x_{2}\geq f(x_{1})+4x_{1}$，①令$g(x)=f(x)+4x$，则$g'(x)=\frac{a + 1}{x}+2ax + 4$，①等价于$g(x)$在$(0,+\infty)$上单调递减，即$\frac{a + 1}{x}+2ax + 4\leq0$，从而$a\leq\frac{-4x - 1}{2x^{2}+1}=\frac{(2x - 1)^{2}-4x^{2}-2}{2x^{2}+1}=\frac{(2x - 1)^{2}}{2x^{2}+1}-2$，故$a$的取值范围为$(-\infty,-2]$。