每日导数35

一道典型但又很难想到思路的双变量问题

已知函数\(f(x)=\ln x-ax-\dfrac{1}{x}\)

\((1)\) 讨论函数\(f(x)\)的单调性

\((2)\) 函数\(f(x)\)有两个零点\(x_1,x_2(x_1<x_2)\)，证明:\(x_1x_2>2e^2\)

解

\((1)\) \(f^{\prime}(x)=\dfrac{1}{x}+\dfrac{1}{x^2}-a=\dfrac{-ax^2+x+1}{x^2}\)

当\(a=0\)时，\(f(x)\)在\(x>0\)上单调递增

当\(a>0\)时，\(\Delta>0\)，从而\(x_1<0,x_2>0\)

则\(f(x)\)在\((0,x_2)\)单调递减，在\((x_2,+\infty)\)上单调递增

当\(a<0\)时，\(f(x)\)单调递增

(2)要证明:\(x_1x_2>2e^2\)，即证：\(\ln x_1+\ln x_2>\ln 2+2\)

由题\(\begin{cases}\ln x_1-ax_1-\dfrac{1}{x_1}=0\\ \\ \ln x_2-ax_2-\dfrac{1}{x_2}=0 \end{cases}\)，即\(\ln x_1+\ln x_2=\dfrac{x_1+x_2}{x_1x_2}+a(x_1+x_2)\)

两式相减有\(a=\dfrac{\ln x_1-\ln x_2}{x_1-x_2}+\dfrac{1}{x_1x_2}\)

从而\(\ln x_1+\ln x_2=\dfrac{x_1+x_2}{x_1x_2}+\dfrac{(\ln x_1-\ln x_2)(x_1+x_2)}{x_1-x_2}+\dfrac{x_1+x_2}{x_1x_2}=\dfrac{\ln x_1-\ln x_2}{x_1-x_2}(x_1+x_2)+\dfrac{2(x_1+x_2)}{x_1x_2}\)
由对数均值不等式：\(\dfrac{a-b}{\ln a-\ln b}<\dfrac{a+b}{2}\)得

\(\ln x_1+\ln x_2>2+\dfrac{2(x_1+x_2)}{x_1x_2}\)

即\(\ln x_1+\ln x_2-\dfrac{2(x_1+x_2)}{x_1x_2}>2\)

而\(\dfrac{2(x_1+x_2)}{x_1x_2}>\dfrac{4}{\sqrt{x_1x_2}}\)

则\(2<\ln x_1+\ln x_2-\dfrac{4}{\sqrt{x_1x_2}}\)

即\(2\ln \sqrt{x_1x_2}-\dfrac{4}{\sqrt{x_1x_2}}>2\)

记\(g(t)=2\ln t-\dfrac{4}{t}-2\)，单调递增

而不难发现\(g(\sqrt{2}e)<0,g(e^2)>0\)

从而\(t>\sqrt{2}e\)

即\(x_1x_2>2e^2\)