简单的但难于计算的二次函数形双参问题

已知函数\(f(x)=x^2-ax+2\ln x\)

(1)讨论\(f(x)\)单调性

(2)已知\(f(x)\)有两个极值点\(x_1,x_2\)且\(x_1<x_2\)，证明:$2f(x_1)-f(x_2)\geq -3\ln 2 $

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

记\(y=2x^2-ax+2,\Delta=a^2-16,x_1+x_2=\dfrac{a}{2},x_1x_2=1,x_{1,2}=\dfrac{a\pm\sqrt{a^2-16}}{4}\)

Case1 \(\Delta\leq 0\)即\(a\in[-4,4]\)时，两根存在

此时\(f(x)\)在\(x>0\)上单调递减

Case2 \(\Delta>0\)即\(a<-4\)或\(a>4\)时，两根存在并且都为正，从而有

\(f(x)\)在\((0,x_1)\)跟\((x_2,+\infty)\)上增，在\((x_1,x_2)\)上减

(2)\(x_1=\dfrac{a-\sqrt{a^2-16}}{4},x_2=\dfrac{a+\sqrt{a^2-16}}{4}\)

\(2f(x_1)-f(x_2)\)

\(=2(x_1^2-ax_1+2\ln x_1)-(x_2^2-ax_2+2\ln x_2)\)

\(=2x_1^2-2ax_1+4\ln x_1-x_2^2+ax_2-2\ln x_2\)

\(=2x_1^2+2(x_1+x_2)(x_2-2x_1)-x_2^2+4\ln x_1-2\ln x_2\)

\(=2x_1^2+2x_1x_2-4x_1^2+2x_2^2-4x_1x_2-x_2^2+4\ln x_1-2\ln x_2\)

\(=-2x_1^2+x_2^2+4\ln x_1-2\ln x_2-2\)

\(=x_2^2-\dfrac{2}{x_2^2}-4\ln x_2-2\ln x_2-2\)

\(=x_2^2-6\ln x_2-\dfrac{2}{x_2^2}-2\)

记\(\gamma(t)=t^2-6\ln t-\dfrac{2}{t^2}-2,t>1\)

\(\gamma^{\prime}(t)=2t-\dfrac{6}{t}+\dfrac{4}{t^3}\)

\(=\dfrac{2t^4-6t^2+4}{t^3}\)

\(=\dfrac{2(t^4-3t^2+2)}{t^3}\)

\(=\dfrac{2(t^2-1)(t^2-2)}{t^3}\)

\(=\dfrac{2(t-1)(t+1)(t-\sqrt{2})(t+\sqrt{2})}{t^3}\)

不难得到\(\gamma(t)\)在\((1,\sqrt{2})\)上递减，在\((\sqrt{2},+\infty)\)上增

从而\(2f(x_1)-f(x_2)=\gamma(t)\geq \gamma(\sqrt{2})=2-6\ln \sqrt{2}-\dfrac{2}{2}-2=-1-3\ln 2\)

得证