类隐零点与切线放缩,越复杂越不怕
已知函数\(f(x)=\ln x-ax(x-1)\)
(1)当\(a\leq 0\)时,探究\(f(x)\)的零点个数
(2)当\(a>0\)时,证明:\(f(x)\leq\dfrac{2+a}{\sqrt{a^2+8a}-a}-\dfrac{3}{2}\)
(1) \(f(x)=\ln x-ax(x-1)= 0\)
考虑\(\dfrac{\ln x}{x}= a(x-1)\)
做出\(\dfrac{\ln x}{x}\)图像,不难得到,\(y=\dfrac{\ln x}{x}\)与\(y=a(x-1)\)总是存在一个交点
从而\(f(x)\)只有一个零点
(2)\(f^{\prime}(x)=\dfrac{1}{x}-2ax+a=\dfrac{-2ax^2+ax+1}{x}\)
记\(g(x)=-2ax^2+ax+1,\Delta=a^2+8a\)
\(x_1=\dfrac{a+\sqrt{a^2+8a}}{4a},x_2=\dfrac{a-\sqrt{a^2+8a}}{4a}=\dfrac{2}{(-a+\sqrt{a^2+8a})}\)
不难得到\(x_1>0,x_2<0\)
从而\(g(x)\)在\((0,x_1)\)上正,在\((x_1,+\infty)\)上负
从而\(f(x)_{\max}=f(x_1)=\ln x_1-ax_1(x_1-1)=\ln x_1-ax_1^2+ax_1\)
因\(-2ax_1^2+ax_1+1=0\),则\(ax_1^2=\dfrac{ax_1+1}{2}\)
从而\(f(x_1)=\ln x_1-\dfrac{ax_1+1}{2}+ax_1=\ln x_1+\dfrac{ax_1-1}{2}\leq x_1-1+\dfrac{ax_1-1}{2}\)
而\(x_1-1+\dfrac{ax_1-1}{2}=\dfrac{x_1(2+a)}{2}-\dfrac{3}{2}\leq \dfrac{2+a}{(-a+\sqrt{a^2+8a})}-\dfrac{3}{2}\)
从而\(f(x)\leq\dfrac{2+a}{\sqrt{a^2+8a}-a}-\dfrac{3}{2}\)