回复 1# 力工
\[f'(x)=\frac{1}{x}+2x-a=0\]
\[x=\frac{1}{4}(a\pm\sqrt{a^2-8})\]
这里要有三个解得保证
\[f\left(\frac{1}{4}(a+\sqrt{a^2-8})\right)=\frac{1}{8}(12-a(a+\sqrt{a^2-8}))+\ln\left(\frac{1}{4}(a+\sqrt{a^2-8})\right)<0\]
这其中如果令中间这块为$g(a)$,有
\[g'(a)=\frac{d}{da}\left[\frac{1}{8}(12-a(a+\sqrt{a^2-8}))+\ln\left(\frac{1}{4}(a+\sqrt{a^2-8})\right)\right]=-\frac{1}{4}(a+\sqrt{a^2-8})<0\]
而$g(3)=0$,因此$a>3$
按中值定理,假设$m<n$,则存在$\xi\in(m,n)$使得$\frac{f(m)-f(n)}{m-n}=f'(\xi)$,这里不妨找一下$f'(x)$的最小值,那么
\[f''(x)=2-\frac{1}{x^2}\]
其最小值在$x=\frac{1}{\sqrt{2}}$取到,此时
\[f'(\frac{1}{\sqrt{2}})=2\sqrt{2}-a<2\sqrt{2}-3\]
那么如果令$a=3+\Delta a$,其中$\Delta a>0$为常数,鉴于
\[\frac{f(m)-f(n)}{m-n}=f'(\xi)>f'(\frac{1}{\sqrt{2}})\]
我们令$m=\frac{1}{\sqrt{2}},n\to \left(\frac{1}{\sqrt{2}}\right)^+$,就会有
\[\lim_{m=\frac{1}{\sqrt{2}},n\to \left(\frac{1}{\sqrt{2}}\right)^+}\frac{f(m)-f(n)}{m-n}=f'(\frac{1}{\sqrt{2}})=2\sqrt{2}-a=2\sqrt{2}-3-\Delta a\]
按极限定义,对任意给定正数$\sigma$,总存在$n=\frac{1}{\sqrt{2}}+\Delta n, \Delta n>0$使得
\[|\frac{f(\frac{1}{\sqrt{2}})-f(\frac{1}{\sqrt{2}}+\Delta n)}{\Delta n}-f'(\frac{1}{\sqrt{2}})|<\sigma\]
那么当$\sigma=\Delta a$时自然也不例外,此时存在$n$使得
\[\frac{f(\frac{1}{\sqrt{2}})-f(n)}{\frac{1}{\sqrt{2}}-n}-(2\sqrt{2}-3-\Delta a)<\Delta a\]
\[\frac{f(\frac{1}{\sqrt{2}})-f(n)}{\frac{1}{\sqrt{2}}-n}<2\sqrt{2}-3\] |