981

设函数 ${\displaystyle f(x)={\displaystyle (\frac{1}{2}{x}^2+ax)(\ln x-\ln a)-\frac{1}{4}{x}^2-ax}}$.

1. 当 ${\displaystyle a=1}$ 时，求曲线 ${\displaystyle f(x)}$ 在点 ${\displaystyle (\mathrm{e},f(\mathrm{e}))}$ 处的切线方程;
2. 证明: 任意 ${\displaystyle 1<a<2}$, 存在 ${\displaystyle x_0\in (0,+\mathrm{\infty })}$, 使得 ${\displaystyle {\displaystyle f(x_0)<-\frac{5}{8}(2+\ln a){\mathrm{e}}^{a-1}}}$.

1. 解 当 ${\displaystyle a=1}$, ${\displaystyle f(x)={\displaystyle (\frac{1}{2}{x}^2+x)\ln x-\frac{1}{4}{x}^2-x,f(\mathrm{e})=\frac{{\mathrm{e}}^2}{4}}}$; ${\displaystyle f^{\prime }(x)=(x+1)\ln x,f^{\prime }(\mathrm{e})=\mathrm{e}+1}$. 则所求切线方程为$$\begin{array}{r}y-\frac{{\mathrm{e}}^2}{4}=(\mathrm{e}+1)(x-\mathrm{e})⟺y=(\mathrm{e}+1)x-\mathrm{e}-\frac{3}{4}{\mathrm{e}}^2.\end{array}$$

2. 证明 考虑 ${\displaystyle x_0=a}$, ${\displaystyle h(a)={\displaystyle \frac{f(a)}{a^2}+\frac{5(2+\ln a){\mathrm{e}}^{a-1}}{8a^2}=-\frac{5}{4}+\frac{5(2+\ln a){\mathrm{e}}^{a-1}}{8a^2}}}$, 则 ${\displaystyle h(1)=0}$, 且$$\begin{array}{r}{h}^{\prime }(a)=\frac{5{\mathrm{e}}^{a-1}(2+\ln a+\frac{1}{a})\cdot 8a^2-16a\cdot 5(2+\ln a){\mathrm{e}}^{a-1}}{64a^4}=\frac{5{\mathrm{e}}^{a-1}}{8a^3}[1+(a-2)(2+\ln a)].\end{array}$$
   记 ${\displaystyle h_1(a)=(a-2)(2+\ln a),1<a<2}$, 则 $$h_1^{\prime }(a)={\displaystyle 3+\ln a-\frac{2}{a},h_1^{″}(a)=\frac{1}{a}+\frac{2}{a^2}>0,}$$ 故 ${\displaystyle h_1^{\prime }(a)>h_1^{\prime }(1)=1>0}$, 故 ${\displaystyle 1+h_1(a)\in (1+h_1(1),1+h_1(2))=(-1,1)}$, 从而 ${\displaystyle h^{\prime }(a)}$ 存在零点 ${\displaystyle 1<a_0<2}$ 且 ${\displaystyle h(a)}$ 在 ${\displaystyle (1,a_0)}$ 上减，在 ${\displaystyle (a_0,2)}$ 上增.

   最后依据 ${\displaystyle \ln 2<{\displaystyle \frac{1}{2}(2-\frac{1}{2})=\frac{3}{4}}}$, 有$$\begin{array}{r}h(2)=\frac{5[(2+\ln 2)\mathrm{e}-8]}{32}<\frac{5}{32}[(2+\frac{3}{4})\cdot 2.72-8]=\frac{5}{32}(7.48-8)<0,\end{array}$$
   故 ${\displaystyle \underset{1<a<2}{max}h(a)<max\{h(1),h(2)\}\le 0}$, 也即任意 ${\displaystyle 1<a<2}$, 可以取 ${\displaystyle x_0=a}$ 使得原不等式成立.