849

已知函数 ${\displaystyle f(x)=ax-\ln x,a\in \mathbb{R}}$.

1. 讨论函数 ${\displaystyle f(x)}$ 的单调性;
2. 当 ${\displaystyle a=1}$ 时, 设 ${\displaystyle g(x)={\displaystyle \frac{(x-1{)}^2}{f(x)}}}$, 求证: 函数 ${\displaystyle g(x)}$ 存在极大值点 ${\displaystyle x_0}$，且 ${\displaystyle {\displaystyle \frac{2}{{\mathrm{e}}^2}<g(x_0)<\frac{2}{3}}}$.

1. ${\displaystyle f^{\prime }(x)={\displaystyle \frac{ax-1}{x}}}$. 当 ${\displaystyle a\le 0}$, ${\displaystyle f(x)}$ 在 ${\displaystyle (0,+\mathrm{\infty })}$ 上单增; 当 ${\displaystyle a>0}$, ${\displaystyle f(x)}$ 在 ${\displaystyle {\displaystyle (0,\frac{1}{a})}}$ 上单减，在 ${\displaystyle {\displaystyle (\frac{1}{a},+\mathrm{\infty })}}$ 上单增.
2. 求导得 ${\displaystyle g^{\prime }(x)={\displaystyle \frac{x-1}{(x-\ln x{)}^2}(x-\frac{1}{x}+2-2\ln x)}}$. 记 ${\displaystyle h(x)={\displaystyle x-\frac{1}{x}+2-2\ln x}}$, 则 ${\displaystyle h^{\prime }(x)={\displaystyle \frac{(x-1{)}^2}{x^2}\ge 0}}$. 又$$\begin{array}{rl}& h\left(\frac{1}{{\mathrm{e}}^2}\right)=6+\frac{1}{{\mathrm{e}}^2}-{\mathrm{e}}^2<7-{\mathrm{e}}^2<0,\\ & h\left(\frac{1}{3}\right)=2\ln 3-\frac{2}{3}>2-\frac{2}{3}>0,\end{array}$$
   从而 ${\displaystyle g^{\prime }(x)}$ 存在唯一零点 ${\displaystyle x_0\in {\displaystyle (\frac{1}{{\mathrm{e}}^2},\frac{1}{3})}}$, 且它是 ${\displaystyle g(x)}$ 的唯一极大值点，进而