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已知函数 ${\displaystyle f(x)=ax-2-\ln x}$, 其中 ${\displaystyle a\in \mathbb{R}}$.

1. 讨论 ${\displaystyle f(x)}$ 的单调性;
2. 若 ${\displaystyle f(x)}$ 有两个零点 ${\displaystyle x_1,x_2}$，且 ${\displaystyle x_2\ge 3x_1}$, 求 ${\displaystyle x_1{x}_2}$ 的最小值.

1. ${\displaystyle f^{\prime }(x)={\displaystyle a-\frac{1}{x}}}$. 当 ${\displaystyle a\le 0}$, ${\displaystyle f(x)}$ 在 ${\displaystyle (0,+\mathrm{\infty })}$ 上单减；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. 由题意$$\begin{array}{r}A=\frac{2+\ln x_1}{x_1}=\frac{2+\ln x_2}{x_2}.\end{array}$$
   记 ${\displaystyle t={\displaystyle \frac{x_2}{x_1}\ge 3}}$, 则上式变为$$\begin{array}{r}T(2+\ln x_1)=2+\ln (t{x}_1),\end{array}$$
   解得 ${\displaystyle {\displaystyle \ln x_2=\frac{\ln t}{t-1}-2,\ln x_1=\frac{t\ln t}{t-1}-2}}$, 从而 ${\displaystyle \ln x_1{x}_2={\displaystyle \frac{t+1}{t-1}\ln t-4}}$. 而 $${\displaystyle {(\frac{t+1}{t-1}\ln t-4)}^{\prime }=\frac{1}{(t-1{)}^2}(t-\frac{1}{t}-2\ln t)\ge 0,}$$ 从而 ${\displaystyle {\displaystyle \ln x_1{x}_2\ge \frac{4}{2}\ln 3-4\Rightarrow x_1{x}_2\ge {\mathrm{e}}^{2\ln 3-4}=\frac{9}{{\mathrm{e}}^4}}}$.