911

已知数列 ${\displaystyle \{a_n\}}$ 满足 ${\displaystyle a_{n+1}={\mathrm{e}}^{1+a_n}}$, 函数 ${\displaystyle f(x)={\displaystyle \frac{\ln x}{x_1}}}$ 的极值点为 ${\displaystyle x_0}$. 若 ${\displaystyle x_0(a_2+1)=\ln a_4}$, 求 ${\displaystyle a_1+a_2}$ 的值.

由题意，${\displaystyle f^{\prime }(x)={\displaystyle \frac{\frac{1}{x}(x+1)-\ln x}{(x+1{)}^2}}}$, 记分子为 ${\displaystyle g(x)={\displaystyle \frac{1}{x}(x+1)-\ln x}}$，则 ${\displaystyle g^{\prime }(x)={\displaystyle -\frac{1}{2x^2}-\frac{1}{x}<0}}$, 又 ${\displaystyle f^{\prime }(1)>0>f^{\prime }({\mathrm{e}}^2)}$, 故 ${\displaystyle f(x)}$ 在 ${\displaystyle (1,{\mathrm{e}}^2)}$ 上唯一存在零点 ${\displaystyle x_0}$ (可以分析出它是函数的唯一最大值点)，且$$\begin{array}{r}{f}^{\prime }(x_0)=0\Rightarrow \frac{1}{x_0}=\frac{\ln x_0}{x_0+1}.\end{array}$$
这样结合条件式和${\displaystyle \{a_n\}}$的定义，$$\begin{array}{r}\frac{\ln a_4}{1+a_2}=\frac{1+a_3}{\ln a_3}=x_0=\frac{x_0+1}{\ln x_0}\Rightarrow \frac{1}{f(a_3)}=\frac{1}{f(x_0)},\end{array}$$
从而${\displaystyle x_0=a_3}$, 进而代入${\displaystyle a_3={\mathrm{e}}^{1+a_2}}$得$$\begin{array}{r}\frac{1}{{\mathrm{e}}^{1+a_2}}=\frac{1+a_2}{{\mathrm{e}}^{1+a_2}+1}\Rightarrow \ln a_2+a_2+1=0,\end{array}$$
从而$$\begin{array}{r}{a}_1+a_2=\ln a_2-1+a_2=-2.\end{array}$$