1733

在平面直角坐标系 ${\displaystyle xOy}$ 中, 曲线 ${\displaystyle {\mathrm{\Gamma }}_1:y=a^x}$ (${\displaystyle a>1}$)与 ${\displaystyle {\mathrm{\Gamma }}_2:xy=m}$ (${\displaystyle m>0}$)交于点 ${\displaystyle A}$.

1. 当 ${\displaystyle a=\mathrm{e}}$ 时, 求曲线 ${\displaystyle {\mathrm{\Gamma }}_1}$ 在 ${\displaystyle (0,1)}$ 的切线方程;
2. 若直线 ${\displaystyle OA}$ 与 ${\displaystyle {\mathrm{\Gamma }}_1}$ 相切于点 ${\displaystyle A}$, 求 ${\displaystyle a^m}$ 的值;
3. 若直线 ${\displaystyle OA}$ 与 ${\displaystyle {\mathrm{\Gamma }}_1}$ 交于另一点 ${\displaystyle B}$, 且 ${\displaystyle |OB|\ge 2|OA|}$, 求 ${\displaystyle a^m}$ 的取值范围.

1. 是 ${\displaystyle y=x+1}$.
2. 记 ${\displaystyle A(x_0,a^{x_0})}$, ${\displaystyle f(x)=a^x}$, 则 ${\displaystyle f(x)}$ 在 ${\displaystyle A}$ 处的切线为 ${\displaystyle y-a^{x_0}=a^{x_0}\ln a(x-x_0)}$, 代入 ${\displaystyle (0,0)}$: ${\displaystyle x_0=\frac{1}{\ln a}}$.
   另一方面 ${\displaystyle A}$ 在 ${\displaystyle {\mathrm{\Gamma }}_2}$ 上, 所以 ${\displaystyle m=a^{x_0}{x}_0}$. 这样 $$m\ln a=a^{x_0}\frac{1}{\ln a}\ln a={\mathrm{e}}^{\ln a\cdot x_0}=\mathrm{e}\Rightarrow a^m={\mathrm{e}}^{\mathrm{e}}.$$
3. 设 ${\displaystyle A(x_1,y_1)}$, ${\displaystyle B(x_2,y_2)}$, ${\displaystyle l_{AB}:y=kx}$. 再记 ${\displaystyle t=\frac{x_2}{x_1}=\frac{OB}{OA}\ge 2}$. 这样 $$k=\frac{y_2}{x_2}=\frac{y_1}{x_1}\Rightarrow \frac{a^{x_2}}{x_2}=\frac{a^{x_1}}{x_1},$$ 消去 ${\displaystyle x_2}$ 得 ${\displaystyle x_1=\frac{\ln t}{(t-1)\ln a}}$, 所以 ${\displaystyle \ln x_1=\ln \ln t-\ln (t-1)-\ln \ln a}$, ${\displaystyle \ln y_1=x_1\ln a=\frac{\ln t}{t-1}}$, ${\displaystyle \ln m=\ln x_1+\ln y_1=\ln \left(\frac{\ln t}{t-1}\right)+\frac{\ln t}{t-1}-\ln \ln a}$. 这样 $$a^m={\mathrm{e}}^{m\ln a}={\mathrm{e}}^{{\mathrm{e}}^{\ln m+\ln \ln a}}=\exp (\exp (\ln \left(\frac{\ln t}{t-1}\right)+\frac{\ln t}{t-1})).$$ 记 ${\displaystyle g(t)=\frac{\ln t}{t-1}}$, 则 ${\displaystyle g^{\prime }(t)=\frac{1-\frac{1}{t}-\ln t}{(t-1{)}^2}\le 0}$, 于是 ${\displaystyle g(t)\le g(2)=\ln 2}$. 这样 $$a^m=\exp (\exp (\ln g(t)+g(t)))\le \exp (\exp (\ln \ln 2+\ln 2))=\exp (2\ln 2)=4.$$ 而 ${\displaystyle m>0}$, ${\displaystyle a>1}$, 所以 ${\displaystyle a^m>1}$.
   综上, ${\displaystyle a^m\in (1,4]}$.