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已知曲线 ${\displaystyle y=\ln x+a(a\in \mathbb{R})}$ 和圆 ${\displaystyle O:x^2+y^2=r^2(r>0)}$ 相交于 ${\displaystyle A,B}$ 两个不同点，记直线 ${\displaystyle AB}$ 的斜率为 ${\displaystyle k}$.

1. 当 ${\displaystyle r=\sqrt{2}}$ 时，证明: ${\displaystyle a>-1}$;
2. 当 ${\displaystyle r={\displaystyle \frac{\sqrt{3}}{2}}}$ 时，证明: ${\displaystyle k>\sqrt{2}}$.

1. 反设 ${\displaystyle a\le -1}$，则$$\begin{array}{r}x-\ln x-a\ge x-(x-1)-(-1)=2>0,\end{array}$$
   因此$$\begin{array}{r}2=x^2+(\ln x+a{)}^2\ge \frac{(x-\ln x-a{)}^2}{2}\ge \frac{2^2}{2}=2,\end{array}$$
   这样等号必须取到，取等时 ${\displaystyle a=-1,x=1}$, 这不能得出两个不同的 ${\displaystyle x}$，与 ${\displaystyle A,B}$ 不同矛盾，因此 ${\displaystyle a>-1}$.

2. 记 ${\displaystyle f(x)=x^2+(\ln x+a{)}^2}$. 不妨设 ${\displaystyle A(x_1,y_1),B(x_2,y_2),x_1<x_2}$. 则 ${\displaystyle x_1,x_2}$ 是 ${\displaystyle {\displaystyle \frac{3}{4}=f(x)}}$ 的解. 当 ${\displaystyle \ln x+a>0}$ (即 ${\displaystyle x>{\mathrm{e}}^{-a}}$)时，${\displaystyle f^{\prime }(x)={\displaystyle \frac{2(x^2+\ln x+a)}{x}>0}}$, 因此 ${\displaystyle x_1,x_2}$ 至多有一个大于 ${\displaystyle {\mathrm{e}}^{-a}}$.

   • 如果 ${\displaystyle x_1,x_2\le {\mathrm{e}}^{-a}}$, 则$$\begin{array}{r}\{\begin{array}{rl}& -\ln x_1-a=\sqrt{\frac{3}{4}-x_1^2},\\ & -\ln x_2-a=\sqrt{\frac{3}{4}-x_2^2}\end{array}\Rightarrow g(x_1)=g(x_2),\end{array}$$
     其中 ${\displaystyle g(x)={\displaystyle \sqrt{\frac{3}{4}-x^2}+\ln x,0<x\le \frac{\sqrt{3}}{4}}}$. ${\displaystyle g^{\prime }(x)={\displaystyle \frac{1}{x}-\frac{x}{\sqrt{\frac{3}{4}-x^2}}}}$,可以解得唯一极大值点 ${\displaystyle {\displaystyle x_0=\frac{\sqrt{2}}{2}}}$. 另一方面要证 ${\displaystyle k>\sqrt{2}}$, 即证 ${\displaystyle h(x_1)<h(x_2),h(x)=\ln x-\sqrt{2}x}$. ${\displaystyle h(x)}$ 同样有唯一极大值点 ${\displaystyle x_0}$.
     首先我们证明${\displaystyle x_0-x_1>x_2-x_0}$, 这是一个传统的极值点偏移问题，只需证明$$\begin{array}{rl}& g(\frac{\sqrt{2}}{2}-x)>g(\frac{\sqrt{2}}{2}+x),\mathrm{\forall }0<x<\frac{\sqrt{2}}{2}\\ ⟺& \sqrt{\frac{1}{4}-x^2+\sqrt{2}x}-\sqrt{\frac{1}{4}-x^2-\sqrt{2}x}>\ln \frac{1+\sqrt{2}x}{1-\sqrt{2}x}.\end{array}$$
     上述不等式$$\begin{array}{rl}\text{左边}=& \frac{2\sqrt{2}x}{\sqrt{\frac{1}{4}-x^2+\sqrt{2}x}+\sqrt{\frac{1}{4}-x^2-\sqrt{2}x}}=\frac{2\sqrt{2}x}{\sqrt{\frac{1}{2}-2x^2+2\sqrt{{(\frac{1}{4}-x^2)}^2-2x^2}}}\\ \ge & \frac{2\sqrt{2}x}{\sqrt{\frac{1}{2}-2x^2+2(\frac{1}{4}-x^2)}}=\frac{2\sqrt{2}x}{\sqrt{1-4x^2}}>\frac{2\sqrt{2}x}{\sqrt{1-4x^2+4x^4}}=\frac{2\sqrt{2}x}{1-2x^2},\\ \text{右边}<& \frac{1}{2}(\frac{1+\sqrt{2}x}{1-\sqrt{2}x}-\frac{1-\sqrt{2}x}{1+\sqrt{2}x})=\frac{2\sqrt{2}x}{1-2x^2},\end{array}$$
     因此得证. 同理，证明$$\begin{array}{r}h(\frac{\sqrt{2}}{2}-x)<h(\frac{\sqrt{2}}{2}+x)⟺\ln \frac{1+\sqrt{2}x}{1-\sqrt{2}x}>2\sqrt{2}x.\end{array}$$
     这由${\displaystyle \ln x>{\displaystyle \frac{2(x-1)}{x+1},x>1}}$立即得到.
     得到了这两个结论，我们可以取${\displaystyle x_3<x_0}$, 使${\displaystyle h(x_3)=h(x_2)}$, 则根据极值点偏移的结论：${\displaystyle x_0-x_3<x_2-x_0<x_0-x_1\Rightarrow x_0>x_3>x_1\Rightarrow h(x_1)<h(x_3)=h(x_2)}$(可以参考如下图示), 问题得证！
   • 如果 ${\displaystyle x_1\le {\mathrm{e}}^{-a}<x_2}$, 则 $$\begin{array}{r}g(x_1)=\ln x_2-\sqrt{\frac{3}{4}-x_2^2}<g(x_2).\end{array}$$
     这样如果 ${\displaystyle x_1<x_2<x_0}$, 则 ${\displaystyle x_1,x_2}$ 也落在 ${\displaystyle h(x)}$ 的单增区间上，${\displaystyle h(x_1)<h(x_2)}$; 如果 ${\displaystyle x_1<x_0<x_2}$, 则 ${\displaystyle \mathrm{\exists }{x}_2^{\prime }>x_2}$, 使得 ${\displaystyle g(x_1)=g(x_2^{\prime })}$. 根据情形一的结论，${\displaystyle h(x_1)<h(x_2^{\prime })<h(x_2)}$.

   综上所述，结论得证！