高考数学？杂题

1

证明：

\[\ln(1+\frac{1}{9})<0.1e^{0.1}<\frac{1}{9} \]

一方面：

\[\begin{aligned} &e^{0.1}<\frac{10}{9} \\ &1<\ln (\frac{10}{9})^{10} \\ &e<(1+\frac{1}{9})^{10} \text{ 显然Q.E.D.} \\ &(1+\frac{1}{n})^n<e<(1+\frac{1}{n})^{n+1} \end{aligned} \]

另一方面：

\[\begin{aligned} &e>(1+\frac{1}{9})^9 \\ &0.1e^{0.1}>0.1(e^{0.9\ln(1+1/9)})>0.1(1+0.9\ln(1+1/9))=0.1+0.09\ln(1+1/9) \\ &10>\ln(1+1/9) \Rightarrow 0.1+0.09\ln(1+1/9)>\ln(1+1/9) \\ &0.1e^{0.1}>\ln(1+1/9) \text{ 显然Q.E.D.} \end{aligned} \]

2

求 \(a\) 的范围，满足 \(e^x+ax \ge 1-\ln(x+1) \quad (x>0)\)

\[\begin{aligned} &h(x)=e^x+ax+\ln(x+1)-1,h'(x)=e^x+a+\frac{1}{x+1} \\ &\forall x>0,\exists \zeta \in (0,x),s.t.h(x)=h(0)+h'(0)\zeta=h'(0)\zeta \\ &h(x) \ge 0(x>0) \Rightarrow h'(0) \ge 0 \Rightarrow a \ge -2 \\ &a \ge -2:h(x)\ge e^x-2x+\ln(x+1)-1 \ge \frac{x^2}{2}-x+\ln(x+1)=g(x) \\ &g'(x)=x-1+\frac{1}{x+1}=\frac{x^2}{x+1}>0 \Rightarrow g(x) \ge g(0)=0 \\ &\Rightarrow h(x) \ge 0 \Rightarrow a \in [-2,+\infty) \end{aligned} \]