853

设函数 ${\displaystyle f(x)=a\ln x-\ln (x+1)+m,a>0,m\in \mathbb{R}}$.

1. 求函数 ${\displaystyle f(x)}$ 的单调区间;
2. 若对任意 ${\displaystyle 0<a<1}$, 函数 ${\displaystyle f(x)}$ 均有 ${\displaystyle 2}$ 个零点, 求实数 ${\displaystyle m}$ 的取值范围;
3. 设 ${\displaystyle n\in {\mathbb{N}}^{\ast }}$ 且 ${\displaystyle n\ge 2}$, 证明: $${\displaystyle \left(\frac{1}{n}\right)\cdot {\left(\frac{2}{n}\right)}^2\cdot {\left(\frac{3}{n}\right)}^3\cdots {\left(\frac{n-1}{n}\right)}^{n-1}>2^{-\frac{n^2}{2}}.}$$

1. 解 ${\displaystyle f^{\prime }(x)={\displaystyle \frac{a}{x}-\frac{1}{x+1}=\frac{x(a-1)+a}{x(x+1)}}}$. 当 ${\displaystyle a\ge 1}$, ${\displaystyle f(x)}$ 在 ${\displaystyle (0,+\mathrm{\infty })}$ 上增；当 ${\displaystyle 0<a<1}$, ${\displaystyle f(x)}$ 在 ${\displaystyle {\displaystyle (0,\frac{a}{1-a})}}$ 上增, 在 ${\displaystyle {\displaystyle (\frac{a}{1-a},+\mathrm{\infty })}}$ 上减.

2. 解 至少要求$$\begin{array}{r}F(x{)}_{max}=f\left(\frac{a}{1-a}\right)=a\ln a+(1-a)\ln (1-a)+m>0.\end{array}$$
   记 ${\displaystyle g(x)=x\ln x+(1-x)\ln (1-x),x\in (0,1)}$, 则 ${\displaystyle g^{\prime }(x)={\displaystyle \ln \frac{x}{1-x}}}$, 易知 ${\displaystyle g(x{)}_{min}={\displaystyle g\left(\frac{1}{2}\right)=-\ln 2}}$, 故 ${\displaystyle m>\ln 2}$.

   此时 $$\begin{array}{rl}& F\left({\mathrm{e}}^{-\frac{m}{a}}\right)<a\ln {\mathrm{e}}^{-\frac{m}{a}}+m=0,\\ & f\left({\mathrm{e}}^{\frac{m}{1-a}}\right)<(a-1)\ln ({\mathrm{e}}^{\frac{m}{1-a}}+1)+m=0.\end{array}$$ 故 ${\displaystyle f(x)}$ 在 ${\displaystyle {\displaystyle ({\mathrm{e}}^{-\frac{m}{a}},\frac{a}{1-a}),(\frac{a}{1-a},{\mathrm{e}}^{\frac{m}{1-a}})}}$ 上确实有两个零点, 综上 ${\displaystyle m>\ln 2}$.

3. 证明 依据2的不等式 ${\displaystyle x\ln x+(1-x)\ln (1-x)\ge -\ln 2,x\in (0,1)}$, 有$$\begin{array}{rl}\sum _{i=1}^{n-1}i\ln \frac{i}{n}=& \frac{1}{2}\sum _{i=1}^{n-1}(i\ln \frac{i}{n}+(n-i)\ln \frac{n-i}{n})=\frac{n}{2}\sum _{i=1}^{n-1}(\frac{i}{n}\ln \frac{i}{n}+\frac{n-i}{n}\ln \frac{n-i}{n})\\ >& \frac{n}{2}\sum _{i=1}^{n-1}(-\ln 2)=-\frac{n(n-1)}{2}\ln 2>-\frac{n^2}{2}\ln 2,\end{array}$$ 两侧作为 ${\displaystyle \mathrm{e}}$ 的底数, 则原不等式得证.