浙江地区出的题太难，上积分了

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

(1)对任意\(0<a<1\)，函数\(f(x)\)有两零点，求\(b\)的取值范围

(2)设\(n\geq 2,n\in\mathbb{N}^{\star}\)，证明:\(\left(\dfrac{1}{n}\right)\cdot\left(\dfrac{2}{n}\right)^2\cdot\left(\dfrac{3}{n}\right)^3\cdots\left(\dfrac{n-1}{n}\right)^{n-1}>2^{-\frac{n^2}{2}}\)

解

(1)\(f^{\prime}(x)=\dfrac{1}{x+1}-\dfrac{a}{x}=\dfrac{x-a(x+1)}{x(x+1)}=\dfrac{x(1-a)-a}{x(x+1)}\)

则得\(f(x)\)在\(\left(0,\dfrac{a}{1-a}\right)\)上减，在\(\left(\dfrac{a}{1-a},+\infty\right)\)上增

则要使得\(f(x)\)有两零点，则必有\(f\left(\dfrac{a}{1-a}\right)<0\)

\(f\left(\dfrac{a}{1-a}\right)=-\ln(1-a)-a\ln a+a\ln(1-a)-b\)

记\(h(a)=-\ln(1-a)-a\ln a+a\ln(1-a)-b\)

\(h^{\prime}(x)=\dfrac{1}{1-a}-\ln a-1+\ln(1-a)-\dfrac{a}{1-a}=\ln(1-a)-\ln a\)递减

而\(h^{\prime}\left(\dfrac{1}{2}\right)=0\)，从而\(h(a)_{\max}=h\left(\dfrac{1}{2}\right)=\ln 2-b\)

从而 \(b>\ln 2\).

现说明此时确实有两个零点\(f(x)=\ln\dfrac{x+1}{x^a}-b\)

\(\lim\limits_{x\to 0}f(x)\to +\infty,\lim\limits_{x\to +\infty}f(x)\to +\infty\)

从而在\(\left(0,\dfrac{a}{1-a}\right)\)上有唯一零点，在\(\left(\dfrac{a}{1-a},+\infty\right)\)上有唯一零点

(2)不等式取对数有

\(\ln\dfrac{1}{n}+2\ln\dfrac{2}{n}+3\ln\dfrac{3}{n}+\cdots+(n-1)\ln\dfrac{n-1}{n}>-\dfrac{n^2}{2}\ln 2\)

即\(\displaystyle\dfrac{1}{n}\sum\limits_{k=1}^{n-1}\dfrac{k}{n}\ln\dfrac{k}{n}>-\dfrac{\ln 2}{2}\)

两边取\(n\to \infty\)，有

即证:\(\displaystyle\int_{0}^{1}x\ln x\mathrm{d}x>-\dfrac{\ln 2}{2}\)

而\(\displaystyle\int_{0}^{1}x\ln x\mathrm{d}x=\dfrac{1}{2}\int_{0}^{1}\ln x\mathrm{d}(x^2)=\dfrac{x^2\ln x}{2}\bigg|_{0}^{1}-\dfrac{1}{2}\int_{0}^{1}x\mathrm{d}x=-\dfrac{1}{4}>-\dfrac{\ln 2}{2}\)

得证.