每日导数41

估值问题

已知函数\(f(x)=(x+1)\ln x-ax+a\)

\((1)\) 若\(a=2\)，判断\(f(x)\)的单调性

\((2)\) 若\(x>1,f(x)>0\)恒成立

(i)求\(a\)的取值范围

(ii)设\(a_n=\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+\cdots+\dfrac{1}{2n},[x]\)表示不超过\(x\)的最大整数，求\(10[a_n](\ln 2\approx0.69)\)

\((1)\) \(a=2,f(x)=(x+1)\ln x-2x+2,f^{\prime}(x)=\ln x+1+\dfrac{1}{x}-2=\ln x+\dfrac{1}{x}-1\)

因\(\ln x\geq 1-\dfrac{1}{x}\)，从而\(f^{\prime}(x)\geq 0\)，则\(f(x)\)单调递增

\((2)\)

(i)不难发现\(f(1)=0,\) \(f^{\prime}(x)=\ln x+\dfrac{1}{x}+1-a\)

因\(\ln x\geq 1-\dfrac{1}{x}\)，则\(f^{\prime}(x)\geq 2-a\)

Case1 当\(a\leq 2\)时，\(f^{\prime}(x)\geq 0\)

则\(f(x)\)单调递增，从而\(f(x)\geq f(1)=0\)，合题

Case2 当\(a>2\)，则\(f^{\prime}(1)=2-a<0\)

\(f^{\prime\prime}(x)=\dfrac{1}{x}-\dfrac{1}{x^2}>0\)

从而\(f^{\prime}(x)\)单调递增，而\(f^{\prime}(1)<0\)

则一定有一段\((1,x_0)\)使得\(f^{\prime}(x)<0\)

在这上面\(f(x)\)单调递减，则\(f(x)<f(1)=0\)，不合题

综上\(a\geq 2\)

(ii)由(1)\(2\cdot \dfrac{x-1}{x+1}\leq \ln x\)

记\(2\cdot\dfrac{x-1}{x+1}=k\)，则\(x=\dfrac{2+k}{2-k}\)

从而\(k<\ln\dfrac{2+k}{2-k}\)

取\(k=\dfrac{1}{n+1},\dfrac{1}{n+2}\cdots \dfrac{1}{2n}\)有

\(\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots +\dfrac{1}{2n}<\ln\dfrac{2n+3}{2n+1}+\ln\dfrac{2n+5}{2n+3}+\cdots\ln\dfrac{4n+1}{4n-1}=\ln\dfrac{4n+1}{2n+1}=\ln\left(2-\dfrac{1}{2n+1}\right)<\ln 2\)

从而\([10a_n]=6\qquad\square\)

耍赖的写法：因\(a_n\)是单调递增的，并且\(a_n<\dfrac{n}{n+1}<1\)是有界的，由单调有界原理极限存在.而

\(\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}\dfrac{1}{n}\left(\dfrac{1}{\dfrac{1}{n}+1}+\dfrac{1}{1+\dfrac{2}{n}}+\dfrac{1}{1+\dfrac{3}{n}}+\cdots+\dfrac{1}{1+\dfrac{n}{n}}\right)=\displaystyle\int_{0}^{1}\ln\dfrac{1}{1+x}\mathrm{d}x=\ln(1+x)\big|_{0}^{1}=\ln 2\approx0.69\)

从而\([10a_n]=6\)