每日导数60

很适合作为改革后的第19题

已知函数\(f(x)=x-\ln(x+a)\)的最小值为\(0,a>0\)

(1)求\(a\)

(2)\(\forall x\geq 0,f(x)\leq kx^2\)成立，求\(k\)最小值

(3) 证明:\(\displaystyle\sum\limits_{i=1}^{n}\dfrac{2}{2i-1}-\ln(2n+1)<2\)

解

(1)\(f^{\prime}(x)=1-\dfrac{1}{x+a}=0\)得\(x=1-a\)

则\(f(1-a)=1-a=0\)得\(a=1\)

(2) \(x-\ln(x+1)\leq kx^2\)

即\(g(x)=x-\ln(x+1)-kx^2\leq 0,g(0)=0\)

\(g^{\prime}(x)=1-\dfrac{1}{x+1}-2kx,g^{\prime}(0)=0\)

Case1 当\(k\leq 0\)时，\(g^{\prime}(x)\)单调递增，则\(g^{\prime}(x)\geq g^{\prime}(0)=0\)

则\(g(x)\)单调递增，从而\(g(x)\geq g(0)=0\)，不合题

Case2当\(k>0\)时，\(g^{\prime}(x)=\dfrac{x+1-2kx(x+1)-1}{x+1}=\dfrac{-2kx^2+x(1-2k)}{x+1}=\dfrac{x(1-2k-2kx)}{x+1}\)

\(g^{\prime}(x)=0\)得\(x_1=0,x_2=\dfrac{1-2k}{2k}=\dfrac{1}{2k}-1>-1\)

Case2.1当\(k\geq \dfrac{1}{2}\)时，\(x_2\leq 0\)

则\(g^{\prime}(x)\)在\(x\geq 0\)上是负

则\(g(x)\)单调递减，则\(g(x)\leq g(0)=0\)合题

Case2.2 当\(0\leq k<\dfrac{1}{2}\)时，\(x_2>0\)

则\(g^{\prime}(x)\)在\([0,x_2)\)上单调递增

则在此区间上\(g(x)\geq (0)=0\)不合题

综上\(k\geq \dfrac{1}{2},k_{\min}=\dfrac{1}{2}\)

(3)取\(k=\dfrac{1}{2},x=\dfrac{2}{2i-1}\)有

\(\dfrac{2}{2i-1}-\ln\left(\dfrac{2}{2i-1}+1\right)\leq \dfrac{1}{2}\left(\dfrac{2}{2i-1}\right)^2=\dfrac{2}{(2i-1)^2}\)

取\(i=1,2,3\cdots n\)累加有

\(\displaystyle\sum\limits_{i=1}^{n}\left[\dfrac{2}{2i-1}-\ln\left(\dfrac{2}{2i-1}+1\right)\right]\leq 2+\displaystyle\sum\limits_{i=2}^{n}\dfrac{2}{(2i-1)^2}\leq 2+\displaystyle\sum\limits_{i=2}^{n}\dfrac{2}{(2i-1)(2i-3)}=2+\dfrac{1}{2}\displaystyle\sum\limits_{i=2}^{n}\left(\dfrac{2}{2i-3}-\dfrac{2}{2i-1}\right)\)
即\(\displaystyle\sum\limits_{i=1}^{n}\dfrac{2}{2i-1}-\ln(2n+1)\leq 2+1-\dfrac{1}{2n-1}=3-\dfrac{1}{2n-1}\)

出现问题，这是因为右边我们从\(i=2\)开始放缩累加

\(i=1,2-\ln 3<2\)

\(i\geq 2,\displaystyle\sum\limits_{i=2}^{n}\dfrac{2}{2i-1}-\displaystyle\sum\limits_{i=2}^{n}\ln\left(\dfrac{2}{2i-1}+1\right)\leq\displaystyle\sum\limits_{i=2}^{n}\dfrac{2}{(2i-1)^2}<\dfrac{1}{2}\displaystyle\sum\limits_{i=2}^{n}\left(\dfrac{2}{2i-3}-\dfrac{2}{2i-1}\right)\)

即\(\displaystyle\sum\limits_{i=2}^{n}\dfrac{2}{2i-1}+\ln 3-\ln(2n+1)<1-\dfrac{1}{2n-1}\)

即\(\displaystyle\sum\limits_{i=2}^{n}\dfrac{2}{2i-1}-\ln(2n+1)<1-\dfrac{1}{2n-1}-\ln 3\)

即\(2+\displaystyle\sum\limits_{i=2}^{n}\dfrac{2}{2i-1}-\ln(2n+1)<3-\ln3-\dfrac{1}{2n-1}\)

即\(\displaystyle\sum\limits_{i=1}^{n}\dfrac{2}{2i-1}-\ln(2n+1)<3-\ln3-\dfrac{1}{2n-1}<2\)