导数几题
$14、已知正实数a,b，满足ae^2(\ln b-\ln a+a-1)\ge be^a,则\cfrac{1}{b}的最小值为(\qquad)$
$类似指对分离,两边{\div} ae^2，\ln b-\ln a+a-1=\ln \cfrac{b}{a}+a-1\ge \cfrac{b}{a}e^{a-2}=e^{\ln \cfrac{b}{a}+a-2}$
${\color{Red} \because \quad } e^x\ge x+1,当且仅当x=0时取=\quad {\color{Red} \therefore \quad } e^{\ln \cfrac{b}{a}+a-2}\ge \ln \cfrac{b}{a}+(a-2)+1$
$\begin{cases} e^{\ln \cfrac{b}{a}+a-2}\ge \ln \cfrac{b}{a}+(a-2)+1\\e^{\ln \cfrac{b}{a}+a-2}\le \ln \cfrac{b}{a}+(a-2)+1\end{cases}\Rightarrow e^{\ln \cfrac{b}{a}+a-2}= \ln \cfrac{b}{a}+(a-2)+1$
$\Rightarrow \ln \cfrac{b}{a}+a-2=0\Rightarrow \ln b=\ln a+2-a，令f(a)=\ln a+2-a$
$ {f}'(a)=\cfrac{1}{a}-1=\cfrac{1-a}{a} $
$a\in (0,1)时，{f}'(a)\gt 0,f(a)\nearrow ;a\in (1,+\infty)时，{f}'(a)\lt 0,f(a)\searrow$
${\color{Red} \therefore \quad } f(a)\le f(1)=1,\Rightarrow \ln b\le 1\Rightarrow b\le e,\cfrac{1}{b}\ge \frac{1}{e}\quad {\color{Red} b_{min}=\cfrac{1}{e}}$
$自贡14题若函数f(x)=k(\ln x)^2-x有3个零点，则实数k的取值范围(\qquad)$
零点变交点，换元法
$令\ln x=t\in R,x=e^t,g(t)=kt^2-e^t,g(t)=0\Leftrightarrow kt^2=e^t,\quad \therefore \cfrac{1}{k}=\cfrac{t^2}{e^t}=h(t)$
${h}'(t)= \cfrac{t(2-t)}{e^t};\qquad t\in (-\infty,0)时，{h}'(t)\lt 0,h(t)\searrow ;$
$a\in (0,+2)时,{h}'(t)\gt 0,h(t)\nearrow;a\in (2,+\infty)时，{h}'(t)\lt 0,h(t)\searrow$
$h(0)=0,h(2)=\cfrac{4}{e^2},故\cfrac{1}{k}\in (0,\cfrac{4}{e^2}),k\in (\cfrac{e^2}{4},+\infty)$
$18、设函数f(x)=x(a+e^x)\quad a\in R,g(x)=1+\ln x,若f(x)\ge g(x)在x\in (0,+\infty)$恒成立，求实数$a$的取值范围。
key:参娈分离，朗博同构，零点变交点，隐零点
$x(a+e^x)\ge 1+\ln x\Rightarrow a\ge \cfrac{1+\ln x}{x}-e^x=\cfrac{1+\ln x-xe^x}{x}$
$=\cfrac{1+\ln x-e^{x+\ln x}}{x}=\cfrac{x+\ln x+1-e^{x+\ln x}-x}{x}\quad自证\quad x+1\le e^x$
$x+\ln x+1\le e^{x+\ln x},当且令当x+\ln x=0时取=$
$a\ge \cfrac{x+\ln x+1-e^{x+\ln x}-x}{x}\ge \cfrac{0-x}{x}=-1\quad 若x+\ln x=0成立，则上式能取得等号成立$
$设g(x)=x+\ln x，x\in (0,+\infty),g(x)单调递增，g(1)=1\gt 0,g(\cfrac{1}{e})=\cfrac{1}{e}-1\lt 0,$
$故存在g(x_0)=0在x_0\in (\cfrac{1}{e},1),即存在x_0\in (\cfrac{1}{e},1)使得a\ge \cfrac{x+\ln x+1-e^{x+\ln x}-x}{x}取＝成立。$
$a\ge -1,a\in (-\infty,1]$
19 、(17分)
$已知函数f(x)=(x-a)\ln x-x$
$(1)当a=0时，①求f(x)的最小值；②设n\in N^*,求证:\cfrac{\ln 1}{2}+\cfrac{\ln 2}{3}+\cfrac{\ln 3}{4}+\cdots+\cfrac{\ln n}{n+1}\le \cfrac{n(n-1)}{4};$
$(2)设x_1\lt x_2,是f(x)的两个极值点,求证:x_1+x_2\gt \cfrac{2}{e}$
$解:a=0,f(x)=x\ln x-x,{f}'(x)=\ln x,令\ln x=0,x=1,f(x)在{\color{Red}x\in(0,1)时 } ，$
${f}'(x)\lt 0,f(x)\searrow;f(x)在{\color{Red}x\in(1,+\infty)},{f}'(x)\gt 0,f(x)\nearrow\qquad f(x)\ge f(1)=-1$
$分析 设S_n=\cfrac{n(n-1)}{4},S_{n-1}=\cfrac{(n-1)(n-2)}{4}\Rightarrow a_n=S_n-S_{n-1}=\cfrac{n-1}{2},即证{\color{Red} \cfrac{\ln n}{n+1}\le \cfrac{n-1}{2}} $
${\color{Red} \Leftrightarrow }\ln n\le \cfrac{1}{2}(n^2-1)\qquad {\color{Red} ①已证x\ln x-x\ge -1}\Rightarrow \ln x \ge 1-\cfrac{1}{x},若令x=\cfrac{1}{n^2}即可得证$
$②证,f(x)\ge -1,即x\ln x-x\ge -1\Rightarrow \ln x\ge 1-\cfrac{1}{x}令x=\cfrac{1}{n^2},得\ln \cfrac{1}{n^2}\ge 1-n^2\Rightarrow -2\ln n\ge 1-n^2$
$2\ln n\le n^2-1\Rightarrow \cfrac{\ln n}{n+1}\le \cfrac{n-1}{2}$
$\cfrac{\ln 1}{2}\le \cfrac{0}{2} $
$\cfrac{\ln 2}{3}\le \cfrac{1}{2} $
$\cfrac{\ln 3}{4}\le \cfrac{2}{2} $
$\cfrac{\ln 4}{5}\le \cfrac{3}{2} $
$\cfrac{\ln 5}{6}\le \cfrac{4}{2} $
$\cdots \cdots$
$\cfrac{\ln (n-1)}{n}\le \cfrac{n-2}{2}$
$\cfrac{\ln n}{n+1}\le \cfrac{n-1}{2}$
$\cfrac{\ln 1}{2}+\cfrac{\ln 2}{3}+\cfrac{\ln 3}{4}+\cdots+\cfrac{\ln n}{n+1}\le \cfrac{0}{2}+\cfrac{1}{2}+\cfrac{2}{2}+\cfrac{3}{2}+\cfrac{4}{2}+\cdots \cfrac{n-2}{2}+\cfrac{n-1}{2}=\cfrac{(0+n-1)n}{2\times 2}$
$(2){\color{Red} {f}'(x)=\ln x-\cfrac{a}{x} \quad }令{f}'(x)=0\Leftrightarrow \ln x=\cfrac{a}{x}\Leftrightarrow {\color{Red} a=x\ln x}$
$f(x)的两个极值点,即y=a与g(x)=x\ln x有两个交点$
$根据g(x)=x\ln x 的图像,可以a\in (-\cfrac{1}{e},0)时,x\in (0,1),有x=\cfrac{1}{e}两边各有一个交点$
${\color{Red} ①}设0\lt x_1\lt \cfrac{1}{e}, \cfrac{2}{e}\lt x_2\lt 1 ,x_1+x_2\gt \cfrac{2}{e}不证自明$
${\color{Red} ②} 设0\lt x_1\lt \cfrac{1}{e} \lt x_2\lt \cfrac{2}{e},欲证x_1+x_2\gt \cfrac{2}{e},即证x_1\gt \cfrac{2}{e}-x_2$
$g(x)=x\ln x求导{g}'(x)=\ln x+1，易知g(x)在x\in(0,\cfrac{1}{e})单调递减；在x\in(\cfrac{1}{e},+\infty)单调递增$
$故即证g(x_1)\lt g(\cfrac{2}{e}-x_2){\color{Red} \Leftrightarrow } g(x_2)\lt g(\cfrac{2}{e}-x_2),构造函数G(x)=g(x)-g(\cfrac{2}{e}-x)\quad (x\in (\cfrac{1}{e},\cfrac{2}{e})$
${G}'(x)={g}'(x)+{g}'(\cfrac{2}{e}-x)=\ln x+1+[\ln (\cfrac{2}{e}-x)+1]$
${\color{Red} {G}''(x)= } \cfrac{1}{x}-\cfrac{1}{\cfrac{2}{e}-x},{G}''(x)\searrow; {G}''(\cfrac{1}{e})=0\Rightarrow {G}''(x)\lt 0\Rightarrow {G}'(x)\searrow$
${G}'(\cfrac{1}{e})=0,\Rightarrow {G}'(x)\lt 0\Rightarrow G(x)\searrow ,G(\cfrac{1}{e})=0,G(x)\lt G(\cfrac{1}{e})=0$
$\Rightarrow G(x)=g(x)-g(\cfrac{2}{e}-x)\lt 0,\Rightarrow g(x)\lt g(\cfrac{2}{e}-x)$