$\log_{a}{b} \lt \log_{a+m}{(b+m)} ,(a\gt b\gt 1,m\gt 0)$

$1、\cfrac{b}{a}\lt \cfrac{b+m}{a+m}\quad (a\gt b\gt 0,m\gt 0)$
a可以认为是糖水质量，b可以认为是糖水中的糖，a克的不饱和糖水里加放m克糖，糖水更甜了
$证明：\cfrac{b+m}{a+m}-\cfrac{b}{a}=\cfrac{a(b+m)-b(a+m)}{am}=\cfrac{m(a-b)}{a(a+m)}\gt 0,得证$

$2、对数糖水不等式:\log_{a}{b} \lt \log_{a+m}{(b+m)} ,(a\gt b\gt 1,m\gt 0)$

$证明：\log_{a}{b}=\cfrac{\lg_{}{b} }{\lg_{}{a} }=\cfrac{\lg{b}+\lg{\cfrac{a+m}{a}} }{\lg{a}+\lg{\cfrac{a+m}{a}}}\lt \cfrac{\lg{\cfrac{ab+am}{a}} }{\lg{(a+m)}} =\cfrac{\lg{(b+m)} }{\lg{(a+m)}} \lt \log_{a+m}{(b+m)}$
重点掌握证明过程，因为对数判断大小常用到此法。

$例1、20年全国三卷，已知5^5\lt 8^4,13^4\lt 8^5.$
$设a=\log_5{3},b=\log_8{5},c=\log_{13}{8},则(\qquad )$
$A.a\lt b\lt c\quad B.b\lt a\lt c\quad C.b\lt c\lt a\quad D.c\lt a\lt b$
$a=\log_5{3}=\cfrac{\lg 3}{\lg 5}\lt \cfrac{\lg 3+\lg \cfrac{8}{5} }{\lg 5+\lg \cfrac{8}{5} }= \cfrac{\lg \cfrac{24}{5} }{\lg 8 }\lt\cfrac{\lg \cfrac{25}{5} }{\lg 8 }= \cfrac{\lg 5 }{\lg 8 }=b=\log_8{5}$

$5^5\lt 8^4两边取8为底的对数,\log_{8}{5^5} \lt \log_{8}{8^4}=4\Rightarrow \log_{8}{5}\lt \cfrac{4}{5}$
$13^4\lt 8^5两边取13为底的对数,\log_{13}{13^4} \lt \log_{13}{8^5}=4\Rightarrow 4\lt 5\log_{13}{8}$
$\log_{8}{5}\lt \cfrac{4}{5}\lt \log_{13}{8}$

$例2、已知a=\log_2{3},b=\log_3{4},c=a=\log_4{5},则(\qquad)$
$A.c\lt b\lt a\quad B.b\lt a\lt c\quad C.a\lt b\lt c\quad D.b\lt c\lt a$
方法一：对数糖水不等式
$ \log_3{2}\lt\log_4{3}\lt \log_5{4}\Rightarrow \cfrac{1}{a}\lt \cfrac{1}{b}\lt \cfrac{1}{c}\Rightarrow a\gt b\gt c$
方法二：$令f(x)=\log_x{(x+1)}=\cfrac{\ln (x+1)}{\ln x},x\gt 1$,必修一课本有此题
${f}' (x)=\cfrac{\cfrac{\ln x}{x+1}-\cfrac{\ln (x+1)}{x}}{\ln ^2x} =\cfrac{x\ln x-(x+1)\ln (x+1)}{x(x+1)\ln ^2x}\lt 0,f(x)\searrow$

$例3、已知9^m=10,a=10^m-11,b=8^m-9,则(\qquad)$
$A.a\gt 0\gt b\quad B.a\gt b\gt 0\quad C.b\gt a\gt 0\quad D.b\gt 0\gt a$
$a=10^m-11=10^m-10-1$
$0=9^m-10=9^m-9-1$
$b=8^m-9=8^m-8-1$
$构造函数f(x)=x^m-x-1\quad x\gt 1,m\gt 1,{f}' (x)=mx^{m-1}-1\gt 0$,
$a\gt 0\gt b$

$1、函数f(x)=\cfrac{2\tan x}{1-\tan^2 x}的最小正周期是(\qquad)$
$A.\cfrac{\pi}{4}\quad B.\cfrac{\pi}{2}\quad C.\pi \quad D.2\pi$
$解:考查f(x)与\tan 2x的定义域，类似y=\cfrac{x^2}{x}=x$
$ 应用倍角公式化简得到 f(x)=\cfrac{2\tan x}{1-\tan^2 x}=\tan 2x ，\tan 2x的周期为\cfrac{\pi}{2},且在x=k\pi \pm\cfrac{\pi}{2}有定义，$
$但f(x)在x=k\pi \pm\cfrac{\pi}{2}处无定义,亦即f(x)化简后扩大了定义域，根据周期函数的对称性，$
故周期为$\pi$

$2、2026年T8联考第15题，在\triangle ABC中，角A,B,C所对边分别为a,b,c,且满足\cos^2\cfrac{A}{2}=\cfrac{b+c}{2c}.$
$①求角C的大小。直角，第二问初中要求$
$②若点D在AB边上，且满足AB=3AD,求\cfrac{\tan A}{\tan \angle ACD}$

--

${\color{Red} 3、} 已知函数f(x)\begin{cases} e^x(2x-1),\quad x\gt 0,\\k(x+1),\quad x\lt 0\end{cases},g(x)=f(x)+f(-x),$
$若函数g(x)恰好有4个零点，则实数k的取值范围是(\qquad)$
$A.(-\infty,1）\quad B.(1,4e^{\cfrac{3}{2}})\quad C.(4e^{\cfrac{3}{2}},+\infty)\quad D.(1,+\infty)$
$令g(x)=0,即f(x)=-f(-x),-f(-x)即是f(x)的中心对称图形。即x\gt 0时e^x(2x-1)与 k(x-1)有两点交点,相切时临界$
$x\gt 0時，{\color{Red}{f}' (x)=e^x(2x+1) } ,设(x_0,y_0)是两曲线的切点，\begin{cases} y_0=k(x_0-1)\qquad \qquad\quad ①\\y_0=e^{x_0}(2x_0-1)\quad\qquad\quad ②\\k={f}'(x_0)=e^{x_0}(2x_0+1)\quad ③\end{cases}$
$ ③代入①,①②左边相等，2x_0-1=(2x_0+1)(x_0-1)\Rightarrow x_0=\cfrac{3}{2},x_0=0(舍去),回代③式k\gt 4e^{\cfrac{3}{2}}$

$4、已知a\gt 0,b\gt 0,\sqrt{ab}=\cfrac{1}{a}+\cfrac{1}{b}，则\cfrac{1}{\log_a 2}+\cfrac{1}{\log_b 2}最小值是(\qquad)$
$A.3\quad B.2\quad C.\sqrt{2}\quad D.1$

$5、已知函数f(x)=(x+a)(2^x-b),若f(x)\ge 0恒成立，则b+2^a的最小值是(\qquad)$
$A.-2\quad B.0\quad C.\sqrt{2}\quad D.2$

$6、等差数列\{a_n\}的前n项和为S_n,已知a_5=8,S_5=20,设b_n=\cfrac{\sin 2}{\cos a_n\cos a_{n+1}},则数列\{b_n\}的前n项和为$_
$解:\because a_n=a_1+(n-1)d=a_n-(n-1)d;S_n=na_1+\cfrac{1}{2}n(n-1)d=na_n-\cfrac{1}{2}n(n-1)d$
$\therefore S_5=na_5-\cfrac{1}{2}\times 5\times 4\times d\Rightarrow 20=40-10d,d=2$
$b_n=\cfrac{\sin 2}{\cos a_n\cos a_{n+1}}=\cfrac{\sin (a_{n+1}-a_n)}{\cos a_n\cos a_{n+1}}=\cfrac{\sin a_{n+1}\cos a_n-\cos a_{n+1}\sin a_n}{\cos a_n\cos a_{n+1}}=\tan a_{n+1}-\tan a_n$
$T_n=b_1+b_2+b_3+\cdots +b_n=\tan a_{n+1}-\tan a_1$

$作业:已知数列\{a_n\}中，a_2=1,设S_n为\{a_n\}的前n项和，2S_n=na_n$
$(1)求\{a_n\}的通项公式；$
$(2)设b_n=\cfrac{\sin 1}{\cos (a_n+1)\cos (a_{n+1}+1)},则数列\{b_n\}的前n项和T_n$

$7、已知数列\{a_n\}的前n项和为S_n.且满足a_1=-5,a_n=\cfrac{S_n}{n}+2(n-1),$
$若对于任意n\in N^*,\lambda\le S_n怛成立，则实数\lambda的取值范围是(\quad )$
$A.(-\infty,-6]\quad B.(-\infty,-5]\quad C.(-\infty,-3]\quad D.(-\infty,-2]$

${\color{Red} 8、}共零点变形题，x(\ln x+2)\le ax^2+\cfrac{2}{a}\ln x对\forall x\ge e恒成立，则实数a的取值范围是(\qquad)$
$a\lt 0(舍去)因为右边等式为负数。a\gt 0时，x\ln x+2x-ax^2-\cfrac{2}{a}\ln x\le 0\Rightarrow (x-\cfrac{2}{a})\ln x+x(2-ax)\le 0$
$\Rightarrow (ax-2)\ln x-ax(ax-2)\le 0\Rightarrow (ax-2)(\ln x-ax)\le 0$
$①容易得到a\ge \cfrac{1}{e}时，ln x\le ax,即ln x-ax\le 0,此时只要ax-2\ge 0即可，\because x\gt e,\Rightarrow a\ge\cfrac{2}{x}\Rightarrow a\ge \cfrac{2}{e}$
$②0\lt a\lt \cfrac{1}{e},\because \quad x\gt e, ⑴e\lt x \lt x_0,\ln x-ax\ge 0;⑵x\gt x_0,\ln x -ax\le 0,此时y=ax-2与y=\ln x-ax共零点x_0。$
$\ln x_0=ax_0=2,\Rightarrow x_0=e^2,x\in [e,e^2],ax-2\le 0,\ln x-ax\ge 0,(ax-2)(\ln x-ax)\le 0;$
$x\gt e^2,ax-x\ge 0,\ln x-ax\le 0,\Rightarrow (ax-2)(\ln x-ax)\le 0$
$综上述，a\in \{e^2\}\cup [\cfrac{1}{e},+\infty)$

${\color{Red} 9、}若关于x的不等式ax^2-ax\ln a-e^x\ln x\gt 0对\forall x\in (0,1)恒成立，则实数a的取值范围为(\qquad)$
$解:ax^2-ax\ln a-e^x\ln x\gt 0\Rightarrow x^2-x\ln a-\cfrac{e^x}{a}\ln x\gt 0\Rightarrow x(x-\ln a)\gt e^{x-\ln a}\ln x\Rightarrow$
$\cfrac{x-\ln a}{e^{x-\ln a}}\gt \cfrac{\ln x}{x}\quad 令f(x)=\cfrac{x}{e^x},上式等价于f(x-\ln a)\gt f(\ln x)\quad \forall x\in (0,1),\ln x\lt 0,,\Rightarrow x-\ln a\gt \ln x$
$x-\ln x\gt \ln a\Rightarrow \ln a\lt x-\ln x\quad (x\in (0,1)),易求导得x-\ln x在(0,1)单调递减$
$(x-\ln x)_{max}=1,a\le 1$

$例1、已知数列\{a_n\}中，a_1=2,a_{n+1}=a_n+n+1.$
$(1)求数列\{a_n\}的通项公式；$
$(2)设b_n=\cfrac{1}{a_n},数列\{b_n\}的前n项和为T_n，证明T_n\lt 2$

$(1)a_{n+1}=a_n+n+1，有递推关系式：$
$a_{n}=a_{n-1}+(n-1)+1$
$a_{n-1}=a_{n-2}+(n-2)+1$
$a_{n-2}=a_{n-3}+(n-3)+1$
$a_{n-3}=a_{n-4}+(n-4)+1$
$\cdots \cdots \cdots\cdots$
$a_3=a_2+2+1$
$a_2=a_1+1+1$
累加之
$a_{n}+a_{n-1}+a_{n-2}+a_{n-3}+\cdots +a_3+a_2=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+\cdots +a_2+a_1$
$+(n-1)+(n-2)+(n-3)+(n-4)+\cdots +1+1+1\cdots +1$
$a_n=a_1+(n-1)+(n-2)+(n-3)+(n-4)+\cdots+2+1(n-1)\times 1$
$=2+(n-1)+(n-2)+(n-3)+(n-4)+\cdots+2+1+(n-1)$
$=n+(n-1)+\cdots +1+1=\cfrac{n(n+1)}{2}+1=\cfrac{n^2+n+2}{2}$
$b_n=\cfrac{1}{a_n}=\cfrac{2}{n^2+n+2}\lt {\color{Red} \cfrac{2}{n(n+1)}=2(\cfrac{1}{n}-\cfrac{1}{n+1})} $
$b_1\lt 2(1-\cfrac{1}{2})$
$b_2\lt 2(\cfrac{1}{2}-\cfrac{1}{3})$
$b_3\lt 2(\cfrac{1}{3}-\cfrac{1}{4})$
$\cdots\cdots\cdots\cdots$
$b_{n-2}\lt 2(\cfrac{1}{n-2}-\cfrac{1}{n-1})$
$b_{n-1}\lt 2(\cfrac{1}{n-1}-\cfrac{1}{n})$
$b_n\lt 2(\cfrac{1}{n}-\cfrac{1}{n+1})$
$T_n\lt 2(1-\cfrac{1}{n+1})\lt 2$

$例2、已知数列\{a_n\}的前n项和为S_n,且a_2=3,a_{n+1}=S_n+n+1$
$(1)证明数列\{a_n+1\}是等比数列$
$递推公式：a_{n+1}=S_n+n+1\quad ①\Rightarrow a_2=S_1+1+1=a_1+2,a_1=1$
$\qquad\qquad a_{n}=S_{n-1}+(n-1)+1\quad ②$
$①-②,得a_{n+1}－a_{n}＝S_n+n+1－[S_{n-1}+(n-1)+1]$
$a_{n+1}－a_{n}=a_n+1\Rightarrow a_{n+1}+1=2(a_n+1)$
$故数列\{a_n+1\}是以a_1+1=2为首项，q=2的等比数列$

$15、d=\cfrac{S_2}{2}-\cfrac{S_1}{1}=\cfrac{a_2+a_1}{2}-\cfrac{a_1}{1}=\cfrac{3-a_1}{2}$
$\cfrac{S_5}{5}=S_1+4d=a_1+4d=a_1+4\times \cfrac{3-a_1}{2}=6-a_1=5\Rightarrow a_1=1,d=1$
$\cfrac{S_n}{n}=n\Rightarrow S_n=n^2\Rightarrow n\ge 2,S_{n-1}=(n-1)^2,S_n-S_{n-1}=a_n=2n-1(n\ge 2)$
$n=1时，a_1=1,故a_n=2n-1在n=1时也成立。$
$(2)、b_n=2(n+1)(a_n+1)=4n(n+1)\Rightarrow \cfrac{1}{b_n}=\cfrac{1}{4n(n+1)}=\cfrac{1}{4}(\cfrac{1}{n}-\cfrac{1}{n+1})$
$\cfrac{1}{b_1}+\cfrac{1}{b_2} +\cdots +\cfrac{1}{b_n}=\cfrac{1}{4}(1-\cfrac{1}{2}+\cfrac{1}{2}-\cfrac{1}{3}+\cfrac{1}{3}-\cfrac{1}{4}+\cdots+\cfrac{1}{n-1}-\cfrac{1}{n}+\cfrac{1}{n}-\cfrac{1}{n+1})$
$=\cfrac{1}{4}(1-\cfrac{1}{n+1})\lt \cfrac{1}{4}$

26年江苏省高三G4联考第17题
$已知数列\{a_n\}满足a_1=1,\cfrac{S_{n+1}}{S_n}=\cfrac{n+2}{n}$
$(1)求数列\{a_n\}的通项公式；$
$(2)设b_n=a_n^2(\cos^2\cfrac{n\pi}{3}-\sin^2\cfrac{n\pi}{3}),求数列\{b_n\}的前n项和T_n$
$解:\cfrac{S_{n+1}}{S_n}=\cfrac{n+2}{n}\Rightarrow \cfrac{S_{n+1}}{S_n}-1=\cfrac{n+2}{n}-1\Rightarrow \cfrac{a_{n+1}}{S_n}=\cfrac{2}{n}\Rightarrow na_{n+1}=2S_n\quad {\color{Red} ①}$
$由递推公式得到(n-1)a_n=2S_{n-1}\quad {\color{Red} ②} \Rightarrow {\color{Red}①-② } ,得na_{n+1}-(n-1)a_n=2a_n$
$na_{n+1}=(n+1)a_n\Rightarrow \cfrac{a_{n+1}}{n+1}=\cfrac{a_{n}}{n}=\cfrac{a_1}{1}=1\Rightarrow a_n=n$
${\color{Red}方法2}:S_1\times \cfrac{S_2}{S_1}\times \cfrac{S_3}{S_2}\times \cfrac{S_4}{S_3}\times \cfrac{S_5}{S_4}\times\cdots\times \cfrac{S_{n+1}}{S_n}=1\times\cfrac{3}{1}\times\cfrac{4}{2} \times\cfrac{5}{3} \times\cfrac{6}{4} \times\cfrac{7}{5}\times \cdots\times \cfrac{n}{n-2}\times\cfrac{n+1}{n-1}\times\cfrac{n+2}{n} $
$S_{n+1}=\cfrac{(n+1)(n+2)}{2}, S_{n}=\cfrac{n(n+1)}{2},n\ge 2时，a_n=S_{n}-S_{n-1}=n,显然n=1也成立。$

$b_n=a_n^2(\cos^2\cfrac{n\pi}{3}-\sin^2\cfrac{n\pi}{3})=n^2\cos \cfrac{2n\pi}{3}$
$b_1=\cos \cfrac{2\pi}{3}=1^2\times(-\cfrac{1}{2})$
$b_2=2^2\times\cos \cfrac{4\pi}{3}=2^2\times(-\cfrac{1}{2})$
$b_3=3^3\times\cos \cfrac{6\pi}{3}=3^2\times(+1)$
$b_4=4^2\times\cos \cfrac{8\pi}{3}=4^2\times(-\cfrac{1}{2})$
$\cdots\cdots\cdots\cdots\cdots$
$①设n=3k时，T_n=-{\color{Red}\cfrac{1}{2}}\times\cfrac{3k(3k+1)(6k+1)}{6} +{\color{Red}\cfrac{3}{2}}\times 9\times \cfrac{k(k+1)(2k+1)}{6}=$
$=-\cfrac{1}{4}k(3k+1)(6k+1)+\cfrac{9}{4}k(k+1)(2k+1)=k[-\cfrac{1}{4}(3k+1)(6k+1)+\cfrac{9}{4}(k+1)(2k+1)]$
$=k[-\cfrac{18}{4}k^2-\cfrac{9}{4}k-\cfrac{1}{4}+\cfrac{18}{4}k^2+\cfrac{9}{4}\times 3k+\cfrac{9}{4}]$
$=k(\cfrac{9}{2}k+2)=\cfrac{9}{2}k^2+2k=\cfrac{9}{2}(\cfrac{n}{3})^2+\cfrac{2n}{3}=\cfrac{3n^2+4n}{6}$
$设②n=3k+1,T_n=\cfrac{9}{2}k^2+2k-{\color{Red}\cfrac{1}{2}(3k+1)^2}=\cfrac{9}{2}k^2+2k-\cfrac{1}{2}(9k^2+6k+1)$
$=-k-\cfrac{1}{2}\quad k=\cfrac{n-1}{3}\Rightarrow T_n=-\cfrac{n-1}{3}-\cfrac{1}{2}=-\cfrac{2n+1}{6}$
$③设n=3k+2,T_n=-k-\cfrac{1}{2}-\cfrac{1}{2}(3k+2)^2=-k-\cfrac{1}{2}-\cfrac{1}{2}\times n^2=-\cfrac{3n^2+2n+1}{6}$
${\color{Green}这里是错的,k值不同于 ②} $
$k=\cfrac{n-2}{3}\quad T_n=-k-\cfrac{1}{2}-\cfrac{1}{2}(3k+2)^2=-\cfrac{n-2}{3}-\cfrac{1}{2}-\cfrac{1}{2}\times n^2=-\cfrac{3n^2+2n-1}{6}$
$综合上述T_n=\begin{cases} \cfrac{3n^2+4n}{6}\qquad\quad n=3k时\\ -\cfrac{2n+1}{6}\qquad\quad n=3k+1时\\-\cfrac{3n^2+2n-1}{6}\quad n=3k+2\end{cases}$

${\color{Red}利用完全立方差或立方差公式推导a_n=n^2的前n项和S_n } $
$(n-1)^3＝n^2-3n^2+3n-1\Rightarrow n^3-(n-1)^3＝3n^2-3n+1$
$n^3-(n-1)^3＝3n^2-3n+1$
$(n-1)^3-(n-2)^3＝3(n-1)^2-3n(n-1)+1$
$(n-2)^3-(n-3)^3＝3(n-2)^2-3n(n-2)+1$
$(n-3)^3-(n-4)^3＝3(n-3)^2-3n(n-3)+1$
$\cdots \cdots \cdots$
$2^3-1^3＝3\cdot 2^2-3\times 2+1$
$1^3-0^3＝3\cdot 1^2-3\times 1+1$
左右两边相加
$n^3=3S_n-3\times \cfrac{n(n+1)}{2}+n$
$3S_n=n^3+3\times \cfrac{n(n+1)}{2}-n=n[n^2+\cfrac{3}{2}(n+1)-1]$
$\cfrac{n}{2}(2n^2+3n+1)=\cfrac{n}{2}(2n+1)(n+1)$
$S_n=\cfrac{1}{6}\times n(n+1)(2n+1)$

数列讨论奇偶求通项－构造相邻项

$21年新高考一卷，已知数列\{a_n\}满足a_1=1,a_{n+1}=\begin{cases}a_n+1,n为奇数\\a_n+2,n为偶数\end{cases}$
$(1)记b_n=a_{2n},写出b_1,b_2,并求数列\{b_n\}的通项公式；$
$(2)求\{a_n\}的前20项和$
$a_2=a_1+1=1+1=2,b_1=a_2=2,a_3=a_2+2=4,b_2=a_4=a_3+1=5$
${\color{Green} \because \quad b_n=a_{2n}}$
$\therefore b_{n+1}=a_ {2(n+1)}=a_ {{\color{Green} 2n+1} +1}=a_ {{\color{Red} 2n}+1}+1=a_{{\color{Red} 2n} }+2+1=b_n+3$
$\Rightarrow b_{n+1}=b_n+3,故b_n为首项为2,公差为3的等差数列,b_n=2+3(n-1)=3n-1$
$(2)求前20项和，需分别计算奇数项和偶数项和；$
$偶数项：共10项(a_2,a_4,a_6\cdots a_{20}),b_{10}=3\times 10-1=29,$
$S_偶=\cfrac{10(b_1+b_{10})}{2}=\cfrac{10(2+29)}{2}=155$
$奇数项：共10项(a_1,a_3,a_5\cdots a_{19})$
$设\{c_k\},(c_k=a_{2k-1},k\in 1,2,3,\cdots 10)$
$c_{k+1}=a_{2k+1}=a_{2k}+2=(a_{2k-1}+1)+2=a_{2k-1}+3=c_k+3$
$\{c_k\}是首项为c_1=1,公差d=3的等差数列，和为S_奇=\cfrac{10(C_1+C_{10})}{2}=145$

$例6.记S_n为正项数列\{a_n\}的前n项和，且a_1=\cfrac{1}{3},2S_n=(3^n-1)a_n;$
$(1)求数列\{a_n\}的通项公式;$
$(2)设b_n=\cfrac{a_{n+1}}{(1-a_n)S_{n+1}},记数列\{b_n\}的前n项和为T_n,证明:T_n\ge \cfrac{3}{8}$
$解:2S_n=(3^n-1)a_n\quad ① n\ge 2时，2S_{n-1}=(3^{n-1}-1)a_{n-1}\quad ②$
$①-②,得2a_n=(3^n-1)a_n-(3^{n-1}-1)a_{n-1}\Rightarrow (3^{n-1}-1)a_{n-1}=(3^n-3)a_n$
$\Rightarrow \cfrac{a_n}{a_{n-1}}=\cfrac{3^{n-1}-1}{3^n-3}=\cfrac{1}{3},所以a_n为首项\cfrac{1}{3}，公式q为\cfrac{1}{3}的等比数列$
$a_n=(\cfrac{1}{3})^n,S_n=\cfrac{\cfrac{1}{3}-(\cfrac{1}{3})^{n+1}}{1-\cfrac{1}{3}}=\cfrac{1-(\cfrac{1}{3})^n}{2}$
$b_n=\cfrac{a_{n+1}}{(1-a_n)S_{n+1}}=\cfrac{(\cfrac{1}{3})^{n+1}}{[1-(\cfrac{1}{3})^{n}]\cfrac{1-(\cfrac{1}{3})^{n+1}}{2}}=\cfrac{2\times (\cfrac{1}{3})^{n+1}}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}$
${\color{Orange} \because \quad \cfrac{1}{1-(\cfrac{1}{3})^{n}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}} =\cfrac{1-(\cfrac{1}{3})^{n+1}-[{1-(\cfrac{1}{3})^{n}}]}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}$
$=\cfrac{-(\cfrac{1}{3})^{n+1}{+(\cfrac{1}{3})^{n}}}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}=\cfrac{{\cfrac{2}{3}\times (\cfrac{1}{3})^{n}}}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}=\cfrac{{2\times (\cfrac{1}{3})^{n+1}}}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}$
${\color{Orange} \therefore \quad b_n=\cfrac{1}{1-(\cfrac{1}{3})^{n}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}} $
$T_n=b_1+b_2+b_3+\cdots +b_n={\color{Orange} \cfrac{1}{1-\cfrac{1}{3}}-\cfrac{1}{1-(\cfrac{1}{3})^2}}+\cfrac{1}{1-(\cfrac{1}{3})^{2}}-\cfrac{1}{1-(\cfrac{1}{3})^{3}}$
$+{\color{Orange} \cfrac{1}{1-(\cfrac{1}{3})^{3}}-\cfrac{1}{1-(\cfrac{1}{3})^{4}}} +\cfrac{1}{1-(\cfrac{1}{3})^{4}}-\cfrac{1}{1-(\cfrac{1}{3})^{5}}+\cdots +\cfrac{1}{1-(\cfrac{1}{3})^{n}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}$
$=\cfrac{1}{1-\cfrac{1}{3}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}},要证\cfrac{3}{2}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}\ge \cfrac{3}{8}即证\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}\le \cfrac{9}{8}$
$即证1-(\cfrac{1}{3})^{n+1}\ge \cfrac{8}{9}=1-\cfrac{1}{9},n=1即可$
$T_n=\cfrac{1}{1-\cfrac{1}{3}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}=\cfrac{3}{2}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}},T_n\nearrow,T_n\ge (T_n)_{min}=T_1=\cfrac{3}{2}-\cfrac{9}{8}=\cfrac{3}{8}$

$例7、数列\{a_n\}的前n项和为S_n,且a_1=１,\{\cfrac{S_n}{a_n}\}是公差为\cfrac{1}{3}的等差数列$
$(1)求\{a_n\}通项公式$
$(2)证明:\cfrac{1}{a_1}+\cfrac{1}{a_2}+\cdots+\cfrac{1}{a_n}\lt 2$
$解:\cfrac{S_n}{a_n}=1+\cfrac{1}{3}(n-1)=\cfrac{n+2}{3}\Rightarrow S_n=\cfrac{n+2}{3}a_n\quad ①$
$n\ge 2时，有S_{n-1}=\cfrac{n-1+2}{3}a_{n-1}=\cfrac{n+1}{3}a_{n-1}\quad②$
$①-②,得\cfrac{a_n}{a_{n-1}}=\cfrac{n+1}{n-1}$
$\cfrac{a_2}{a_1}=\cfrac{2+1}{2-1}=\cfrac{3}{1}$
$\cfrac{a_3}{a_2}=\cfrac{3+1}{3-1}=\cfrac{4}{2}$
$\cfrac{a_4}{a_3}=\cfrac{4+1}{4-1}=\cfrac{5}{3}$
$\cdots \cdots\cdots\cdots $
$\cfrac{a_{n-1}}{a_{n-2}}=\cfrac{n-1+1}{n-1-1}=\cfrac{n}{n-2}$
$\cfrac{a_n}{a_{n-1}}=\cfrac{n+1}{n-1}$
$\cfrac{a_n}{a_{n-1}}\times\cfrac{a_{n-1}}{a_{n-2}}\cdots \cfrac{a_4}{a_3}\times \cfrac{a_3}{a_2}\times \cfrac{a_2}{a_1}=a_n=\cfrac{n+1}{n-1}\times\cfrac{n}{n-2}\cdots\cfrac{5}{3}\times \cfrac{4}{2}\times \cfrac{3}{1}=\cfrac{n(n+1)}{2}$
${\color{Red} \because \quad} \cfrac{1}{a_n}=\cfrac{2}{n(n+1)}=2(\cfrac{1}{n}-\cfrac{1}{n+1})$
${\color{Red} \therefore\quad} \cfrac{1}{a_1}+\cfrac{1}{a_2}+\cdots+\cfrac{1}{a_n}=2(1-\cfrac{1}{2}+\cfrac{1}{2}-\cfrac{1}{3}+\cdots +\cfrac{1}{n}-\cfrac{1}{n+1})=2(1-\cfrac{1}{n+1})\lt 2$

$例8、已知正项数列\{a_n\}的前n项和S_n，满足S_n=(\cfrac{a_n+1}{2})^2.$
$(1)求数列\{a_n\}的通项公式;$
$(2)记b_n=\cfrac{n+1}{S_nS_{n+2}},设数列\{b_n\}的前n项和为T_n.求证T_n\lt \cfrac{5}{16}$
$解：(1).a_1=S_1=(\cfrac{a_1+1}{2})^2\Rightarrow a_1=1$
$n\ge 2时，S_{n-1}=(\cfrac{a_{n-1}+1}{2})^2$
$a_n=S_n-S_{n-1}=(\cfrac{a_n+1}{2})^2-(\cfrac{a_{n-1}+1}{2})^2=\cfrac{a_n+a_{n-1}+2}{2} \times\cfrac{a_n-a_{n-1}+2}{2}$
$\Rightarrow 4a_n=a_n^2-a_{n-1}^2+2(a_n-a_{n-1})\Rightarrow a_n^2-a_{n-1}^2-2(a_n+a_{n-1})=0$
$\Rightarrow (a_n+a_{n-1})(a_n-a_{n-1}-2)=0,a_n+a_{n-1}=0(舍去),a_n-a_{n-1}=2$
$a_{n-1}-a_{n-2}=2$
$a_{n-2}-a_{n-3}=2$
$a_2-a_1=2$
$a_n-a_1=2(n-1)\Rightarrow a_n=2n-1\quad(n\ge 2),a=1$
$(2){\color{Red} \because \quad } S_n=(\cfrac{a_n+1}{2})^2=n^2,所以S_{n+2}=(n+2)^2$
$b_n=\cfrac{n+1}{S_nS_{n+2}}=\cfrac{n+1}{n^2(n+2)^2}$
${\color{Red} \because \quad } \cfrac{1}{n^2(n+2)^2} =\cfrac{1}{4(n+1)}[\cfrac{1}{n^2}-\cfrac{1}{(n+2)^2} ]$
${\color{Red} \therefore \quad } b_n=\cfrac{n+1}{n^2(n+2)^2} =\cfrac{1}{4}[\cfrac{1}{n^2}-\cfrac{1}{(n+2)^2} ]$
$T_n=\cfrac{1}{4}(1-\cfrac{1}{3^2}+\cfrac{1}{2^2}-\cfrac{1}{4^2}+\cfrac{1}{3^2}-\cfrac{1}{5^2}+\cdots+\cfrac{1}{n^2}-\cfrac{1}{(n+2)^2}$
$=\cfrac{1}{4}[1+\cfrac{1}{2^2}-\cfrac{1}{(n+1)^2}-\cfrac{1}{(n+2)^2}]\lt \cfrac{5}{16}$

https://uu.890222.xyz/usr/uploads/2025/12/705270186.[5]: https://uu.890222.xyz/usr/uploads/2025/12/2791294339.png

1、

2、
3、

$4、\cfrac{b}{\sin B}=\cfrac{c}{\sin C}\Rightarrow\cfrac{3\sin C}{\sin B}=\cfrac{3c}{b}\Rightarrow \cfrac{\cos A}{a}+\cfrac{\cos B}{b}=\cfrac{3\sin C}{\sin B}=\cfrac{3c}{b}\quad 两边\times ab \Rightarrow b\cos A+a\cos B=3ac$
$\sin B\cos A+\cos B\sin A=3a\sin C\Rightarrow \sin (A+B)=\sin C=3a\sin C\Rightarrow 3a=1$

$5、\begin{cases} e=\cfrac{c}{a}=\sqrt{10}\\ c^2=a^2+b^2 \end{cases}\Rightarrow 10a^2=a^2+b^2\Rightarrow y=\pm \cfrac{b}{a}x=\pm 3x$
$圆心(-2,-1),r=2\Rightarrow 圆与直线y=-3x无交点。圆心到渐近线3x-y=0的距离d=\cfrac{|-6+1|}{\sqrt{10}}=\cfrac{5}{\sqrt{10}}$
$|MN|=2\sqrt{4-(\cfrac{5}{\sqrt{10}})^2}=2\sqrt{\cfrac{3}{2}}=\sqrt{6}$

$6、\cfrac{\cos \theta(1-\sin 2\theta)}{\sqrt{2}\sin(\theta-\cfrac{\pi}{4})}=\cfrac{\cos\theta(\sin \theta-\cos\theta)^2}{\sin \theta-\cos\theta}=\cos\theta(\sin \theta-\cos\theta)$
$=\cfrac{\cos\theta(\sin \theta-\cos\theta)}{\sin^2\theta+\cos ^2\theta }=\cfrac{\tan\theta-1}{\tan^2\theta-1}$

7、

$8、设x_1\gt x_2,\cfrac{2x_2f(x_1)-2x_1f(x_2)}{x_1x_2(x_1-x_2)}=\cfrac{2[\cfrac{f(x_1)}{x_1}-\cfrac{f(x_2)}{x_2}]}{x_1-x_2}\lt 0\qquad 构造函数g(x)=\cfrac{f(x)}{x},$
$故上式\cfrac{\cfrac{f(x_1)}{x_1}-\cfrac{f(x_2)}{x_2}}{x_1-x_2}=\cfrac{g(x_1)-g(x_2)}{x_1-x_2}\lt 0，即g(x)在x\gt 0为递减函数，f(x)为奇函数\Rightarrow g(x)为偶函数$
$f(2025)=4050\Leftrightarrow \cfrac{f(2025)}{2025}=2=g(2025)\quad f(x)\lt 2x\Leftrightarrow g(x)=\cfrac{f(x)}{x}\lt 2=g(2025)\Rightarrow g(x)\lt g(2025)$
$|x|\gt 2025\Rightarrow x\gt 2025,或x\lt -2025,故选A$

$9、l=\sqrt{2^2+(4\sqrt{2})^2}=6,S_侧=\pi rl=12\pi;S_\triangle =\cfrac{1}{2}\times 4\times 4\sqrt{2}=8\sqrt{2},半周长\cfrac{1}{2}C=4+12=16,$
$内切圆半径r_{in}=\cfrac{2S}{C}=\sqrt{2},C，刚好落在高的中点，故母线为3;选ABC;r_t=\cfrac{4}{3},故D不对$

$10、-\cfrac{3}{4}=e^2-1=k_{PM}k_{PN}\Rightarrow a=^2=4,c^2=1,b^2=3,故选A;$
$\quad B显然不成立,P为上下顶点时角最大,只有当b=1时才成立。Q点为上顶点，$
$|PQ|^2=x^2+(y-1)^2=4(1-\cfrac{1}{3}y^2)+y^2-2Y+1=-\cfrac{1}{3}y^2-2y+5=-\cfrac{1}{3}(y+3)^2+8$
$y\in [-\sqrt{3},\sqrt{3}],最大值在y=--\sqrt{3}处取得，|PQ|^2=4+2\sqrt{3}\Rightarrow |PQ|=\sqrt{3}+1$
$曲线\Gamma 上存在点P,使得|PD|=e,|PD|^2=(x-1)^2+y^2=x^2-2x+1+3(1-\cfrac{x^2}{4})=\cfrac{x^2}{4}-2x+4$
$=\cfrac{1}{4}(x-4)^2, 令它=e^2\Rightarrow x-4=\pm 2e\in (-2,2),故D正确$

11、

$12、C_{5}^{1} (\cfrac{\sqrt{x}}{3})^4(\cfrac{3}{x^2})^1=5\times \cfrac{1}{3^3}=\cfrac{5}{27}$

$13、O与B、C共线，且为BC的中心。\angle A=\cfrac{\pi}{2},\overrightarrow{CA} 在\overrightarrow{CB}的投影向量为 \cfrac{3}{4}\overrightarrow{CB},用射影定理可求BC边上的高为\sqrt{3},B=\cfrac{\pi}{3}$

$14、已知a\gt 0,函数f(x)=ae^x+\ln \cfrac{a}{x-1}+1的定义域为x\gt 1(因为\cfrac{a}{x-1}\gt 0且a\gt 0)。$
$要求存在x\gt 1使得f(x)\lt 0,至少有一个f(x)\lt 0$
法一：$步骤1、求导数与极值点：$
${f}' (x)=ae^x-\cfrac{1}{x-1}.$
$由于a\gt 0且x\gt 1，ae^x严格递增-\cfrac{1}{x-1}也严格递增，故{f}' (x)在(1,+\infty)上严格递增$
$当x\to 1^+时，{f}' (x)\to -\infty;当x\to +\infty时，{f}' (x)\to +\infty。$由介值定理，存在唯一$x_0\in (1,+\infty)$使得${f}'(x_0)=0,$即：$ae^{x_0}=\cfrac{1}{x_0-1}\Rightarrow a=\cfrac{1}{(x_0-1)}e^{x_0}$
$步骤2、求最小值f(x_0)$
$f(x)在(1,x_0)\quad \nearrow;f(x)在(x_0,+\infty)\quad \searrow;故f(x)在x=x_0处取得最小值：f(x_0)=ae^{x_0}+\ln \cfrac{a}{x_0-1}+1.$
$将a=\cfrac{1}{(x_0-1)}e^{x_0}代入,化简:f(x_0)=\cfrac{1}{x_0-1}=(x_0-1)-2\ln (x_0-1)$
$步骤3、换元与单调性分析：令t=x_0-1\quad (t\gt 0),f(x_0)=g(t)=\cfrac{1}{t}-t -2\ln t.$
${g}' (t)=-\cfrac{1}{t^2}-1- \cfrac{2}{t}=-\cfrac{(t+1)^2}{t^2}\lt 0,故g(t)在(0,+\infty)严格递减。$
$又g(1)=0,故t\gt 1時，g(t)\lt 0$
$步骤4、轉化為a的範圍。t\gt 1\Leftrightarrow x_0\gt 2.將a=\cfrac{1}{(x_0-1)}e^{x_0}視作關於x_0的函數$
$令h(x)=(x-1)e^x,則a=\cfrac{1}{h(x)},{h}' (x)=xe^x\gt 0,h(x)\nearrow \Rightarrow a=\cfrac{1}{h(x)} \searrow$
$當t=1,即x_0=2時，a=\cfrac{1}{(2-1)e^2}=\cfrac{1}{e^2};當x_0\to +\infty時，a\to 0^+.因此當x_0\gt 2時，$
$a\in (0,\cfrac{1}{e^2}),此時f(x_0)\lt 0,滿足條件。$
綜上，$a的取值範圍是(0,\cfrac{1}{e^2})$
法二：用同构法
$f(x)\lt 0\Rightarrow ae^x+\ln \cfrac{a}{x-1}+1\lt 0\quad \because a\gt 0\quad\therefore x\gt 1$
$\Rightarrow ae^x+\ln a-\ln(x-1)+1\lt 0, 指对分家ae^x+\ln a\lt \ln (x-1)-1$
$\Rightarrow ae^x+x+\ln a\lt \ln (x-1)+(x-1)\Rightarrow e^{x+\ln a}+x+\ln a\lt \ln (x-1)+(x-1)$
$\Rightarrow e^{x+\ln a}+x+\ln a\lt \ln (x-1)+e^{\ln (x-1)}$
$构造函数g(x)=e^x+x,g(x+\ln a)\lt g(\ln (x-1))\Rightarrow x+\ln a\lt \ln (x-1)\Rightarrow \ln a\lt \ln (x-1)-x$
$存在性问题，\ln a\lt [\ln (x-1)-x]_{max}\quad x\gt 1,令h(x)=\ln (x-1)-x,{h}'(x)=\cfrac{1}{x-1}-1$
$=\cfrac{2-x}{x-1},故x\in (1,2),{h}'(x)\gt 0,h(x)\nearrow ;x\in (2,+\infty),{h}'(x)\lt 0,h(x)\searrow,$
$故h(x)在x=2有最大值h(2)=-2,\ln a\lt -2\Rightarrow 0\lt a \lt e^{-2}$

$15、d=\cfrac{S_2}{2}-\cfrac{S_1}{1}=\cfrac{a_2+a_1}{2}-\cfrac{a_1}{1}=\cfrac{3-a_1}{2}$
$\cfrac{S_5}{5}=S_1+4d=a_1+4d=a_1+4\times \cfrac{3-a_1}{2}=6-a_1=5\Rightarrow a_1=1,d=1$
$\cfrac{S_n}{n}=n\Rightarrow S_n=n^2\Rightarrow n\ge 2,S_{n-1}=(n-1)^2,S_n-S_{n-1}=a_n=2n-1(n\ge 2)$
$n=1时，a_1=1,故a_n=2n-1在n=1时也成立。$
$(2)、b_n=2(n+1)(a_n+1)=4n(n+1)\Rightarrow \cfrac{1}{b_n}=\cfrac{1}{4n(n+1)}=\cfrac{1}{4}(\cfrac{1}{n}-\cfrac{1}{n+1})$
$\cfrac{1}{b_1}+\cfrac{1}{b_2} +\cdots +\cfrac{1}{b_n}=\cfrac{1}{4}(1-\cfrac{1}{2}+\cfrac{1}{2}-\cfrac{1}{3}+\cfrac{1}{3}-\cfrac{1}{4}+\cdots+\cfrac{1}{n-1}-\cfrac{1}{n}+\cfrac{1}{n}-\cfrac{1}{n+1})$
$=\cfrac{1}{4}(1-\cfrac{1}{n+1})\lt \cfrac{1}{4}$

16、

$17、(1)设圆心坐标为(x,y),x+\cfrac{1}{2}=\sqrt{(x-1)^2+y^2}-\cfrac{1}{2}\Rightarrow y^2=4x$
$(2)设A(x_1,y_1),C(x_2,y_2),B(x_3,y_3),D(x_4,y_4),S_\Box =\cfrac{1}{2}\times|AC|\times|BD|$
$|AC|=|AF|+|CF|=x_1-(-1)+x_2-(-1)=x_1+x_2+2,同理|BD|=x_3+x_4+2$
$显然k_{AC},k_{BD}的斜率不为0，设k_{AC}=k,则k_{BD}=-\cfrac{1}{k},l_{AC}:y=k(x-1),l_{BD}:y=-\cfrac{1}{k}(x-1)$
$\begin{cases} y^2=4x\\ y=kx-k\end{cases}\Rightarrow k^2(x-1)^2-4x=0\Rightarrow k^2x^2-(2k^2+4)x+k^2=0$
$x_1+x_2=\cfrac{2k^2+4}{k^2}\quad同理x_3+x_4=\cfrac{2(\cfrac{1}{k})^2+4}{(\cfrac{1}{k})^2}=2+4k^2$
$x_1+x_2+2=4+\cfrac{4}{k^2},x_3+x_4+2=4+4k^2$
$S=\cfrac{1}{2}\times(x_1+x_2+2)(x_3+x_4+2)=8(1+\cfrac{1}{k^2})\times (1+k^2)=8(2+k^2+\cfrac{1}{k^2})\ge 32$
$当且仅当\cfrac{1}{k^2}=k^2,k=\pm 1时取等号$
所以四边形面积的最小值为32

$18、(1)三棱锥D-MAB,即三棱锥M-ABD, 当M在弧AMD的中点时体积最大。$
$以底面圆心O为坐标原点，\overrightarrow{OM};\overrightarrow{OD};\overrightarrow{OO'}分别为x,y,z轴正向。$
$A(0,-2,0),B(0,-2,4),C(0,2,4),M(2,0,0);设平面BCM的法向量为\vec{n_1},平面ABM的法向量为\vec{n_2}$
$\overrightarrow{BC}=(0,4,0),\overrightarrow{BM}=(2,2,-4),\vec{n_1}\cdot \overrightarrow{BC}＝0和\vec{n_1}\cdot \overrightarrow{BM}＝0;解得\vec{n_1}=(2,0,1)$
$\overrightarrow{AM}=(2,2,0),\vec{n_2}\cdot \overrightarrow{AM}＝0和\vec{n_2}\cdot \overrightarrow{BM}＝0;解得\vec{n_2}=(1,-1,0),\cos \theta=\sin <\vec{n_1},\vec{n_2}>=\cfrac{\vec{n_1}\cdot \vec{n_2}}{|\vec{n_1}|\times|\vec{n_1}|}=\cfrac{2}{\sqrt{5}\sqrt{2}}=\cfrac{\sqrt{10}}{5}$

$(2)\triangle AMD为直角三角形，M点在AD的垂足E，\angle MBE为直线MB与平面ABCD所成的角。$
$\tan \angle MBE=\cfrac{ME}{AB}，当ME为半径时，\tan \angle MBE最大$

$19、(1)a=0,ff(x)=2\ln (1-x)+2x\quad (x\lt 1),{f}'(x)=-\cfrac{2}{1-x}+2=\cfrac{-2x}{1-x}$
$x\in (-\infty,0), {f}'(x)\gt 0,f(x)\nearrow ;\quad x\in (0,+\infty), {f}'(x)\lt 0,f(x)\searrow;$
$故f(x)在x=0处有极大值，f(0)=0；无极小值$
$(2)当x\le 0时，f(x)\ge 0恒成立，求a的取值范围。$
$f(x)=(2+2ax)\ln (1-x)+2x,x\le 0.$
$x=0时，f(0)=0,满足条件。$
$x\lt 0时，\ln (1-x)\gt 0,2x\lt 0$
$①若a\gt 0,当x=-\cfrac{2}{a}\lt 0时，f(-\cfrac{2}{a})\lt 0,不合题意。$
$②若a= 0,根据(1)结论可知，x\le 0,f(x)\le 0,不合题意。$
$③若a\lt 0，当x\le 0,f(x)\ge 0等价于 \ln (1-x)+\cfrac{2x}{2+ax}\ge 0,令g(x)=\ln (1-x)+\cfrac{2x}{2+ax}$
${f}'(x)=\cfrac{-1}{1-x}+\cfrac{4}{(2+ax)^2}=\cfrac{-x(a^2x+4a+4)}{(1-x)(2+ax)^2}$
$若a\le -1,则当x\lt 0时，{g}'(x)\lt 0,g(x)\searrow ;所以当x\le 0时g(x)\ge 0, 故f(x)\ge 0.$

$14、已知正实数a,b，满足ae^2(\ln b-\ln a+a-1)\ge be^a,则\cfrac{1}{b}的最小值为(\qquad)$
$类似指对分离,两边{\div} ae^2，\ln b-\ln a+a-1=\ln \cfrac{b}{a}+a-1\ge \cfrac{b}{a}e^{a-2}=e^{\ln \cfrac{b}{a}+a-2}$
${\color{Red} \because \quad } e^x\ge x+1,当且仅当x=0时取=\quad {\color{Red} \therefore \quad } e^{\ln \cfrac{b}{a}+a-2}\ge \ln \cfrac{b}{a}+(a-2)+1$
$\begin{cases} e^{\ln \cfrac{b}{a}+a-2}\ge \ln \cfrac{b}{a}+(a-2)+1\\e^{\ln \cfrac{b}{a}+a-2}\le \ln \cfrac{b}{a}+(a-2)+1\end{cases}\Rightarrow e^{\ln \cfrac{b}{a}+a-2}= \ln \cfrac{b}{a}+(a-2)+1$
$\Rightarrow \ln \cfrac{b}{a}+a-2=0\Rightarrow \ln b=\ln a+2-a，令f(a)=\ln a+2-a$
$ {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} $
$\cdots \cdots$
$\cfrac{\ln (n-1)}{n}\le \cfrac{n-2}{2}$
$\cfrac{\ln n}{n+1}\le \cfrac{n-1}{2}$
$\cfrac{\ln 1}{2}+\cfrac{\ln 2}{3}+\cfrac{\ln 3}{4}+\cdots+\cfrac{\ln n}{n+1}\le \cfrac{0}{2}+\cfrac{1}{2}+\cfrac{2}{2}+\cfrac{3}{2}+\cfrac{4}{2}+\cdots \cfrac{n-2}{2}+\cfrac{n-1}{2}=\cfrac{(0+n-1)n}{2\times 2}$
$(2){\color{Red} {f}'(x)=\ln x-\cfrac{a}{x} \quad }令{f}'(x)=0\Leftrightarrow \ln x=\cfrac{a}{x}\Leftrightarrow {\color{Red} a=x\ln x}$
$f(x)的两个极值点,即y=a与g(x)=x\ln x有两个交点$
$根据g(x)=x\ln x 的图像,可以a\in (-\cfrac{1}{e},0)时,x\in (0,1),有x=\cfrac{1}{e}两边各有一个交点$
${\color{Red} ①}设0\lt x_1\lt \cfrac{1}{e}, \cfrac{2}{e}\lt x_2\lt 1 ,x_1+x_2\gt \cfrac{2}{e}不证自明$
${\color{Red} ②} 设0\lt x_1\lt \cfrac{1}{e} \lt x_2\lt \cfrac{2}{e},欲证x_1+x_2\gt \cfrac{2}{e},即证x_1\gt \cfrac{2}{e}-x_2$
$g(x)=x\ln x求导{g}'(x)=\ln x+1，易知g(x)在x\in(0,\cfrac{1}{e})单调递减；在x\in(\cfrac{1}{e},+\infty)单调递增$
$故即证g(x_1)\lt g(\cfrac{2}{e}-x_2){\color{Red} \Leftrightarrow } g(x_2)\lt g(\cfrac{2}{e}-x_2),构造函数G(x)=g(x)-g(\cfrac{2}{e}-x)\quad (x\in (\cfrac{1}{e},\cfrac{2}{e})$
${G}'(x)={g}'(x)+{g}'(\cfrac{2}{e}-x)=\ln x+1+[\ln (\cfrac{2}{e}-x)+1]$
${\color{Red} {G}''(x)= } \cfrac{1}{x}-\cfrac{1}{\cfrac{2}{e}-x},{G}''(x)\searrow; {G}''(\cfrac{1}{e})=0\Rightarrow {G}''(x)\lt 0\Rightarrow {G}'(x)\searrow$
${G}'(\cfrac{1}{e})=0,\Rightarrow {G}'(x)\lt 0\Rightarrow G(x)\searrow ,G(\cfrac{1}{e})=0,G(x)\lt G(\cfrac{1}{e})=0$
$\Rightarrow G(x)=g(x)-g(\cfrac{2}{e}-x)\lt 0,\Rightarrow g(x)\lt g(\cfrac{2}{e}-x)$