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

$(1)a_{n+1}=a_n+n+1，有递推关系式：$
$a_{n}=a_{n-1}+(n-1)+1$
$a_{n-1}=a_{n-2}+(n-2)+1$
$a_{n-2}=a_{n-3}+(n-3)+1$
$a_{n-3}=a_{n-4}+(n-4)+1$
$\cdots \cdots \cdots\cdots$
$a_3=a_2+2+1$
$a_2=a_1+1+1$
累加之
$a_{n}+a_{n-1}+a_{n-2}+a_{n-3}+\cdots +a_3+a_2=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+\cdots +a_2+a_1$
$+(n-1)+(n-2)+(n-3)+(n-4)+\cdots +1+1+1\cdots +1$
$a_n=a_1+(n-1)+(n-2)+(n-3)+(n-4)+\cdots+2+1(n-1)\times 1$
$=2+(n-1)+(n-2)+(n-3)+(n-4)+\cdots+2+1+(n-1)$
$=n+(n-1)+\cdots +1=\cfrac{n(n+1)}{2}$

$b_n=\cfrac{1}{a_n}=\cfrac{1}{\cfrac{n(n+1)}{2}}=\cfrac{2}{n(n+1)}=2(\cfrac{1}{n}-\cfrac{1}{n+1})$
$b_1=2(1-\cfrac{1}{2})$
$b_2=2(\cfrac{1}{2}-\cfrac{1}{3})$
$b_3=2(\cfrac{1}{3}-\cfrac{1}{4})$
$\cdots\cdots\cdots\cdots$
$b_{n-2}=2(\cfrac{1}{n-2}-\cfrac{1}{n-1})$
$b_{n-1}=2(\cfrac{1}{n-1}-\cfrac{1}{n})$
$b_n=2(\cfrac{1}{n}-\cfrac{1}{n+1})$
$T_n=2(1-\cfrac{1}{n+1})\lt 2$

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

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

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、

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

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

$14、已知a\gt 0,函数f(x)=ae^x+\ln \cfrac{a}{x-1}+1的定义域为x\gt 1(因为\cfrac{a}{x-1}\gt 0且a\gt 0)。$
$要求存在x\gt 1使得f(x)\lt 0,至少有一个f(x)\lt 0$
$步骤1、求导数与极值点：$
${f}' (x)=ae^x-\cfrac{1}{x-1}.$
$由于a\gt 0且x\gt 1，ae^x严格递增-\cfrac{1}{x-1}也严格递增，故{f}' (x)在(1,+\infty)上严格递增$
$当x\to 1^+时，{f}' (x)\to -\infty;当x\to +\infty时，{f}' (x)\to +\infty。$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，$a的取值範圍是(0,\cfrac{1}{e^2})$

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

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
$x(a+e^x)\ge 1+\ln x\Rightarrow a\ge \cfrac{1+\ln x}{x}-e^x=\cfrac{1+\ln x-xe^x}{x}$
$=\cfrac{1+\ln x-e^{x+\ln x}}{x}=\cfrac{x+\ln x+1-e^{x+\ln x}-x}{x}\quad自证\quad x+1\le e^x$
$x+\ln x+1\le e^{x+\ln x},当且令当x+\ln x=0时取=$
$a\ge \cfrac{x+\ln x+1-e^{x+\ln x}-x}{x}\ge \cfrac{0-x}{x}=-1\quad 若x+\ln x=0成立，则上式能取得等号成立$
$设g(x)=x+\ln x，x\in (0,+\infty),g(x)单调递增，g(1)=1\gt 0,g(\cfrac{1}{e})=\cfrac{1}{e}-1\lt 0,$
$故存在g(x_0)=0在x_0\in (\cfrac{1}{e},1),即存在x_0\in (\cfrac{1}{e},1)使得a\ge \cfrac{x+\ln x+1-e^{x+\ln x}-x}{x}取＝成立。$
$a\ge -1,a\in (-\infty,1]$

$1、在复平面内，复数z=(2+i)i对应的点位于第几象限；$
$2、已知集合A=\{x|-1\le x\le 2,x\in Z\},B=\{x|\cfrac{1}{x}\le 1\},A\cap B=$
$5、 已知\cfrac{\cos 2\theta}{\cos (\theta-\cfrac{\pi}{4})}=\cfrac{\sqrt{2}}{2},则\sin 2\theta$
$解：\cfrac{\cos^2\theta -\sin^2\theta}{\cos\theta +\sin\theta}=(\cfrac{\sqrt{2}}{2})^2\Rightarrow \cos \theta-\sin \theta=0.5\Rightarrow 1-\sin 2\theta=\cfrac{1}{4}$

$15、在\triangle ABC中,角A,B,C的对边分别为a,b,c,ab\cos C+ac\cos B=bc\sin A.$
$(1)证明:\sin A=\sin B\sin C$
$(2)若\tan B=2\tan C，求\tan A$
$证明:(1)\cfrac{2S_\triangle}{\sin C}\cos C+\cfrac{2S_\triangle}{\sin B}\cos B=2S_\triangle\Rightarrow \cfrac{\cos C}{\sin C}+\cfrac{\cos B}{\sin B}=1$
$\Rightarrow \cfrac{\sin B\cos C+\cos B\sin C}{\sin B\sin C}=1\Rightarrow \sin A=\sin (B+C)=\sin B\sin C$
$(2)-\tan A=\tan(\pi- A)=\tan(B+C)=\cfrac{\tan B +\tan C}{1-\tan B\tan C}$
$\Rightarrow \tan A+\tan B+\tan C=\tan A\tan B\tan C\quad 正切恒等式$
$\because \sin A=\sin (B+C)=\sin B\cos C+\cos B\sin C=\sin B\sin C$
$\times \cos B\cos C,得\tan B+\tan C=\tan B\tan C\Rightarrow 3\tan C=2\tan^2 C\Rightarrow \tan C=\cfrac{3}{2}$
$\tan A+3\tan C=\tan A\times 2\tan^2C\Rightarrow \tan A=\cfrac{7}{2}$

$解法二:\cfrac{a}{\sin A}=\cfrac{b}{\sin B}=\cfrac{c}{\sin C}\Rightarrow b\sin A=a\sin B
\quad bc\sin A=ac\sin B=ab\sin C左边$
$ac\sin B=ab\cos C+ac\cos B \Rightarrow c\sin B=b\cos C+c\cos B\Rightarrow\sin B\cos C+\cos B\sin C=\sin B\sin C$
$\Rightarrow \sin(B+C)=\sin B\sin C\Rightarrow \sin A=\sin(B+C)=\sin B\sin C$
$a\sin B\cos C+a\cos B\sin C=a\sin B\sin C \Rightarrow \sin B\cos C+\cos B\sin C=\sin B\sin C$
$两边{\div} \cos B\cos C,得\tan B+\tan C=\tan B\tan C\quad \tan B=2 \tan C\Rightarrow 3\tan C=2\tan^2C$
$\Rightarrow \tan C=\cfrac{3}{2},\tan B=3,-\tan A=\tan (\pi-A)=\tan (B+C)=\cfrac{\tan B+\tan C}{1-\tan B\tan C}= $

$16、已知椭圆C:\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1(a\gt b\gt 0)的左顶点为(-2,0)，且离心率为\cfrac{\sqrt{3}}{2}.$
$(1)求椭圆方程:\cfrac{x^2}{4}+y^2=1$
$(2)过点(6,0)的直线l交于不同的两点PQ(异于A），记直线AP,AQ的斜率分别为k_1,k_2,求k_1k_2$
$解：点乘双根法能大大减少此题的计算量。$
$设P(x_1,y_1)Q(x_2,y_2),求k_1k_2\cfrac{y_1y_2}{(x_1+2)(x_2+2)},x_1,x_2,y_1,y_2是直线l$方程与椭圆方程的公共解。
$\begin{cases}直线l:y=k(x-6)\\ \qquad \\椭圆C:x^2+4y^2-4=0\end{cases}\Rightarrow x^2+4k^2(x-6)^2-4=0\quad ①$
$因为x_1,x_2是①式的根，因此，上式可改写成:x^2+4k^2(x-6)^2-4＝(1+4k^2)(x_1-x)(x_2-x)$
$令x=-2,可得(x_1+2)(x_2+2)=\cfrac{64\times 4k^2}{1+4k^2}$
$\begin{cases} 直线l:\qquad x=my+6\\ 椭圆C:\quad x^2+4y^2-4=0\end{cases}\Rightarrow (my+6 )^2+4y^2-4=0\quad ②$
$因为y_1,y_2是②式的根，因此，上式可改写成:(my+6 )^2+4y^2-4＝(m^2+4)(y_1-y)(y_2-y)$
$令y=0,可得y_1y_2=\cfrac{32}{m^2+4}$
$\because k_1k_2=\cfrac{y_1y_2}{(x_1+2)(x_2+2)}=\cfrac{\cfrac{32}{m^2+4}}{\cfrac{64\times 4k^2}{1+4k^2}}\quad \because m=\cfrac{1}{k}\Rightarrow k_1k_2=\cfrac{1}{8}$

$17、已知函数f(x)=ax^2-x\ln x,a\in R$
$(1)若函数f(x)在定义域内单调递增，求的取值范围$
$(2)若过点(0,1)可作曲线y=f(x)的两条切线l_1，l_2，记直线l_1，l_2的斜率分别为k_1,k_2,求k_1+k_2的取值范围$
$解:{f}' (x)=2ax-1-\ln x\ge0\Rightarrow 2a\ge \cfrac{1+\ln x}{x}\quad $
$令g(x)= \cfrac{1+\ln x}{x}\Rightarrow {g}' (x)=\cfrac{-\ln x}{x^2}\quad$
$x\in (0,1) \quad {g}' (x)\gt 0,g(x)\nearrow ;\qquad x\in (1,+\infty ) \quad {g}' (x)\lt 0,g(x)\searrow ;$
$\therefore g(x)在x=1处有最大值g(1)=1,2a\ge 1\Rightarrow a\ge \cfrac{1}{2}$
$解:(2)设切点坐标为(t,t(at-\ln t)),切线斜率为{f}' (t)=2at-1-\ln t,故切线方程为:$
$y-t(at-\ln t)=(2at-1-\ln t)(x-t),切线经过(0,1)，故有1-t(at-\ln t)=-t(2at-1-\ln t)\Rightarrow$
$1-at^2+t\ln t=-2at^2+t+t\ln t,\Rightarrow at^2-t+1=0$
$由于x\gt 0，即切点坐标t\gt 0\therefore t_1+t_2=t_1t_2=\cfrac{1}{a},且a\gt 0\Rightarrow \Delta =1-4a\gt 0$
$0\lt a\lt \cfrac{1}{4},k_1+k_2=2a(t_1+t_2)-2-\ln(t_1+t+2)=2a\cfrac{1}{a}-2-\ln \cfrac{1}{a}=\ln a$
$0\lt a\lt \cfrac{1}{4},\therefore k_1+k_2=\ln a \in (-\infty,-2\ln2)$

$18、如图所示，在长方体ABCD-A_1B_1C_1D_1中，点M,N分别是直线AB_1,BC_1上的动点。$
$(1)若M,N分别为线段AB_1,BC_1的中点，证明：MN//平面ABCD;$
$(2)若AB=BC=3,且二面角A-B_1D_1－C的斜弦值为\cfrac{1}{3}.$
$①求AA_1;$
$②若直线MN与平面ABCD所成角为\cfrac{\pi}{3}，求线段MN长度的最小值。$
$解:(1)证明：以D为坐标原点，DA,DC,DD_1分别为x,y,z$轴正方向，建立空间直角坐标系,令$DA=a,DC=b,DD_1=C,$
$则A(a,0,0),B_1(a,b,c),B(a,b,0),C(0,b,c),M(a,\cfrac{b}{2},\cfrac{c}{2}),N(\cfrac{a}{2},b,\cfrac{c}{2}),\overrightarrow{MN}=(-\cfrac{a}{2},\cfrac{b}{2},0)$
$显然平面ABCD的法向量为\vec{n} =(0,0,1),\vec{n}\cdot \overrightarrow{MN}=0,所以MN//平面ABCD $
$(2)解①A(3,0,0),B_1(3,3,c),C(0,3,0),D_1(0,0,C),\overrightarrow{AB_1}=(0,3,0),\overrightarrow{D_1B_1}=(3,3,0),\overrightarrow{CB_1}=(3,0,0)$
$平面AB_1D_1的法向量为\vec{n_1}:得\vec{n_1}\cdot \overrightarrow{AB_1}=0,\vec{n_1}\cdot \overrightarrow{D_1B_1}=0，解得\vec{n_1}=(c,-c,3)$
$平面CB_1D_1的法向量为\vec{n_2}:得\vec{n_2}\cdot \overrightarrow{CB_1}=0,\vec{n_2}\cdot \overrightarrow{D_1B_1}=0，解得\vec{n_2}=(-c,c,3)$
$\cos <\vec{n_1},\vec{n_2}>=\cfrac{\vec{n_1}\cdot\vec{n_2}}{|\vec{n_1}|\times|\vec{n_2}|}=\cfrac{|9-2c^2|}{2c^2+9}=\cfrac{1}{3},3(2c^2-9)=2c^2+9\Rightarrow c=3$
$3(9-2c^2)=2c^2+9\Rightarrow c=\cfrac{3}{2},此夹角为钝角，舍去,故可得AA_1=3$
$解②,A(3,0,0),B(3,3,0),B_1(3,3,3),C_1(0,3,3),\quad\therefore \overrightarrow{AB}=(0,3,0),\overrightarrow{AB_1}=(0,3,3),\overrightarrow{BC_1}=(-3,0,3)$
$设\overrightarrow{AM}=\lambda\overrightarrow{AB_1},\overrightarrow{BN}=\mu\overrightarrow{BC_1},\overrightarrow{MN}=\overrightarrow{AB}+\overrightarrow{BN}-\overrightarrow{AM}=(0,3,0)+\mu (-3,0,3)-\lambda (0,3,3)$
$\overrightarrow{MN}=(-3\mu,3=3\lambda,-3\lambda +3\mu)=-3(\mu,\lambda -1,\mu -\lambda),平面ABCD的法向量\vec{n}=(0,0,1)$
$\sin\cfrac{\pi}{3}=\cfrac{\sqrt{3}}{2}=\cfrac{\vec{n}\cdot \overrightarrow{MN}}{|\vec{n}||\overrightarrow{MN}|}=\cfrac{3|\mu -\lambda|}{3\sqrt{\mu^2+(\lambda-1)^2+(\mu-\lambda)^2}}$
$分析：求MN的最小值，即求3\sqrt{\mu^2+(\lambda-1)^2+(\mu-\lambda)^2}=\cfrac{3|\mu -\lambda|}{\cfrac{\sqrt{3}}{2}}=2\sqrt{3}|\mu -\lambda|的最小值，$
$由上式可知即求|\mu-\lambda|的最小值。上等式包含\mu与\lambda之间的关系，要求最值，可用\Delta法,即令t=\mu-\lambda$
$\cfrac{3}{4}=\cfrac{(\mu-\lambda)^2}{\mu^2+(\lambda-1)^2+(\mu -\lambda)^2}\Rightarrow 4\mu ^2-8\mu\lambda +4\lambda ^2 =3\mu ^2+3\lambda ^2-6\lambda +3+3\mu ^2-6\mu \lambda +3\lambda ^2$
$\Rightarrow 2\mu^2+2\lambda ^2 +2\mu \lambda -6\lambda +3=0,{\color{Red} t=\mu-\lambda,\mu=t+\lambda代入上式} $
$2(t+\lambda)^2+2\lambda ^2 +2(t+\lambda) \lambda -6\lambda +3, 展开整理为关于\lambda的一元二次方程：$
$6\lambda^2+6(t-1)\lambda+t^2+3=0,因为\lambda\in R,\Delta\ge 0,可得36(t-1)^2-4\times 6\times (t^2+3)\ge 0$
$t^2+6t+3\le 0,-3-\sqrt{6}\le t\le -3+\sqrt{6},{\color{Red} \quad\therefore 2\sqrt{3}|t|_{min}=} 2\sqrt{3}(3-\sqrt{6})=6\sqrt{3}-6\sqrt{2}$
${\color{Red}解②法二： } 设M(3,m,m),N(n,3,3-n);A(3,0,0),B(3,3,0),\overrightarrow{AM}=(0,m,m),$
$\overrightarrow{BN}=(n-3,0,3-n),平面ABCD的法向量\vec{n}=(0,0,1)$
$\overrightarrow{MN}=\overrightarrow{AB}+\overrightarrow{BN}-\overrightarrow{AM}=(0,3,0)+(n-3,0,3-n)-(0,m,m)=(n-3,3-m,3-n-m),$
$\sin \cfrac{\pi}{3}=\cfrac{\sqrt{3}}{2}=\cfrac{\vec{n}\cdot\overrightarrow{MN}}{|\vec{n}|\times|\overrightarrow{MN}|}=\cfrac{|3-n-m|}{\sqrt{(n-3)^2+(3-m)^2+(3-n-m)^2}}$
$\cfrac{3}{4}=\cfrac{(3-n-m)^2}{(n-3)^2+(3-m)^2+(3-n-m)^2}\Rightarrow 3[(n-3)^2+(3-m)^2+(3-n-m)^2]=4(3-n-m)^2$
$3(n-3)^2+3(m-3)^2=(3-n-m)^2,(n-3)^2+(m-3)^2\ge \cfrac{(6-n-m)^2}{2}$
$\Rightarrow (3-n-m)^2\ge \cfrac{3}{2}\times(3-n-m)^2,换元，令t=n+m$
$2(3-t)^2\ge 3(6-t)^2\Rightarrow t^2-24t+90\le 0\Rightarrow 12-3\sqrt{6}\le t\le 12+3\sqrt{6}$
$|\overrightarrow{MN}|=\cfrac{2\sqrt{3}}{3}|3-t|\Rightarrow 9-3\sqrt{6}\le t-3\le 9+3\sqrt{6}$
$\Rightarrow \cfrac{2\sqrt{3}}{3}(9-3\sqrt{6})\le \cfrac{2\sqrt{3}}{3}(t-3)\le \cfrac{2\sqrt{3}}{3}(9+3\sqrt{6})\Rightarrow 6(\sqrt{3}-\sqrt{2})\le |\overrightarrow{MN}|\le 6(\sqrt{3}+\sqrt{2})$
$故|\overrightarrow{MN}|最小值为6(\sqrt{3}-\sqrt{2})$