$1、已知函数f(x)=\ln x-ax+1,其中a\in R.$
$(1)若函数f(x)的图像恒不在x轴上方，求实数a的取值范围;$
$(2)证明:1+\cfrac{1}{2}+\cfrac{1}{3}+\cdots+\cfrac{1}{n}\gt \ln (n+1),其中n\in N*.$
$(1)定义域x\in R,\ln x-ax +1\le 0,\Rightarrow a\ge \cfrac{\ln x+1}{x}$
$令g(x)=\cfrac{\ln x+1}{x},求g(x)的最大值。{g}' (x)=\cfrac{-\ln x}{x^2} $
$x\in (0,1),{g}' (x)\gt 0,g(x) \nearrow ;x\in (1,+\infty),{g}' (x)\lt 0,g(x) \searrow;$
$g(x)在x=1时有最大值g(x)_{max}=1,故a\ge 1$
$(2)分析，\cfrac{1}{k}\ge \ln (k+1)-\ln k=\ln (1+\cfrac{1}{k})$
$\ln x\le x-1,令x=1+\cfrac{1}{x}\Rightarrow \ln (1+\cfrac{1}{x})\le \cfrac{1}{x},当且仅当x=1时取=$
$1+\cfrac{1}{2}+\cfrac{1}{3}+\cdots+\cfrac{1}{n}\gt \ln 2-\ln 1+\ln 3-\ln 2+\ln4-\ln3 +\cdots +\ln (n+1)-\ln n=\ln (n+1)$

$2、已知函数f(x)=2x\ln x,g(x)=ax^2-1.$
$(1)求f(x)的单调区间;$
$(2)当x\ge 1时,f(x)\le g(x),求a的取值范围;$
$(3)证明:{\textstyle \sum_{n}^{k=2}} \cfrac{1}{k\ln k} \gt \cfrac{3}{2}-\cfrac{1}{n}-\cfrac{1}{n+1},n\in N^*且n\ge 2.$
$解:(1) f(x)=2x\ln x,{f}' (x)=2\ln x+2=2(\ln x+1)$
$x\in (0,\cfrac{1}{e}), {f}' (x)\lt 0,f(x)\searrow;x\in (\cfrac{1}{e},+\infty ), {f}' (x)\gt 0,f(x)\nearrow$
$(2) x\gt 1，f(x)\le g(x)\Rightarrow 2x\ln x\le ax^2-1\Rightarrow a\ge \cfrac{2x\ln x+1}{x^2}$
$令h(x)=\cfrac{2x\ln x+1}{x^2},{h}' (x)=\cfrac{2(\ln x+1)x^2-2x(2x\ln x+1)}{x^4}=$
$\cfrac{2x(\ln x+1)-2(2x\ln x+1)}{x^3}=\cfrac{2x\ln x+2x-4x\ln x-2}{x^3}=\cfrac{-2x\ln x+2x-2}{x^3}$
$令\varphi (x)=-2x\ln x+2x-2,{\varphi }' (x)=2-2\ln x-2=-2\ln x$
$当 x\gt 1时,{\varphi }' (x)\lt 0,\varphi (x)\searrow,且\varphi (1)=0\Rightarrow\varphi (x)\lt 0\Rightarrow h'(x)\lt 0$
$h(x)在x\gt 1时，h(x)\searrow,且有h(1)=0,h(x)\lt 0，h(x)_{max}=h(1)=0$
$故a\gt 1，亦可用必要性探路得到此结论,请自行证明$/
$(a)证明必要性:x\ge 1时，2x\ln x\le ax^2-1，令x=1，0\le a-1\Rightarrow a\ge 1.$
$(b)证明充分性，即证a\ge 1,x\ge 1时，证明 ax^2-1\ge 2x\ln x.$
$即证x^2-1\ge 2x\ln x\Leftrightarrow x^2-1-2x\ln x \ge 0 $
$令h(x)=x^2-1-2x\ln x,h'(x)=2x-2(\ln x+1)=\varphi (x)$
$\varphi '(x)=2-\cfrac{2}{x},x\ge 1时，\varphi '(x)\gt 0,\varphi (x)=h'(x)\nearrow $
$h'(1)=0\Rightarrow h'(x)\ge 0\Rightarrow h(x)\nearrow,h(1)=0,h(x)\ge h(1)=0$

$(3)2x\ln x \le x^2-1,x\ge 1，当且仅当x=1时取等号，当x\gt 1时，左右两边均为正数$
$令x=k,k\ge 2,k\in N^*，有2k\ln k\lt k^2-1,两边取倒数 ，则有\cfrac{1}{k\ln k}\gt \cfrac{2}{k^2-1}=\cfrac{1}{k-1}-\cfrac{1}{k+1}$
$\sum_{k=2}^{n} \cfrac{1}{k\ln k}\gt 1-\cfrac{1}{3}+\cfrac{1}{2}-\cfrac{1}{4}+\cfrac{1}{3}-\cfrac{1}{5}+\cfrac{1}{n-2}-\cfrac{1}{n}+\cfrac{1}{n-1}-\cfrac{1}{n+1}$
$\sum_{k=2}^{n} \cfrac{1}{k\ln k}\gt \cfrac{3}{2}-\cfrac{1}{n}-\cfrac{1}{n+1}$

$3、已知函数f(x)=\cfrac{ae^x-1}{x}的图像在(1,f(1))处的切线经过点(2,e).$
$(1)求a的值及函数f(x)的单调区间;$
$(2)若关于x的不等式e^x\ln x-\cfrac{\ln x +x^2+(\lambda -1)x-\lambda }{e^\lambda }\ge 0在区间(1,+\infty)上恒成立，求正实数\lambda 的取值范围。$
$解:(1)定义域为x\ne 0\qquad f'(x)=\cfrac{axe^x-ae^x+1}{x^2}$
$k=\cfrac{e-f(1)}{2-1}=f'(1)\Rightarrow e-ae+1=1\Rightarrow a=1$
$f(x)=\cfrac{e^x-1}{x}\quad f'(x)=\cfrac{xe^x-e^x+1}{x^2},令h(x)=xe^x-e^x+1$
$h'(x)=xe^x,x\in (-\infty,0),h'(x)\lt 0,h(x)\searrow;x\in (0,+\infty),h'(x)\gt 0,h(x)\nearrow$
$(2)x\in (1,+\infty),\Rightarrow \ln x \gt 0且x+\lambda\gt 0,e^x\ln x-\cfrac{\ln x +x^2+(\lambda -1)x-\lambda }{e^\lambda }\ge 0$
$e^x\ln x-\cfrac{\ln x +(x+\lambda)(x -1)}{e^\lambda }\ge 0\Rightarrow e^{x+\lambda}-1-\cfrac{(x+\lambda)(x -1)}{\ln x}\ge 0$
$\cfrac{e^{x+\lambda}-1}{x+\lambda}-\cfrac{x -1}{\ln x}\ge 0\Rightarrow \cfrac{e^{x+\lambda}-1}{x+\lambda}\ge \cfrac{e^{\ln x} -1}{\ln x}$
$已知f(x)=\cfrac{e^x-1}{x}在x\in (0,+\infty)上单调递增,又{\color{Red} \because } f(x+\lambda)\ge f(\ln x)$
$x+\lambda\ge \ln x\Rightarrow \lambda\ge \ln x-x$
$\lambda\ge -1$

$4、已知函数f(x)＝\ln x+\sin x.$
$(1)求曲线y=f(x)在点(1,f(1))处的切线方程;$
$(2)求函数f(x)在区间[1,e]上的最小值;$
$(3)证明函数f(x)只有一个零点.$
$解:(1)f'(x)=\cfrac{1}{x}+\cos x,f'(1)=1+\cos 1,f(1)=\sin 1$
$y-\sin 1=(1+\cos 1)(x-1)$
$(2)f'(x)=\cfrac{1}{x}+\cos x,在x\in(0,\pi )\cos x单调递减,$
$所以[1,e]\subset (0,\pi)，所以{f}' (x)在[1,e]单调递减。$
${f}' (\cfrac{\pi}{2} )=\cfrac{2}{\pi}+\cos \cfrac{\pi}{2}=\cfrac{2}{\pi}\gt 0;\qquad {f}' (\cfrac{2\pi}{3} )=\cfrac{3}{2\pi} -\cfrac{1}{2} \lt 0$
$故存在唯一的x_0\in (\cfrac{\pi}{2},\cfrac{2\pi}{3})，使得{f}' (x_0 )=0,$
$即x\in[1,x_0),{f}' (x)\gt 0,f(x)\nearrow;x\in(x_0,e],{f}' (x)\lt 0,f(x)\searrow$
$f(x)在[1,e]存在最大值f(x)_{max}=f(x_0),最小值在左右端点$
$f(1)=\sin 1 \lt 1;f(e)=1+\sin e,e\in (0,\pi)\Rightarrow \sin e\gt 0,即f(1)\lt f(e)$
$故函数f(x)在区间[1,e]上的最小值为\sin 1$
$(3)$

$解：(2)令x_0=t,设A(x_1,y_1),B(x_2,y_2),F(1,0);利用两点式的变形及对偶式$
$\because \angle BFA的角平分线垂直于x轴,\therefore k_{BF}+k_{AF}=0\Rightarrow \cfrac{y_1}{x_1-1}+\cfrac{y_2}{x_2-1}=0$
$\Rightarrow y_1(x_2-1)+y_2(x_1-1)=0\Rightarrow y_1x_2+y_2x_1={\color{Red}y_1+y_2 } $
$由两点式直线方程：\cfrac{y-y_1}{x-x_1} =\cfrac{y_2-y_1}{x_2-x_1}$
$交叉相乘合并同类项，得 y_1x_2-y_2x_1=y(x_2-x_1)+x(y_1-y_2)$
$(x_1,y_1),(x_2,y_2),(t,0)三点共线,得y_1x_2-y_2x_1={\color{Red}t(y_1-y_2) } $
$构造对偶式：y_1x_2+y_2x_1=\cfrac{(y_1x_2)^2-(y_2x_1)^2}{y_1x_2-y_2x_1}=\cfrac{y_1^2x_2^2-y_2^2x_1^2}{t(y_1-y_2)}$
$(x_1,y_1),(x_2,y_2)在椭圆上,有\cfrac{x_1^2}{4}+\cfrac{y_1^2}{3}=1,\Rightarrow \cfrac{x_1^2}{4}=1-\cfrac{y_1^2}{3}$
$\cfrac{x_2^2}{4}+\cfrac{y_2^2}{3}=1,\Rightarrow \cfrac{x_2^2}{4}=1-\cfrac{y_2^2}{3}$
$y_1x_2+y_2x_1=\cfrac{4y_1^2(1-\cfrac{y_2^2}{3})-4y_2^2(1-\cfrac{y_1^2}{3})}{t(y_1-y_2)}=\cfrac{4y_1^2-4y_2^2}{t(y_1-y_2)}=\cfrac{t}{4}(y_1+y_2)$
$即\cfrac{t}{4}=1,t=4$

$19、已知函数f(x)=x+a\cos ax,a\gt 0$
$(1)当a=\sqrt{2}时，求f(x)在区间(0,\cfrac{\pi}{2})上的最大值$
$解：a=\sqrt{2},f(x)x+\sqrt{2}\cos \sqrt{2}x,{f}' (x)=1-2\sin \sqrt{2}x $
$令{f}' (x)=0,1-2\sin \sqrt{2}x =0,解得x=\cfrac{\sqrt{2}\pi}{12},$
$当x\in (0,\cfrac{\sqrt{2}\pi}{12}),{f}' (x)\gt 0,f(x)\nearrow;x\in (\cfrac{\sqrt{2}\pi}{12},\cfrac{\pi}{2}),{f}' (x)\lt 0;f(x)\searrow$
$f(x)在x=\cfrac{\sqrt{2}\pi}{12}有最大值，f(\cfrac{\sqrt{2}\pi}{12})=\cfrac{\sqrt{2}\pi}{12}+\sqrt{2}\cos (\cfrac{\pi}{6})=\cfrac{\sqrt{2}\pi}{12}+\cfrac{\sqrt{6}}{2}$

$15、\triangle ABC中, 角A,B,C所对的边为a,b,c,A=\cfrac{3\pi}{4}.$
$(1)若c=\sqrt{2},b=2,求\sin C$
解：$a^2=b^2+c^2-2bc\cos A=4+2-4\sqrt{2}\cdot (-\cfrac{\sqrt{2}}{2})=10$
$\cfrac{a}{\sin A}=\cfrac{c}{\sin C}\Rightarrow \sin C=\cfrac{c\sin A}{a}=\cfrac{\sqrt{2}\cfrac{\sqrt{2}}{2}}{\sqrt{10}}=\cfrac{\sqrt{10}}{10}$
$(2)\sin B+\sqrt{3}\cos B=2,b=4,求\triangle ABC面积$
$解：\cfrac{1}{2}\sin B+\cfrac{\sqrt{3}}{2}\cos B=1\Rightarrow \sin(B+\cfrac{\pi}{3})=1\Rightarrow B+\cfrac{\pi}{3}=\cfrac{\pi}{2},B=\cfrac{\pi}{6}$

$例1.雅礼中学26届高三月考卷五18.已知函数f(x)=\ln x,a\in \mathbb{R}$
$(2)当a\gt 0时 ,\forall x\in [1,+\infty),均有af(x)+(x+1)^2-\cfrac{x^2}{a}\le 0恒成立,$
$求实数a的取值范围。$
$解:(2) 对\forall x\in [1,+\infty),均有g(x)=af(x)+(x+1)^2-\cfrac{x^2}{a}$
$=a\ln x+(x+1)^2-\cfrac{x^2}{a}\le 0恒成立。$
${\color{Red} \therefore \quad g(1)}\le 0即4-\cfrac{1}{a}\le 0,又a\gt 0\quad {\color{Red} \therefore \quad 0\lt a\le \cfrac{1}{4}}$

$当0\lt a\le \cfrac{1}{4}时,-\cfrac{1}{a}\le －4，又x\ge 1,\therefore \ln x\ge 0,x^2\ge 1$
$\therefore a\ln x \le \cfrac{1}{4}\ln x,-\cfrac{x^2}{a}\le -4x^2.$
${\color{Red} \therefore} \quad g(x)\le \cfrac{1}{4}\ln x+(x+1)^2-4x^2=\cfrac{1}{4}\ln x-3x^2+2x+1$
$令h(x)=\cfrac{1}{4}\ln x-3x^2+2x+1 \quad (x\ge 1)$
${h}' (x)=\cfrac{1}{4x}-6x+2\le {h}' (1)=-\cfrac{15}{4}\lt 0$
${\color{Red} \therefore} \quad h(x)在[1,+\infty)上\searrow ,{\color{Red} \therefore} \quad x\ge 1时,h(x)\le h(1)＝0$
$所以当0\lt a\le \cfrac{1}{4}时,g(x)\le 0对\forall x\ge 1恒成立。$
$故实数a的取值范围为(0,\cfrac{1}{4}]$

${\color{Red} 必要性探路不需要证明a取其他值，不成立。}$

例2.2022年新高考二卷，用必要性探路。https://uu.890222.xyz/index.php/archives/282/

$例3.云南师大附中2026届高考适应性月考卷六$
$19.已知函数f(x)=2xe^x-mx^2+1\quad (m\in \mathbb{R}$
$(3)对\forall x\in [-1,+\infty),f(x)\le \cfrac{1-m}{m}e^{2x}恒成立，求m的取值范围。$
$(3)解:由题意知f(0)\le \cfrac{1-m}{m}\cdot e^0,即1\le \cfrac{1-m}{m},解得0\lt m\le \cfrac{1}{2},$
$即0\lt m\le \cfrac{1}{2}为題中不等式恆成立的必要條件。$
${\color{Red} 注:必要性探路,令\begin{cases} f(x_0)-\cfrac{1-m}{m}\cdot e^{2x_0}=0\\{ f}' (x_0)-(\cfrac{1-m}{m}\cdot e^{2x_0})'=0\end{cases}} ，可得 \begin{cases} x_0=0\\m=\cfrac{1}{2}\end{cases} 是方程組的唯一解，$

$1.若x_1满足2x+2^x=5,x_2满足2x+2\log_{2}{(x-1)} =5,x_1+x_2=(\qquad)$
$解:2^x与\log_{2}{x}是反函数,关于y=x对称.\quad {\color{Red} \because }\quad 2x+2\log_{2}{(x-1)} =5$
$\Rightarrow x+\log_{2}{(x-1)} =\cfrac{5}{2}\Rightarrow \log_{2}{(x-1)} =\cfrac{5}{2}-x{\color{Green} \Rightarrow\begin{cases} y=\log_{2}{(x-1)}\\y=\cfrac{5}{2}-x\end{cases}} $
$同理2x+2^x=5\Rightarrow x+2^{x-1}=\cfrac{5}{2}\Rightarrow 2^{x-1} =\cfrac{5}{2}-x{\color{Green}\Rightarrow \begin{cases} y=2^{x-1}\\y=\cfrac{5}{2}-x\end{cases}}$
${\color{Red}\because\quad} y=2^{x-1}和y=\log_{2}{(x-1) 关于y=x-1对称},x_1,x_2的中心坐标为直线y=x-1与y=\cfrac{5}{2}-x$

$2.已知x\gt1,方程x-(x-1)2^x=0,x-(x-1)\log _2x=0在区间(1,+\infty)的根分别$
$为a,b,以下结论正确的有:$
$A.b-a=2^a-\log _2b\quad B.\cfrac{1}{a}+\cfrac{1}{b}=1\quad C.a+b\lt 4\quad D.a+b\gt 4$
$解:\begin{cases}b=2^a\\a=\log_2 b\end{cases}\Rightarrow b-a=2^a-a=2^a-\log_2b,故A正确。$
$x=(x-1)2^x\Rightarrow \cfrac{x}{x-1}=2^x\Rightarrow 2^a=\cfrac{a}{a-1}{\color{Red} =b} $
$\cfrac{1}{a}+\cfrac{1}{b}=\cfrac{1}{a}+\cfrac{1}{2^a}=\cfrac{1}{a}+\cfrac{a-1}{a}=1$
$a+b=a+\cfrac{a}{a-1}=a-1+\cfrac{a-1+1}{a-1}+1=2+a-1+\cfrac{1}{a-1}\ge 4$

$3.求\cfrac{2\sin 40+\sin 20}{\cos 30}$

$已知P(x_0,y_0) 是圆C:x^2+y^2-2x-2y+1=0上任意一点，则\cfrac{y_0+1}{x_0-3}的最小值为(\qquad )$
$A.\cfrac{4+\sqrt{7}}{3}\quad B.\cfrac{-4-\sqrt{7}}{3}\quad C.\cfrac{4-\sqrt{7}}{3}\quad D.\cfrac{-4+\sqrt{7}}{3}\quad$
$令\cfrac{y_0+1}{x_0-3}=k\Rightarrow y_0+1=k(x_0-3),即转换成直线与圆C相交时，直线斜率k的最小值$
$\begin{cases} y=k(x-3)-1\\x^2+y^2-2x-2y+1=0\end{cases}\Rightarrow x^2+[k(x-3)-1]^2-2x-2[k(x-3)-1]+1=0$
$\Delta =0,解得k 值。计算量巨大$
$我们应用圆心(1,1)到切线的距离不大于半径,(x-1)^2+(y-1)^2=1,d=\cfrac{|k(1-3)-1-1|}{\sqrt{1+k^2}}\le 1$
$4(k+1)^2\le 1+k^2\Rightarrow 3k^2+8k+3\le 0\Rightarrow \cfrac{-4-\sqrt{7}}{3}\le k\le \cfrac{-4+\sqrt{7}}{3},C选$

$设实数x,y满足x^2+(y-1)^2=1,则\cfrac{y-x}{x-2}的最大值为(\qquad)$
$A.\cfrac{1}{6}\quad B.\cfrac{1}{4}\quad C.\cfrac{1}{3}\quad D.\cfrac{1}{2}\quad$
$分离常数\cfrac{y-x}{x-2}=-1+\cfrac{y-2}{x-2},故转换成上题一样。先令\cfrac{y-2}{x-2}=k,k(x-2)-y+2=0$
$圆心(0,1)到切线y-2=k(x-2)的距离不大于半径1,d=\cfrac{|k(0-2)-1+2|}{\sqrt{1+k^2}}\le 1$
$(2k+1)^\le 1+k^2\Rightarrow 3k^2-4k\le 0\quad \Rightarrow 0\le k\le \cfrac{4}{3}, \cfrac{4}{3}-1=C$