2023 3月衢丽湖二模

23单位尺只需标6个刻度（两端不用标）便可测量所有23以内整单位，则8单位尺最少要标几个刻度？ 答案：3
错误点：省题，第8单位不用标

原题干误导人，题意应该是，三个问题三个不同答对概率，能选择一个先答剩下两个随机，问选哪个能使连续答对两个的概率最大，那么应该先答概率小点
11.
知识点：椭圆上一点P和椭圆上关于原点对称的点对A,B有\(K_{PA} \centerdot K_{PB}=-\frac{b^2}{a^2}\)

\[x>0,y>0;x^3+y^3=x-y; \]

\[x+y\in ?\ \ \ x^2+y^2\in ? \]

考场做法：令\(x\rightarrow y\rightarrow0\)，可排除\(A.x+y>\sqrt{\frac{2\sqrt{2}}{3}}-1\)和\(D.x^2+y^2>\frac{1}{2}\)那么直接选BC就行了
解法1：齐次化

\[\begin{aligned} \because &x^3+y^3=x-y \\ \therefore &1=\frac{x-y}{x^3+y^3} \\ &令t=\frac{x}{y}>1 \\ \therefore &(x+y)^2=\frac{(x+y)^2(x-y)}{x^3+y^3}=\frac{(t+1)^2(t-1)}{t^3+1}\in(0,\frac{2\sqrt{2}}{3}) \\ &即(x+y)\in(0,\sqrt{\frac{2\sqrt{2}}{3}}) \\ &x^2+y^2=\frac{(x^2+y^2)(x-y)}{x^3+y^3}=\frac{(t^2+1)(t-1)}{t^3+1}\in(0,1) \end{aligned} \]

解法2：比值代换

\[\begin{aligned} \because &x^3+y^3=x-y \Leftrightarrow (x^2-xy+y^2)=\frac{x-y}{x+y} \\ &即[\frac{3}{4}(x-y)^2+\frac{1}{4}(x+y)^2]=\frac{x-y}{x+y} \\ \therefore &令t=\frac{x-y}{x+y}\in(0,1) \\ &则\frac{3t^2+1}{4}(x+y)^2=t \\ \therefore &(x+y)^2=\frac{4t}{3t^2+1}\in(0,\frac{2\sqrt{2}}{3}) \\ &即(x+y)\in(0,\sqrt{\frac{2\sqrt{2}}{3}}) \\ &又(x-y)^2=t^2 (x+y)^2 \\ \therefore &x^2+y^2=\frac{1}{2}[(x+y)^2+(x-y)^2]=\frac{2t^2+2t}{3t^2+1}\in(0,1) \end{aligned} \]

19. 看清楚条件给出的角
20. 暴算
21. 双曲线\(U:\frac{x^2}{4}-y^2=1,A(-2,1),B(2,1)\)，点C在AB线段上（不含端点），过C作双曲线两切线，切点PQ，取PQ中点F，过F作双曲线两切线，切点DE，求\(S_{\Delta DEF min}\)

\[\begin{aligned} &设P(x_1,y_1),Q(x_2,y_2),C(t,1),F(x_0,y_0) \\ &由题意容易知道CP斜率存在且不为0，设CP:x=my+n (其中 n=x_1-my_1) \\ &\begin{cases} CP\\ U \end{cases} \Rightarrow (m^2-4)y^2+2mny+n^2-4=0 \\ \because &\Delta =4m^2n^2-4(m^2-4)(n^2-4)=0 \\ \therefore &m^2+n^2=4 即 {x_1}^2-2mx_1y_1+m^2({y_1}^2+1)=4 \\ &又 {x_1}^2=4({y_1}^2+1) \therefore \frac{{x_1}^2}{4}m^2-2x_1y_1m+4{y_1}^2=0 \\ \therefore &m=\frac{4y_1}{x_1} \therefore CP:x_1x-4y_1=4 \\ 同理 &CQ:x_2x-4y_2=4 \\&带入C点 得 \begin{cases} (x_1)(t)-(4y_1)(1)=4\\ (x_2)(t)-(4y_2)(1)=4 \end{cases} \Rightarrow PQ:\frac{t}{4}x-y=1 \\ &类似的 DE:\frac{x_0}{4}x-y_0y=1 \\ &\begin{cases} U\\ PQ \end{cases} \Rightarrow (\frac{t^2}{16}-\frac{1}{4})x^2-\frac{t}{2}x+2=0 \\ &\Rightarrow F(\frac{4t}{t^2-4},\frac{4}{t^2-4}) \therefore t=\frac{x_0}{y_0} \\ &将t=\frac{x_0}{y_0}代回F坐标 \\ &得{x_0}^2=4y_0(y_0+1),(y_0<-1) \\ &\begin{cases} DE\\ U \end{cases} \Rightarrow (4{y_0}^2-{x_0}^2)y^2+8y_0y+4-{x_0}^2=0 \\ &即 y_0y^2-2y_0y+{y_0}^2+y_0-1=0 \\ \therefore &\begin{cases} y_D+y_E=2\\ y_Dy_E=y_0-\frac{1}{y_0}+1 \end{cases} \\ \therefore &|DE|=\sqrt{(\frac{4}{y_0}-4y_0)((\frac{4y_0}{x_0})^2+1)} \\ &又 d_{F\rightarrow DE}=\frac{|{x_0}^2-4{y_0}^2-4|}{\sqrt{{x_0}^2+(4y_0)^2}} \\ \therefore S_{\Delta DEF}&=\frac{1}{2}|DE|d_{F\rightarrow DE}\\ &=\frac{1}{2}|{x_0}^2-4{y_0}^2-4|\sqrt{\frac{4(1-{y_0}^2)}{y_0{x_0}^2}}\\ &=\frac{1}{2}|4y_0-4|\sqrt{\frac{4(1-{y_0}^2)}{y_0(4{y_0}^2+4y_0)}}\\ &=\frac{2}{-y_0}(1-y_0)^{\frac{3}{2}}\\ &\geq 3\sqrt{3} (求导可知，当y_0=-2取等) \end{aligned} \]

补充：
配极原理：当F在C的极线(切点弦)上，可得C在F的极线上；且当F为弦中点时可得C也为弦中点
22.\(f(x)=a(e^x-e^{-x})-2x\)
(1)略（2）当\(x>0\)时有\(f(x)>0\)，求\(a\)范围

\[\begin{aligned} &\because x>0 \therefore 令t=e^x>1 则 f(x)>0\Leftrightarrow F(t)=a(t-t^{-1})-2\ln{t}>0 \\ &由重要不等式 \ln{t}<\frac{1}{2}(t-t^{-1}) (t>1) 可看出a\geq 1，以下进行证明 \\ &F'(t)=a(1+\frac{1}{t^2})-\frac{2}{t} \\ &必要性探路：\because F(1)=0 \therefore F'(1)=2a-2\geq 0 即a\geq 1 \\&下证a\geq 1的充分性\\ &\because a\geq 1 \therefore F'(t)\geq 1+\frac{1}{t^2}-\frac{2}{t}=(1-\frac{1}{t})^2 \geq 0 \\ &\therefore F(t) 在(1,+\infty)\uparrow \therefore F(t)>F(1)=0 \\ &\therefore a\geq 1 充分 \therefore a\in [1,+\infty) \end{aligned} \]

(2)\(m>n,\ m,n\in N^*\)证明：\(\ln{\frac{m}{n}}-\sum_{k=n+1}^{m}{\frac{1}{k}}<\frac{m-n}{2mn}\)

\[\begin{aligned} 原不等式&\Leftrightarrow \forall m>n且m,n\in N^*\ \ln{m}-\sum_{k=1}^{m}{\frac{1}{k}}+\frac{1}{2m}<\ln{n}-\sum_{k=1}^{n}{\frac{1}{k}}+\frac{1}{2n} \\ &\Leftrightarrow \forall n\in N^*\ \ln{(n+1)}-\sum_{k=1}^{n+1}{\frac{1}{k}}+\frac{1}{2(n+1)}<\ln{n}-\sum_{k=1}^{n}{\frac{1}{k}}+\frac{1}{2n} \\ &\Leftrightarrow \forall n\in N^*\ \ln{\frac{n+1}{n}}-\frac{1}{2(n+1)}-\frac{1}{2n}<0 \\ &\Leftrightarrow \forall n\in N^*\ \ln{(1+\frac{1}{n})}-\frac{2n+1}{2n(n+1)}<0 \\ &\Leftrightarrow \forall n\in N^*\ \ln{(1+\frac{1}{n})}-\frac{(\frac{1}{n})^2+\frac{2}{n}}{2(1+\frac{1}{n})}<0 \\ &\Leftrightarrow \forall n\in N^*\ \ln{(1+\frac{1}{n})}-\frac{(1+\frac{1}{n})^2-1}{2(1+\frac{1}{n})}<0 \\ &\Leftrightarrow \forall n\in N^*\ \ln{t}<\frac{1}{2}(t-t^{-1}) （其中t=1+\frac{1}{n}>1） \\ &由（2）可知 \forall t>1有 \ln{t}<\frac{1}{2}(t-t^{-1}) ，证毕 \end{aligned} \]