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[{"title":"2019全国高中数学联赛加试T4引理证明","url":"//post/2019%E5%85%A8%E5%9B%BD%E9%AB%98%E4%B8%AD%E6%95%B0%E5%AD%A6%E8%81%94%E8%B5%9B%E5%8A%A0%E8%AF%95T4%E5%BC%95%E7%90%86%E8%AF%81%E6%98%8E/","content":"<center></center>\n<span id=\"more\"></span>\n\n<p>引理: 若一简单连通简单图$G(V, E)$中有$\\vert E\\vert$条边, 则该图中有$\\lfloor \\frac{E}{2} \\rfloor$个两两无公共边的共点边对(下文称为"角"). [^1]</p>\n<p>对$\\vert V\\vert$进行数学归纳(标答是对$\\vert E\\vert$)</p>\n<p>$\\vert V\\vert=1$时显然成立</p>\n<p>对于任意正整数$k$, 若$\\vert V\\vert<k$时结论成立, 下证$\\vert V \\vert=k$时结论成立</p>\n<p>开始之前, 我们先分析一种简单的情况</p>\n<p>对于任意点$V_0$, 若$degree(V_0)$为偶数</p>\n<p>那么将$V_0$及与$V_0$相连的边两两去掉并计数, 剩下的图$G’(V’, E’)$满足$\\vert V\\vert<k$, 由归纳假设, 结论"成立"(为什么有引号)</p>\n<p>这个反例还是挺容易构造的</p>\n<!-- <Graph indexType=\"custom\" height=\"400\" width=\"400\" nodes={[{label:\"0\",center:{x:206.3,y:204.9}},{label:\"1\",center:{x:221.8,y:164.8}},{label:\"2\",center:{x:262.1,y:185.5}},{label:\"3\",center:{x:243.8,y:226.9}},{label:\"4\",center:{x:197.2,y:247.4}},{label:\"5\",center:{x:154.3,y:233}},{label:\"6\",center:{x:166.2,y:189.3}}]} edges={[{source:1,target:2},{source:2,target:3},{source:3,target:1},{source:4,target:5},{source:5,target:6},{source:6,target:4},{source:0,target:1},{source:0,target:2},{source:0,target:3},{source:0,target:4},{source:0,target:5},{source:0,target:6}]} /> -->\n\n<p>如图<img src=\"/file/2019%E5%85%A8%E5%9B%BD%E9%AB%98%E4%B8%AD%E6%95%B0%E5%AD%A6%E8%81%94%E8%B5%9B%E5%8A%A0%E8%AF%95T4%E5%BC%95%E7%90%86%E8%AF%81%E6%98%8E/graph1.png\" alt=\"graph1.png\"></p>\n<p>$degree(V_0)$为偶数, 但显然我们不能这么做(想一想为什么)</p>\n<p>因为剩下的$G’$不是一个连通图, 更要命的是$G’$的连通分量集中还有大于1个有奇数条边的连通分量</p>\n<!-- <Graph indexType=\"custom\" height=\"400\" width=\"400\" nodes={[{label:1,center:{x:207.5,y:141.8}},{label:2,center:{x:232.7,y:184.2}},{label:3,center:{x:181.3,y:183.7}},{label:4,center:{x:182.1,y:230.6}},{label:5,center:{x:233.5,y:231.1}},{label:\"6\",center:{x:207.3,y:273}}]} edges={[{source:0,target:1},{source:1,target:2},{source:4,target:5},{source:2,target:0},{source:3,target:4},{source:5,target:3}]} /> -->\n\n<p>如图<img src=\"/file/2019%E5%85%A8%E5%9B%BD%E9%AB%98%E4%B8%AD%E6%95%B0%E5%AD%A6%E8%81%94%E8%B5%9B%E5%8A%A0%E8%AF%95T4%E5%BC%95%E7%90%86%E8%AF%81%E6%98%8E/graph2.png\" alt=\"graph2.png\">显然不成立</p>\n<p>这种情况下再去"限制$G’$的连通分量集中有小于1个有奇数条边的连通分量"已经没有多大用处了</p>\n<p>这种做法似乎走到了尽头…</p>\n<p>除非…</p>\n<p>$G$中没有割点!</p>\n<p>没错, 没有割点就不会出现去掉一个点后剩下的图不连通(这是割点的定义! 同时, 我们由上面的推导中可以发现可作限制"剩下的$G’$是一个连通图", 进而联想到割点)的情况</p>\n<p>这时再进行如上操作便合情合理, 即"存在$V_0$使$degree(V_0)$为偶数且$G$为点双连通分量时"结论成立</p>\n<p>若$G$中没有度数为偶数的点呢? </p>\n<p>这个好办, 你把任意一个"角"去掉不就有了吗? (这样就可以化归为上面或下面情况中的一种啦) [^2]</p>\n<p>接下来考虑剩下的情况, 即"$G$中存在割点"的情况</p>\n<!-- <Graph indexType=\"custom\" height=\"400\" width=\"400\" nodes={[{label:\"0a\",center:{x:152.2,y:172.1}},{label:\"0b\",center:{x:237.6,y:180.1}},{label:\"1\",center:{x:192,y:187.7}},{label:\"2\",center:{x:137.5,y:214.5}},{label:\"3\",center:{x:177.9,y:231.3}},{label:\"4\",center:{x:277.6,y:198.2}},{label:\"5\",center:{x:222.7,y:223}},{label:\"6\",center:{x:261.7,y:240.3}}]} edges={[{source:2,target:3},{source:3,target:4},{source:4,target:2},{source:5,target:6},{source:6,target:7},{source:7,target:5},{source:0,target:2},{source:0,target:3},{source:0,target:4},{source:1,target:5},{source:1,target:6},{source:1,target:7}]} /> -->\n\n<p>如图<img src=\"/file/2019%E5%85%A8%E5%9B%BD%E9%AB%98%E4%B8%AD%E6%95%B0%E5%AD%A6%E8%81%94%E8%B5%9B%E5%8A%A0%E8%AF%95T4%E5%BC%95%E7%90%86%E8%AF%81%E6%98%8E/graph3.png\" alt=\"graph3.png\"></p>\n<p>我们去掉割点$V_0$后, 不妨设$G$被分为了$m$个连通分量$G_1(V_1, E_1), G_2(V_2, E_2), \\dots , G_m(V_m, E_m)$, 设$V_0$与这些连通分量相连的边集分别为$E_{01}, E_{02}, \\dots E_{0m}$</p>\n<p>考虑$G_1’(V_1\\cup V_0, E_1\\cup E_{01}), G_2’(V_2\\cup V_0, E_2\\cup E_{02}), \\dots , G_m’(V_m\\cup V_0, E_m\\cup E_{0m})$, 若$G_1’, G_2’, \\dots , G_m’$中有$n$个含有奇数条边的图, 设为$G_{j1}, G_{j2}, \\dots , G_{jn}$取n条边$E_1\\in G_{j1}, E_2\\in G_{j2}, \\dots , E_n\\in G_{jn}$可配成$\\lfloor\\frac{n}{2}\\rfloor$个"角"(这时有人要问, 若$n$为奇数那又怎么办呢? 不是会剩下一条边无法配对吗? $n$为奇数说明原图$G$中有奇数条边, 本来就会剩下一条边, 又何必担心呢? ), 将这$n$条边全部去掉后$G_1’, G_2’, \\dots , G_m’$均含有偶数条边, 且顶点数均小于$k$, 结论成立</p>\n<p>综上所述, 得证</p>\n<p>$Q.E.D.$</p>\n<hr>\n<p>[^1]: 为什么只有引理? 因为后面我也不会了. 这个引理也是看标答才知道的…</p>\n<p>[^2]: 其实有割点的情况应该放在前面写的, 这样就不会出现前面引用后面结论破坏连贯性的情况了, 但是为了体现思考过程, 避免看完以后出现"看懂了, 想不到"的情况, 姑且这么写吧. </p>\n<hr>\n<p>本来计划2023-06-16 19:18:39开写的, 但实际上咕到了2023-06-23 22:59:56才写完…</p>\n","tags":["全国高中数学联赛","Maths","高中"]},{"title":"20211103|数海钓鱼每日一题","url":"//post/20211103%7C%E6%95%B0%E6%B5%B7%E9%92%93%E9%B1%BC%E6%AF%8F%E6%97%A5%E4%B8%80%E9%A2%98/","content":"<center></center>\n<span id=\"more\"></span>\n\n<p><a href=\"https://mp.weixin.qq.com/s?__biz=MzkyMjI2MzAyNA==&mid=2247484115&idx=1&sn=d9079669adb14acefc8ba2d823ff900e\">20211103|数海钓鱼每日一题</a></p>\n<p><img src=\"/file/20211103%7C%E6%95%B0%E6%B5%B7%E9%92%93%E9%B1%BC%E6%AF%8F%E6%97%A5%E4%B8%80%E9%A2%98/640.png\" alt=\"640.png\"></p>\n<p>$22. (本题满分15分)已知函数f(x)=\\frac{x^{2}+ax+1}{x-1}\\ln{x}, a\\in{\\mathbb{R}}. $</p>\n<p>$(I)求f(x)的极值点个数; $</p>\n<p>$(II)若x\\geq 1时恒有2f(x)\\geq (a+2)(x+1)成立, 求a的取值范围; $</p>\n<p>$(III)设a=-2, 若函数y=f(x)-k有两个零点m, n且(m-1)(n-1)<\\lambda k^{2}-k恒成立, 求\\lambda 的最小值. $</p>\n<hr>\n<p>(I)(II) 没导明白</p>\n<hr>\n<p>(III)</p>\n<p>重量寄的来了, 说实话确实没导明白这题, 如果放考场上2个小时全做这题也大概率被创飞</p>\n<p>先换元</p>\n<p>$x_{1}=\\ln{m}, x_{2}=\\ln{n}, x_{1}<x_{2}, h_{1}(x)=x(e^{x}-1), h_{1}(x_{1})=h_{1}(x_{2})=k$</p>\n<p>构造函数</p>\n<p>$h_{2}(x)=\\lambda (x^{2}e^{x}-x^{2})-((\\frac{1}{2}x-1)e^{x}+\\frac{1}{2}x+1)$</p>\n<p>求导</p>\n<p>$h_2’(x)=\\lambda ((x^{2}+2x)e^{x}-2x)-\\frac{1}{2}((x-1)e^{x}+1)$</p>\n<p>$h_{2}’’(x)=\\lambda ((x^{2}+4x+2)e^{x}-2)-\\frac{1}{2}xe^{x}$</p>\n<p>$h_{2}’’’(x)=\\lambda (x^{2}+6x+6)e^{x}-\\frac{1}{2}(x+1)e^{x}$</p>\n<p>先证$\\lambda <\\frac{1}{12}$时不成立</p>\n<p>若$\\lambda <\\frac{1}{12}$</p>\n<p>则$h_{2}’’’(0)=6\\lambda -\\frac{1}{2}<0$</p>\n<p>又$h_{2}’’’(x)$为$\\mathbb{R}$上连续函数</p>\n<p>故存在包含$0$的一个连续开区间$U_{1}$</p>\n<p>满足$\\forall x\\in{U_{1}}$均有$h_{2}’’’(x)<0$</p>\n<p>即$h_{2}’’’(x)$在$U_{1}$上恒小于$0$</p>\n<p>$h_{2}’’(x)$在$U_{1}$上单调递减</p>\n<p>$x\\in{(-\\infty, 0)\\cap U_{1}}$, $h_{2}’’(x)>h_{2}’’(0)=0$, $h_{2}’(x)$单调递增, $h_{2}’(x)<h_2’(0)=0$, $h_{2}$单调递减; </p>\n<p>$x\\in{U_{1}\\cap (0, +\\infty)}$, $h_{2}’’(x)<h_{2}’’(0)=0$, $h_{2}’(x)$单调递减, $h_{2}’(x)<h_2’(0)=0$, $h_{2}$单调递减; </p>\n<p>因此$h_{2}(x)$在$U_{1}$上单调递减</p>\n<p>又$k\\to 0^{+}$时$x_{1}\\to 0^{-}$, $x_{2}\\to 0^{+}$</p>\n<p>故存在$k$使得$x_{1}\\in{U_{1}}$, $x_{2}\\in{U_{1}}$</p>\n<p>$x_{1}<x_{2}$</p>\n<p>$\\Leftrightarrow h_{2}(x_{1})>h_{2}(x_{2})$</p>\n<p>$\\Leftrightarrow \\lambda ({x_{1}}^{2}e^{x_{1}}-{x_{1}}^{2})-((\\frac{1}{2}x-1)e^{x_{1}}+\\frac{1}{2}x_{1}+1)>\\lambda ({x_{2}}^{2}e^{x}-{x_{2}}^{2})-((\\frac{1}{2}x_{2}-1)e^{x_{2}}+\\frac{1}{2}x_{2}+1)$</p>\n<p>$\\Leftrightarrow \\lambda {x_{1}}^{2}(e^{x_{1}}-1)-(\\frac{1}{2}x_{1}(e^{x_{1}}-1)-e^{x_{1}}+x_{1}+1)>\\lambda {x_{2}}^{2}(e^{x_{2}}-1)-(\\frac{1}{2}x_{2}(e^{x_{2}}-1)-e^{x_{2}}+x_{2}+1)$</p>\n<p>$\\Leftrightarrow \\lambda x_{1}k-x_{1}-\\frac{1}{2}k+(e^{x_{1}}-1)>\\lambda x_{2}k-x_{2}-\\frac{1}{2}k+(e^{x_{2}}-1)$</p>\n<p>$\\Leftrightarrow \\frac{x_{1}}{k}(\\lambda k^{2}-k)+\\frac{x_{2}}{k}(e^{x_{1}}-1)(e^{x_{2}}-1)>\\frac{x_{2}}{k}(\\lambda k^{2}-k)+\\frac{x_{1}}{k}(e^{x_{1}}-1)(e^{x_{2}}-1)$</p>\n<p>$\\Leftrightarrow \\frac{x_{2}}{k}(e^{x_{1}}-1)(e^{x_{2}}-1)-\\frac{x_{1}}{k}(e^{x_{1}}-1)(e^{x_{2}}-1)>\\frac{x_{2}}{k}(\\lambda k^{2}-k)-\\frac{x_{1}}{k}(\\lambda k^{2}-k)$</p>\n<p>$\\Leftrightarrow (\\frac{x_{2}}{k}-\\frac{x_{1}}{k})(e^{x_{1}}-1)(e^{x_{2}}-1)>(\\frac{x_{2}}{k}-\\frac{x_{1}}{k})(\\lambda k^{2}-k)$</p>\n<p>$\\Leftrightarrow (e^{x_{1}}-1)(e^{x_{2}}-1)>\\lambda k^{2}-k$</p>\n<p>与$(m-1)(n-1)<\\lambda k^{2}-k恒成立$即$(e^{x_{1}}-1)(e^{x_{2}}-1)<\\lambda k^{2}-k恒成立$矛盾</p>\n<p>故$\\lambda \\geq \\frac{1}{12}$</p>\n<p>再证$\\lambda = \\frac{1}{12}$时$(e^{x_{1}}-1)(e^{x_{2}}-1)<\\lambda k^{2}-k恒成立$</p>\n<p>若$\\lambda = \\frac{1}{12}$</p>\n<p>$h_{2}’’’(x)=\\frac{1}{12}x^{2}e^{x}\\geq 0$, 当且仅当$x=0$时取等</p>\n<p>$h_{2}’’(x)$单调递增</p>\n<p>$x\\in{(-\\infty, 0)}$时, $h_{2}’’(x)<h_{2}’’(0)=0$,$h_{2}’(x)$单调递减, $h_{2}’(x)>h_2’(0)=0$, $h_{2}(x)$单调递增; </p>\n<p>$x\\in{(0, +\\infty)}$时, $h_{2}’’(x)>h_{2}’’(0)=0$,$h_{2}’(x)$单调递增, $h_{2}’(x)>h_2’(0)=0$, $h_{2}(x)$单调递增; </p>\n<p>故$h_{2}(x)$在$\\mathbb{R}$上单调递增</p>\n<p>$x_{1}<x_{2}$</p>\n<p>$\\Leftrightarrow h_{2}(x_{1})<h_{2}(x_{2})$</p>\n<p>$\\Leftrightarrow \\lambda ({x_{1}}^{2}e^{x_{1}}-{x_{1}}^{2})-((\\frac{1}{2}x-1)e^{x_{1}}+\\frac{1}{2}x_{1}+1)<\\lambda ({x_{2}}^{2}e^{x}-{x_{2}}^{2})-((\\frac{1}{2}x_{2}-1)e^{x_{2}}+\\frac{1}{2}x_{2}+1)$</p>\n<p>$\\Leftrightarrow \\lambda {x_{1}}^{2}(e^{x_{1}}-1)-(\\frac{1}{2}x_{1}(e^{x_{1}}-1)-e^{x_{1}}+x_{1}+1)<\\lambda {x_{2}}^{2}(e^{x_{2}}-1)-(\\frac{1}{2}x_{2}(e^{x_{2}}-1)-e^{x_{2}}+x_{2}+1)$</p>\n<p>$\\Leftrightarrow \\lambda x_{1}k-x_{1}-\\frac{1}{2}k+(e^{x_{1}}-1)<\\lambda x_{2}k-x_{2}-\\frac{1}{2}k+(e^{x_{2}}-1)$</p>\n<p>$\\Leftrightarrow \\frac{x_{1}}{k}(\\lambda k^{2}-k)+\\frac{x_{2}}{k}(e^{x_{1}}-1)(e^{x_{2}}-1)<\\frac{x_{2}}{k}(\\lambda k^{2}-k)+\\frac{x_{1}}{k}(e^{x_{1}}-1)(e^{x_{2}}-1)$</p>\n<p>$\\Leftrightarrow \\frac{x_{2}}{k}(e^{x_{1}}-1)(e^{x_{2}}-1)-\\frac{x_{1}}{k}(e^{x_{1}}-1)(e^{x_{2}}-1)<\\frac{x_{2}}{k}(\\lambda k^{2}-k)-\\frac{x_{1}}{k}(\\lambda k^{2}-k)$</p>\n<p>$\\Leftrightarrow (\\frac{x_{2}}{k}-\\frac{x_{1}}{k})(e^{x_{1}}-1)(e^{x_{2}}-1)<(\\frac{x_{2}}{k}-\\frac{x_{1}}{k})(\\lambda k^{2}-k)$</p>\n<p>$\\Leftrightarrow (e^{x_{1}}-1)(e^{x_{2}}-1)<\\lambda k^{2}-k$</p>\n<p>即$\\lambda$最小值为$\\frac{1}{12}$</p>\n<hr>\n<p>现在我们来处理两个问题, 一小一大</p>\n<p>先看小问题</p>\n<p>$h_{2}(x)$是如何构造<del>xjb凑</del>出来的</p>\n<p>回到$(e^{x_{1}}-1)(e^{x_{2}}-1)<\\lambda k^{2}-k$</p>\n<p>同乘$(\\frac{x_{2}}{k}-\\frac{x_{1}}{k})$即$(\\frac{1}{e^{x_{2}}-1}-\\frac{1}{e^{x_{1}}-1})$制造同构式</p>\n<p>得到$\\lambda {x_{1}}^{2}(e^{x_{1}}-1)-x_{1}+(e^{x_{1}}-1)<\\lambda {x_{2}}^{2}(e^{x_{2}}-1)-x_{2}+(e^{x_{2}}-1)$</p>\n<p>令$h_{3}(x)=\\lambda x^{2}(e^{x}-1)-x+(e^{x}-1)$</p>\n<p>但是我们发现$h_{3}(x)$不是一个$\\mathbb{R}$上的单调递增函数, 与我们期望的不同</p>\n<details class=\"note info no-icon\"><summary><p>为什么</p>\n</summary>\n<p>$h_{3}’(x)=\\lambda ((x^{2}+2x)e^{x}-2x)-(1-e^{x})$</p>\n<p>$h_{3}’’(x)=\\lambda ((x^{2}+4x+2)e^{x}-2)+e^{x}$</p>\n<p>有$h_{3}’’(0)=1>0$</p>\n<p>又$h_{3}’’(x)$为$\\mathbb{R}$上连续函数</p>\n<p>故存在包含$0$的一个连续开区间$U_{2}$</p>\n<p>满足$\\forall x\\in{U_{2}}$均有$h_{3}’’(x)>0$</p>\n<p>即$h_{3}’’(x)$在$U_{2}$上恒大于$0$</p>\n<p>$h_{3}’(x)$在$U_{2}$上单调递增</p>\n<p>$x\\in{(-\\infty, 0)\\cap U_{2}}$, $h_{3}’(x)<h_{2}’’(0)=0$, $h_{3}(x)$单调递减; </p>\n<p>$x\\in{U_{2}\\cap (0, +\\infty)}$, $h_{2}’’(x)>h_{2}’’(0)=0$, $h_{3}(x)$单调递增; </p>\n<p>又$k\\to 0^{+}$时$x_{1}\\to 0^{-}$, $x_{2}\\to 0^{+}$</p>\n<p>故存在$k$使得$x_{1}\\in{U_{2}}$, $x_{2}\\in{U_{2}}$</p>\n<p>因此我们不能只通过$x_{1}<x_{2}$</p>\n<p>得出$h_{3}(x_{1})<h_{3}(x_{2})$</p>\n\n</details>\n\n<p>那怎么办? </p>\n<p>这部分思想有点类似拉格朗日乘数法, 直线系, 曲线系之类的</p>\n<p>如果我们构造两个函数$h_{4}(x, y)$, $h_{5}(x)=h_{4}(h_{3}(x), h_{1}(x))$均是关于$x$单调递增的</p>\n<p>那么有$h_{5}(x_{1})=h_{4}(h_{3}(x_{1}), h_{1}(x_{1}))<h_{5}(x_{2})=h_{4}(h_{3}(x_{2}), h_{1}(x_{2}))$</p>\n<p>不就有$h_{3}(x_{1})<h_{3}(x_{2})$了吗? </p>\n<p>实际上一般构造的是$h_{4}(x, y)=x+\\mu y$</p>\n<p>本题中是依据$\\lambda \\geq \\frac{1}{12}$取的$\\mu =-\\frac{1}{2}$</p>\n<p>具体细节不详细描述, 可参看</p>\n<p><a href=\"https://zhuanlan.zhihu.com/p/542468423\">【导数】从增强函数透析“极值点偏移”本质（1/2）</a></p>\n<p>现在来看大问题</p>\n<p>$\\frac{1}{12}$是怎么来的</p>\n<p>前置知识: 洛必达法则, 隐函数求导</p>\n<p>$(e^{x_{1}}-1)(e^{x_{2}}-1)<\\lambda k^{2}-k$</p>\n<p>$\\Leftrightarrow \\frac{(e^{x_{1}}-1)(e^{x_{2}}-1)}{k^{2}}<\\frac{\\lambda k^{2}-k}{k^{2}}$</p>\n<p>$\\Leftrightarrow \\frac{1}{x_{1}x_{2}}+\\frac{1}{k}<\\lambda$</p>\n<p>$h_{6}(k)=\\frac{1}{x_{1}x_{2}}+\\frac{1}{k}$</p>\n<p>一个直觉是$k\\to 0^{+}$时$h_{6}(k)$最大(我也不知道为什么这是对的)</p>\n<p>$\\lim_{k\\to 0^{+}}h_{6}(k)$怎么求? </p>\n<p>$h_{7}(x)=\\sqrt{h_{1}(x)}=\\sqrt{x(e^{x}-1)}$</p>\n<p><del>取根号的原因自己算算就知道, 后面极限最起码不是$0$或$\\infty$你就偷着乐吧</del></p>\n<p>$h_{7}’(x)=\\frac{(x+1)e^{x}-1}{2 \\sqrt{x(e^{x}-1)}}$</p>\n<p>$h_{7}’’(x)=\\frac{(x^{2}+2x-1)e^{2x}-2(x^{2}+x-1)e^{x}-1}{4(x(e^{x}-1))^{3/2}}$</p>\n<p>$h_{7}’’’(x)=\\frac{(x^{3}+3x^{2}-3x+3)e^{3x}-(2x^{3}+9x^{2}-6x+9)e^{2x}+(4x^{3}+6x^{2}-3x+9)e^{x} -3}{8(x(e^{x}-1))^{5/2}}$</p>\n<p>$h_{7}’’’’(x)=\\frac{((x^{4}+4x^{3}-6x^{2}+12x-15)e^{4x}-4(x^{4}+3x^{3}-6x^{2}+9x-15)e^{3x}-(4x^{4}-24x^{3}+30x^{2}-36x+90)e^{2x}-4(2x^{4}+4x^{3}-3x^{2}+3x-15)e^{x}-15)}{16(x(e^{x}-1))^{7/2}}$</p>\n<p>$h_{7}(x_{1})=\\sqrt{h_{1}(x_{1})}=\\sqrt{x_{1}(e^{x_{1}}-1)}=\\sqrt{k}=t$</p>\n<p>$h_{7}(x_{2})=\\sqrt{h_{1}(x_{2})}=\\sqrt{x_{2}(e^{x_{2}}-1)}=\\sqrt{k}=t$</p>\n<p>$h_{8}(t)=\\frac{1}{x_{1}x_{2}}+\\frac{1}{t^{2}}$</p>\n<p>$h_{9}(t)=x_1$</p>\n<p>$h_{10}(t)=x_2$</p>\n<p>则</p>\n<p>$h_{7}(h_{9}(t))=t$</p>\n<p>$h_{7}(h_{10}(t))=t$</p>\n<p>对$t$求导得</p>\n<p>$h_{9}’(t)h_{7}’(h_{9}(t))=1$</p>\n<p>$h_{10}’(t)h_{7}’(h_{10}(t))=1$</p>\n<p>即</p>\n<p>$h_{9}’(t)h_{7}’(x_{1})=1$</p>\n<p>$h_{10}’(t)h_{7}’(x_{2})=1$</p>\n<p>可得</p>\n<p>$h_{9}’(t)=\\frac{1}{h_{7}’(x_{1})}$</p>\n<p>$h_{10}’(t)=\\frac{1}{h_{7}’(x_{2})}$</p>\n<p>重复对$t$求导可得</p>\n<p>$h_{9}’’(t)=-\\frac{h_{7}’’(x_{1})}{(h_{7}’(x_{1}))^{3}}$</p>\n<p>$h_{10}’’(t)=-\\frac{h_{7}’’(x_{2})}{(h_{7}’(x_{2}))^{3}}$</p>\n<p>$h_{9}’’’(t)=\\frac{3(h_{7}’’(x_{1}))^{2}-h_{7}’’’(x_{1})h_{7}’(x_{1})}{(h_{7}’(x_{1}))^{5}}$</p>\n<p>$h_{10}’’’(t)=\\frac{3(h_{7}’’(x_{2}))^{2}-h_{7}’’’(x_{2})h_{7}’(x_{2})}{(h_{7}’(x_{2}))^{5}}$</p>\n<p>$h_{9}’’’’(t)=\\frac{10h_{7}’’’(x_{1})h_{7}’’(x_{1})h_{7}’(x_{1})-h_{7}’’’’(x_{1})(h_{7}’(x_{1}))^{2}-15(h_{7}’’(x_{1}))^{3}}{(h_{7}’(x_{1}))^{7}}$</p>\n<p>$h_{10}’’’’(t)=\\frac{10h_{7}’’’(x_{2})h_{7}’’(x_{2})h_{7}’(x_{2})-h_{7}’’’’(x_{2})(h_{7}’(x_{2}))^{2}-15(h_{7}’’(x_{2}))^{3}}{(h_{7}’(x_{2}))^{7}}$</p>\n<p>$h_{11}(t)=h_{9}(t)h_{10}(t)$</p>\n<p>$h_{11}’(t)=h_{9}’(t)h_{10}(t)+h_{9}(t)h_{10}’(t)$</p>\n<p>$h_{11}’’(t)=h_{9}’’(t)h_{10}(t)+2h_{9}’(t)h_{10}’(t)+h_{9}(t)h_{10}’’(t)$</p>\n<p>$h_{11}’’’(t)=h_{9}’’’(t)h_{10}(t)+3h_{9}’’(t)h_{10}’(t)+3h_{9}’(t)h_{10}’’(t)+h_{9}(t)h_{10}’’’(t)$</p>\n<p>$h_{11}’’’’(t)=h_{9}’’’’(t)h_{10}(t)+4h_{9}’’’(t)h_{10}’(t)+6h_{9}’’(t)h_{10}’’(t)+4h_{9}’(t)h_{10}’’’(t)+h_{9}(t)h_{10}’’’’(t)$</p>\n<p>$\\lim_{x\\to 0^{-}}h_{7}’(x)=-1$</p>\n<p>$\\lim_{x\\to 0^{+}}h_{7}’(x)=1$</p>\n<p>$\\lim_{x\\to 0^{-}}h_{7}’’(x)=-\\frac{1}{2}$</p>\n<p>$\\lim_{x\\to 0^{+}}h_{7}’’(x)=\\frac{1}{2}$</p>\n<p>$\\lim_{x\\to 0^{-}}h_{7}’’’(x)=-\\frac{5}{16}$</p>\n<p>$\\lim_{x\\to 0^{+}}h_{7}’’’(x)=\\frac{5}{16}$</p>\n<p>$\\lim_{x\\to 0^{-}}h_{7}’’’’(x)=-\\frac{3}{16}$</p>\n<p>$\\lim_{x\\to 0^{+}}h_{7}’’’’(x)=\\frac{3}{16}$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{9}(t)=0$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{10}(t)=0$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{9}’(t)=-1$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{10}’(t)=1$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{9}’’(t)=-\\frac{1}{2}$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{10}’’(t)=-\\frac{1}{2}$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{9}’’’(t)=-\\frac{7}{16}$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{10}’’’(t)=\\frac{7}{16}$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{9}’’’’(t)=-\\frac{1}{2}$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{10}’’’’(t)=-\\frac{1}{2}$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{11}(t)=0$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{11}’(t)=0$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{11}’’(t)=-2$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{11}’’’(t)=0$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{11}’’’’(t)=-2$</p>\n<p>$\\lim_{t\\to 0^{+}}h_{8}(t)=\\frac{1}{h_{11}(t)}+\\frac{1}{t^{2}}=\\frac{t^{2}+h_{11}(t)}{h_{11}(t)t^{2}}=\\frac{2t+h_{11}’(t)}{h_{11}(t)t^{2}+2h_{11}(t)t}=\\frac{h_{11}’’(t)+2}{4h_{11}’(t)t+h_{11}’’(t)t^{2}+2h_{11}(t)}=\\frac{h_{11}’’’(t)}{6h_{11}’’(t)t+6h_{11}’(t)+h_{11}’’’(t)t^{2}}=\\frac{h_{11}’’’’(t)}{12h_{11}’’(t)+8h_{11}’’’(t)t+h_{11}’’’’(t)t^{2}}=\\frac{-2}{-24}=\\frac{1}{12}$</p>\n<p>$Q.E.D$</p>\n","tags":["Maths","高中","高考"]},{"title":"2022年普通高等学校招生全国统一考试数学(北京卷)T21","url":"//post/2022%E5%B9%B4%E6%99%AE%E9%80%9A%E9%AB%98%E7%AD%89%E5%AD%A6%E6%A0%A1%E6%8B%9B%E7%94%9F%E5%85%A8%E5%9B%BD%E7%BB%9F%E4%B8%80%E8%80%83%E8%AF%95%E6%95%B0%E5%AD%A6(%E5%8C%97%E4%BA%AC%E5%8D%B7)T21/","content":"<center></center>\n<span id=\"more\"></span>\n\n<p>题目: </p>\n<p>待补充<del>(求发个$\\LaTeX$的, 懒得码)</del></p>\n<hr>\n<p>昨晚上刷到这题, 看了一眼就摸鱼去了<del>反正不是京爷考不到我</del>, 今早(12:00)起来做了一下</p>\n<hr>\n<p>搜了一下没发现什么好的解答, 不是<a href=\"https://edu.sina.com.cn/gaokao/2022-06-08/doc-imizirau7279844.shtml\">伪证</a>就是<a href=\"https://blog.csdn.net/weixin_47790125/article/details/125201280\">太复杂</a></p>\n<p>但还是感谢一下<a href=\"https://zhuanlan.zhihu.com/p/525808497\">2022年北京高考数学压轴题赏析</a>, 提到了富比尼原理, 指明了方向</p>\n<hr>\n<p>(1)(2)自己做</p>\n<hr>\n<p>(3): </p>\n<p>实际上本题有个条件是无用的<del>自己猜是哪个</del></p>\n<p>只证$k=6$不成立, $k<6$的情况补$0$即为$k=6$的情况, 同时得知$k=6$时$\\forall i\\in{\\{1,2,3,4,5,6\\}}, a_{i}\\neq 0$</p>\n<p>$k=6$时</p>\n<p>设$S$为所有连续和组成的可重集合</p>\n<p>由</p>\n<p>$|S|=21$</p>\n<p>$\\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\\}\\subseteq S$</p>\n<p>知</p>\n<p>$\\exists t\\in{S}, \\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\\}=S/\\{t\\}$</p>\n<p>算两次</p>\n<p>$\\sum \\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\\}=210=6\\sum_{i\\in{\\{1,2,3,4,5,6\\}}}a_{i}-t=\\sum S/\\{t\\}$</p>\n<p>$\\sum_{i\\in{\\{1,2,3,4,5,6\\}}}a_{i}=\\frac{210+t}{6}$</p>\n<p>若$\\exists j\\in{\\{1,2,3,4,5,6\\}}, a_{j}<0$</p>\n<p>则$t=a_{j}<0$</p>\n<p>$\\sum_{i\\in{\\{1,2,3,4,5,6\\}}, i\\neq j}a_{i}=\\frac{210-5t}{6}\\geq\\frac{210-5(-6)}{6}=40$</p>\n<p>又$a_{j}$至多将$\\{a_{n}\\}$分为两部, 或有一部连续和大于$20$, 或有两部连续和均大于等于$20$</p>\n<p>均矛盾</p>\n<p>若$\\forall i\\in{\\{1,2,3,4,5,6\\}}, a_{i}>0$</p>\n<p>则$t>0$</p>\n<p>$\\sum_{i\\in{\\{1,2,3,4,5,6\\}}}a_{i}=\\frac{210+t}{6}\\geq\\frac{210+6}{6}=36$</p>\n<p>考察等于$20$的连续和</p>\n<p>$l,r\\in{\\mathbb{Z}^{+}},l,r-1\\in{\\{1,2,3,4,5,6\\}},\\sum_{i\\in{\\mathbb{Z}^{+}},i\\in{[l,r-1]}}a_{i}=20$</p>\n<p>若$r-l<5$</p>\n<p>则由$(r-l)+1<6$知$\\exists L,R\\in{\\mathbb{Z}^{+}},L,R-1\\in{\\{1,2,3,4,5,6\\}},[l,r-1]\\subset[L,R-1],R-L=5$</p>\n<p>$\\sum_{i\\in{\\mathbb{Z}^{+}},i\\in{[L,R-1]}}a_{i}>\\sum_{i\\in{\\mathbb{Z}^{+}},i\\in{[l,r-1]}}a_{i}=20$</p>\n<p>又</p>\n<p>$\\sum_{i\\in{\\{1,2,3,4,5,6\\}}}a_{i}\\geq 36>20$</p>\n<p>即</p>\n<p>$\\sum_{i\\in{\\mathbb{Z}^{+}},i\\in{[L,R-1]}}a_{i}\\notin \\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\\}$</p>\n<p>$\\sum_{i\\in{\\{1,2,3,4,5,6\\}}}a_{i}\\notin \\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\\}$</p>\n<p>$\\{i|i\\in{\\mathbb{Z}^{+}},i\\in{[L,R-1]}\\}\\neq \\{i|i\\in{\\{1,2,3,4,5,6\\}}\\}$</p>\n<p>矛盾</p>\n<p>且由</p>\n<p>$\\sum_{i\\in{\\{1,2,3,4,5,6\\}}}a_{i}\\notin \\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\\}$</p>\n<p>知$r-l\\neq 6$又$r-l\\leq 6$故$r-l=5$</p>\n<p>考察$j\\notin{[l,r-1]}$</p>\n<p>$a_{j}=\\sum_{i\\in{\\{1,2,3,4,5,6\\}}}a_{i}-\\sum_{i\\in{\\mathbb{Z}^{+}},i\\in{[l,r-1]}}a_{i}\\geq 36-20=16$</p>\n<p>$\\exists L,R\\in{\\mathbb{Z}^{+}},L,R-1\\in{\\{1,2,3,4,5,6\\}},j\\in{[L,R-1]},R-L=5$</p>\n<p>$\\sum_{i\\in{\\mathbb{Z}^{+}},i\\in{[L,R-1]}}a_{i}\\geq 16+1+1+1+1=20$</p>\n<p>$\\{i|i\\in{\\mathbb{Z}^{+}},i\\in{[L,R-1]}\\}\\neq \\{i|i\\in{\\{1,2,3,4,5,6\\}}\\}$</p>\n<p>$\\{i|i\\in{\\mathbb{Z}^{+}},i\\in{[L,R-1]}\\}\\neq \\{i|i\\in{\\mathbb{Z}^{+}},i\\in{[l,r-1]}\\}$</p>\n<p>或有两个连续和等于$20$, 或有两个连续和不属于$\\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\\}$</p>\n<p>均矛盾</p>\n<p>$Q.E.D.$</p>\n","tags":["Maths","高中","高考"]},{"title":"2024年阿里巴巴全球数学竞赛决赛分析赛道T4","url":"//post/2024%E5%B9%B4%E9%98%BF%E9%87%8C%E5%B7%B4%E5%B7%B4%E5%85%A8%E7%90%83%E6%95%B0%E5%AD%A6%E7%AB%9E%E8%B5%9B%E5%86%B3%E8%B5%9B%E5%88%86%E6%9E%90%E8%B5%9B%E9%81%93T4/","content":"<center></center>\n<span id=\"more\"></span>\n\n<p>Problem 4. 定义序列</p>\n<p>$$a_{n+1}=a_{n}+\\frac{ {a_{n}}^{2} }{n^{2}}. 0\\leq a_{1}<1. $$</p>\n<p>证明极限 $\\lim\\limits_{n\\to +\\infty}a_{n}$ 存在并且有限. </p>\n<hr>\n<p>$$(n+1)-a_{n+1}=(n-a_{n})(1+\\frac{a_{n}+n}{n^{2}})$$</p>\n<p>先证不存在 $\\lambda\\geq 1$, 使得 $\\exists N_{0}\\in\\mathbb{N}, k>0$ 满足 $\\frac{a_{n}}{(n+1)^{\\lambda}}\\geq k$ 对 $\\forall n\\in\\mathbb{N}, n>N_{0}$ 恒成立</p>\n<p>若存在， 则 $\\lambda=1$ 时一定符合要求</p>\n<p>下证 $\\lambda=1$ 时不成立</p>\n<p>对 $n>N_{0}$</p>\n<p>$$1+\\frac{a_{n}+n}{n^{2}}\\geq\\frac{n^{2}+n+k(n+1)}{n^2}=\\frac{(n+1)(n+k)}{n^2}$$</p>\n<p>$$\\begin{equation}\\begin{split}(n+1)-a_{n+1}&=((N_{0}+1)-a_{N_{0}+1})\\prod\\limits_{i=N_{0}+1}^{n}(1+\\frac{a_{i}+i}{i^{2}})\\\\& \\geq((N_{0}+1)-a_{N_{0}+1})\\prod\\limits_{i=N_{0}+1}^{n}\\frac{(i+1)(i+k)}{i^2}\\\\& =((N_{0}+1)-a_{N_{0}+1})\\frac{n+1}{N_{0}+1}\\prod\\limits_{i=N_{0}+1}^{n}(1+\\frac{k}{i})\\\\& >((N_{0}+1)-a_{N_{0}+1})\\frac{n+1}{N_{0}+1}(1+k\\sum\\limits_{i=N_{0}+1}^{n}(\\frac{1}{i}))\\\\& >((N_{0}+1)-a_{N_{0}+1})\\frac{n+1}{N_{0}+1}(1+k\\sum\\limits_{i=N_{0}+1}^{n}(\\ln(\\frac{i+1}{i})))\\\\& =((N_{0}+1)-a_{N_{0}+1})\\frac{n+1}{N_{0}+1}(1+k\\ln(\\frac{n+1}{N_{0}+1}))\\end{split}\\end{equation}$$</p>\n<p>当 $n$ 充分大时, $((N_{0}+1)-a_{N_{0}+1})\\frac{n+1}{N_{0}+1}(1+k\\ln(\\frac{n+1}{N_{0}+1}))>n+1$, 即 $(n+1)-a_{n+1}>(n+1)$, 矛盾. </p>\n<p>由此知, 对 $\\forall N_{0}\\in\\mathbb{N}, k>0$, $\\exists n>N_{0}$, 使得 $\\frac{a_{n}}{n+1}<k$ 成立</p>\n<p>再证不存在 $1>\\lambda>0$, 使得 $\\exists N_{0}\\in\\mathbb{N}, k>0$ 满足 $\\frac{a_{n}}{(n+1)^{\\lambda}}\\geq k$ 对 $\\forall n\\in\\mathbb{N}, n>N_{0}$ 恒成立</p>\n<p>若存在, 则有</p>\n<p>$$1+\\frac{a_{n}+n}{n^{2}}\\geq\\frac{n^{2}+n+k(n+1)^{\\lambda}}{n^2}=\\frac{(n+1)(n+\\frac{k}{(n+1)^{1-\\lambda}})}{n^2}$$</p>\n<p>$$\\begin{equation}\\begin{split}(n+1)-a_{n+1}&=((N_{0}+1)-a_{N_{0}+1})\\prod\\limits_{i=N_{0}+1}^{n}(1+\\frac{a_{i}+i}{i^{2}})\\\\& \\geq((N_{0}+1)-a_{N_{0}+1})\\prod\\limits_{i=N_{0}+1}^{n}\\frac{(i+1)(i+\\frac{k}{(i+1)^{1-\\lambda}})}{i^2}\\\\& =((N_{0}+1)-a_{N_{0}+1})\\frac{n+1}{N_{0}+1}\\prod\\limits_{i=N_{0}+1}^{n}(1+\\frac{k}{i(i+1)^{1-\\lambda}})\\\\& >((N_{0}+1)-a_{N_{0}+1})\\frac{n+1}{N_{0}+1}(1+k\\sum\\limits_{i=N_{0}+1}^{n}(\\frac{1}{i(i+1)^{1-\\lambda}}))\\\\& >((N_{0}+1)-a_{N_{0}+1})\\frac{n+1}{N_{0}+1}(1+\\frac{k}{1-\\lambda}\\sum\\limits_{i=N_{0}+1}^{n}(\\frac{1}{i^{1-\\lambda}}-\\frac{1}{(i+1)^{1-\\lambda}}))\\\\& =((N_{0}+1)-a_{N_{0}+1})\\frac{n+1}{N_{0}+1}(1+\\frac{k}{1-\\lambda}(\\frac{1}{(N_{0}+1)^{1-\\lambda}}-\\frac{1}{(n+1)^{1-\\lambda}}))\\end{split}\\end{equation}$$</p>\n<p>仿照上面, 尝试构造使得</p>\n<p>$$(1-\\frac{a_{N_{0}+1}}{N_{0}+1})(1+\\frac{k}{1-\\lambda}(\\frac{1}{(N_{0}+1)^{1-\\lambda}}-\\frac{1}{(n+1)^{1-\\lambda}}))\\geq 1$$</p>\n<p>即</p>\n<p>$$\\frac{\\frac{k}{1-\\lambda}(\\frac{1}{(N_{0}+1)^{1-\\lambda}}-\\frac{1}{(n+1)^{1-\\lambda}})}{1+\\frac{k}{1-\\lambda}(\\frac{1}{(N_{0}+1)^{1-\\lambda}}-\\frac{1}{(n+1)^{1-\\lambda}})}\\geq\\frac{a_{N_{0}+1}}{N_{0}+1}$$</p>\n<p>设 $t=N_{0}+1, \\epsilon=\\frac{a_{N_{0}+1}}{k(N_{0}+1)}-(\\frac{1}{(N_{0}+1)^{1-\\lambda}}-\\frac{1}{(n+1)^{1-\\lambda}})=\\frac{a_{N_{0}+1}}{mk(N_{0}+1)}$</p>\n<p>化简得</p>\n<p>$$m\\geq\\frac{1-\\frac{a_{t}}{t}}{\\lambda-\\frac{a_{t}}{t}}$$</p>\n<p>即</p>\n<p>$$\\epsilon\\leq\\frac{\\lambda-\\frac{a_{t}}{t}}{1-\\frac{a_{t}}{t}}\\frac{a_{t}}{kt}$$</p>\n<p>又因为 $\\frac{1}{(n+1)^{1-\\lambda}}$ 可以充分小, 故只需 $k>\\frac{a_{t}(1-\\lambda)}{t^{\\lambda}(1-\\frac{a_{t}}{t})}$ 即可</p>\n<p>$$\\frac{\\frac{a_{t}(1-\\lambda)}{t^{\\lambda}(1-\\frac{a_{t}}{t})}}{\\frac{a_{t}}{(t+1)^{\\lambda}}}=\\frac{(1-\\lambda)(t+1)^{\\lambda}}{(1-\\frac{a_{t}}{t})t^{\\lambda}}<\\frac{(1-\\lambda)(t+1)}{(1-\\frac{a_{t}}{t})t}$$</p>\n<p>当 $\\frac{a_{t}}{t}<\\lambda^{2}$ 且 $t>\\frac{1}{\\lambda}$ 时 $\\frac{(1-\\lambda)(t+1)}{(1-\\frac{a_{t}}{t})t}<1$</p>\n<p>又由对 $\\forall N_{0}\\in\\mathbb{N}, k>0$, $\\exists n>N_{0}$, 使得 $\\frac{a_{n}}{n+1}<k$ 成立</p>\n<p>知 $\\forall N_{0}\\in\\mathbb{N}, k>0$, $\\exists n>N_{0}$, 使得 $\\frac{a_{n}}{n}<k$ 成立</p>\n<p>知 $\\forall N_{0}\\in\\mathbb{N}$, $\\exists t>N_{0}$, 使得 $\\frac{\\frac{a_{t}(1-\\lambda)}{t^{\\lambda}(1-\\frac{a_{t}}{t})}}{\\frac{a_{t}}{(t+1)^{\\lambda}}}=\\frac{(1-\\lambda)(t+1)^{\\lambda}}{(1-\\frac{a_{t}}{t})t^{\\lambda}}<\\frac{(1-\\lambda)(t+1)}{(1-\\frac{a_{t}}{t})t}<1$ 成立</p>\n<p>接下来, 我们对数列${\\frac{a_{n}}{(n+1)^{\\lambda}}}$进行考察</p>\n<p>$$\\frac{a_{n+1}}{((n+1)+1)^{\\lambda}}<\\frac{a_{n}}{(n+1)^{\\lambda}}$$</p>\n<p>$$\\Leftrightarrow \\frac{a_{n}}{(n+1)^{\\lambda}}<\\frac{n^{2}((n+2)^{\\lambda}-(n+1)^{\\lambda})}{(n+1)^{2\\lambda}}$$</p>\n<p>又</p>\n<p>$$\\frac{n^{2}((n+2)^{\\lambda}-(n+1)^{\\lambda})}{(n+1)^{2\\lambda}}<\\frac{(n+1)^{2}(((n+1)+2)^{\\lambda}-((n+1)+1)^{\\lambda})}{((n+1)+1)^{2\\lambda}}$$</p>\n<p>若</p>\n<p>$$\\frac{a_{n+1}}{((n+1)+1)^{\\lambda}}<\\frac{a_{n}}{(n+1)^{\\lambda}}$$</p>\n<p>则</p>\n<p>$$\\frac{a_{n+1}}{((n+1)+1)^{\\lambda}}<\\frac{a_{n}}{(n+1)^{\\lambda}}<\\frac{n^{2}((n+2)^{\\lambda}-(n+1)^{\\lambda})}{(n+1)^{2\\lambda}}<\\frac{(n+1)^{2}(((n+1)+2)^{\\lambda}-((n+1)+1)^{\\lambda})}{((n+1)+1)^{2\\lambda}}$$</p>\n<p>即</p>\n<p>$$\\frac{a_{(n+1)+1}}{(((n+1)+1)+1)^{\\lambda}}<\\frac{a_{(n+1)}}{((n+1)+1)^{\\lambda}}$$</p>\n<p>故数列${\\frac{a_{n}}{(n+1)^{\\lambda}}}$为 递增数列 或 先递增后递减数列 或 递减数列</p>\n<p>若 $\\exists N_{0}\\in\\mathbb{N}$ 使得 $\\forall n\\in\\mathbb{N}, n>N_{0}$ 均有 $\\frac{a_{n+1}}{((n+1)+1)^{\\lambda}}<\\frac{a_{n}}{(n+1)^{\\lambda}}$ 成立</p>\n<p>则知极限 $\\lim\\limits_{n\\to +\\infty}\\frac{a_{n}}{(n+1)^{\\lambda}}$ 存在并且大于$0$</p>\n<p>设$k_{max}=\\lim\\limits_{n\\to +\\infty}\\frac{a_{n}}{(n+1)^{\\lambda}}$</p>\n<p>显然 $k=k_{max}$ 符合定义</p>\n<p>因此对 $\\forall\\sigma>0$, $\\exists t_{0}\\in\\mathbb{N}, t_{0}>N_{0}$ 使得 $\\forall t>t_{0}$ 时有 $\\frac{a_{t}}{(t+1)^{\\lambda}}<(1+\\sigma)k_{max}$</p>\n<p>又由 $\\frac{\\frac{a_{t}(1-\\lambda)}{t^{\\lambda}(1-\\frac{a_{t}}{t})}}{\\frac{a_{t}}{(t+1)^{\\lambda}}}=\\frac{(1-\\lambda)(t+1)^{\\lambda}}{(1-\\frac{a_{t}}{t})t^{\\lambda}}<\\frac{(1-\\lambda)(t+1)}{(1-\\frac{a_{t}}{t})t}$</p>\n<p>知</p>\n<p>存在无穷个 $t$ 使得 $k_{max}>\\frac{a_{t}(1-\\lambda)}{t^{\\lambda}(1-\\frac{a_{t}}{t})}$ 成立</p>\n<p>取$k=k_{max}$, 符合条件的$t$, $N_{0}=t-1$, 此时有</p>\n<p>$$(1-\\frac{a_{N_{0}+1}}{N_{0}+1})(1+\\frac{k}{1-\\lambda}(\\frac{1}{(N_{0}+1)^{1-\\lambda}}-\\frac{1}{(n+1)^{1-\\lambda}}))\\geq 1$$</p>\n<p>对任意充分大的 $n$ 成立, 矛盾</p>\n<p>若数列${\\frac{a_{n}}{(n+1)^{\\lambda}}}$为递增数列</p>\n<p>取 $k_{max}=\\frac{a_{t}}{(t+1)^{\\lambda}}$ 即可, 同理存在无穷个 $t$ 使得 $k_{max}>\\frac{a_{t}(1-\\lambda)}{t^{\\lambda}(1-\\frac{a_{t}}{t})}$ 成立</p>\n<p>取$k=k_{max}$, 符合条件的$t$, $N_{0}=t-1$, 此时有</p>\n<p>$$(1-\\frac{a_{N_{0}+1}}{N_{0}+1})(1+\\frac{k}{1-\\lambda}(\\frac{1}{(N_{0}+1)^{1-\\lambda}}-\\frac{1}{(n+1)^{1-\\lambda}}))\\geq 1$$</p>\n<p>对任意充分大的 $n$ 成立, 矛盾</p>\n<p>得证! </p>\n<p>综上对 $\\forall\\lambda>0, k>0$, $\\exists N_{0}\\in\\mathbb{N}$ 使得对 $\\forall n\\in\\mathbb{N}, n>N_{0}$ 均有 $\\frac{a_{n}}{(n+1)^{\\lambda}}<k$ 成立</p>\n<p>(从 "$\\forall\\lambda>0, k>0$, $\\exists n\\in\\mathbb{N}$ 有 $\\frac{a_{n}}{(n+1)^{\\lambda}}<k$ 成立" 到 "$\\forall\\lambda>0, k>0$, $\\exists N_{0}\\in\\mathbb{N}$ 使得对 $\\forall n\\in\\mathbb{N}, n>N_{0}$ 均有 $\\frac{a_{n}}{(n+1)^{\\lambda}}<k$ 成立" 可能有点绕(提示: 仿照上面证明中最后一步讨论), 但是无所谓, 我们可以使用一个更弱的结论, $\\forall\\lambda>0$, $\\exists N_{0}\\in\\mathbb{N}, k>0$ 使得对 $\\forall n\\in\\mathbb{N}, n>N_{0}$ 均有 $\\frac{a_{n}}{(n+1)^{\\lambda}}<k$ 成立)</p>\n<p>$$a_{n+1}=a_{N_{0}+1}+\\sum\\limits_{i=N_{0}+1}^{n}(\\frac{a_{i}}{i^{2}})<a_{N_{0}+1}+\\sum\\limits_{i=N_{0}+1}^{n}(\\frac{k^2(i+1)^{2\\lambda}}{i^{2}})$$</p>\n<p>取 $0<\\lambda<\\frac{1}{2}$ 则原题显然</p>\n<p>$$Q.E.D.$$</p>\n","tags":["Maths"]},{"title":"FRTI64","url":"//post/FRTI64/","content":"<p>居然破天荒的有简介了欸</p>\n<p>嗯, 讲一下个人经历吧. </p>\n<p>本文按照时间顺序讲述了FRTI64的形成, 演变过程</p>\n<center></center>\n<span id=\"more\"></span>\n\n<hr>\n<p>前幼儿园时代: 不记得惹\\</p>\n<hr>\n<p>幼儿园——初见端倪</p>\n<p>这个时期的FRTI64过的还是相对比较幸福的</p>\n<p>嗯Fr02enT在打游戏(RA3)</p>\n<p>Frantinesy在发情, 试着把某些东西塞回去, 试着在脸上涂点什么东西(虽然由于极其低下的执行力并没有这样做)</p>\n<p>FROZEN-TIME在外面陪朋友们玩, 不过这个时候已经出现了: 虽然是你的球被别的小孩弄到水坑里了, 很脏, 但你要自己去捡, 没有教养的东西, 看我衣架!</p>\n<p>这种很不愉快的事情</p>\n<hr>\n<p>小学1-3年级——不太记得惹</p>\n<p>嗯Fr02enT还在打游戏</p>\n<p>Frantinesy还在发情</p>\n<p>FROZEN-TIME还在外面陪朋友们玩</p>\n<p>不过吃胖了好多现在都没减下来ww</p>\n<hr>\n<p>小学4-6年级——Frantinesy没有发情的生活, 会是什么样呢? </p>\n<p>这个时期主要是Fr02enT还在打游戏</p>\n<p>FROZEN-TIME还在外面陪朋友们玩</p>\n<p>不过他和Frantinesy一起承担了一部分学习任务</p>\n<p>父母已经变成这样惹: 小学四年级教你解二元一次方程组, 但是教不明白, 故急眼, 使用衣架狠狠注入武大优秀基因</p>\n<hr>\n<p>初中——光与暗的变幻</p>\n<p>莫名其妙, FROZEN-TIME与Frantinesy就带着Fr02enT走进了有品(我也是天才少年力)</p>\n<p>不过父母已经开始吵架</p>\n<p>每次父亲要打母亲时, 母亲都会躲到我房间里, 和我一起, 但父亲还是冲进来</p>\n<p>成为oier(az其实是因为选信息课的时候没选到程序设计, 机器人之类的, 莫名奇妙就这样了)</p>\n<p>直到初二, 第一次因为沉迷oi不学文化课挨打, 心里有什么东西悄悄的碎掉惹</p>\n<p>从此只会抄题解放弃思考了</p>\n<p>这也是接下来两次csp失利的一部分原因(嗯主要还是Frantinesy在考场上发情gd惹)</p>\n<p>然后就是初三</p>\n<p>初三就是爆父母米, 玩手机, 和父母吵架的两年(因为休学惹)了</p>\n<p>Fr02enT在打游戏</p>\n<p>然后就是Frantinesy发情gd躲被子里哭</p>\n<p>也没有找到渠道买糖所以就拖成了现在这样呜呜呜</p>\n<p>真的同学都非常好, 那时还和xzq, zcc几个上课传字条, 下课跟在xxr后面当小跟班, 好想回到那个时候啊</p>\n<hr>\n<p>高中——新环境新起点</p>\n<p>嗯就感觉挺好的, 糖糖什么的也就放了一段时间</p>\n<p>直到…有个姐姐停糖, 想把剩下的糖卖出去</p>\n<p>嗯, 但是最后我也没买成</p>\n<p>不过也就此真的开始含糖惹(但是好像来不及惹ww)</p>\n<p>未完待续</p>\n"},{"title":"Hello World","url":"//post/Hello%20World/","content":"<center></center>\n<span id=\"more\"></span>\n\n<p>Welcome to <a href=\"https://hexo.io/\">Hexo</a>! This is your very first post. Check <a href=\"https://hexo.io/docs/\">documentation</a> for more info. If you get any problems when using Hexo, you can find the answer in <a href=\"https://hexo.io/docs/troubleshooting.html\">troubleshooting</a> or you can ask me on <a href=\"https://github.com/hexojs/hexo/issues\">GitHub</a>.</p>\n<h2 id=\"Quick-Start\"><a href=\"#Quick-Start\" class=\"headerlink\" title=\"Quick Start\"></a>Quick Start</h2><h3 id=\"Create-a-new-post\"><a href=\"#Create-a-new-post\" class=\"headerlink\" title=\"Create a new post\"></a>Create a new post</h3><figure class=\"highlight bash\"><table><tr><td class=\"code\"><pre><span class=\"line\">$ hexo new <span class=\"string\">"My New Post"</span></span><br></pre></td></tr></table></figure>\n\n<p>More info: <a href=\"https://hexo.io/docs/writing.html\">Writing</a></p>\n<h3 id=\"Run-server\"><a href=\"#Run-server\" class=\"headerlink\" title=\"Run server\"></a>Run server</h3><figure class=\"highlight bash\"><table><tr><td class=\"code\"><pre><span class=\"line\">$ hexo server</span><br></pre></td></tr></table></figure>\n\n<p>More info: <a href=\"https://hexo.io/docs/server.html\">Server</a></p>\n<h3 id=\"Generate-static-files\"><a href=\"#Generate-static-files\" class=\"headerlink\" title=\"Generate static files\"></a>Generate static files</h3><figure class=\"highlight bash\"><table><tr><td class=\"code\"><pre><span class=\"line\">$ hexo generate</span><br></pre></td></tr></table></figure>\n\n<p>More info: <a href=\"https://hexo.io/docs/generating.html\">Generating</a></p>\n<h3 id=\"Deploy-to-remote-sites\"><a href=\"#Deploy-to-remote-sites\" class=\"headerlink\" title=\"Deploy to remote sites\"></a>Deploy to remote sites</h3><figure class=\"highlight bash\"><table><tr><td class=\"code\"><pre><span class=\"line\">$ hexo deploy</span><br></pre></td></tr></table></figure>\n\n<p>More info: <a href=\"https://hexo.io/docs/one-command-deployment.html\">Deployment</a></p>\n"},{"title":"利用GitHub Actions进行Hexo的多镜像站部署","url":"//post/%E5%88%A9%E7%94%A8GitHub%20Actions%E8%BF%9B%E8%A1%8CHexo%E7%9A%84%E5%A4%9A%E9%95%9C%E5%83%8F%E7%AB%99%E9%83%A8%E7%BD%B2/","content":"<center></center>\n<span id=\"more\"></span>\n\n<!-- 其实早就有解决方法了\n\n前几天的[LOG[2021-09-05]](/LOG[2021-09-05]/)就是个实例, 同时部署到GitHub, Cloudflare, Vercel\n\n本文主要是做个总结\n\n------ -->\n\n<p>最简单的做法就是为每个镜像站额外维护一个配置文件</p>\n<p>但这样做的缺点很明显, 一旦有更改, 就必须手动同步, 像git, svn这些版本控制系统都难以自动完成"对于部分项保留当前分支, 对于其它项合并其它分支"的工作<del>我知道有<code>git cherry-pick</code>, 但我不会</del></p>\n<p>好吧还是给一个思路, 从主站分支创建镜像站分支, 将需要保留且长期的项修改并提交到当前分支</p>\n<p>每次主站commit, 自动记录commit编号, cherry-pick到分支站中</p>\n<p>差不多就这样</p>\n<hr>\n<p>接下来讲正解</p>\n<p>首先要清楚镜像站之间的配置不同在哪里</p>\n<p>大部分时候都是这样</p>\n<figure class=\"highlight yaml\"><table><tr><td class=\"code\"><pre><span class=\"line\"><span class=\"comment\"># URL</span></span><br><span class=\"line\"><span class=\"comment\">## Set your site url here. For example, if you use GitHub Page, set url as 'https://username.github.io/project'</span></span><br><span class=\"line\"><span class=\"attr\">url:</span></span><br><span class=\"line\"></span><br><span class=\"line\"><span class=\"comment\"># Deployment</span></span><br><span class=\"line\"><span class=\"comment\">## Docs: https://hexo.io/docs/one-command-deployment</span></span><br><span class=\"line\"><span class=\"attr\">deploy:</span></span><br><span class=\"line\"> <span class=\"attr\">type:</span> <span class=\"string\">git</span></span><br><span class=\"line\"> <span class=\"attr\">repo:</span></span><br><span class=\"line\"> <span class=\"attr\">branch:</span></span><br><span class=\"line\"></span><br></pre></td></tr></table></figure>\n\n<p>那么我们很明显需要一个东西来自动修改这些项</p>\n<p>nodejs的解决方案是<code>js-yaml</code></p>\n<p>当然你也可以拿C/C++, shell或者其它的东西写, 本文就只讲nodejs解决方案</p>\n<figure class=\"highlight javascript\"><table><tr><td class=\"code\"><pre><span class=\"line\"><span class=\"keyword\">const</span> fs = <span class=\"built_in\">require</span>(<span class=\"string\">'fs'</span>);</span><br><span class=\"line\"><span class=\"keyword\">const</span> yaml = <span class=\"built_in\">require</span>(<span class=\"string\">'js-yaml'</span>);</span><br><span class=\"line\"></span><br><span class=\"line\"><span class=\"keyword\">try</span> {</span><br><span class=\"line\"> <span class=\"keyword\">let</span> hexo_config = yaml.<span class=\"title function_\">load</span>(fs.<span class=\"title function_\">readFileSync</span>(<span class=\"string\">'./_config.yml'</span>, <span class=\"string\">'utf8'</span>)); <span class=\"comment\">// 主题配置文件也可以这样修改</span></span><br><span class=\"line\"> <span class=\"comment\">// 例如deploy:</span></span><br><span class=\"line\"> <span class=\"comment\">// repo:</span></span><br><span class=\"line\"> <span class=\"comment\">// 可以写成deploy.repo = ''</span></span><br><span class=\"line\"> fs.<span class=\"title function_\">writeFileSync</span>(<span class=\"string\">'./hexo-config.yml'</span>, yaml.<span class=\"title function_\">dump</span>(hexo_config), <span class=\"string\">'utf8'</span>); <span class=\"comment\">// 将修改后的文件存在当前目录的hexo-config.yml中</span></span><br><span class=\"line\">}</span><br><span class=\"line\"><span class=\"keyword\">catch</span> (e) {</span><br><span class=\"line\"> <span class=\"variable language_\">console</span>.<span class=\"title function_\">log</span>(e);</span><br><span class=\"line\">}</span><br></pre></td></tr></table></figure>\n\n<p>接下来讲一下GitHub Actions的配置要点</p>\n<figure class=\"highlight yaml\"><table><tr><td class=\"code\"><pre><span class=\"line\"><span class=\"attr\">on:</span> </span><br><span class=\"line\"> <span class=\"comment\"># push:</span></span><br><span class=\"line\"> <span class=\"comment\"># branches:</span></span><br><span class=\"line\"> <span class=\"comment\"># - pages # pages分支有push时触发</span></span><br><span class=\"line\"> <span class=\"attr\">watch:</span></span><br><span class=\"line\"> <span class=\"attr\">types:</span> [<span class=\"string\">started</span>] <span class=\"comment\"># 有人Star时触发</span></span><br></pre></td></tr></table></figure>\n<figure class=\"highlight yaml\"><table><tr><td class=\"code\"><pre><span class=\"line\"><span class=\"bullet\">-</span> <span class=\"attr\">name:</span> <span class=\"string\">Checkout</span> <span class=\"string\">Repository</span> <span class=\"string\">master</span> <span class=\"string\">branch</span></span><br><span class=\"line\"> <span class=\"attr\">uses:</span> <span class=\"string\">actions/checkout@master</span></span><br><span class=\"line\"> <span class=\"attr\">with:</span></span><br><span class=\"line\"> <span class=\"attr\">ref:</span> <span class=\"comment\"># 自行填写分支</span></span><br><span class=\"line\"></span><br><span class=\"line\"><span class=\"bullet\">-</span> <span class=\"attr\">name:</span> <span class=\"string\">Setup</span> <span class=\"string\">Node.js</span> <span class=\"number\">16.</span><span class=\"string\">x</span></span><br><span class=\"line\"> <span class=\"attr\">uses:</span> <span class=\"string\">actions/setup-node@master</span></span><br><span class=\"line\"> <span class=\"attr\">with:</span></span><br><span class=\"line\"> <span class=\"attr\">node-version:</span> <span class=\"string\">"16.x"</span></span><br><span class=\"line\"></span><br><span class=\"line\"><span class=\"bullet\">-</span> <span class=\"attr\">name:</span> <span class=\"string\">Setup</span> <span class=\"string\">Hexo</span> <span class=\"string\">Dependencies</span></span><br><span class=\"line\"> <span class=\"attr\">run:</span> <span class=\"string\">|</span></span><br><span class=\"line\"><span class=\"string\"> npm install hexo-cli -g</span></span><br><span class=\"line\"><span class=\"string\"> npm install</span></span><br><span class=\"line\"><span class=\"string\"></span></span><br><span class=\"line\"><span class=\"bullet\">-</span> <span class=\"attr\">name:</span> <span class=\"string\">Setup</span> <span class=\"string\">Deploy</span> <span class=\"string\">Key</span></span><br><span class=\"line\"> <span class=\"attr\">run:</span> <span class=\"string\">|</span></span><br><span class=\"line\"><span class=\"string\"> mkdir ~/.ssh/</span></span><br><span class=\"line\"><span class=\"string\"> cp ./.ssh/id_rsa ~/.ssh/id_rsa # 复制ssh密钥, 可以选择存储在环境变量中</span></span><br><span class=\"line\"><span class=\"string\"> cp ./.ssh/id_rsa.pub ~/.ssh/id_rsa.pub</span></span><br><span class=\"line\"><span class=\"string\"> cp ./.ssh/known_hosts ~/.ssh/known_hosts</span></span><br><span class=\"line\"><span class=\"string\"> sudo chmod 0600 ~/.ssh/id_rsa # 权限不这样设会报错</span></span><br><span class=\"line\"><span class=\"string\"> sudo chmod 0600 ~/.ssh/id_rsa.pub</span></span><br><span class=\"line\"><span class=\"string\"> sudo chmod 0600 ~/.ssh/known_hosts</span></span><br><span class=\"line\"><span class=\"string\"></span></span><br><span class=\"line\"><span class=\"bullet\">-</span> <span class=\"attr\">name:</span> <span class=\"string\">Setup</span> <span class=\"string\">Git</span> <span class=\"string\">Information</span></span><br><span class=\"line\"> <span class=\"attr\">run:</span> <span class=\"string\">|</span></span><br><span class=\"line\"><span class=\"string\"> git config --global user.name '' # 登录用户名</span></span><br><span class=\"line\"><span class=\"string\"> git config --global user.email '' # 登录邮箱</span></span><br><span class=\"line\"><span class=\"string\"></span></span><br><span class=\"line\"><span class=\"bullet\">-</span> <span class=\"attr\">name:</span> <span class=\"string\">Generate</span> <span class=\"string\">For</span> <span class=\"string\">FT</span></span><br><span class=\"line\"> <span class=\"attr\">run:</span> <span class=\"string\">|</span></span><br><span class=\"line\"><span class=\"string\"> node ft-pages.js</span></span><br><span class=\"line\"><span class=\"string\"> cp "./hexo-config.yml" ./_config.yml # 如果修改了主题配置文件也要一起复制</span></span><br><span class=\"line\"><span class=\"string\"></span></span><br><span class=\"line\"><span class=\"bullet\">-</span> <span class=\"attr\">name:</span> <span class=\"string\">Deploy</span> <span class=\"string\">Hexo</span> <span class=\"string\">For</span> <span class=\"string\">FT</span></span><br><span class=\"line\"> <span class=\"attr\">run:</span> <span class=\"string\">|</span></span><br><span class=\"line\"><span class=\"string\"> hexo clean</span></span><br><span class=\"line\"><span class=\"string\"> hexo generate</span></span><br><span class=\"line\"><span class=\"string\"> hexo deploy</span></span><br><span class=\"line\"><span class=\"string\"></span></span><br></pre></td></tr></table></figure>\n\n<p>晚安</p>\n","tags":["Blog","GitHub","GitHub Actions","Hexo"]},{"title":"圆梦导数杯T5一个不完整的证明","url":"//post/%E5%9C%86%E6%A2%A6%E5%AF%BC%E6%95%B0%E6%9D%AFT5%E4%B8%80%E4%B8%AA%E4%B8%8D%E5%AE%8C%E6%95%B4%E7%9A%84%E8%AF%81%E6%98%8E/","content":"<center></center>\n<span id=\"more\"></span>\n\n<p><img src=\"/file/%E5%9C%86%E6%A2%A6%E5%AF%BC%E6%95%B0%E6%9D%AFT5%E4%B8%80%E4%B8%AA%E4%B8%8D%E5%AE%8C%E6%95%B4%E7%9A%84%E8%AF%81%E6%98%8E/v2-4c9cbb3c1ff5da0a820d59a1992b8cd3_r.jpg\" alt=\"v2-4c9cbb3c1ff5da0a820d59a1992b8cd3_r.jpg\"></p>\n<p>已知函数$f(x)=a \\mathbb{e}^{x}-2x+\\ln{\\frac{x}{a}}(a>0)$, 且$f(x_{1})=f(x_{2})=0(1<x_{1}<x_{2})$. </p>\n<p>(I)求$a$的取值范围; </p>\n<p>(II)讨论函数$r(x)=f(x-x_{1})+f(x-x_{2})$的零点个数. </p>\n<hr>\n<p>目前没有找到什么能完整写完的思路</p>\n<hr>\n<p>(I)</p>\n<p>$f’(x)=a e^x+\\frac{1}{x}-2$</p>\n<p>令$g(x)=\\frac{2-\\frac{1}{x}}{e^{x}}$</p>\n<p>$g’(x)=\\frac{(2 x+1)(1-x)}{x^2e^{x}}$</p>\n<p>$x\\in (0, 1)$时, $g’(x)>0$, $g(x)$单调递增; </p>\n<p>$x\\in (1, +\\infty)$时, $g’(x)<0$, $g(x)$单调递减; </p>\n<p>$g(\\frac{1}{2})=0$</p>\n<p>$\\lim_{x\\to +\\infty}g(x)=0$</p>\n<p>$g(1)=\\frac{1}{e}$</p>\n<p>若$a\\in (0, \\frac{1}{e})$</p>\n<p>$g(x)=a$在$(\\frac{1}{2}, 1)$与$(1, +\\infty)$上各恰有一解, 不妨设为$x_{3}$, $x_{4}$, $(x_{3}<x_{4})$</p>\n<p>$x\\in (0, x_{3})$时, $g(x)<a$, $f’(x)>0$, $f(x)$单调递增; </p>\n<p>$x\\in (x_{3}, x_{4})$时, $g(x)>a$, $f’(x)<0$, $f(x)$单调递减; </p>\n<p>$x\\in (x_{4}, +\\infty)$时, $g(x)<a$, $f’(x)>0$, $f(x)$单调递增; </p>\n<p>$\\lim_{x\\to 0^{+}}f(x)=-\\infty$</p>\n<p>$f(1)=e a-\\ln a-2>0$</p>\n<p>$f(\\ln\\frac{1}{a})=1-\\ln\\frac{1}{a}+\\ln\\ln\\frac{1}{a}<0$</p>\n<p>$\\lim_{x\\to +\\infty}f(x)=+\\infty$</p>\n<p>$x_{3}<1<\\ln\\frac{1}{a}<x_{4}$</p>\n<p>$f(x)$在$(0, 1)$, $(1, \\ln\\frac{1}{a})$, $(\\ln\\frac{1}{a}, +\\infty)$上各有一根</p>\n<p>故$a\\in (0, \\frac{1}{e})$符合题意</p>\n<p>若$a\\in (\\frac{1}{e}, +\\infty)$</p>\n<p>$g(x)<a$对$x\\in (0, +\\infty)$恒成立, $f(x)>0$恒成立, $f(x)$单调递增, 至多有一根, 不符题意</p>\n<p>综上, $a\\in (0, \\frac{1}{e})$</p>\n<hr>\n<p>(II)</p>\n<p>注意到</p>\n<p>$r(x_{1}+x_{2})=f(x_{2})+f({x_1})=0$</p>\n<p>我们证明</p>\n<p>$r’(x_{1}+x_{2})=f’(x_{2})+f’({x_1})>0$</p>\n<p>及</p>\n<p>$f(x_{3})+f(x_{4})>0$</p>\n<p>即可</p>\n<p>对于后者, 我们设$\\frac{2-\\frac{1}{x_{4}}}{2-\\frac{1}{x_{3}}}=k$</p>\n<p>则</p>\n<p>$x_{3}=\\frac{1}{4} \\left(-2 \\ln (k)+\\frac{\\sqrt{k+4 \\ln (k) (k+(k-1) \\ln (k)+1)-1}}{\\sqrt{k-1}}+1\\right)$</p>\n<p>$x_{4}=\\frac{1}{4} \\left(2 \\ln (k)+\\frac{\\sqrt{k+4 \\ln (k) (k+(k-1) \\ln (k)+1)-1}}{\\sqrt{k-1}}+1\\right)$</p>\n<p>$x_{3}+x_{4}=\\frac{1}{2} \\left(\\frac{\\sqrt{k+4 \\ln (k) (k+(k-1) \\ln (k)+1)-1}}{\\sqrt{k-1}}+1\\right)$</p>\n<p>$x_{3}x_{4}=\\frac{1}{8} \\left(\\frac{2 (k+1) \\ln (k)}{k-1}+\\frac{\\sqrt{k+4 \\ln (k) (k+(k-1) \\ln (k)+1)-1}}{\\sqrt{k-1}}+1\\right)$</p>\n<p>$f(x_{3})+f(x_{4})=4+2\\ln(\\frac{x_{3}x_{4}}{4x_{3}x_{4}-2(x_{3}+x_{4})+1})-(x_{3}+x_{4})(1+\\frac{1}{x_{3}x_{4}})$</p>\n<p>我们可以得到一个关于$k$的函数</p>\n<p>由$Mathematica$, 得证…</p>\n","tags":["Maths","高中","高考"]},{"title":"数海导数闯关练","url":"//post/%E6%95%B0%E6%B5%B7%E5%AF%BC%E6%95%B0%E9%97%AF%E5%85%B3%E7%BB%83/","content":"<center></center>\n<span id=\"more\"></span>\n\n<p><a href=\"https://mp.weixin.qq.com/s?__biz=MzkyMjI2MzAyNA==&mid=2247484138&idx=1&sn=7404e371b456753505cb820625a47c66\">【原创试卷】数海导数闯关练</a></p>\n<p><img src=\"/file/%E6%95%B0%E6%B5%B7%E5%AF%BC%E6%95%B0%E9%97%AF%E5%85%B3%E7%BB%83/1.jpg\" alt=\"1.jpg\"></p>\n<p><img src=\"/file/%E6%95%B0%E6%B5%B7%E5%AF%BC%E6%95%B0%E9%97%AF%E5%85%B3%E7%BB%83/2.jpg\" alt=\"2.jpg\"></p>\n<p>持续更新ing…</p>\n<hr>\n"},{"title":"武汉市2023届高中毕业生四月调研考试数学T22","url":"//post/%E6%AD%A6%E6%B1%89%E5%B8%822023%E5%B1%8A%E9%AB%98%E4%B8%AD%E6%AF%95%E4%B8%9A%E7%94%9F%E5%9B%9B%E6%9C%88%E8%B0%83%E7%A0%94%E8%80%83%E8%AF%95%E6%95%B0%E5%AD%A6T22/","content":"<center></center>\n<span id=\"more\"></span>\n\n<script src=\"/file/GeoGebra/deployggb.js\"></script>\n\n<p>$22. (12分)$</p>\n<p>$已知函数f(x)=x\\ln{x}-\\frac{k}{x} , 其中k>0. $</p>\n<p>$(1)证明: f(x)恒有唯一零点; $</p>\n<p>$(2)记(1)中零点为x_{0} , 当0<k<\\frac{e}{2}时, 证明: f(x)图像上存在关于(x_{0}, 0)对称的两点. $</p>\n<hr>\n<p>(1)</p>\n<p>$g(x)=\\ln{x}-\\frac{k}{x^{2}}$</p>\n<p>$f(x)=0\\Leftrightarrow g(x)=0$</p>\n<p>$g’(x)=\\frac{1}{x}+\\frac{2k}{x^{3}}>0$</p>\n<p>因此$g(x)$在$\\mathbb{R}^{+}$上单调递增, 故$g(x)$在$\\mathbb{R}^{+}$上至多有一零点</p>\n<p>又</p>\n<p>$\\lim_{x \\to 0^{+}}g(x)=-\\infty$</p>\n<p>$\\lim_{x \\to +\\infty}g(x)=+\\infty$</p>\n<details class=\"note info no-icon\"><summary><p>取点</p>\n</summary>\n<p>$g(1)=-k<0$</p>\n<p>$g(e^k)=k-\\frac{k}{e^{2k}}>0$</p>\n<details class=\"note info no-icon\"><summary><p>Geogebra</p>\n</summary>\n<div>\n <center><div id=\"ggb-element1\"></div> </center>\n <script> \n var params1 = {\"appName\": \"graphing\", \"width\": 800, \"height\": 600, \"showToolBar\": true, \"showAlgebraInput\": true, \"showMenuBar\": true, \"filename\":\"/file/武汉市2023届高中毕业生四月调研考试数学T22/geogebra-export1.ggb\"};\n var applet1 = new GGBApplet(params1, true);\n window.addEventListener(\"load\", function() { \n applet1.setHTML5Codebase('/file/GeoGebra/HTML5/5.0/web3d/');\n applet1.inject('ggb-element1');\n });\n </script>\n</div>\n</details>\n</details>\n\n<p>且$g(x)$在$\\mathbb{R}^{+}$上连续</p>\n<p>故$g(x)$在$\\mathbb{R}^{+}$上至少有一零点</p>\n<p>综上$g(x)$在$\\mathbb{R}^{+}$上有唯一零点, 即$f(x)$有唯一零点</p>\n<hr>\n<p>(2)</p>\n<p>显然$1<x_{0}<e^{\\frac{1}{2}}$</p>\n<p>不妨设两点为</p>\n<p>$(x_{0}(1-t), f(x_{0}(1-t))), (x_{0}(1+t), f(x_{0}(1+t))), 0<t<1$</p>\n<p>则有</p>\n<p>$f(x_{0}(1-t))+f(x_{0}(1+t))=0$</p>\n<p>即</p>\n<p>$((x_{0}(1-t))\\ln{(x_{0}(1-t))}-\\frac{k}{x_{0}(1-t)}) + ((x_{0}(1+t))\\ln{(x_{0}(1+t))}-\\frac{k}{x_{0}(1+t)})=0$</p>\n<p>参变分离</p>\n<p>$\\frac{1-t^{2}}{t^{2}}((1-t)\\ln{(1-t)}+(1+t)\\ln{(1+t)})=2\\ln{x_{0}}$</p>\n<p>令$h(t)=\\frac{1-t^{2}}{t^{2}}((1-t)\\ln{(1-t)}+(1+t)\\ln{(1+t)})$</p>\n<p>$\\lim_{t \\to 0^{+}}h(t)=1>2\\ln{x_{0}}$</p>\n<p>$\\lim_{t \\to 1^{-}}h(t)=0<2\\ln{x_{0}}$</p>\n<details class=\"note info no-icon\"><summary><p>Geogebra</p>\n</summary>\n<div>\n <center><div id=\"ggb-element2\"></div> </center>\n <script> \n var params2 = {\"appName\": \"graphing\", \"width\": 800, \"height\": 600, \"showToolBar\": true, \"showAlgebraInput\": true, \"showMenuBar\": true, \"filename\":\"/file/武汉市2023届高中毕业生四月调研考试数学T22/geogebra-export2.ggb\"};\n var applet2 = new GGBApplet(params2, true);\n window.addEventListener(\"load\", function() { \n applet2.setHTML5Codebase('/file/GeoGebra/HTML5/5.0/web3d/');\n applet2.inject('ggb-element2');\n });\n </script>\n</div>\n</details>\n\n<p>又$h(t)$在$(0,1)$上连续</p>\n<p>由连续介值定理得证</p>\n<details class=\"note info no-icon\"><summary><p>取点</p>\n</summary>\n<p>由不等式$\\ln{x}\\leq x-1$</p>\n<p>$h(t)=\\frac{1-t^{2}}{t^{2}}((1-t)\\ln{(1-t)}+(1+t)\\ln{(1+t)})\\leq 2(1-t^{2})$</p>\n<details class=\"note info no-icon\"><summary><p>Geogebra</p>\n</summary>\n<div>\n <center><div id=\"ggb-element3\"></div> </center>\n <script> \n var params3 = {\"appName\": \"graphing\", \"width\": 800, \"height\": 600, \"showToolBar\": true, \"showAlgebraInput\": true, \"showMenuBar\": true, \"filename\":\"/file/武汉市2023届高中毕业生四月调研考试数学T22/geogebra-export3.ggb\"};\n var applet3 = new GGBApplet(params3, true);\n window.addEventListener(\"load\", function() { \n applet3.setHTML5Codebase('/file/GeoGebra/HTML5/5.0/web3d/');\n applet3.inject('ggb-element3');\n });\n </script>\n</div>\n</details>\n\n<p>$\\forall t\\in [\\sqrt{1-\\ln{x_{0}}}, 1), h(t)\\leq 2\\ln{x_{0}}$</p>\n<p>又</p>\n<p>$h(t)=\\frac{1-t^{2}}{t^{2}}((1-t)\\ln{(1-t)}+(1+t)\\ln{(1+t)})=\\frac{1-t^{2}}{t^{2}}\\ln{(1-t^{2})}+\\frac{1-t^{2}}{t}\\ln{(\\frac{1+t}{1-t})}$</p>\n<details class=\"note info no-icon\"><summary><p>Geogebra</p>\n</summary>\n<div>\n <center><div id=\"ggb-element4\"></div> </center>\n <script> \n var params4 = {\"appName\": \"graphing\", \"width\": 800, \"height\": 600, \"showToolBar\": true, \"showAlgebraInput\": true, \"showMenuBar\": true, \"filename\":\"/file/武汉市2023届高中毕业生四月调研考试数学T22/geogebra-export4.ggb\"};\n var applet4 = new GGBApplet(params4, true);\n window.addEventListener(\"load\", function() { \n applet4.setHTML5Codebase('/file/GeoGebra/HTML5/5.0/web3d/');\n applet4.inject('ggb-element4');\n });\n </script>\n</div>\n</details>\n\n<p>由不等式$\\ln{x}\\geq 1-\\frac{1}{x}$</p>\n<p>$h(t)=\\frac{1-t^{2}}{t^{2}}\\ln{(1-t^{2})}+\\frac{1-t^{2}}{t}\\ln{(\\frac{1+t}{1-t})}\\geq 1-2t$</p>\n<details class=\"note info no-icon\"><summary><p>Geogebra</p>\n</summary>\n<div>\n <center><div id=\"ggb-element5\"></div> </center>\n <script> \n var params5 = {\"appName\": \"graphing\", \"width\": 800, \"height\": 600, \"showToolBar\": true, \"showAlgebraInput\": true, \"showMenuBar\": true, \"filename\":\"/file/武汉市2023届高中毕业生四月调研考试数学T22/geogebra-export5.ggb\"};\n var applet5 = new GGBApplet(params5, true);\n window.addEventListener(\"load\", function() { \n applet5.setHTML5Codebase('/file/GeoGebra/HTML5/5.0/web3d/');\n applet5.inject('ggb-element5');\n });\n </script>\n</div>\n</details>\n\n<p>$\\forall t\\in (0, \\frac{1-2\\ln{x_{0}}}{2}], h(t)\\geq 2\\ln{x_{0}}$</p>\n<p>依此取点即可</p>\n</details>","tags":["Maths","高中","高考"]},{"title":"湖南省2023届高三九校联盟第二次联考数学T22","url":"//post/%E6%B9%96%E5%8D%97%E7%9C%812023%E5%B1%8A%E9%AB%98%E4%B8%89%E4%B9%9D%E6%A0%A1%E8%81%94%E7%9B%9F%E7%AC%AC%E4%BA%8C%E6%AC%A1%E8%81%94%E8%80%83%E6%95%B0%E5%AD%A6T22/","content":"<center></center>\n<span id=\"more\"></span>\n\n\n<p>$22. (本小题满分 12 分)$</p>\n<p>$已知f(x)=\\frac{1}{2}x^2-x-a\\ln(x-a), a\\in R$</p>\n<p>$(1)判断函数f(x)的单调性; $</p>\n<p>$(2)若 x_1, x_2是函数 g(x)=f(x+a)-a(x+\\frac{1}{2}a)的两个极值点, 且 x_1< x_2, 求证: 0 < f (x_1)-f(x_2) < \\frac{1}{2}. $</p>\n<hr>\n<p>(1)</p>\n<p>自己做</p>\n<hr>\n<p>(2)</p>\n<p>$f(x)=\\frac{1}{2}x^2-x-a\\ln(x-a)$</p>\n<p>$g(x)=\\frac{1}{2}x^2-x-a\\ln(x)$</p>\n<p>$f’(x)=\\frac{x(x-(a+1))}{x-a}$</p>\n<p>$g’(x)=\\frac{(x-\\frac{1+\\sqrt{1+4a}}{2})(x-\\frac{1-\\sqrt{1+4a}}{2})}{x}$</p>\n<p>此题中$x_1$与$x_2$均为$g(x)$极值点, 但是最后需证的不等式中却没有$g(x)$的身影, 出题人不会就只是想考一下我们多项式化简吧(或者还有韦达), 不会吧不会吧</p>\n<p>题目改为$0 < f (\\frac{1-\\sqrt{1+4a}}{2})-f(\\frac{1+\\sqrt{1+4a}}{2}) < \\frac{1}{2}$不是更好?</p>\n<p>这个不等式就非常好, 显然告诉我们不可能通过展开计算, 当然不在考场上的你可以试试</p>\n<p>中间的看不出眉目, 不妨看向两边</p>\n<p>$0, \\frac{1}{2}$都是常数</p>\n<p>这意味着原不等式其实非常松, 我们可以大胆放缩</p>\n<p>怎么放?</p>\n<p>$0 < f (\\frac{1-\\sqrt{1+4a}}{2})-f(\\frac{1+\\sqrt{1+4a}}{2}) < \\frac{1}{2}$这个式子为什么恶心</p>\n<p>恶心在其中的$\\frac{1\\pm\\sqrt{1+4a}}{2}$, 这不得给它整掉?</p>\n<p>来小亮给他整个活</p>\n<p>还记得(1)问中$f(x)$有一个单调递减区间$(0, a + 1)$吗?</p>\n<p>再回到(2)问, $x_1< x_2, 求证: 0 < f(x_1)-f(x_2)$</p>\n<p>看不明白? 移一下项$x_1< x_2, 求证: f(x_1) < f(x_2)$</p>\n<p>等一下, 如果, 我是说如果啊, 如果$x_1,x_2\\in(0,a+1)$, 这个式子是不是显然成立</p>\n<p>然后你真的发现$\\frac{1\\pm\\sqrt{1+4a}}{2}\\in(0,a+1)$</p>\n<p>那剩下什么, 放啊!</p>\n<p>$f(x_1)<f(0), f(x_2)>f(a+1)\\Rightarrow f(x_1)-f(x_2)<f(0)-f(a+1)$(其实更紧一点的是$f(x_1)<f(-a), f(x_2)>f(a+1)\\Rightarrow f(x_1)-f(x_2)<f(-a)-f(a+1)$但本题均够用)</p>\n<p>我们只需证$f(0)-f(a+1)<\\frac{1}{2}$即可</p>\n<p>自己写吧</p>\n<hr>\n<p>总结, 本题实际上观察不等式中间与两边复杂度的不对等性可以很明确的给出思路</p>\n<ol>\n<li>利用单调性</li>\n<li>寻找$x_1$与$x_2$的上下界</li>\n</ol>\n","tags":["Maths","高中","2023"]}]