Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2014年浙江全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "充分非必要条件 " } ], [ { "aoVal": "B", "content": "必要非充分条件 " } ], [ { "aoVal": "C", "content": "充分必要条件 " } ], [ { "aoVal": "D", "content": "既非充分又非必要条件 " } ] ]

[ "竞赛->知识点->逻辑->常用逻辑用语" ]

[ "在$$\\triangle ABC$$中$$\\angle A\\ne \\angle B\\Leftrightarrow \\sin \\angle A\\ne \\sin \\angle B$$． " ]

[ "2017年AMC10竞赛A第13题" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "$$10$$ " } ] ]

[ "美国AMC10/12->Knowledge Point->Combination->Reasoning->Recurrence and Recursion" ]

[ "A patten starts to emerge as the function is continued. The repeating patten is $$0$$, $$1$$, $$1$$, $$2$$, $$0$$, $$2$$, $$2$$, $$1$$ $$\\ldots$$. The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which is $$(\\rm D) 9$$. " ]

[ "2012年黑龙江全国高中数学联赛竞赛初赛第6题5分", "2016~2017学年6月陕西西安莲湖区西安市第一中学高二下学期月考理科第9题3分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ] ]

[ "课内体系->素养->逻辑推理", "课内体系->知识点->计数原理->排列与组合->组合->隔板法" ]

[ "根据题意， 先在编号为$$2$$、$$3$$的三个盒子中分别放入$$1$$、$$2$$个小球，编号为$$1$$的盒子里不放， 再将剩下的$$7$$个小球放入$$3$$个盒子里，每个盒子里至少一个， 分析可得，$$7$$个小球排好，有$$6$$个空位，在$$6$$个空位中任选$$2$$个， 插入挡板，共$$\\text{C}_{6}^{2}=15$$种装法， 即可得符合题目要求的装法共有$$15$$种， 故选：$$\\text{C}$$． " ]

[ "2018年陕西全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{\\sqrt{2}}{6}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{\\sqrt{2}}{12}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{\\sqrt{3}}{6}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{\\sqrt{3}}{12}$$ " } ] ]

[ "竞赛->知识点->立体几何与空间向量" ]

[ "解：四面体以等腰直角$$\\triangle ABD$$为底，高为$$\\frac{\\sqrt{2}}{2}$$，故体积为$$\\frac{1}{3}\\times \\frac{\\sqrt{2}}{2}\\times \\frac{1}{2}=\\frac{\\sqrt{2}}{12}$$． " ]

[ "2015年黑龙江全国高中数学联赛竞赛初赛第7题5分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "竞赛->知识点->不等式->线性规划", "竞赛->知识点->不等式->几个重要的不等式->均值" ]

[ "$$ab+4=8$$，$$ab=4$$，所以$$a+b\\geqslant 2\\sqrt{ab}=4$$． " ]

[ "2011年浙江全国高中数学联赛竞赛初赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$2\\sin \\theta $$ " } ], [ { "aoVal": "B", "content": "$$-2\\sin \\theta $$ " } ], [ { "aoVal": "C", "content": "$$-2\\cos \\theta $$ " } ], [ { "aoVal": "D", "content": "$$2\\cos \\theta $$ " } ] ]

[ "竞赛->知识点->三角函数->三角恒等变换" ]

[ "因为$$\\theta \\in [\\frac{5 \\pi }{4},\\frac{3 \\pi }{2}]$$，所以$$\\sqrt{1-\\sin 2\\theta }-\\sqrt{1+\\sin 2\\theta }$$ $$=\\left\\textbar{} \\cos \\theta -\\sin \\theta \\right\\textbar-\\left\\textbar{} \\cos \\theta +\\sin \\theta \\right\\textbar=2\\cos \\theta $$． " ]

[ "2018年吉林全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "因为 $$y=\\frac{{{({{2}^{x}}+1)}^{2}}}{{{2}^{x}}\\cdot x}=\\left( {{2}^{x}}+\\frac{1}{{{2}^{x}}}+2 \\right)\\cdot \\frac{1}{x}$$ 为$$[-\\infty ,0)\\cup (0,+\\infty ]$$上的奇函 数，故 $$f(x)=\\frac{{{({{2}^{x}}+1)}^{2}}}{{{2}^{x}}\\cdot x}+1$$ 的图象关于点$$(0,1)$$对称，所以$$M+N=2$$． 故选$$\\text{B}$$． " ]

[ "2012年辽宁全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{3\\sqrt{2}}{4}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{9}{8}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3\\sqrt{5}}{5}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{3\\sqrt{2}}{2}$$ " } ] ]

[ "竞赛->知识点->复数与平面向量->平面向量的概念与运算", "竞赛->知识点->解析几何->双曲线" ]

[ "将$$A\\left( c,\\frac{bc}{a} \\right), B\\left( c, -\\frac{bc}{a} \\right)$$代入$$\\overrightarrow{OP}=m\\overrightarrow{OA}+n\\overrightarrow{OB}$$， 得$$P\\left( \\left( m+n \\right)c,\\left( m-n \\right)\\frac{bc}{a} \\right)$$，代入双曲线方程，得 $$\\frac{{{\\left( m+n \\right)}^{2}}{{c}^{2}}}{{{a}^{2}}}-\\frac{{{\\left( m-n \\right)}^{2}}{{b}^{2}}{{c}^{2}}}{{{a}^{2}}{{b}^{2}}}=1$$， 化简得$$4mn{{c}^{2}}={{a}^{2}}$$， $$e=\\frac{c}{a}=\\frac{1}{\\sqrt{4mn}}=\\sqrt{\\frac{9}{8}}=\\frac{3\\sqrt{2}}{4}$$． " ]

[ "1991年全国高中数学联赛竞赛一试第4题" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "若$$3+\\alpha $$是$$f(x)=0$$的一个根，则由已知 $$f\\left( 3-\\alpha \\right)=f\\left( 3+\\alpha \\right)=0$$ 即$$3-\\alpha $$也是一个根．因此可设方程$$f\\left( x \\right)=0$$的六个根为 $$3\\mathsf{\\pm }{{\\alpha }_{1}}$$，$$3\\mathsf{\\pm }{{\\alpha }_{2}}$$，$$3\\mathsf{\\pm }{{\\alpha }_{3}}$$． 于是它们的行等于$$18$$． " ]

[ "2013年四川全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "最多有一个 " } ], [ { "aoVal": "B", "content": "最多有两个 " } ], [ { "aoVal": "C", "content": "最多有四个 " } ], [ { "aoVal": "D", "content": "可以有无穷多个 " } ] ]

[ "竞赛->知识点->解析几何->圆与方程", "竞赛->知识点->逻辑->逻辑推理" ]

[ "设$$\\left( a,b \\right),\\left( c,d \\right)$$为这个圆上的两个有理点，则 $${{\\left( a-1 \\right)}^{2}}+{{\\left( b-\\sqrt{2} \\right)}^{2}}={{\\left( c-1 \\right)}^{2}}+{{\\left( d-\\sqrt{2} \\right)}^{2}}$$， 整理得$$2\\sqrt{2}\\left( b-d \\right)={{a}^{2}}+{{b}^{2}}-{{c}^{2}}-{{d}^{2}}+2\\left( c-a \\right)$$. 显然等式右边是有理数，因此当且仅当$$b=d$$时，等式的左边是有理数，此时$$a\\ne c,a+c=2$$.由此得这两个有理点关于直线$$x=1$$对称，而圆上与$$\\left( a,b \\right)$$关于$$x=1$$对称的点最多只有一个. " ]

[ "高一上学期单元测试《二次与对勾函数》竞赛第7题" ]

[ [ { "aoVal": "A", "content": "$$A\\subseteq B\\Leftrightarrow {{f}_{A}}(x)\\leqslant {{f}_{B}}(x)，x\\in X$$ " } ], [ { "aoVal": "B", "content": "$${{f}_{{{C}_{X}}A}}(x)=1-{{f}_{A}}(x)，x\\in X$$ " } ], [ { "aoVal": "C", "content": "$${{f}_{A\\cap B}}(x)={{f}_{A}}(x){{f}_{B}}(x)，x\\in X$$ " } ], [ { "aoVal": "D", "content": "$${{f}_{A\\cup B}}(x)={{f}_{A}}(x)+{{f}_{B}}(x)，x\\in X$$ " } ] ]

[ "竞赛->知识点->函数->二次函数", "竞赛->知识点->函数->函数的图像与性质" ]

[ "若$$x\\in A\\cap B$$，则$${{f}_{A\\cup B}}(x)={{f}_{A}}(x)={{f}_{B}}(x)=1$$，故$\\text{D}$不正确． " ]

[ "2012年吉林全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$7$$个 " } ], [ { "aoVal": "B", "content": "$$8$$个 " } ], [ { "aoVal": "C", "content": "$$9$$个 " } ], [ { "aoVal": "D", "content": "$$10$$个 " } ] ]

[ "竞赛->知识点->函数->函数的概念" ]

[ "$$1\\times 3\\times 3=9$$． " ]

[ "2009年浙江全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$${{S}_{10}}$$ " } ], [ { "aoVal": "B", "content": "$${{S}_{11}}$$ " } ], [ { "aoVal": "C", "content": "$${{S}_{20}}$$ " } ], [ { "aoVal": "D", "content": "$${{S}_{21}}$$ " } ] ]

[ "知识标签->素养->数学运算", "知识标签->题型->数列->等差数列->等差数列的性质问题->求等差数列前n项和的最值  ", "知识标签->知识点->数列->等差数列->等差数列的前n项和", "知识标签->知识点->数列->等差数列->等差数列的概念通项公式", "知识标签->知识点->数列->等差数列->等差数列的性质及应用" ]

[ "由$$3{{a}_{8}}=5{{a}_{13}}$$可得$${{a}_{20}}\\textgreater0$$，$${{a}_{21}} ~\\textless{} ~0$$，∴$${{S}_{20}}$$最大． " ]

[ "2014年AMC12竞赛A第21题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{ \\log 2015}{ \\log 2014}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{ \\log 2014}{ \\log 2013}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2014}{2013}$$ " } ], [ { "aoVal": "E", "content": "$$2014^{\\frac{1}{2014}}$$ " } ] ]

[ "美国AMC10/12->Knowledge Point->Number Theory->Gaussian Function", "课内体系->知识点->基本初等函数->对数的概念及其运算->对数的运算" ]

[ "记$\\left { x\\right }=x-\\left\\lfloor x\\right\\rfloor\\textgreater0$. 则$f(x)\\leq 1\\Leftrightarrow 2014^{\\left {x\\right }}-1\\leq \\frac{1}{\\left\\lfloor x\\right\\rfloor}$ $\\Leftrightarrow \\left {x\\right }\\leq\\log_{2014}{\\frac{\\left\\lfloor x\\right\\rfloor+1}{\\left\\lfloor x\\right\\rfloor}}$ 于是单个满足$f(x)\\leq 1$的区间长度为$\\log_{2014}{\\frac{\\left\\lfloor x\\right\\rfloor+1}{\\left\\lfloor x\\right\\rfloor}}$ 区间总长度$=\\sum_{x=1}^{2013}\\log_{2014}{\\frac{\\left\\lfloor x\\right\\rfloor+1}{\\left\\lfloor x\\right\\rfloor}}=\\log_{2014}{\\frac{2}{1}\\times\\frac{3}{2}\\times\\cdots\\times\\frac{2014}{2013}}=\\log_{2014}{2014}=1$, 选$$\\text{A}$$. " ]

[ "2012年天津全国高中数学联赛竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$35$$ " } ], [ { "aoVal": "C", "content": "$$34$$ " } ], [ { "aoVal": "D", "content": "$$33$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "$${{a}_{n}}=\\begin{cases}{{S}_{1}},n=1 {{S}_{n}}-{{S}_{n-1}},n\\geqslant 2 \\end{cases}$$当$$n\\textgreater1$$时，$${{a}_{n}}={{S}_{\\text{n}}}-{{S}_{\\text{n-1}}}=2\\text{n}-3$$，因此$${{a}_{3}}+{{a}_{17}}=34$$． " ]

[ "2019年吉林全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{4}{5}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{3}{4}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2}{3}$$ " } ] ]

[ "竞赛->知识点->复数与平面向量->复数的概念与运算" ]

[ "注意到， $$\\left\\textbar{} \\overline{z}+\\frac{1}{z} \\right\\textbar=\\left\\textbar{} \\frac{\\overline{z}z+1}{z} \\right\\textbar=\\frac{{{\\left\\textbar{} z \\right\\textbar}^{2}}+1}{\\left\\textbar{} z \\right\\textbar}=\\frac{5}{2}$$， 解得$$\\left\\textbar{} z \\right\\textbar=2$$ （舍去）或$$\\left\\textbar{} z \\right\\textbar=\\frac{1}{2}$$． 故选$$\\text{C}$$． " ]

[ "2022年江苏徐州贾汪区江苏省贾汪中学高一竞赛（下学期春季）第8题" ]

[ [ { "aoVal": "A", "content": "$\\left[ \\frac{2}{3},\\frac{4}{3} \\right]$ " } ], [ { "aoVal": "B", "content": "$\\left[ \\frac{2}{3},\\frac{3}{4} \\right]$ " } ], [ { "aoVal": "C", "content": "$\\left( 0,\\frac{2}{3} \\right]$ " } ], [ { "aoVal": "D", "content": "$\\left( 0,\\frac{3}{2} \\right]$ " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 由题意可得$t=\\omega x+\\frac{\\text{ }\\pi\\text{ }}{\\text{6}}\\in (\\frac{\\text{ }\\pi\\text{ }\\omega }{2}+\\frac{\\text{ }\\pi\\text{ }}{\\text{6}},\\text{ }\\pi\\text{ }\\omega +\\frac{\\text{ }\\pi\\text{ }}{\\text{6}})$，再根据$y=\\sin t$的单调区间，列出不等式组求解即可.\\\\ 【详解】\\\\ 解：因为$\\omega \\textgreater0$，$x\\in \\left( \\frac{\\text{ }\\pi\\text{ }}{\\text{2}}\\text{, }\\pi\\text{ } \\right)$，\\\\ 所以$t=\\omega x+\\frac{\\text{ }\\pi\\text{ }}{\\text{6}}\\in (\\frac{\\text{ }\\pi\\text{ }\\omega }{2}+\\frac{\\text{ }\\pi\\text{ }}{\\text{6}},\\text{ }\\pi\\text{ }\\omega +\\frac{\\text{ }\\pi\\text{ }}{\\text{6}})$，\\\\ 又因为$y=\\sin t$在$(\\frac{\\text{ }\\pi\\text{ }}{2}+2k\\text{ }\\pi\\text{ },\\frac{3\\text{ }\\pi\\text{ }}{2}+2k\\text{ }\\pi\\text{ }),k\\in \\text{Z}$上单调递减，\\\\ 所以$\\text{ } {\\text{ }\\begin{array}{*{35}{l}} \\frac{\\text{ }\\pi\\text{ }\\omega }{2}+\\frac{\\text{ }\\pi\\text{ }}{\\text{6}}\\ge \\frac{\\text{ }\\pi\\text{ }}{2}+2k\\text{ }\\pi\\text{ } \\text{ }\\pi\\text{ }\\omega +\\frac{\\text{ }\\pi\\text{ }}{\\text{6}}\\le \\frac{3\\text{ }\\pi\\text{ }}{2}+2k\\text{ }\\pi\\text{ } \\end{array}\\text{ }$，$k\\in \\text{Z}$,\\\\ 解得：$\\frac{2}{3}+2k\\text{ }\\pi\\text{ }\\le \\omega \\le \\frac{4}{3}+2k\\text{ }\\pi\\text{ },k\\in \\text{Z}$.\\\\ 当$k=0$时，$\\frac{2}{3}\\le \\omega \\le \\frac{4}{3}$.\\\\ 故选：A " ]

[ "2017年AMC10竞赛B第14题" ]

[ [ { "aoVal": "A", "content": "$$\\frac 15$$ " } ], [ { "aoVal": "B", "content": "$$\\frac 25$$ " } ], [ { "aoVal": "C", "content": "$$\\frac 35$$ " } ], [ { "aoVal": "D", "content": "$$\\frac 45$$ " } ], [ { "aoVal": "E", "content": "$$1$$ " } ] ]

[ "美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Geometric Models of Probabilities" ]

[ "By Fermat\\textquotesingle s Little Theorem, $$N^{16}=\\left( N^{4}\\right)^{4}\\equiv1$$(mod $$5$$), when $$N$$ is relatively prime to $$5$$. Hence, this happens with probability $$\\frac 45$$. Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits $$0-9$$. The pattern for $$0$$ is $$0$$, no matter what power, so $$0$$ doesn\\textquotesingle t work. Likewise, the pattern for $$5$$ is always $$5$$. Doing the same for the rest of the digits, we find that the units digits of $$1^{16}$$,$$2^{16}$$,$$3^{16}$$,$$4^{16}$$,$$6^{16}$$,$$7^{16}$$,$$8^{16}$$ and $$9^{16}$$ all have the remainder of $$1$$ when divided by $$5$$, so $$\\frac 45$$. We can use modular arithmetic for each residue of $$n$$ (mod $$5$$). If $$n\\equiv0$$ (mod $$5$$), then $$n^{16}\\equiv0^{16}\\equiv0$$(mod $$5$$), If $$n\\equiv1$$ (mod $$5$$), then $$n^{16}\\equiv1^{16}\\equiv0$$(mod $$5$$), If $$n\\equiv2$$ (mod $$5$$), then $$n^{16}\\equiv\\left( n^{2}\\right)^{8}\\equiv\\left( 2^{2}\\right)^{8}\\equiv4^{8}\\equiv\\left( -1\\right)^{8}\\equiv1$$ (mod $$5$$). If $$n\\equiv3$$ (mod $$5$$), then $$n^{16}\\equiv\\left( n^{2}\\right)^{8}\\equiv\\left( 3^{2}\\right)^{8}\\equiv9^{8}\\equiv\\left( -1\\right)^{8}\\equiv1$$ (mod $$5$$). If $$n\\equiv4$$ (mod $$5$$), then $$n^{16}\\equiv4^{16}\\equiv\\left( -1\\right)^{16}=1$$ (mod $$5$$). In $$4$$ out of the $$5$$ cases, the result was $$1$$(mod $$5$$), and since each case occurs equally as $$2020\\equiv0$$ (mod $$5$$), the answer is $$\\frac 45$$. " ]

[ "2008年黑龙江全国高中数学联赛竞赛初赛第12题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{5}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{25}$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "竞赛->知识点->不等式->几个重要的不等式->均值" ]

[ "因为$$1\\leqslant 1+{{\\cos }^{2}}(2x+3y-1)\\leqslant 2$$， 且$$\\frac{{{x}^{2}}+{{y}^{2}}+2(x+1)(1-y)}{x-y+1}=\\frac{1}{x-y+1}+x-y+1\\geqslant 2$$， 所以$$\\begin{cases}x-y+1=1 2x+3y-1=k \\pi ~\\end{cases}$$， 即$$x=y=\\frac{k \\pi +1}{5}$$， 其中$$k\\in Z$$．所以$$xy$$的最小值为$${{\\left( \\frac{1}{5} \\right)}^{2}}=\\frac{1}{25}$$．故选$$\\text{B}$$． " ]

[ "2009年第二十届全国希望杯高二竞赛复赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$132$$ " } ], [ { "aoVal": "B", "content": "$$138$$ " } ], [ { "aoVal": "C", "content": "$$252$$ " } ], [ { "aoVal": "D", "content": "$$258$$ " } ] ]

[ "竞赛->知识点->解析几何->椭圆", "竞赛->知识点->解析几何->直线与圆锥曲线" ]

[ "过两个焦点的弦为长轴，长度是$$12$$．仅过左焦点的好弦中，长度最长为$$11$$，最短为$$3$$（垂直于长轴的焦点弦），且长度为$$45\\cdots 11$$的弦均有$$2$$条，故只过左焦点的好弦长度之和为$$3+2\\times (4+5+\\cdots +11)=123$$，因此所有好弦的长度之和是$$123\\times 2+12=258$$． " ]

[ "2008年天津全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{4}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "2 " } ] ]

[ "竞赛->知识点->函数->基本初等函数" ]

[ "方程$${{a}^{x}}+2x-4=0$$等价于$$\\frac{{{a}^{x}}}{2}=2-x$$， 其根即为$$y=\\frac{{{a}^{x}}}{2}$$与$$y=2-x$$的交点的横坐标． $${{\\log }_{a}}^{2x}+x-2=0$$等价于$${{\\log }_{a}}^{2x}=2-x$$， 其根即为$$y={{\\log }_{a}}^{2x}$$与$$y=2-x$$的交点的横坐标． 因为$$y=\\frac{{{a}^{x}}}{2}$$与$$y={{\\log }_{a}}^{2x}$$互为反函数， 所以此它们的图像关于$$y=x$$对称， 因此所有根的算术平均就是$$y=x$$与$$y=2-x$$交点的横坐标$$1$$． 故选$$\\text{C}$$． " ]

[ "2010年河南全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$76$$种 " } ], [ { "aoVal": "B", "content": "$$100$$种 " } ], [ { "aoVal": "C", "content": "$$132$$种 " } ], [ { "aoVal": "D", "content": "$$150$$种 " } ] ]

[ "课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->知识点->计数原理->排列与组合->组合->分组分配法", "课内体系->知识点->计数原理->排列与组合->排列组合综合" ]

[ "分两步：第一步，分配甲，有${\\rm{A}}_{\\rm{2}}^{\\rm{1}}$种方法，设 甲被分配到$$B$$宿舍； 第二步，分配其余$$4$$名同学，分两类：第一类，其余$$4$$名 同学都不分配到$$B$$宿舍，有${\\rm{C}}_{\\rm{4}}^{\\rm{3}}{\\rm{A}}_{\\rm{2}}^{\\rm{2}} + {\\rm{C}}_{\\rm{4}}^{\\rm{1}}{\\rm{C}}_{\\rm{2}}^{\\rm{2}}$种方法； 第二类，其余$$4$$名同学有人分配到$$B$$宿舍，有${\\rm{C}}_{\\rm{4}}^{\\rm{2}}{\\rm{A}}_{\\rm{3}}^{\\rm{3}}$种 方法．故分配其余$$4$$名同学有${\\rm{C}}_{\\rm{3}}^{\\rm{2}}{\\rm{A}}_{\\rm{2}}^{\\rm{2}} + {\\rm{C}}_{\\rm{4}}^{\\rm{2}}{\\rm{C}}_{\\rm{2}}^{\\rm{2}} + {\\rm{C}}_{\\rm{4}}^{\\rm{2}}{\\rm{C}}_{\\rm{3}}^{\\rm{3}}$种方法 所以不同的分配方案种数是$${N = {\\rm{A}}_{\\rm{2}}^{\\rm{1}}{\\rm{(C}}_{\\rm{4}}^{\\rm{3}}{\\rm{A}}_{\\rm{2}}^{\\rm{2}} + {\\rm{C}}_{\\rm{4}}^{\\rm{2}}{\\rm{C}}_{\\rm{2}}^{\\rm{2}} + {\\rm{C}}_{\\rm{4}}^{\\rm{2}}{\\rm{A}}_{\\rm{3}}^{\\rm{3}}{\\rm{)}} = 100}$$．故选$${\\rm{B}}$$． " ]

[ "2009年浙江全国高中数学联赛竞赛初赛第3题5分", "2017~2018学年6月河北石家庄辛集市河北辛集中学高一下学期月考第14题5分" ]

[ [ { "aoVal": "A", "content": "$${{S}_{10}}$$ " } ], [ { "aoVal": "B", "content": "$${{S}_{11}}$$ " } ], [ { "aoVal": "C", "content": "$${{S}_{20}}$$ " } ], [ { "aoVal": "D", "content": "$${{S}_{21}}$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->知识点->数列->等差数列->等差数列的前n项和->求等差数列前n项和的最值  " ]

[ "由$$3{{a}_{8}}=5{{a}_{13}}$$可得$${{a}_{20}}\\textgreater0$$，$${{a}_{21}} ~\\textless{} ~0$$，∴$${{S}_{20}}$$最大． " ]

[ "2023年江苏连云港灌南县灌南县中学高三竞赛（下学期3月解题能力竞赛）第7题" ]

[ [ { "aoVal": "A", "content": "$p=1$ " } ], [ { "aoVal": "B", "content": "$p=2$ " } ], [ { "aoVal": "C", "content": "$p=3$ " } ], [ { "aoVal": "D", "content": "$p=4$ " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 根据数列的递推公式得出$b_{n}=\\frac{a_{n}}{3^{n}}=\\frac{n}{3^{n}}$，然后根据等差数列的性质进项求解即可得出结果.\\\\ 【详解】\\\\ 数列$\\left {a_{n}\\right }$满足$a_{1}=1$，$a_{n}a_{n+1}=2S_{n}$，\\\\ 当$n=1$时，$a_{1}a_{2}=2S_{1}=2a_{1}$，解得：$a_{2}=2$；\\\\ 当$n\\geq 2$时，$2a_{n}=2(S_{n}-S_{n-1})=a_{n}(a_{n+1}-a_{n-1})$，\\\\ 因为$a_{n}\\neq 0$，所以$a_{n+1}-a_{n-1}=2$，所以数列$\\left {a_{n}\\right }$是首项为1，公差为1的等差数列，\\\\ 所以$a_{n}=1+(n-1)=n$，$b_{n}=\\frac{a_{n}}{3^{n}}=\\frac{n}{3^{n}}$，\\\\ 若存在正整数$p,q\\left(p\\textless{} q\\right)$，使得$b_{1}$，$b_{p}$，$b_{q}$成等差数列，\\\\ 则$2b_{p}=b_{1}+b_{q}$，所以$\\frac{2p}{3^{p}}=\\frac{1}{3}+\\frac{q}{3^{q}}$~~~~~~~~①\\\\ 因为数列$ {b_{n} }$是单调递减数列，\\\\ 当$p=1$时，由$\\frac{2}{3}=\\frac{1}{3}+\\frac{q}{3^{q}}$，解得：$q=1$，舍去；\\\\ 当$2\\leq p\\textless{} q$时，则$\\frac{1}{3}\\geq \\frac{p-1}{3^{p-1}}$，$\\frac{p-1}{3^{p-1}}-\\frac{2p}{3^{p}}=\\frac{p-3}{3^{p}}$；\\\\ 当$3\\leq p$时，$\\frac{1}{3}\\geq \\frac{p-1}{3^{p-1}}\\geq \\frac{2p}{3^{p}}$，$\\frac{q}{3^{q}}\\textgreater{} 0$，所以$\\frac{2p}{3^{p}}\\textless{} \\frac{1}{3}+\\frac{q}{3^{q}}$，①式不成立，\\\\ 所以$p=2$，则有$\\frac{4}{9}=\\frac{1}{3}+\\frac{q}{3^{q}}$，解得：$q=3$，\\\\ 故选：$\\mathrm{B}$. " ]

[ "2017~2018学年山东青岛崂山区青岛第二中学高一上学期期末第7题5分", "2008年吉林全国高中数学联赛竞赛初赛第1题6分", "2017~2018学年12月山西太原迎泽区太原市第五中学高一上学期月考第9题3分" ]

[ [ { "aoVal": "A", "content": "向右平移$$\\frac{\\pi }{6}$$个单位 " } ], [ { "aoVal": "B", "content": "向右平移$$\\frac{\\pi }{3}$$个单位 " } ], [ { "aoVal": "C", "content": "向左平移$$\\frac{\\pi }{6}$$个单位 " } ], [ { "aoVal": "D", "content": "向左平移$$\\frac{\\pi }{3}$$个单位 " } ] ]

[ "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->诱导公式", "课内体系->知识点->三角函数->三角函数的图象与性质->正余弦型、正切型函数图象变换", "课内体系->素养->逻辑推理" ]

[ "$$y=\\sin \\left( 2x-\\frac{\\pi }{6} \\right)=\\cos \\left( \\frac{2\\pi }{3}-2x \\right)=\\cos \\left( 2x-\\frac{2\\pi }{3} \\right)=\\cos \\left( 2\\left( x-\\frac{\\pi }{3} \\right) \\right)$$． 故选择$$B$$选项． " ]

[ "2011年浙江全国高中数学联赛竞赛初赛第2题5分", "2018年广东深圳龙岗区深圳科学高中高三四模文科第2题5分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$2\\sqrt{2}$$ " } ], [ { "aoVal": "C", "content": "$$\\pm 2$$ " } ], [ { "aoVal": "D", "content": "$$\\pm 2\\sqrt{2}$$ " } ] ]

[ "竞赛->知识点->复数与平面向量->复数的概念与运算" ]

[ "由题意得$$\\sqrt{2}\\cdot \\sqrt{{{a}^{2}}+4}=4\\Rightarrow a=\\pm 2$$． " ]

[ "全国全国高中数学联赛竞赛一试" ]

[ [ { "aoVal": "A", "content": "$$\\textbar k\\textbar\\textgreater1$$ " } ], [ { "aoVal": "B", "content": "$$\\textbar k\\textbar\\ne 1$$ " } ], [ { "aoVal": "C", "content": "$$-1\\textless{}k\\textless{}1$$ " } ], [ { "aoVal": "D", "content": "$$0\\textless{}\\textbar k\\textbar\\textless{}1$$ " } ] ]

[ "课内体系->知识点->圆锥曲线->椭圆->椭圆的定义、标准方程->椭圆的定义", "课内体系->素养->数学抽象" ]

[ "椭圆外部的点可以离原点很远， 它的坐标$$x$$、$$y$$的绝对值可以很大， 使得方程左边大于$$0$$， 所以内部点的坐标使方程左边小于$$0$$， 用原点的坐标代入方程左边得$${{k}^{2}}-1\\textless{}0$$ 又$$k\\ne 0$$，（否则方程不表示椭圆）， 所以$$0\\textless{}\\textbar k\\textbar\\textless{}1$$． 故选$$\\text{D}$$． " ]

[ "2020~2021学年河南郑州中原区郑州市第十九中学高二上学期期中理科第11题5分", "2020~2021学年河南郑州郑东新区郑州市第四十七中学高二上学期期中理科第11题5分", "2020~2021学年河南郑州高新区郑州中学高二上学期期中理科第11题5分", "2020~2021学年河南郑州二七区郑州市第二中学高二上学期期中理科第11题5分", "2009年辽宁全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$\\left( 1,1+\\frac{\\sqrt{3}}{2} \\right]$$ " } ], [ { "aoVal": "B", "content": "$$\\left[ \\frac{1}{2},1+\\frac{\\sqrt{3}}{2} \\right]$$ " } ], [ { "aoVal": "C", "content": "$$(1,\\sqrt{2}]$$ " } ], [ { "aoVal": "D", "content": "$$\\left[ \\frac{1}{2},\\sqrt{2} \\right]$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列", "竞赛->知识点->三角函数->三角形中的问题->解三角形", "竞赛->知识点->三角函数->三角函数的图像与性质", "竞赛->知识点->三角函数->三角恒等变换" ]

[ "$$ac={{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\\cos B\\geqslant 2ac-2ac\\cos B\\Rightarrow \\cos B\\geqslant \\frac{1}{2}\\Rightarrow 0\\textless{}B\\leqslant \\frac{ \\pi }{3}$$， 于是$$\\sin B+\\cos B=\\sqrt{2}\\sin \\left( B+\\frac{ \\pi }{4} \\right)\\in (1,\\sqrt{2}]$$． " ]

[ "2012年黑龙江全国高中数学联赛竞赛初赛第11题5分" ]

[ [ { "aoVal": "A", "content": "$$\\pm \\frac{2}{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\pm \\frac{3}{2}$$ " } ], [ { "aoVal": "C", "content": "$$\\pm \\frac{3}{4}$$ " } ], [ { "aoVal": "D", "content": "$$\\pm \\frac{4}{3}$$ " } ] ]

[ "竞赛->知识点->解析几何->直线与圆锥曲线" ]

[ "设$$A\\left( {{x}_{1}}, {{y}_{1}} \\right), B\\left( {{x}_{2}}, {{y}_{2}} \\right)$$，由$$\\overrightarrow{FA}=-4\\overrightarrow{FB}$$，得 $${{y}_{1}}=-4{{y}_{2}}$$．~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ① 设$$AB$$直线为$$y=k\\left( x-1 \\right)$$，与抛物线联立得$$k{{y}^{2}}-4y-4k=0$$．则 $${{y}_{1}}+{{y}_{2}}=4/k$$，~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ② $${{y}_{1}}\\cdot {{y}_{2}}=-4$$.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ③ 由①②③联立得出$$k$$值． " ]

[ "2011年吉林全国高中数学联赛竞赛初赛第6题5分" ]

[ [ { "aoVal": "A", "content": "必为$$0$$ " } ], [ { "aoVal": "B", "content": "必为$$1$$ " } ], [ { "aoVal": "C", "content": "是奇数 " } ], [ { "aoVal": "D", "content": "是偶数 " } ] ]

[ "竞赛->知识点->数列与数学归纳法->数列的综合应用" ]

[ "假设$$M$$是奇数，则$${{a}_{i}}-{{b}_{i}}(i=1, 2,\\cdots , 2011)$$必定都是奇数，从而这$$2011$$个奇数的和也是奇数；另一方面，其和 $$({{a}_{1}}-{{b}_{1}})+({{a}_{2}}-{{b}_{2}})+({{a}_{3}}-{{b}_{3}})+\\cdots +({{a}_{2011}}-{{b}_{2011}})=({{a}_{1}}+{{a}_{2}}+{{a}_{3}}+\\cdots +{{a}_{2011}})-({{b}_{1}}+{{b}_{2}}+{{b}_{3}}+\\cdots +{{b}_{2011}})=0$$为偶数，矛盾，所以假设不成立，即$$M$$是偶数． " ]

[ "1986年全国高中数学联赛竞赛一试第5题" ]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$28$$ " } ] ]

[ "竞赛->知识点->组合->图论（二试）" ]

[ "要使$$M$$中的点的个数最少，各圆要尽可能多地共同经过$$M$$中的点．因为不相同的两圆最多只有两个公共点，而$${{C}_{7}}$$经过$$M$$中的$$7$$个点，所以$${{C}_{6}}$$至少要经过$$M$$中的另外四个点，而$${{C}_{5}}$$除了和$${{C}_{7}}$$、$${{C}_{6}}$$各有两个公共点外，还要经过$$M$$中的一个点，则$$M$$中至少要有$$12$$个点． 另一方面，如果有$$12$$个点，就能满足要求： 先作两两相交的三个圆，记为$${{C}_{7}}$$、$${{C}_{6}}$$、$${{C}_{5}}$$．两作圆$${{C}_{4}}$$与两圆$${{C}_{6}}$$、$${{C}_{7}}$$都相交于另外的点（但与$${{C}_{5}}$$不相交），又过$${{C}_{5}}$$与$${{C}_{7}}$$的一个交点作圆$${{C}_{3}}$$，使与$${{C}_{5}}$$及$${{C}_{7}}$$各交一个新的点，将所有的交点都算作$$M$$中的点，此时$$M$$中的点为$$12$$个．$${{C}_{7}}$$与$${{C}_{6}}$$、$${{C}_{5}}$$、$${{C}_{4}}$$各有两个交点，又与$${{C}_{3}}$$有一个新交点，共$$7$$个点；$${{C}_{6}}$$与$${{C}_{7}}$$、$${{C}_{5}}$$、$${{C}_{4}}$$各有两个交点，共$$6$$个点；$${{C}_{5}}$$与$${{C}_{7}}$$、$${{C}_{6}}$$各有两个交点，且与$${{C}_{3}}$$有一个新交点，共$$5$$个点；$${{C}_{4}}$$与$${{C}_{7}}$$、$${{C}_{6}}$$各有两个交点，共$$4$$个点；$${{C}_{3}}$$是过$${{C}_{7}}$$与$${{C}_{5}}$$的一个交点，且与其各交得一个新点，共有$$3$$个点．最后在$$M$$的$$12$$个点中，选取两个点、一个点分别作满足要求的圆$${{C}_{2}}$$、$${{C}_{1}}$$是容易的事． 故选$$\\text{B}$$． " ]

[ "2018年上海杨浦区高三二模第16题5分", "2017年辽宁全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$\\arccos \\frac{1}{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\arccos \\frac{\\sqrt{2}}{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\arccos \\frac{\\sqrt{3}}{9}$$ " } ], [ { "aoVal": "D", "content": "$$\\arccos \\frac{\\sqrt{6}}{9}$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->知识点->立体几何初步->基本立体图形->空间几何体的概念->棱柱、棱锥、棱台的结构特征", "课内体系->知识点->立体几何初步->基本图形位置关系->空间中的基本事实与定理->点、直线、平面之间的位置关系", "课内体系->知识点->立体几何初步->基本图形位置关系->探索性问题->几何法求空间角" ]

[ "设三条棱$$a\\leqslant b\\leqslant c$$， ∴$$ab+ac+bc=\\frac{45}{4}$$，$$a+b+c=6$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=\\frac{27}{2}$$， $${{a}^{2}}+{{b}^{2}}+{{c}^{2}}\\geqslant {{a}^{2}}+2bc={{a}^{2}}+2\\left[ \\frac{45}{4}-a\\left( 6-a \\right) \\right]$$， 整理得$${{a}^{2}}-4a+3\\leqslant 0$$， ∴$$1\\leqslant a\\leqslant 3$$， ∴最短棱长为$$1$$，体对角线长为$$\\frac{3\\sqrt{6}}{2}$$，$$\\cos \\theta =\\frac{2}{3\\sqrt{6}}=\\frac{\\sqrt{6}}{9}$$， 故选$$\\text{D}$$． " ]

[ "2014年湖南全国高中数学联赛竞赛初赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$9\\in M$$，$$10\\in M$$ " } ], [ { "aoVal": "B", "content": "$$9\\notin M$$，$$10\\in M$$ " } ], [ { "aoVal": "C", "content": "$$9\\in M$$，$$10\\notin M$$ " } ], [ { "aoVal": "D", "content": "$$9\\notin M$$，$$10\\notin M$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "因为$$9={{3}^{2}}-{{0}^{2}}$$，所以$$9\\in M$$；假设$$10\\in M$$，则存在整数$$m$$、$$n$$使得 $$10={{m}^{2}}-{{n}^{2}}=\\left( m+n \\right)\\left( m-n \\right)$$， 则因为将$$10$$分解为两个整数因子之积，必定一个因子为奇数，另一个因子为偶数，而$$\\left( m+n \\right)$$与$$\\left( m-n \\right)$$同奇同偶，矛盾，所以$$10\\notin M$$． " ]

[ "1990年全国高中数学联赛竞赛一试第2题", "2006年上海复旦大学自主招生千分考第24题" ]

[ [ { "aoVal": "A", "content": "$$x+4$$ " } ], [ { "aoVal": "B", "content": "$$2-x$$ " } ], [ { "aoVal": "C", "content": "$$3-\\left\\textbar{} x+1 \\right\\textbar$$ " } ], [ { "aoVal": "D", "content": "$$2+\\left\\textbar{} x+1 \\right\\textbar$$ " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "当$$x\\in \\left[ -2,-1 \\right]$$时，$$f\\left( x \\right)=f\\left( x+4 \\right)=x+4$$， 当$$x\\in \\left[ -1,0 \\right]$$时，$$f\\left( x \\right)=f\\left( -x \\right)=f\\left( 2-x \\right)=2-x$$． " ]

[ "2019~2020学年10月广东深圳南山区深圳市南山区育才中学高三上学期月考理科第10题5分", "2017~2018学年广东深圳龙岗区深圳科学高中高二上学期期中第10题5分", "2013年黑龙江全国高中数学联赛竞赛初赛第12题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{3}{2}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{5}{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{9}{4}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{25}{6}$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->知识点->数列->等比数列->等比数列的概念与通项公式", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的实际应用", "课内体系->知识点->等式与不等式->不等式->基本不等式->利用基本不等式求最值", "课内体系->知识点->基本初等函数->实数指数幂运算", "课内体系->知识点->基本初等函数->指数函数->指数函数的图象及性质" ]

[ "由各项均为正数的等比数列$$\\left { {{a}_{n}} \\right }$$满足$${{a}_{7}}={{a}_{6}}+2{{a}_{5}}$$， 可得$${{a}_{1}}{{q}^{6}}={{a}_{1}}{{q}^{5}}+2{{a}_{1}}{{q}^{4}}$$， 所以$${{q}^{2}}-q-2=0$$，解得$$q=2$$或$$q=-1$$（舍去）． 因为$$\\sqrt{{{a}_{m}}{{a}_{n}}}=4{{a}_{1}}$$，所以$${{q}^{m+n-2}}=16$$， 所以$${{2}^{m+n-2}}={{2}^{4}}$$，所以$$m+n=6$$． 所以$$\\frac{1}{m}+\\frac{4}{n}=\\frac{1}{6}\\left( m+n \\right)\\left( \\frac{1}{m}+\\frac{4}{n} \\right)$$ $$=\\frac{1}{6}\\left( 5+\\frac{n}{m}+\\frac{4m}{n} \\right)$$ $$\\geqslant \\frac{1}{6}\\left( 5+2\\sqrt{\\frac{n}{m}\\cdot \\frac{4m}{n}} \\right)$$ $$=\\frac{3}{2}$$． 当且仅当$$\\frac{n}{m}=\\frac{4m}{n}$$，即$$n=2m$$时，等号成立， 又$$m+n=6$$，所以$$m=2$$，$$n=4$$，符合题意． 故最小值为$$\\frac{3}{2}$$． " ]

[ "2022~2023学年湖南永州宁远县高一上学期月考（明德湘南中学基础知识竞赛）第4题" ]

[ [ { "aoVal": "A", "content": "$y=-\\frac{1}{x}$ " } ], [ { "aoVal": "B", "content": "$y=-{{x}^{3}}$ " } ], [ { "aoVal": "C", "content": "$y=x+1$ " } ], [ { "aoVal": "D", "content": "$y=x\\left\\textbar{} x \\right\\textbar$ " } ] ]

[ "课内体系->知识点->函数的概念与性质->函数的性质->奇偶性->函数奇偶性的应用" ]

[ "\\hfill\\break 【分析】\\\\ 根据基本函数的奇偶性、单调性逐项判断即可.\\\\ 【详解】\\\\ 对于A：$y=-\\frac{1}{x}$为反比例函数，为奇函数，在区间$\\left( -\\infty ,0 \\right)$以及$\\left( 0,+\\infty \\right)$上都是增函数，但在定义域内不是增函数，故A错误；\\\\ 对于B：$y=-{{x}^{3}}$为幂函数，既是奇函数又是减函数，不符合题意，故B错误；\\\\ 对于C：$y=x+1$为一次函数，不是奇函数，不符合题意，故C错误；\\\\ 对于D： $y=x\\left\\textbar{} x \\right\\textbar=\\left { \\begin{array}{*{35}{l}} {{x}^{2}}, \\& x\\ge 0 -{{x}^{2}}, \\& x\\textless{} 0 \\end{array} \\right.$为奇函数；$x\\ge 0$时， $y={{x}^{2}}$为增函数，$x\\textless{} 0$时，$y=-{{x}^{2}}$ 为增函数，且${{x}^{2}}\\ge -{{x}^{2}}$该函数在\\emph{R}上为增函数，故D正确；\\\\ 故选： D " ]

[ "2009年高二竞赛广州市第1题6分" ]

[ [ { "aoVal": "A", "content": "$$3+\\text{i}$$ " } ], [ { "aoVal": "B", "content": "$$1+\\frac{1}{3}\\text{i}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2}{3}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{4}{3}$$ " } ] ]

[ "课内体系->知识点->复数->复数的概念及几何意义->复数的基本概念", "课内体系->知识点->复数->复数的运算->复数的四则运算综合", "课内体系->知识点->复数->复数的运算->复数的乘法和除法", "课内体系->素养->数学运算" ]

[ "$$z=1+\\frac{\\text{i}}{3}$$，故实部与虚部之和为$$1+\\frac{1}{3}=\\frac{4}{3}$$． " ]

[ "2017~2018学年广东广州白云区艺术中学高二下学期期中文科第6题5分", "2010年黑龙江全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{8}{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{3}{2}$$ " } ], [ { "aoVal": "C", "content": "$$-\\frac{8}{3}$$ " } ], [ { "aoVal": "D", "content": "$$-\\frac{3}{2}$$ " } ] ]

[ "课内体系->知识点->复数->复数的运算->复数的乘法和除法", "课内体系->知识点->复数->复数的概念及几何意义->复数的基本概念->实部与虚部", "课内体系->素养->数学运算" ]

[ "$$\\frac{{{z}_{1}}}{{{z}_{2}}}=\\frac{m+2\\text{i}}{3-4\\text{i}}=\\frac{(m+2\\text{i})(3+4\\text{i})}{(3-4\\text{i})(3+4\\text{i})}=\\frac{3m-8+(6+4m)\\text{i}}{25}$$， ∵$$\\frac{{{z}_{1}}}{{{z}_{2}}}$$为实数， ∴$$6+4m=0$$即$$m=-\\frac{3}{2}$$． 故选$$\\text{D}$$． " ]

[ "2012年四川全国高中数学联赛竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$\\left { x\\textbar-5\\leqslant x\\textless{}-1 \\right }$$ " } ], [ { "aoVal": "B", "content": "$$\\left { x\\textbar-5\\leqslant x\\textless{}5 \\right }$$ " } ], [ { "aoVal": "C", "content": "$$\\left { x\\textbar-1\\textless{}x\\leqslant 1 \\right }$$ " } ], [ { "aoVal": "D", "content": "$$\\left { x\\textbar1\\leqslant x\\textless{}5 \\right }$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "$$S=\\left { x\\textbar-1\\textless{}x\\textless{}6 \\right }, T=\\left { x\\textbar-5\\leqslant x\\leqslant 1 \\right }$$，故$$S\\cap T=\\left { x\\textbar-1\\textless{}x\\leqslant 1 \\right }$$． " ]

[ "2020年江苏全国高中数学联赛竞赛初赛第9题7分" ]

[ [ { "aoVal": "A", "content": "$$17$$ " } ], [ { "aoVal": "B", "content": "$$37$$ " } ], [ { "aoVal": "C", "content": "$$57$$ " } ], [ { "aoVal": "D", "content": "以上都不对 " } ] ]

[ "课内体系->知识点->函数的概念与性质", "竞赛->知识点->数论模块->整除->质数（算数基本定理）" ]

[ "无 " ]

[ "2011年吉林全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "有且仅有一个 " } ], [ { "aoVal": "B", "content": "有时有两个，有时有一个 " } ], [ { "aoVal": "C", "content": "恰有两个 " } ], [ { "aoVal": "D", "content": "有时不存在 " } ] ]

[ "竞赛->知识点->立体几何与空间向量->空间中的平行和垂直" ]

[ "在面$$ABCDE$$内过$$A$$作$$AE$$的垂线交$$DC$$延长线于$$F$$，在面$${{A}_{1}}{{B}_{1}}{{C}_{1}}{{D}_{1}}{{E}_{1}}$$内过$${{E}_{1}}$$作$${{E}_{1}}{{A}_{1}}$$的垂线交$${{B}_{1}}{{C}_{1}}$$的延长线与点$${{F}_{1}}$$，则面$$AF{{F}_{1}}{{E}_{1}}$$与$$C{{C}_{1}}$$的交点为$$P$$，由正五边形的图形特征知只有一个交点，即$$P$$恰有一个． " ]

[ "2004年全国高中数学联赛竞赛一试第3题6分" ]

[ [ { "aoVal": "A", "content": "$$[2,3)$$ " } ], [ { "aoVal": "B", "content": "$$(2,3]$$ " } ], [ { "aoVal": "C", "content": "$$[2,4)$$ " } ], [ { "aoVal": "D", "content": "$$(2,4]$$ " } ] ]

[ "竞赛->知识点->不等式->不等式的解法" ]

[ "原不等式等价于$$\\begin{cases}\\sqrt{{{\\log }_{2}}x-1}-\\dfrac{3}{2}{{\\log }_{2}}x+\\dfrac{3}{2}+\\dfrac{1}{2}\\textgreater0 {{\\log }_{2}}x-1\\geqslant 0 \\end{cases}$$， 设$$\\sqrt{{{\\log }_{2}}x-1}=t$$，则有$$\\begin{cases}t-\\dfrac{3}{2}{{t}^{2}}+\\dfrac{1}{2}\\textgreater0 t\\geqslant 0 \\end{cases}$$，解得$$0\\leqslant t\\textless{}1$$， 即$$0\\leqslant {{\\log }_{2}}x-1\\textless{}1$$，$$2\\leqslant x\\textless{}4$$． 故选$$\\text{C}$$． " ]

[ "1998年全国高中数学联赛竞赛一试第1题6分" ]

[ [ { "aoVal": "A", "content": "等于$$\\text{lg2}$$ " } ], [ { "aoVal": "B", "content": "等于$$1$$ " } ], [ { "aoVal": "C", "content": "等于$$0$$ " } ], [ { "aoVal": "D", "content": "不是与$$a$$，$$b$$无关的常数 " } ] ]

[ "竞赛->知识点->函数->基本初等函数" ]

[ "$$a+b=ab$$，$$\\left( a-1 \\right)\\left( b-1 \\right)=1$$，由$$a-1\\textgreater0$$，$$b-1\\textgreater0$$，故$$\\text{lg}\\left( a-1 \\right)\\left( b-1 \\right)=0$$． 故选$$\\text{C}$$． " ]

[ "2014年浙江全国高中数学联赛竞赛初赛第10题5分" ]

[ [ { "aoVal": "A", "content": "$$\\left[ 0,\\frac{ \\pi }{4} \\right)$$ " } ], [ { "aoVal": "B", "content": "$$\\left[ 0,\\frac{{{ \\pi }^{2}}}{2} \\right)$$ " } ], [ { "aoVal": "C", "content": "$$\\left[ 0,\\frac{{{ \\pi }^{2}}}{4} \\right)$$ " } ], [ { "aoVal": "D", "content": "$$\\left[ 0,\\frac{ \\pi }{2} \\right)$$ " } ] ]

[ "竞赛->知识点->三角函数->三角函数的图像与性质" ]

[ "由已知得$2k \\pi +b\\cos x=\\frac{ \\pi }{2}-a\\sin x$，$\\left( 2k+1 \\right) \\pi -b\\cos x=\\frac{ \\pi }{2}-a\\sin x\\left( k\\in \\mathbf{Z} \\right)$， 无解，所以方程$\\sqrt{{{a}^{2}}+{{b}^{2}}}\\sin \\left( x+\\varphi~ \\right)=\\frac{ \\pi }{2}+2k \\pi $，或 $\\sqrt{{{a}^{2}}+{{b}^{2}}}$$\\sqrt{{{a}^{2}}+{{b}^{2}}}\\sin \\left( x-\\varphi~ \\right)=\\frac{ \\pi }{2}-\\left( 2k+1 \\right) \\pi $ 其中$\\sin \\varphi =\\frac{b}{\\sqrt{{{a}^{2}}+{{b}^{2}}}}$，$\\cos \\varphi =\\frac{a}{\\sqrt{{{a}^{2}}+{{b}^{2}}}}$，无解． 所以$\\sqrt{{{a}^{2}}+{{b}^{2}}}\\textless\\left\\textbar{} \\frac{ \\pi }{2}+2k \\pi \\right\\textbar$或$\\left( \\left\\textbar{} \\frac{ \\pi }{2}-\\left( 2k+1 \\right) \\pi \\right\\textbar{} \\right)\\Rightarrow {{a}^{2}}+{{b}^{2}}$取值范围为$\\left[ 0 \\frac{{{ \\pi }^{2}}}{4} \\right)$． $$.$$ " ]

[ "2022~2023学年12月安徽滁州定远县安徽省定远县民族中学高一上学期月考第5题", "2022~2023学年湖南邵阳隆回县湖南省隆回县第二中学高一上学期月考（竞赛）第3题", "2021~2022学年江苏南京鼓楼区金陵中学高一上学期期中第4题" ]

[ [ { "aoVal": "A", "content": "$\\sqrt{2}$ " } ], [ { "aoVal": "B", "content": "2$\\sqrt{2}$ " } ], [ { "aoVal": "C", "content": "$\\frac{\\sqrt{10}}{2}$ " } ], [ { "aoVal": "D", "content": "2 " } ] ]

[ "课内体系->知识点->等式与不等式->不等式->基本不等式" ]

[ "\\hfill\\break 应用对数运算得到$ab=10$，由目标式结合基本不等式有$\\frac{2}{a}+\\frac{5}{b}\\ge 2\\sqrt{\\frac{2}{a}\\cdot \\frac{5}{b}}$即可求其最小值.\\\\ 【详解】\\\\ ∵$\\lg a+\\lg b=1$，即$\\lg ab=1$，\\\\ ∴$ab=10$，而$a\\textgreater0,b\\textgreater0$，\\\\ ∴$\\frac{2}{a}+\\frac{5}{b}\\ge 2\\sqrt{\\frac{2}{a}\\cdot \\frac{5}{b}}=2$当且仅当$a=2,b=5$时等号成立.\\\\ ∴$\\frac{2}{a}+\\frac{5}{b}$的最小值为2.\\\\ 故选：D\\\\ 【点睛】\\\\ 易错点睛：利用基本不等式求最值时，须满足的三个条件：\\\\ （1）``一正二定三相等''\\,``一正''就是各项必须为正数；\\\\ （2）``二定''就是要求和的最小值，必须把构成和的二项之积转化成定值；要求积的最大值，则必须把构成积的因式的和转化成定值；\\\\ （3）``三相等''是利用基本不等式求最值时，必须验证等号成立的条件，若不能取等号则这个定值就不是所求的最值，这也是最容易发生错误的地方 " ]

[ "2008年安徽全国高中数学联赛竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "30 " } ], [ { "aoVal": "B", "content": "40 " } ], [ { "aoVal": "C", "content": "50 " } ], [ { "aoVal": "D", "content": "60 " } ] ]

[ "竞赛->知识点->数论模块->同余->几个重要的定理（欧拉定理、费尔马小定理、威尔逊定理）" ]

[ "$$\\frac{1}{2008}=\\frac{1}{8\\cdot 251}=\\frac{1}{1000}\\cdot \\frac{125}{251}$$，则$$\\frac{125}{251}=0.\\underbrace{498\\ldots 625}_{{}}\\underbrace{498\\ldots 625}_{{}}\\ldots $$． 设循环节长度为$$k$$，则 $$\\frac{125}{251}\\cdot {{10}^{k}}=\\underbrace{498\\ldots 625}_{{}}+0.\\underbrace{498\\ldots 625}_{{}}\\underbrace{498\\ldots 625}_{{}}\\ldots =\\underbrace{498\\cdots 625}_{{}}+\\frac{125}{251}=n+\\frac{125}{251}$$，$$n\\in {{\\mathbf{N}}^{+}}$$． 于是，$$251n=125\\left( {{10}^{k}}-1 \\right)$$，则$$251\\left\\textbar{} {{10}^{k}}-1 \\right.$$． 由费马小定理，$$251\\left\\textbar{} {{10}^{250}}-1 \\right.$$，则$$k\\left\\textbar{} 250 \\right.$$，只能选$$\\text{C}$$． " ]

[ "2021~2022学年安徽阜阳太和县安徽省太和中学高一下学期月考（竞赛考试）第3~3题" ]

[ [ { "aoVal": "A", "content": "－5 " } ], [ { "aoVal": "B", "content": "－3 " } ], [ { "aoVal": "C", "content": "3 " } ], [ { "aoVal": "D", "content": "5 " } ] ]

[ "课内体系->知识点->平面向量->平面向量基本定理及其坐标表示" ]

[ "\\hfill\\break 【分析】\\\\ 根据向量的线性坐标运算求得向量$$\\vec{a}-\\vec{b}$$，再由向量垂直的坐标表示建立方程，求解即可.\\\\ 【详解】\\\\ 解：由题意得$$\\vec{a}-\\vec{b}=(-2，m\\text{+}2)$$.\\\\ 又$$(\\vec{a}-\\vec{b})\\bot \\vec{b}$$，∴$$(\\vec{a}-\\vec{b})\\cdot \\vec{b}=-6-2(m\\text{+}2)=0$$ ，解得\\emph{m}＝－5.\\\\ 故选：A. " ]

[ "2015年辽宁全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$660$$ " } ], [ { "aoVal": "B", "content": "$$664$$ " } ], [ { "aoVal": "C", "content": "$$668$$ " } ], [ { "aoVal": "D", "content": "$$672$$ " } ] ]

[ "竞赛->知识点->函数->二次函数", "竞赛->知识点->函数->函数的图像与性质" ]

[ "$${{x}^{2}}-53x+196=\\left( x-4 \\right)\\left( x-49 \\right)$$， 当$$x\\textless{}4$$或$$x\\textgreater49$$时，$${{x}^{2}}-53x+196\\textgreater0$$，$$f\\left( x \\right)=2\\left( {{x}^{2}}-53x+196 \\right)$$； 当$$4\\leqslant x\\leqslant 49$$时，$${{x}^{2}}-53x+196\\leqslant 0$$，$$f\\left( x \\right)=0$$． $$f\\left( 1 \\right)+f\\left( 2 \\right)+f\\left( 3 \\right)+f\\left( 50 \\right)=330$$． " ]

[ "2008年安徽全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "椭圆 " } ], [ { "aoVal": "B", "content": "双曲线的一部分 " } ], [ { "aoVal": "C", "content": "抛物线的一部分 " } ], [ { "aoVal": "D", "content": "矩形 " } ] ]

[ "竞赛->知识点->解析几何->曲线与方程" ]

[ "以两条定直线所成的角的两条角平分线所在的直线分别为$$x,y$$轴建立平面直角坐标系，设两条直线方程为$$y=kx$$和$$y=-kx$$，$$k\\textgreater0$$． 点$$\\left( x,y \\right)$$到$$y=\\pm kx$$的距离和为$$1$$，则$$\\frac{\\left\\textbar{} y-kx \\right\\textbar+\\left\\textbar{} y+kx \\right\\textbar}{\\sqrt{{{k}^{2}}+1}}=1$$． 轨迹由$$4$$条线段组成，构成一个矩形． " ]

[ "2015年天津全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "一个圆 " } ], [ { "aoVal": "B", "content": "两个半圆 " } ], [ { "aoVal": "C", "content": "一个椭圆 " } ], [ { "aoVal": "D", "content": "以上结论都不对 " } ] ]

[ "竞赛->知识点->解析几何->曲线与方程" ]

[ "$$y\\textgreater0$$时，$${{\\left( y-1 \\right)}^{2}}+{{\\left( x-1 \\right)}^{2}}=1$$，$$y\\geqslant 1$$，表示一个半圆； $$y\\textless{}0$$时，$${{\\left( y+1 \\right)}^{2}}+{{\\left( x-1 \\right)}^{2}}=1$$，$$y\\leqslant -1$$，表示一个半圆． " ]

[ "2019~2020学年江西新余高二上学期期末理科第6题5分", "2009年高考真题湖北卷理科第5题5分", "2018~2019学年云南昆明高二下学期期中理科昆明三中、滇池中学第9题5分", "2013年黑龙江全国高中数学联赛竞赛初赛第5题5分", "2016~2017学年12月广东广州天河区天河中学高三上学期月考理科第6题5分", "2017~2018学年8月四川成都龙泉驿区成都龙泉一中高三上学期月考文科第3题5分" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

[ "课内体系->知识点->计数原理->排列与组合->排列", "课内体系->知识点->计数原理->排列与组合->组合->分组分配法", "课内体系->素养->逻辑推理" ]

[ "如丙、丁分到同一个班级，则方法数就是三个元素的一个全排列，即$$\\text{A}_{3}^{3}$$；若丙分到甲或乙所在的班级，则丁只能独自一个班级，方法数是$$2\\text{A}_{3}^{3}$$；同理，若丁分到甲或乙所在的班级，则丙独自一个班级，方法数是$$2\\text{A}_{3}^{3}$$．根据分类加法计数原理，得总的方法数是$$5\\text{A}_{3}^{3}=30$$． ", "<p>总的方法数是$$\\text{C}_{4}^{2}\\text{A}_{3}^{3}=36$$，甲、乙被分到同一个班级的方法数是$$\\text{A}_{3}^{3}=6$$，故甲、乙不分到同一个班级的方法数是$$36-6=30$$．</p>" ]

[ "1993年全国全国高中数学联赛竞赛一试第3题" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$26$$ " } ], [ { "aoVal": "D", "content": "$$27$$ " } ] ]

[ "竞赛->知识点->排列组合与概率->排列与组合" ]

[ "已知$$A\\cup B= {{{a}_{1}},{{a}_{2}},{{a}_{3}} }$$，则作为其子的$$A$$，$$B$$最多只有$$3$$元素． （$$1$$）若$$A= {{{a}_{1}},{{a}_{2}},{{a}_{3}} }$$则满足题意得$$B$$可以使空集，或者是单元素的集合，或是二元素的集合，或是三元素的集合，这样的$$B$$有$$\\text{C}_{3}^{0}+\\text{C}_{3}^{1}+\\text{C}_{3}^{2}+\\text{C}_{3}^{3}={{2}^{3}}$$个，这时$$(A,B)$$有$$\\text{C}_{3}^{3}\\cdot {{2}^{3}}$$对． （$$2$$）若$$A$$为二元素的集合，则有$$\\text{C}_{3}^{2}$$种，其对应的$$B$$的$${{2}^{3}}$$个（$$\\text{C}_{2}^{0}+\\text{C}_{2}^{1}+\\text{C}_{2}^{2}={{2}^{2}}$$），这时$$(A,B)$$有$$C_{3}^{2}\\cdot {{2}^{2}}$$时． （$$3$$）若$$A$$为单元素的集合，则有$$\\text{C}_{3}^{1}$$种，其对应的$$B$$有$$2$$个，这时$$(A,B)\\text{C}_{3}^{1}\\cdot 2$$对． （$$4$$）若$$A$$是空集，则有$$\\text{C}_{3}^{0}$$种，其对应的$$B$$有一个，这时$$(A,B)$$有$$\\text{C}_{3}^{0}\\cdot 1$$对． ∴ 这样的$$(A,B)$$共有 $$\\text{C}_{3}^{3}\\cdot {{2}^{3}}+\\text{C}_{3}^{2}\\cdot {{2}^{2}}+\\text{C}_{3}^{1}\\cdot 2+\\text{C}_{3}^{0}\\cdot {{2}^{0}}={{3}^{3}}=27$$． 故选$$\\text{D}$$． " ]

[ "2022~2023学年广东深圳南山区深圳实验学校高二上学期段考（三）第8题", "2010年四川全国高中数学联赛竞赛初赛第7题5分" ]

[ [ { "aoVal": "A", "content": "$$\\left( 0,\\frac{1}{2} \\right)$$ " } ], [ { "aoVal": "B", "content": "$$\\left( 0,\\frac{\\sqrt{2}}{2} \\right)$$ " } ], [ { "aoVal": "C", "content": "$$\\left( \\frac{1}{2}, 1 \\right)$$ " } ], [ { "aoVal": "D", "content": "$$\\left( \\frac{\\sqrt{2}}{2}, 1 \\right)$$ " } ] ]

[ "课内体系->知识点->圆锥曲线->椭圆->椭圆的简单几何性质" ]

[ "由题设知$$\\angle OP{{A}_{2}}=90{}^{}\\circ $$，设$$P(x, y)(x\\textgreater0)$$， 以$$O{{A}_{2}}$$为直径的圆方程为$${{\\left( x-\\frac{a}{2} \\right)}^{2}}+{{y}^{2}}=\\frac{{{a}^{2}}}{4}$$， 与椭圆方程联立得$$\\left( 1-\\frac{{{b}^{2}}}{{{a}^{2}}} \\right){{x}^{2}}-ax+{{b}^{2}}=0$$． 由题设知，要求此方程在$$(0, a)$$上有实根． 由此得$$0\\textless{}\\frac{a}{2\\left( 1-\\frac{{{b}^{2}}}{{{a}^{2}}} \\right)}\\textless{}a$$化简得$${{e}^{2}}\\textgreater\\frac{1}{2}$$， 所以$$e$$的取值范围为$$\\left( \\frac{\\sqrt{2}}{2}, 1 \\right)$$．故选$$\\text{D}$$． " ]

[ "2012年河北全国高中数学联赛高二竞赛初赛第7题8分", "2015~2016学年天津和平区天津市耀华中学高二上学期期中文科第18题4分", "2017~2018学年天津和平区天津市耀华中学高二上学期期中理科第14题4分", "2017~2018学年天津和平区天津市耀华中学高二上学期期中文科第14题4分", "2017~2018学年天津和平区天津市耀华中学高一下学期期中第14题4分" ]

[ [ { "aoVal": "A", "content": "$\\sqrt{2}\\pi $ " } ], [ { "aoVal": "B", "content": "$\\sqrt{3}\\pi $ " } ], [ { "aoVal": "C", "content": "$\\sqrt{6}\\pi $ " } ], [ { "aoVal": "D", "content": "$3\\pi $ " } ] ]

[ "竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题" ]

[ "在三棱锥$$A-BCD$$中，侧棱$$AB$$、$$AC$$、$$AD$$两两垂直， 补成长方体，两者的外接球是同一个，长方体的体对角线即为球的直径， 设长方体的三度分别为$$a$$、$$b$$、$$c$$， 则有$$\\frac{1}{2}ab=\\frac{\\sqrt{6}}{2}$$，$$\\frac{1}{2}bc=\\frac{\\sqrt{3}}{2}$$，$$\\frac{1}{2}bc=\\frac{\\sqrt{2}}{2}$$， 解得：$$a=\\sqrt{3}$$，$$b=\\sqrt{2}$$，$$c=1$$， 所以球的直径$$d=\\sqrt{3+2+1}=\\sqrt{6}$$， 球的半径$$r=\\frac{d}{2}=\\frac{\\sqrt{6}}{2}$$， ∴三棱锥$$A-BCD$$的外接球的体积为： $$V=\\frac{4}{3} \\pi \\cdot {{\\left( \\frac{\\sqrt{6}}{2} \\right)}^{3}}=\\sqrt{6} \\pi $$． " ]

[ "2016~2017学年9月天津静海区天津市静海区第一中学高三上学期月考理科第7题5分", "2010年浙江全国高中数学联赛竞赛初赛第8题5分" ]

[ [ { "aoVal": "A", "content": "$$a\\textgreater b\\textgreater c$$ " } ], [ { "aoVal": "B", "content": "$$b\\textgreater c\\textgreater a$$ " } ], [ { "aoVal": "C", "content": "$$c\\textgreater a\\textgreater b$$ " } ], [ { "aoVal": "D", "content": "$$b\\textgreater a\\textgreater c$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->知识点->基本初等函数->指对幂比较大小", "课内体系->知识点->基本初等函数->对数的概念及其运算->对数的运算", "课内体系->知识点->函数的概念与性质->函数的性质->单调性->函数单调性的应用->用单调性比较大小" ]

[ "$$a=f\\left( {{\\log }_{e}}\\frac{1}{ \\pi } \\right)=-{{\\log }_{e}} \\pi \\in (-2,-1)$$，$$b=f({{\\log }_{ \\pi }}\\frac{1}{e})=-{{\\log }_{ \\pi }}e\\in (-1,0)$$，$$c=f\\left( {{\\log }_{\\frac{1}{e}}}\\frac{1}{{{ \\pi }^{2}}} \\right)=-2{{\\log }_{2}} \\pi \\in (-2,-4)$$，由此能求出结果． 【解答】解：∵$$f(x)=-\\textbar x\\textbar$$， ∴$$a=f(lo{{g}_{e}}\\frac{1}{ \\pi })=-\\left\\textbar{} {{\\log }_{e}}\\frac{1}{ \\pi } \\right\\textbar=-{{\\log }_{e}} \\pi \\in (-2,-1)$$， $$b=f\\left( {{\\log }_{ \\pi }}\\frac{1}{e} \\right)=-\\left\\textbar{} {{\\log }_{ \\pi }}\\frac{1}{e} \\right\\textbar=-{{\\log }_{ \\pi }}e\\in (-1,0)$$， $$c=f\\left( {{\\log }_{\\frac{1}{e}}}\\frac{1}{{{ \\pi }^{2}}} \\right)=-\\left\\textbar{} {{\\log }_{\\frac{1}{e}}}\\frac{1}{{{ \\pi }^{2}}} \\right\\textbar=-{{\\log }_{e}}{{ \\pi }^{2}}=-2{{\\log }_{e}} \\pi \\in (-2,-4)$$， ∴$$b\\textgreater a\\textgreater c$$． 故选$$\\text{D}$$． " ]

[ "2018年辽宁全国高中数学联赛竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "$$45$$ " } ], [ { "aoVal": "B", "content": "$$81$$ " } ], [ { "aoVal": "C", "content": "$$165$$ " } ], [ { "aoVal": "D", "content": "$$216$$ " } ] ]

[ "竞赛->知识点->排列组合与概率->排列与组合" ]

[ "依题意，等边三角形共有$$9$$个，等腰非等边三角形共有$$3\\times \\left( 8+8+8+8+8+6+4+2 \\right)=156$$个， 因此，所求为$$9+156=165$$． " ]

[ "2017~2018学年江西南昌西湖区南昌市第八中学高一上学期期末第10题5分", "2018~2019学年广东深圳高一上学期期末高中联考联盟第8题5分", "2018~2019学年浙江杭州江干区杭州第四中学下沙校区高二下学期期中第12题4分", "2018~2019学年江苏盐城高一上学期期末第5题5分", "2017~2018学年北京海淀区中央民族大学附属中学高一下学期期中第7题5分", "2008年黑龙江全国高中数学联赛竞赛初赛第4题5分", "2015年北京海淀区高三三模第4题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{19}{25}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{16}{25}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{14}{25}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{7}{25}$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式->正余弦和差积相互转化求值", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正弦", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->两角和与差的正弦->利用正弦和差角公式凑角求值" ]

[ "$$\\sin \\left( \\frac{ \\pi }{4}-x \\right)=\\frac{\\sqrt{2}}{2}(\\cos x-\\sin x)=\\frac{3}{5}$$，所以$$\\cos x-\\sin x=\\frac{3\\sqrt{2}}{5}$$， 所以$${{(\\text{cos }x-\\text{sin }x)}^{2}}=1-\\sin 2x=\\frac{18}{25}$$，所以$$\\sin 2x=\\frac{7}{25}$$，故选$$\\text{D}$$． " ]

[ "1995年全国高中数学联赛竞赛一试第5题" ]

[ [ { "aoVal": "A", "content": "$${{\\log }_{\\sin 1}}\\cos 1\\textless{}{{\\log }_{\\cos 1}}\\sin 1\\textless{}{{\\log }_{\\sin 1}}\\text{tg1lo}{{\\text{g}}_{\\cos 1}}\\text{tg1}$$ " } ], [ { "aoVal": "B", "content": "$${{\\log }_{\\cos 1}}\\sin 1\\textless{}{{\\log }_{\\cos 1}}\\text{tg1lo}{{\\text{g}}_{\\sin 1}}\\cos 1\\textless{}{{\\log }_{\\sin 1}}\\text{tg1}$$ " } ], [ { "aoVal": "C", "content": "$${{\\log }_{\\sin 1}}\\text{tg}1\\textless{}{{\\log }_{\\cos 1}}\\text{tg1lo}{{\\text{g}}_{\\cos 1}}\\sin 1\\textless{}{{\\log }_{\\sin 1}}\\cos 1$$ " } ], [ { "aoVal": "D", "content": "$${{\\log }_{\\cos 1}}\\text{tg}1\\textless{}{{\\log }_{\\sin 1}}\\text{tg1lo}{{\\text{g}}_{\\sin 1}}\\cos 1\\textless{}{{\\log }_{\\cos 1}}\\sin 1$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->三角函数线", "课内体系->知识点->基本初等函数->指对幂比较大小", "课内体系->知识点->基本初等函数->对数的概念及其运算", "课内体系->知识点->基本初等函数->实数指数幂运算" ]

[ "由$$\\frac{ \\pi }{4}\\textless{}1$$知$$\\cos 1\\textless{}\\sin 1\\textless{}1\\textless{}\\text{tg}1$$，从而$${{\\log }_{\\sin 1}}\\text{tg1}\\textless{}0$$且$${{\\log }_{\\cos 1}}\\text{tg1}\\textless{}0$$，而$${{\\log }_{\\sin 1}}\\text{cos 1}\\textgreater0$$且$${{\\log }_{\\cos 1}}\\text{sin 1}\\textgreater0$$，于是可排除$$\\text{A}$$和$$\\text{B}$$，只剩$$\\text{C}$$和$$\\text{D}$$又由$${{\\log }_{\\cos 1}}\\text{sin 1}\\textless{}{{\\log }_{\\cos 1}}\\text{cos 1=}{{\\log }_{\\sin 1}}\\text{sin 1}{{\\log }_{\\sin 1}}\\text{cos 1}$$即知应． 故选$$\\text{C}$$． " ]

[ "2016年AMC10竞赛A第20题" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$17$$ " } ], [ { "aoVal": "E", "content": "$$19$$ " } ] ]

[]

[ "All the desired terms are in the form $$a^{}xb^{}yc^{}zd^{}w1^{}t$$, where $$x+y+z+w+t=N$$ (the $$1^{}t$$ part is necessary to make stars and bars work better.) Since $$x$$, $$y$$, $$z$$, and~ $$w$$ must be at least $$1$$ ($$t$$ can be $$0$$), let $$x^{}\\prime =x-1$$, $$y^{}\\prime =y-1$$, $$z^{}\\prime =z-1$$, and $$w^{}\\prime =w-1$$, so $$x^{}\\prime +y^{}\\prime+z^{}\\prime+w^{}\\prime+t=N-4$$. Now, we use stars and bars to see that there are $$\\left(\\frac {(N-4)+4}4\\right)$$ or $$\\left(\\frac N4\\right)$$ solutions to this equation. We notice that $$1001=7\\cdot 11\\cdot 13$$, which leads us to guess that $$N$$ is around these numbers. This suspicion proves to be correct, as we see that $$\\left(\\frac {14}4\\right)=1001$$, giving us our answer of $$N=14$$. An alternative is to instead make the transformation $$6^{}\\prime =t+1$$, so $$x+y+z+w+t^{}\\prime =N+1$$, and all variables are positive integers. The solution to this, by Stars and Bars is $$\\left(\\frac {(N+1)-1}4\\right)=\\left(\\frac N4\\right)$$ and we can proceed as above. ", "<p>By Hockey Stick Identity, the number of terms that have all $$a$$, $$b$$, $$c$$, $$d$$ raised to a positive power is $$\\left(\\frac {N-1}3\\right)+\\left(\\frac {N-2}3\\right)+\\cdots +\\left(\\frac 43\\right)+\\left(\\frac 33\\right)=\\left(\\frac N4\\right)$$. We now want to find some $$N$$ such that $$\\left(\\frac N4\\right)=1001$$. As mentioned above, after noticing that $$1001=7\\cdot 11\\cdot 13$$, and some trial and error, we find that $$\\left(\\frac {14}4\\right)=1001$$, giving us our answer of $$N=14$$.</p>" ]

[ "2010年辽宁全国高中数学联赛竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算", "竞赛->知识点->集合->集合的划分与覆盖" ]

[ "$$s$$是奇数，且最小值是$$3$$，最大值是$$27$$，中间缺少$$17$$、$$19$$，故集合$$C$$共有$$11$$个元素． " ]

[ "2018年天津全国高中数学联赛竞赛初赛第6题6分" ]

[ [ { "aoVal": "A", "content": "$$4\\sqrt{2}$$ " } ], [ { "aoVal": "B", "content": "$$4\\sqrt{3}$$ " } ], [ { "aoVal": "C", "content": "$$5\\sqrt{2}$$ " } ], [ { "aoVal": "D", "content": "$$5\\sqrt{3}$$ " } ] ]

[ "竞赛->知识点->复数与平面向量->复数的应用" ]

[ "$$\\left( z+1 \\right)+\\text{i}\\left( 7-z \\right)=\\left( 1-\\text{i} \\right)z+\\left( 1+7\\text{i} \\right)=\\left( 1-\\text{i} \\right)\\left( z-3+4\\text{i} \\right)$$， 即$$\\left\\textbar{} \\left( z+1 \\right)+\\text{i}\\left( 7-z \\right) \\right\\textbar=\\left\\textbar{} 1-\\text{i} \\right\\textbar\\cdot \\left\\textbar{} z-3+4\\text{i} \\right\\textbar=\\sqrt{2}\\cdot \\left\\textbar{} z-\\left( 3-4\\text{i} \\right) \\right\\textbar$$， 即可以表示点$$z$$到$$3-4\\text{i}$$的距离的$$\\sqrt{2}$$倍， ∵$$z$$在单位圆上， ∴$$\\left\\textbar{} \\left( z+1 \\right)+\\text{i}\\left( 7-z \\right) \\right\\textbar$$的取值范围是$$\\left[ 4\\sqrt{2},6\\sqrt{2} \\right]$$． 故选：$$\\text{D}$$． " ]

[ "2008年全国高中数学联赛竞赛一试第6题6分", "2016年辽宁全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "$$\\left( 0,+\\infty \\right)$$ " } ], [ { "aoVal": "B", "content": "$$\\left( 0,\\frac{\\sqrt{5}+1}{2} \\right)$$ " } ], [ { "aoVal": "C", "content": "$$\\left( \\frac{\\sqrt{5}-1}{2},\\frac{\\sqrt{5}+1}{2} \\right)$$ " } ], [ { "aoVal": "D", "content": "$$\\left( \\frac{\\sqrt{5}-1}{2},+\\infty \\right)$$ " } ] ]

[ "课内体系->知识点->数列->等比数列->等比数列的性质及应用", "课内体系->知识点->解三角形->解三角形的实际应用", "课内体系->素养->数据分析", "课内体系->素养->数学运算" ]

[ "$$\\frac{\\sin A\\cot C+\\cos A}{\\sin B\\cot C+\\cos B}=\\frac{\\sin A\\cos C+\\sin C\\cos A}{\\sin B\\cos C+\\sin C\\cos B}=\\frac{\\sin \\left( A+C \\right)}{\\sin \\left( B+C \\right)}=\\frac{\\sin B}{\\sin A}=\\frac{b}{a}$$， 由题设有$${{b}^{2}}=ac$$，则$$a\\left\\textbar{} a-b \\right\\textbar\\textless{}{{b}^{2}}\\textless{}a\\left( a+b \\right)$$，解得$$\\frac{\\sqrt{5}-1}{2}\\textless{}\\frac{b}{a}\\textless{}\\frac{\\sqrt{5}+1}{2}$$． 故选$$\\text{C}$$． " ]

[ "2011年天津全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$-4$$ " } ], [ { "aoVal": "B", "content": "$$-2$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "竞赛->知识点->导数模块->导数", "竞赛->知识点->函数->基本初等函数" ]

[ "设切点的横坐标为$${{x}_{0}}$$．在$$x={{x}_{0}}$$处，曲线$$y={{\\text{e}}^{x+a}}$$的斜率为$${{\\text{e}}^{{{x}_{0}}+a}}$$．而直线$$y=x-3$$的斜率为$$1$$．因此$${{\\text{e}}^{{{x}_{0}}+a}}=1$$，得$${{x}_{0}}=-a$$．因此，切点的纵坐标$${{\\text{e}}^{{{x}_{0}}+a}}=1={{x}_{0}}-3$$，即$$1=-a-3$$，所以$$a=-4$$． " ]

[ "2008年浙江全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "直角三角形 " } ], [ { "aoVal": "B", "content": "等边三角形 " } ], [ { "aoVal": "C", "content": "等腰直角三角形 " } ], [ { "aoVal": "D", "content": "等腰三角形 " } ] ]

[ "竞赛->知识点->三角函数->三角形中的问题->解三角形" ]

[ "因为$$\\overrightarrow{PB}-\\overrightarrow{PA}=\\overrightarrow{AB}$$，$$\\overrightarrow{PB}+\\overrightarrow{PA}-2\\overrightarrow{PC}=\\overrightarrow{CB}+\\overrightarrow{CA}$$， 所以已知条件可改写为$$\\overrightarrow{AB}\\cdot (\\overrightarrow{CB}+\\overrightarrow{CA})=0$$． 容易得到此三角形为等腰三角形．故选$$\\text{D}$$． " ]

[ "2002年AMC12竞赛B第22题" ]

[ [ { "aoVal": "A", "content": "$$-2$$ " } ], [ { "aoVal": "B", "content": "$$-1$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{2002}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{1001}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{1}{2}$$ " } ] ]

[ "课内体系->知识点->基本初等函数->对数的概念及其运算->对数的运算", "美国AMC10/12->Knowledge Point->Algebra->Sequence->Sum of the First n Terms for Geometric Sequences" ]

[ "$\\frac{1}{\\log_a{b}}=\\log_b{a}$ " ]

[ "2010年河北全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$31$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$33$$ " } ] ]

[ "竞赛->知识点->集合->集合的划分与覆盖" ]

[ "2010是数列$${{a}_{n}}=2n$$的第$$1005$$项， 设$$2010$$位于第$$n$$组， 则$$\\begin{cases}1+3+5+\\cdots +(2n-1)\\geqslant 1005 1+3+5+\\cdots +(2n-3)\\textless{}1005 \\end{cases}$$， 即$$\\begin{cases}{{n}^{2}}\\geqslant 1005 {{(n-1)}^{2}}\\textless{}1005 \\end{cases}$$， 所以$$n=32$$，故$$2010$$位于第$$32$$组． " ]

[ "2008年山东全国高中数学联赛竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{31}{65}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{32}{65}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{33}{65}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{34}{65}$$ " } ] ]

[ "竞赛->知识点->复数与平面向量->平面向量的概念与运算", "竞赛->知识点->三角函数->三角恒等变换" ]

[ "由题意知$$\\textbar\\vec{a}-\\vec{b}{{\\textbar}^{2}}={{(\\cos \\alpha -\\cos \\beta )}^{2}}+{{(\\sin \\alpha -\\sin \\beta )}^{2}}$$ $$=2-2(\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta )=2-2\\cos (\\alpha -\\beta )$$， 则$$2-2\\cos (\\alpha -\\beta )=\\frac{4}{5}$$，$$\\cos (\\alpha -\\beta )=\\frac{3}{5}$$． 由$$\\alpha =(\\alpha -\\beta )+\\beta $$，且$$0\\textless{}\\alpha \\textless{}\\frac{ \\pi }{2}$$，$$-\\frac{ \\pi }{2}\\textless{}\\beta \\textless{}0$$及$$\\sin \\beta =-\\frac{5}{13}$$，$$\\cos (\\alpha -\\beta )=\\frac{3}{5}$$， 知$$\\cos \\beta =\\frac{12}{13}$$，$$\\sin (\\alpha -\\beta )=\\frac{4}{5}$$， 从而得$$\\sin \\alpha =\\sin [\\beta +(\\alpha -\\beta )]=\\sin \\beta \\cos (\\alpha -\\beta )+\\cos \\beta \\sin (\\alpha -\\beta )$$ $$=-\\frac{5}{13}\\times \\frac{3}{5}+\\frac{12}{13}\\times \\frac{4}{5}=\\frac{33}{65}$$， 故选$$\\text{C}$$． " ]

[ "2009年河北全国高中数学联赛竞赛初赛第6题6分" ]

[ [ { "aoVal": "A", "content": "焦点在$$x$$轴上的双曲线 " } ], [ { "aoVal": "B", "content": "双曲线，其焦点所在的轴与$$n$$有关 " } ], [ { "aoVal": "C", "content": "焦点在$$y$$轴上的椭圆 " } ], [ { "aoVal": "D", "content": "椭圆，其焦点所在的轴与$$n$$有关 " } ] ]

[ "竞赛->知识点->解析几何->椭圆" ]

[ "当$$n=1$$时，$$\\sin 19{}^{}\\circ \\textless{}\\cos 19{}^{}\\circ $$； $$n=2$$时， $$\\sin ({{19}^{2}}){}^{}\\circ =\\sin 361{}^{}\\circ =\\sin 1{}^{}\\circ \\textless{}\\cos 1{}^{}\\circ $$； $$n=2k(k\\in N*)$$时， $$\\sin ({{19}^{2k}}){}^{}\\circ =\\sin ({{361}^{k}}){}^{}\\circ =\\sin 1{}^{}\\circ \\textless{}\\cos 1{}^{}\\circ $$； $$n=2k+1(k\\in N*)$$时， $$\\sin ({{19}^{2k+1}}){}^{}\\circ =\\sin [19\\times {{(361)}^{k}}]{}^{}\\circ =\\sin 19{}^{}\\circ \\textless{}\\cos 19{}^{}\\circ $$． 所以方程表示焦点在$$y$$轴上的椭圆．选$$\\text{C}$$． " ]

[ "2011年浙江全国高中数学联赛竞赛初赛第9题5分" ]

[ [ { "aoVal": "A", "content": "$$\\left( \\frac{1}{2},1 \\right)$$ " } ], [ { "aoVal": "B", "content": "$$\\left[ \\frac{1}{2},1 \\right]$$ " } ], [ { "aoVal": "C", "content": "$$\\left[ \\frac{1}{2},1 \\right)$$ " } ], [ { "aoVal": "D", "content": "$$\\left( \\frac{1}{2},1 \\right]$$ " } ] ]

[ "课内体系->知识点->三角函数->三角函数的图象与性质->正弦型函数的图象与性质", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦函数的图象和性质", "课内体系->知识点->函数的应用->函数的零点->零点的个数问题->已知零点情况求参数的取值范围", "课内体系->知识点->函数的应用->函数的零点->函数零点的概念", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的定义域", "课内体系->素养->直观想象", "课内体系->素养->数学运算", "课内体系->思想->数形结合思想" ]

[ "$$\\because x\\in \\left[ 0,\\frac{\\pi }{2} \\right]$$，$$\\therefore 2x-\\frac{\\pi }{6}\\in \\left[ -\\frac{\\pi }{6},\\frac{5\\pi }{6} \\right]$$，$$\\therefore \\sin \\left( 2x-\\frac{\\pi }{6} \\right)\\in \\left[ -\\frac{1}{2},1 \\right]$$，令$$t=2x-\\frac{\\pi }{6}$$，$$y=m$$， 在同一直角坐标系中作出$$y=\\sin t$$（$$t\\in \\left[ -\\frac{\\pi }{6},\\frac{5\\pi }{6} \\right]$$）与$$y=m$$的图象，由图象可知， 当$$\\frac{1}{2}\\leqslant m \\textless{} 1$$时，$$y=\\sin t$$（$$t\\in \\left[ -\\frac{\\pi }{6},\\frac{5\\pi }{6} \\right]$$）的图象与直线$$y=m$$有两个交点， 即函数$$f(x)=\\sin \\left( 2x-\\frac{\\pi }{6}-m \\right)$$在$$\\left[ 0,\\frac{\\pi }{2} \\right]$$上有两个零点． " ]

[ "2009年湖南全国高中数学联赛竞赛初赛第11题7分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{2\\sqrt{3}}{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{4\\sqrt{3}}{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\sqrt{3}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{5\\sqrt{3}}{3}$$ " } ] ]

[ "竞赛->知识点->解析几何->双曲线" ]

[ "易知$$a=2, c=\\sqrt{4+{{b}^{2}}}, e=\\frac{\\sqrt{4+{{b}^{2}}}}{2}$$． 渐近线方程为$$\\frac{x}{2}\\pm \\frac{y}{b}=0$$，即$$bx\\pm 2y=0$$． 右焦点$$(\\sqrt{4+{{b}^{2}}}, 0)$$到渐近线的距离$$d=b$$，从而得方程$$b=\\frac{\\sqrt{4+{{b}^{2}}}}{2}$$． 解方程得，$$b=\\frac{2}{3}\\sqrt{3}, e=\\frac{\\sqrt{4+\\frac{4}{3}}}{2}=\\frac{2\\sqrt{3}}{3}$$．故填$$\\frac{2\\sqrt{3}}{3}$$． " ]

[ "2016年吉林全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$-\\frac{7}{25}$$ " } ], [ { "aoVal": "B", "content": "$$-\\frac{24}{25}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{24}{25}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{7}{25}$$ " } ] ]

[ "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式的化简和求值", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->半角公式", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正弦", "课内体系->素养->数学运算" ]

[ "$$\\cos 2x=\\cos \\left( 2x-\\frac{\\mathrm{ }\\pi }{2}+\\frac{\\mathrm{ }\\pi }{2} \\right)$$ $$=-\\sin \\left( 2x-\\frac{\\mathrm{ }\\pi }{2} \\right)$$ $$=-2\\sin \\left( x-\\frac{\\mathrm{ }\\pi }{4} \\right)\\cos \\left( x-\\frac{\\mathrm{ }\\pi }{4} \\right)$$． 故选$$\\text{B}$$． " ]

[ "2017~2018学年3月天津河西区天津市新华中学高一下学期月考第8题", "2009年湖南全国高中数学联赛竞赛初赛第6题5分", "2018~2019学年陕西西安碑林区西安建筑科技大学附属中学高一下学期期中（必修4）第10题5分", "2016~2017学年湖北黄冈黄州区湖北省黄冈中学高一上学期期末第7题5分", "2018~2019学年山东济南历城区济南外国语学校高二上学期开学考试第9题5分", "2019~2020学年山东济南历城区济南历城一中高二上学期开学考试第9题5分" ]

[ [ { "aoVal": "A", "content": "三边均不相等的三角形 " } ], [ { "aoVal": "B", "content": "直角三角形 " } ], [ { "aoVal": "C", "content": "等腰非等边三角形 " } ], [ { "aoVal": "D", "content": "等边三角形 " } ] ]

[ "竞赛->知识点->复数与平面向量->平面向量的概念与运算", "竞赛->知识点->复数与平面向量->平面向量的应用" ]

[ "设$$\\frac{\\overrightarrow{AB}}{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}=\\overrightarrow{AE}$$，$$\\frac{\\overrightarrow{AC}}{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}=\\overrightarrow{AF}$$， 则原式$$=\\left( \\overrightarrow{AE}+\\overrightarrow{AF} \\right)\\cdot \\overrightarrow{BC}=0$$， 即$$\\overrightarrow{AD}\\cdot \\overrightarrow{BC}=0$$， ∴$$AD\\bot BC$$， ∴四边形$$AEDF$$为菱形． ∵$$\\overrightarrow{AE}\\cdot \\overrightarrow{AF}=\\left\\textbar{} \\overrightarrow{AE} \\right\\textbar\\cdot \\left\\textbar{} \\overrightarrow{AF} \\right\\textbar\\cos \\angle BAC=\\frac{1}{2}$$， ∴$$\\cos \\angle BAC=\\frac{1}{2}$$， ∴$$\\angle BAC=60{}^{}\\circ $$， ∴$$\\angle BAD=\\angle DAC=30{}^{}\\circ $$， ∴$$\\triangle ABH$$ ≌ $$\\triangle AHC$$， ∴$$AB=AC$$， ∴$$\\triangle ABC$$为等边三角形． 故选$$\\text{D}$$． 设$$\\frac{\\overrightarrow{AB}}{\\textbar\\overrightarrow{AB\\textbar}}$$为$$\\overrightarrow{AB}$$上的单位向量，$$\\frac{\\overrightarrow{AC}}{\\left\\textbar{} AC \\right\\textbar}$$为$$\\overrightarrow{AC}$$上的单位向量． 则$$\\frac{\\overrightarrow{AB}}{\\textbar\\overrightarrow{AB}\\textbar}+\\frac{\\overrightarrow{AC}}{\\textbar\\overrightarrow{AC}\\textbar}$$的方向为$$\\angle BAC$$的角平分线$$\\overrightarrow{AO}$$的方向． 而$$\\overrightarrow{AO}\\cdot \\overrightarrow{BC}=0$$，∴$$\\overrightarrow{AO}\\bot \\overrightarrow{BC}$$． ∴$$\\triangle ABC$$为等腰三角形，$$AB=AC$$． $$\\frac{\\overrightarrow{AB}}{\\textbar\\overrightarrow{AB}\\textbar}\\cdot \\frac{\\overrightarrow{AC}}{\\textbar\\overrightarrow{AC}\\textbar}=\\left\\textbar{} \\frac{\\overrightarrow{AB}}{\\textbar\\overrightarrow{AB}\\textbar} \\right\\textbar\\left\\textbar{} \\frac{\\overrightarrow{AC}}{\\textbar\\overrightarrow{AC}\\textbar} \\right\\textbar\\cos \\angle BAC=\\frac{1}{2}$$． ∵$$\\left\\textbar{} \\frac{\\overrightarrow{AB}}{\\textbar\\overrightarrow{AB}\\textbar} \\right\\textbar=\\left\\textbar{} \\frac{\\overrightarrow{AC}}{\\textbar\\overrightarrow{AC}\\textbar} \\right\\textbar=1$$，∴$$\\angle BAC=60{}^{}\\circ $$． 综上所述，$$\\triangle ABC$$为等边三角形． ", "<p>由$$\\left( \\frac{\\overrightarrow{AB}}{|\\overrightarrow{AB}|}+\\frac{\\overrightarrow{AC}}{|\\overrightarrow{AC}|} \\right)\\cdot \\overrightarrow{BC}=0$$，得$$\\angle BAC$$的平分线垂直于$$BC$$，</p>\n<p>&there4;$$AB=AC$$，</p>\n<p>∵$$\\frac{\\overrightarrow{AB}}{|\\overrightarrow{AB}|}\\cdot \\frac{\\overrightarrow{AC}}{|\\overrightarrow{AC}|}=\\cos \\left\\langle \\overrightarrow{AB},\\overrightarrow{AC} \\right\\rangle =\\frac{1}{2}$$，$$\\left\\langle \\overrightarrow{AB},\\overrightarrow{AC} \\right\\rangle =\\in (0{}^\\circ ,180{}^\\circ )$$，</p>\n<p>&there4;$$\\angle BAC=60{}^\\circ $$，$$\\triangle ABC$$为等边三角形．</p>\n<p>故选$$\\text{D}$$．</p>\n" ]

[ "2015年辽宁全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2}{3}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{3}{2}$$ " } ] ]

[ "竞赛->知识点->立体几何与空间向量->空间向量" ]

[ "用向量法. " ]

[ "2010年浙江全国高中数学联赛竞赛初赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$${{a}_{2}}\\textgreater{{b}_{2}}$$ " } ], [ { "aoVal": "B", "content": "$${{a}_{3}}\\textless{}{{b}_{3}}$$ " } ], [ { "aoVal": "C", "content": "$${{a}_{5}}\\textgreater{{b}_{5}}$$ " } ], [ { "aoVal": "D", "content": "$${{a}_{6}}\\textgreater{{b}_{6}}$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "设等差数列的公差为$$d$$，等比数列公比为$$q$$，由$${{a}_{1}}={{b}_{1}}=4,{{a}_{4}}={{b}_{4}}=1,$$ 得$$d=-1, q=\\frac{\\sqrt[3]{2}}{2},$$得$${{a}_{2}}=3, {{b}_{2}}=2\\sqrt[3]{2}$$；$${{a}_{3}}=2, {{b}_{3}}=\\sqrt[3]{4}$$；$${{a}_{5}}=0,{{b}_{5}}=\\frac{\\sqrt[3]{2}}{2}$$；$${{a}_{6}}=-1,,$$ $$ {{b}_{6}}=\\frac{\\sqrt[3]{4}}{4}$$ " ]

[ "2010年高考真题江苏卷理科第10题5分", "2018~2019学年3月河南郑州金水区郑州市第七中学高一下学期月考第16题5分", "2018~2019学年11月河北石家庄辛集市河北辛集中学高一上学期月考第22题5分", "2010年湖南全国高中数学联赛竞赛初赛第7题7分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{4}{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{6}{5}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{2}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2}{3}$$ " } ] ]

[ "竞赛->知识点->三角函数->三角函数的图像与性质", "课内体系->知识点->三角函数" ]

[ "由$$\\begin{cases}y=6\\cos x y=5\\tan x \\end{cases}$$消$$y$$得$$6\\cos x=5\\tan x$$．整理得$$6{{\\cos }^{2}}x=5\\sin x$$， $$6{{\\sin }^{2}}x+5\\sin x-6=0$$．$$\\left( 3\\sin x-2 \\right)\\left( 2\\sin x+3 \\right)=0$$，所以$$\\sin x=\\frac{2}{3}$$或$$\\sin x=-\\frac{3}{2}$$（舍去）． 故点$${{P}_{2}}$$的纵坐标为$${{y}_{2}}=\\frac{2}{3}$$．所以$$\\left\\textbar{} {{P}_{1}}{{P}_{2}} \\right\\textbar=\\frac{2}{3}$$． " ]

[ "2015年天津全国高中数学联赛竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "充分条件但不是必要条件 " } ], [ { "aoVal": "B", "content": "必要条件但不是充分条件 " } ], [ { "aoVal": "C", "content": "充分必要条件 " } ], [ { "aoVal": "D", "content": "既非充分也非必要条件 " } ] ]

[ "竞赛->知识点->逻辑->常用逻辑用语", "竞赛->知识点->集合->集合的概念与运算" ]

[ "如果$$B$$和$$C$$都是$$A$$的子集，则$$A\\cap B=B$$，$$A\\cap C=C$$，从而$$\\left( A\\cap B \\right)\\cup \\left( A\\cap C \\right)=B\\cup C$$成立；反过来，如果$$\\left( A\\cap B \\right)\\cup \\left( A\\cap C \\right)=B\\cup C$$成立，则由$$A\\cap B$$和$$A\\cap C$$都是$$A$$的子集知$$B\\cup C$$是$$A$$的子集，即$$B$$和$$C$$都是$$A$$的子集．故选$$\\text{C}$$． " ]

[ "1985年全国高中数学联赛竞赛一试第2题" ]

[ [ { "aoVal": "A", "content": "$${{S}_{1}}\\textgreater{{S}_{2}}$$ " } ], [ { "aoVal": "B", "content": "$${{S}_{1}}\\textless{}{{S}_{2}}$$ " } ], [ { "aoVal": "C", "content": "$${{S}_{1}}\\mathsf{\\geqslant }{{S}_{2}}$$ " } ], [ { "aoVal": "D", "content": "有时$${{S}_{1}}\\textgreater{{S}_{2}}$$，有时$${{S}_{1}}={{S}_{2}}$$，有时$${{S}_{1}}\\textless{}{{S}_{2}}$$． " } ] ]

[ "竞赛->知识点->解析几何->抛物线" ]

[ "在$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$、$$\\text{D}$$四个答案中，$$\\text{C}$$包含着$$\\text{A}$$，即若$$\\text{A}$$正确，则$$\\text{C}$$必定正确，因此不必考虑$$\\text{A}$$． 当过焦点的弦$$PQ$$的倾角（与$$x$$轴的夹角）很小时，$$PQ$$绕$$l$$旋转一周所得旋转体（圆台）的母线很长，而相应的高却很短，亦即球的直径很小，因此$$\\text{B}$$不成立． 我们知道，圆台的侧面积由母线$$PQ$$和两底半径$$PMQN$$的长确定．根据抛物线的定义， $$\\left\\textbar{} PM \\right\\textbar=\\left\\textbar{} PF \\right\\textbar\\mathsf{}\\left\\textbar{} QN \\right\\textbar=\\left\\textbar{} QF \\right\\textbar$$ （其中$$F$$为焦点） 若取$$PQ$$的倾角$$\\alpha $$为参数，设 $$\\left\\textbar{} PF \\right\\textbar={{\\rho }_{1}}\\mathsf{} \\left\\textbar{} QF \\right\\textbar={{\\rho }_{2}}$$ 则$$\\left\\textbar{} PM \\right\\textbar={{\\rho }_{1}}\\mathsf{} \\left\\textbar{} QN \\right\\textbar={{\\rho }_{2}}\\mathsf{} \\left\\textbar{} PQ \\right\\textbar={{\\rho }_{1}}+{{\\rho }_{2}}$$ 于是 $${{S}_{1}}= \\pi ({{\\rho }_{1}}+{{\\rho }_{2}}{{)}^{2}}$$， $${{S}_{2}}= \\pi \\centerdot {{\\left\\textbar{} MN \\right\\textbar}^{2}}= \\pi {{\\sin }^{2}}\\alpha {{({{\\rho }_{1}}+{{\\rho }_{2}})}^{2}}$$ 显然$${{S}_{1}}\\mathsf{\\geqslant }{{S}_{2}}$$，且当$$a=\\frac{ \\pi }{2}$$时取到等号． " ]

[ "2009年浙江全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$(-16,16)$$ " } ], [ { "aoVal": "B", "content": "$$\\left[ -16, 16 \\right]$$ " } ], [ { "aoVal": "C", "content": "$$(-\\infty ,-8)$$ " } ], [ { "aoVal": "D", "content": "$$(8,+\\infty )$$ " } ] ]

[ "竞赛->知识点->导数模块->导数" ]

[ "令$$f(x)={{x}^{3}}-12x+a$$，则$$f\\prime (x)=3{{x}^{2}}-12$$， $$f\\prime (x)=3{{x}^{2}}-12=0\\Rightarrow x=\\pm 2$$． 要使$$f(x)=0$$有三个不同的实数根，则必须有$$f(2)f(-2)\\textless{}0$$，即$$(a-16)(a+16)\\textless{}0$$，也即有$$-16\\textless a\\textless16$$． " ]

[ "2018年四川全国高中数学联赛竞赛初赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->数列的通项与求和" ]

[ "由条件设$$S_n=kn(2n+6)$$，$$T_n=kn(n+1)$$． 当$$m=1$$时，$$\\frac{a_{1}}{b_{1}}= \\frac{S_{1}}{T_{1}}= \\frac{8}{2}=4$$ ，不满足题中条件，舍去． 当$$m\\geqslant 2$$时， $$\\frac{a_{m}}{b_{m}}= \\frac{S_{m}-S_{m-1}}{T_{m}-T_{m-1}}= \\frac{4m+4}{2m}=2+ \\frac{2}{m}$$． 于是，仅在$$m=2$$时， $$\\frac{a_{m}}{b_{m}}=3$$ 为素数． 因此，所求正整数$$m=2$$． 故选$$\\text{A}$$． " ]

[ "2007年全国全国高中数学联赛竞赛一试第4题6分" ]

[ [ { "aoVal": "A", "content": "$$-\\frac{1}{2}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "C", "content": "−1 " } ], [ { "aoVal": "D", "content": "1 " } ] ]

[ "课内体系->知识点->三角函数->三角恒等变换->和差角公式->辅助角公式", "课内体系->知识点->三角函数->三角恒等变换->倍角、和差角公式综合", "课内体系->素养->数学运算" ]

[ "令$$c= \\pi $$，则对任意的$$x\\in \\mathbf{R}$$，都有$$f(x)+f(x-c)=2$$，于是取$$a=b=\\frac{1}{2}$$，$$c= \\pi $$，则对任意的$$x\\in \\mathbf{R}$$，$$af(x)+bf(x-c)=1$$，由此得$$\\frac{b\\cos c}{a}=-1$$． 一般地，由题设可得$$f(x)=\\sqrt{13}\\sin (x+\\phi )+1$$，$$f(x-c)=\\sqrt{13}\\sin (x+\\phi -c)+1$$，其中$$0 ~\\textless{} ~\\phi ~~\\textless{} ~\\frac{\\pi }{2}$$且$$\\tan \\phi =\\frac{2}{3}$$，于是$$af(x)+bf(x-c)=1$$可化为 $$\\sqrt{13}a\\sin (x+\\phi )+\\sqrt{13}b\\sin (x+\\phi -c)+a+b=1$$，即 $$\\sqrt{13}a\\sin (x+\\phi )+\\sqrt{13}b\\sin (x+\\phi )\\cos c-\\sqrt{13}b\\sin c\\cos (x+\\phi )+(a+b-1)=0$$，所以 $$\\sqrt{13}(a+b\\cos c)\\sin (x+\\phi )-\\sqrt{13}b\\sin c\\cos (x+\\phi )+(a+b-1)=0$$． 由已知条件，上式对任意$$x$$∈$$R$$恒成立，故必有$$\\begin{cases} a+b\\cos c=0 （1） b\\sin c=0 （2） a+b-1=0 （3） \\end{cases}$$， 若$$b$$=0，则由（$$1$$）知$$a$$=0，显然不满足（$$3$$）式，故$$b\\ne 0$$．所以，由（$$2$$）知$$\\sin c$$=0，故$$c=2k \\pi + \\pi $$或$$c=2k \\pi $$（$$k\\in \\mathbf{Z}$$）．当$$c=2k \\pi $$时，$$\\cos c=1$$，则（$$1$$）、（$$3$$）两式矛盾．故$$c=2k \\pi + \\pi $$（$$k\\in \\mathbf{Z}$$），$$\\cos c$$ =−1．由（$$1$$）、（$$3$$）知$$a=b=\\frac{1}{2}$$，所以$$\\frac{b\\cos c}{a}=-1$$． 故选$$\\text{C}$$． " ]

[ "2005年全国高中数学联赛竞赛一试第2题6分" ]

[ [ { "aoVal": "A", "content": "只有一个 " } ], [ { "aoVal": "B", "content": "有二个 " } ], [ { "aoVal": "C", "content": "有四个 " } ], [ { "aoVal": "D", "content": "有无穷多个 " } ] ]

[ "竞赛->知识点->立体几何与空间向量->空间向量" ]

[ "注意到$${{3}^{2}}+{{11}^{2}}=1130={{7}^{2}}+{{9}^{2}}$$，由于$$\\overrightarrow{AB}+\\overrightarrow{BC}+\\overrightarrow{CD}+\\overrightarrow{DA}=\\vec{0}$$， 则$$D{{A}^{2}}={{\\overrightarrow{DA}}^{2}}={{(\\overrightarrow{AB}+\\overrightarrow{BC}+\\overrightarrow{CD})}^{2}}=A{{B}^{2}}+B{{C}^{2}}+C{{D}^{2}}+2(\\overrightarrow{AB}\\cdot \\overrightarrow{BC}+\\overrightarrow{BC}\\cdot \\overrightarrow{CD}+\\overrightarrow{CD}\\cdot \\overrightarrow{AB})$$ $$=A{{B}^{2}}-B{{C}^{2}}+C{{D}^{2}}+2({{\\overline{BC}}^{2}}+\\overrightarrow{AB}\\cdot \\overrightarrow{BC}+\\overrightarrow{BC}\\cdot \\overrightarrow{CD}+\\overrightarrow{CD}\\cdot \\overrightarrow{AB})$$ $$=A{{B}^{2}}-B{{C}^{2}}+C{{D}^{2}}+2(\\overrightarrow{AB}+\\overrightarrow{BC})\\cdot (\\overrightarrow{BC}+\\overrightarrow{CD})$$， 即$$2\\overrightarrow{AC}\\cdot \\overrightarrow{BD}=A{{D}^{2}}+B{{C}^{2}}-A{{B}^{2}}-C{{D}^{2}}=0$$，∴$$\\overrightarrow{AC}\\cdot \\overrightarrow{BD}$$只有一个值得$$0$$， 故选$$\\text{A}$$． " ]

[ "2017年福建全国高中数学联赛竞赛初赛第9题6分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

[ "竞赛->知识点->不等式->几个重要的不等式->柯西", "课内体系->知识点->等式与不等式->不等式->柯西不等式" ]

[ "由柯西不等式，$${{\\left( f\\left( x \\right) \\right)}^{2}}\\leqslant \\left( 3+2+6 \\right)\\left( \\frac{2x-7}{3}+\\frac{12-x}{2}+\\frac{44-x}{6} \\right)={{11}^{2}}$$，所以$$f\\left( x \\right)\\leqslant 11$$，当且仅当$$x=8$$时等号成立． " ]

[ "2009年AMC10竞赛A第19题" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$50$$ " } ], [ { "aoVal": "E", "content": "$$90$$ " } ] ]

[ "课内体系->知识点->圆锥曲线", "美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Circle" ]

[ "The circumference of circle $$A$$ is $$200\\pi$$ , and the circumference of circle $$B$$ with radius r is $$2r\\pi$$. Since circle $$B$$ makes a complete revolution and ends up on the same point, the circumference of $$A$$ must be a multiple of the circumference of $$B$$, therefore the quotient must be an integer. Thus, $$\\dfrac{200 \\pi}{2 \\pi \\cdot r}= \\dfrac{100}{r}$$. Therefore $$r$$ must then be a factor of $$100$$, excluding $$100$$ (because then circle $$B$$ would be the same size as circle $$A$$). $$100 =2^{2}\\cdot5^{2}$$. Therefore $$100$$ has $$(2+1)\\cdot (2+1)$$ factors*. But you need to subtract $$1$$ from $$9$$, in order to exclude $$100$$. Therefore the answer is $$8$$. " ]

[ "2008年山东全国高中数学联赛竞赛初赛第17题12分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "前三个答案都不对 " } ] ]

[ "竞赛->知识点->不等式->换元技巧->代数换元", "竞赛->知识点->不等式->不等式的证明", "课内体系->方法->换元法" ]

[ "任取$$a\\textgreater0$$，令$$b=ax，c=by$$，由$$xyz=1$$ 得$$x=\\frac{b}{a}，y=\\frac{c}{b}，z=\\frac{a}{c}$$， 从而有$$\\frac{1}{1+x}+\\frac{1}{1+y}+\\frac{1}{1+z}=\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{a+c}$$ $$\\textgreater\\frac{a}{a+b+c}+\\frac{b}{a+b+c}+\\frac{c}{a+b+c}=1$$， 又 $$\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{a+c} ~\\textless{} ~\\frac{a+c}{a+b+c}+\\frac{a+b}{a+b+c}+\\frac{b+c}{a+b+c}=2$$， 所以 $$1 ~\\textless{} ~\\frac{1}{1+x}+\\frac{1}{1+y}+\\frac{1}{1+z} ~\\textless{} ~2$$． " ]

[ "2011年天津全国高中数学联赛竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$\\left( -\\infty ,\\frac{\\pi }{2} \\right)$$ " } ], [ { "aoVal": "B", "content": "$$\\left( -\\infty ,\\left. \\frac{ \\pi }{2} \\right] \\right.$$ " } ], [ { "aoVal": "C", "content": "$$\\left( -\\infty ,\\frac{2}{\\pi } \\right)$$ " } ], [ { "aoVal": "D", "content": "$$\\left( -\\infty ,\\left. \\frac{2}{ \\pi } \\right] \\right.$$ " } ] ]

[ "竞赛->知识点->三角函数->三角函数的图像与性质" ]

[ "作出$$y=\\sin x$$和$$y=kx$$的图象，易知选$$\\text{C}$$． " ]

[ "2007年全国全国高中数学联赛竞赛一试第6题6分", "2019~2020学年上海浦东新区上海市建平中学高一上学期期中第16题3分" ]

[ [ { "aoVal": "A", "content": "$$62$$ " } ], [ { "aoVal": "B", "content": "$$66$$ " } ], [ { "aoVal": "C", "content": "$$68$$ " } ], [ { "aoVal": "D", "content": "$$74$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "先证$$\\textbar A\\cup B\\textbar\\leqslant 66$$，只须证$$\\left\\textbar{} A \\right\\textbar\\leqslant 33$$，为此只须证若$$A$$是{$$1$$，$$2$$，\\ldots，$$49$$}的任一个$$34$$元子集，则必存在$$n\\in A$$，使得$$2n+2\\in B$$，证明如下： 将{$$1$$，$$2$$，\\ldots，$$49$$}分成如下$$33$$个集合：{$$1$$，$$4$$}，{$$3$$，$$8$$}，{$$5$$，$$12$$}，\\ldots，{$$23$$，$$48$$}共$$12$$个；{$$2$$，$$6$$}，{$$10$$，$$22$$}，{$$14$$，$$30$$}，{$$18$$，$$38$$}共$$4$$个；{$$25$$}，{$$27$$}，{$$29$$}，\\ldots，{$$49$$}共$$13$$个；{$$26$$}，{$$34$$}，{$$42$$}，{$$46$$}共$$4$$个．由于$$A$$是{$$1$$，$$2$$，\\ldots，$$49$$}的$$34$$元子集，从而由抽屉原理可知上述$$33$$个集合中至少有一个$$2$$元集合中的数均属于$$A$$，即存在$$n\\in A$$，使得$$2n+2\\in B$$， 如取$$A$$={$$1$$，$$3$$，$$5$$，\\ldots，$$23$$，$$2$$，$$10$$，$$14$$，$$18$$，$$25$$，$$27$$，$$29$$，\\ldots，$$49$$，$$26$$，$$34$$，$$42$$，$$46$$}，$$B$$=$$ {2n+2\\textbar n\\in A }$$，则$$AB$$满足题设且$$\\textbar A\\cup B\\textbar\\leqslant 66$$． 故选$$\\text{B}$$． " ]

[ "2011年辽宁全国高中数学联赛竞赛初赛第3题6分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{3}{5}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{4}{7}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{3}{7}$$ " } ] ]

[ "竞赛->知识点->排列组合与概率->概率初步" ]

[ "$$7$$个球全被取出的方式共有$$\\text{C}_{7}^{3}=\\text{C}_{7}^{4}=35$$（种），$$3$$个黑球全部被取出的方式有$$\\text{C}_{6}^{3}=20$$（种），故所求概率为$$\\frac{20}{35}=\\frac{4}{7}$$． " ]

[ "2011年黑龙江全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{4}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{8}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{16}$$ " } ] ]

[ "竞赛->知识点->解析几何->直线与圆综合" ]

[ "直线$$mx+ny=4$$和圆$${{x}^{2}}+{{y}^{2}}=4$$相离，则$$\\frac{\\left\\textbar{} 4 \\right\\textbar}{\\sqrt{{{m}^{2}}+{{n}^{2}}}}\\textgreater2$$，得$${{m}^{2}}+{{n}^{2}}\\textless{}4$$． 方程$${{x}^{2}}+2mx+{{n}^{2}}=0$$有实根时，判别式$$\\Delta =4{{m}^{2}}-4{{n}^{2}}\\geqslant 0\\Rightarrow \\left\\textbar{} m \\right\\textbar\\geqslant \\left\\textbar{} n \\right\\textbar$$． 所以有实根的概率为$$\\frac{1}{2}$$． " ]

[ "2016~2017学年10月北京西城区北京市第十四中学高三上学期月考理科第1题5分", "2014年北京海淀区高三一模理科第1题", "2016年山西太原高三三模文科第1题5分", "2014年黑龙江全国高中数学联赛竞赛初赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$\\left { \\frac{1}{2}\\right }$$ " } ], [ { "aoVal": "B", "content": "$$\\left { 2 \\right }$$ " } ], [ { "aoVal": "C", "content": "$$\\left { 1 \\right }$$ " } ], [ { "aoVal": "D", "content": "$$\\varnothing $$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->知识点->集合->集合的基本运算->交集", "课内体系->知识点->集合->集合的基本运算->交、并、补集混合运算" ]

[ "由题可知：$$B=\\left { 1,4,\\frac{1}{4} \\right }$$，所以 $$A\\cap B=\\left {1 \\right }$$． " ]

[ "2016年北京海淀区中国人民大学附属中学高三零模理科第6题5分", "2014年吉林全国高中数学联赛竞赛初赛第4题6分", "2016~2017学年2月湖南长沙岳麓区湖南师范大学附属中学高三上学期月考理科第6题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{16}{33}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{33}{128}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{32}{33}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{4}{11}$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->知识点->统计与概率->概率->事件与概率->古典概型->古典概型的概率计算（涉及计数原理）" ]

[ "由题意知本题是一个古典概型， 试验发生的总事件是从$$12$$名选手中选出$$4$$个优胜者，共有$$\\text{C}_{12}^{4}$$种结果， 而满足条件的是选出的$$4$$名选手中恰有且只有两个人 是同一省份的歌手表示从$$6$$个省中选一个省， 它的两名选手都获奖，同时从余下的$$10$$名选手中选一个， 再从剩下的$$4$$个省中选一个，共有$$\\text{C}_{6}^{1}\\text{C}_{10}^{1}\\text{C}_{4}^{1}$$种选法， ∴$$P=\\frac{\\text{C}_{6}^{1}\\text{C}_{10}^{1}\\text{C}_{4}^{1}}{\\text{C}_{12}^{4}}=\\frac{16}{33}$$． 故选$$\\text{A}$$． " ]

[ "2015年吉林全国高中数学联赛竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "$$(-\\infty ,-1)$$ " } ], [ { "aoVal": "B", "content": "$$(-\\infty ,-1]$$ " } ], [ { "aoVal": "C", "content": "$$(-\\infty ,-2)$$ " } ], [ { "aoVal": "D", "content": "$$(-\\infty ,-2]$$ " } ] ]

[ "课内体系->知识点->函数的概念与性质->函数的性质->单调性->函数单调性的应用->利用函数单调性解不等式", "竞赛->知识点->函数->函数的图像与性质" ]

[ "显然$$m\\textless{}0$$，所以$$f\\left( x+m \\right)\\textless{}-mf\\left( x \\right)=f\\left( \\sqrt{-m}x \\right)$$． 因为$$f\\left( x \\right)$$是单调增的奇函数，所以$$x+m\\textless{}\\sqrt{-m}x$$，即$$\\left( \\sqrt{-m}-1 \\right)x\\textgreater m$$． 所以必须$$\\sqrt{-m}-1\\geqslant 0$$，$$m\\leqslant -1$$． " ]

[ "2011年AMC10竞赛A第20题" ]

[ [ { "aoVal": "A", "content": "$$\\dfrac{1}{6}$$ " } ], [ { "aoVal": "B", "content": "$$\\dfrac{1}{5}$$ " } ], [ { "aoVal": "C", "content": "$$\\dfrac{1}{4}$$ " } ], [ { "aoVal": "D", "content": "$$\\dfrac{1}{3}$$ " } ], [ { "aoVal": "E", "content": "$$\\dfrac{1}{2}$$ " } ] ]

[ "课内体系->知识点->统计与概率->概率->事件与概率->几何概型->与角度有关的几何概率的计算", "课内体系->知识点->直线和圆的方程->圆与方程->直线与圆的位置关系->圆的弦长的相关问题", "美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Geometric Models of Probabilities" ]

[ "Fix a point $$A$$ from which we draw a clockwise chord. In order for the clockwise chord from another point $$B$$ to intersect that of point $$A$$, $$A$$ and $$B$$ must be no more than $$r$$ units apart.By drawing the circle, we quickly see that $$B$$ can be on $$\\dfrac{120}{360}=\\dfrac{1}{3}(\\rm D)$$ of the perimeter of the crcle. (lmagine a regular hexagon inscribed in the circle). " ]

[ "2022年浙江宁波竞赛第2题" ]

[ [ { "aoVal": "A", "content": "充分不必要条件 " } ], [ { "aoVal": "B", "content": "必要不充分条件 " } ], [ { "aoVal": "C", "content": "充要条件 " } ], [ { "aoVal": "D", "content": "既不充分也不必要条件 " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 判断条件间的推出关系，根据充分必要性的定义判断即可.\\\\ 【详解】\\\\ 当$a\\textgreater b$：\\\\ 若$a,b$异号，即$a\\textgreater0\\textgreater b$，显然$\\left\\textbar{} a \\right\\textbar\\textgreater b$成立；\\\\ 若$a\\textgreater b\\ge 0$或$0\\ge a\\textgreater b$，均有$\\left\\textbar{} a \\right\\textbar\\textgreater b$成立；\\\\ 所以充分性成立；\\\\ 当$\\left\\textbar{} a \\right\\textbar\\textgreater b$：若$a=-2$，$b=1$，显然$a\\textgreater b$不成立，故必要性不成立.\\\\ 所以``$a\\textgreater b$''是``$\\left\\textbar{} a \\right\\textbar\\textgreater b$''的充分不必要条件.\\\\ 故选：A " ]

[ "2015年吉林全国高中数学联赛竞赛初赛第6题6分" ]

[ [ { "aoVal": "A", "content": "$$(-\\frac{5}{2},-1)$$ " } ], [ { "aoVal": "B", "content": "$$(-\\frac{5}{2},-1]$$ " } ], [ { "aoVal": "C", "content": "$$(-\\frac{5}{2},-\\frac{1}{2})$$ " } ], [ { "aoVal": "D", "content": "$$(-\\frac{5}{2},-\\frac{1}{2}]$$ " } ] ]

[ "竞赛->知识点->导数模块->导数" ]

[ "$${f}'\\left( x \\right)=12{{x}^{2}}-3=3\\left( 2x+1 \\right)\\left( 2x-1 \\right)$$，$$f\\left( x \\right)$$在$$\\left( -\\infty ,-\\frac{1}{2} \\right)$$单调增，在$$\\left( -\\frac{1}{2},\\frac{1}{2} \\right)$$单调减，在$$\\left( \\frac{1}{2},+\\infty \\right)$$单调增，所以$$f\\left( x \\right)$$在$$\\left( a,a+2 \\right)$$上的最大值只能是$$f\\left( -\\frac{1}{2} \\right)=1$$． 因此，$$a\\textless{}-\\frac{1}{2}\\textless{}a+2$$，另外$$f\\left( 1 \\right)=1$$，所以$$a+2\\leqslant 1$$，综合有$$-\\frac{5}{2}\\textless{}a\\leqslant -1$$． " ]

[ "2016年AMC10竞赛第23题5分" ]

[ [ { "aoVal": "A", "content": "$$109$$ " } ], [ { "aoVal": "B", "content": "$$201$$ " } ], [ { "aoVal": "C", "content": "$$301$$ " } ], [ { "aoVal": "D", "content": "$$3049$$ " } ], [ { "aoVal": "E", "content": "33,601 " } ] ]

[]

[ "We see that $$a$$◇$$a=1$$, and think of division. Testing, we see that the first condition $$a$$◇$$(b$$◇$$c)=(a$$◇$$b)\\cdot c$$ is satisfied, because $$\\frac {a}{\\dfrac bc}=\\frac ab\\cdot c$$. Therefore, division can be the operation ◇. Solving the equation, $$\\frac {2016}{\\dfrac 6x}=\\frac {2016}6\\cdot x=336x=100\\Rightarrow x=\\frac {100}{336}=\\frac {25}{84}$$, so the answer is $$25+84=109$$. We can manipulate the given identities to arrive at a conclusion about the binary operator ◇. Substituting $$b=c$$ into the first identity yields $$(a$$◇$$b)\\cdot b=a$$◇$$(b$$◇$$b)=a$$◇$$1=a$$◇$$(a$$◇$$a)=(a$$◇$$a)\\cdot a=a$$. Hence, $$(a$$◇$$b)\\cdot b=a$$, or, dividing both sides of the equation by $$b$$$, $$(a$$◇$$b)=\\frac ab$$. Hence, the given equation becomes $$\\frac {2016}{\\dfrac 6x}=100$$. Solving yields $$x=\\frac {100}{336}=\\frac {25}{84}$$, so the answer is $$25+84=109$$. One way to eliminate the ◇ in this equation is to make $$a=b$$ so that $$a$$◇$$(b$$◇$$c)=c$$. In this case, we can make $$b=2016$$. $$2016$$◇$$(6$$◇$$x)=100\\Rightarrow (2016$$◇$$6)\\cdot x=100$$ By multiplying both sides by $$x$$, we get: $$(2016$$◇$$6)\\cdot 6=\\frac {600}x\\Rightarrow 2016$$◇$$(6$$◇$$6)=\\frac {600}x$$ Because $$6$$◇$$6=2016$$◇$$2016=1$$: $$2016$$◇$$(2016$$◇$$2016)=\\frac {600}x\\Rightarrow (2016$$◇$$2016)\\cdot 2016=\\frac {600}x\\Rightarrow 2016=\\frac {600}x$$ Therefore, $$x=\\frac {600}{2016}=\\frac {25}{84}$$, so the answer is $$25+84=109$$. " ]

[ "2014年吉林全国高中数学联赛竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "论断正确的有①②③④． 因为$${{a}_{1}}-{{a}_{1}}=0\\in \\left { {{a}_{n}} \\right }$$，所以$${{a}_{1}}=0$$； 因为 $$a={{a}_{1}}\\textless{}{{a}_{3}}-{{a}_{2}}\\textless{}{{a}_{4}}-{{a}_{2}}\\textless{}{{a}_{5}}-{{a}_{2}}\\textless{}{{a}_{5}}$$， 且 $${{a}_{3}}-{{a}_{2}}$$，$${{a}_{4}}-{{a}_{2}}$$，$${{a}_{5}}-{{a}_{2}}\\in \\left { {{a}_{n}} \\right }$$， 所以 $${{a}_{3}}-{{a}_{2}}={{a}_{2}}$$，$${{a}_{4}}-{{a}_{2}}={{a}_{3}}$$，$${{a}_{5}}-{{a}_{2}}={{a}_{4}}$$； 于是 $${{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}}={{a}_{5}}-{{a}_{4}}$$． 所以$$\\left { {{a}_{n}} \\right }$$为等差数列，且$$\\left { {{a}_{n}} \\right }$$：$$0$$,d,2d,3d,4d，因此$${{a}_{5}}=4{{a}_{2}}$$；集合$$A=\\left { {{a}_{i}}+{{a}_{j}}\\left\\textbar{} 1\\leqslant i\\leqslant j\\leqslant 5 \\right. \\right }$$含$$9$$个元素． " ]

[ "2009年新疆全国高中数学联赛竞赛初赛第6题6分" ]

[ [ { "aoVal": "A", "content": "$$2+\\sqrt{3}$$ " } ], [ { "aoVal": "B", "content": "$$2-\\sqrt{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\sqrt{3}$$ " } ], [ { "aoVal": "D", "content": "$$-\\sqrt{3}$$ " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->函数综合" ]

[ "当$$x=1$$时， $$f(3)\\left[ 1-f(x) \\right]=1+f(1)$$ 所以$$-\\sqrt{3}\\left[ 1-f(1) \\right]=1+f(1)$$ 所以$$f(1)=\\frac{1+\\sqrt{3}}{\\sqrt{3}-1}=2+\\sqrt{3}$$ 又因为$$f(x+2)=\\frac{1+f(x)}{1-f(x)}$$ 所以$$f(x+4)=\\frac{1+f(x+2)}{1-f(x+2)}$$ $$=\\frac{1+\\frac{1+f(x)}{1-f(x)}}{1-\\frac{1+f(x)}{1-f(x)}} $$ $$=\\frac{2}{-2f(x)}$$ $$=-\\frac{1}{f(x) }$$ 所以$$f(x+8)=-\\frac{1}{f(x+4)}=-\\frac{1}{-\\frac{1}{f(x)}}=f(x)$$ 所以$$f(2009)=f(251\\times 8+1)=f(1)=2+\\sqrt{3}$$ 故选$$\\text{A}$$． " ]

[ "2021年吉林全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$${{a}_{1}}$$ " } ], [ { "aoVal": "B", "content": "$${{a}_{9}}$$ " } ], [ { "aoVal": "C", "content": "$${{a}_{10}}$$ " } ], [ { "aoVal": "D", "content": "$${{a}_{12}}$$ " } ] ]

[ "课内体系->思想->函数思想", "课内体系->素养->逻辑推理", "课内体系->素养->数学抽象", "课内体系->方法->分离常数法", "课内体系->知识点->数列->数列的概念->数列的函数特性->数列中最大项与最小项的求解问题", "课内体系->知识点->数列->数列的概念->数列的函数特性->数列单调性问题" ]

[ "通项公式$${{a}_{n}}=1+\\frac{1}{n-9.5}$$， 当$$n\\leqslant 9$$时，有$${{a}_{n}}\\textless{}1$$， 当$$n\\geqslant 10$$时，有$${{a}_{n}}\\textgreater1$$，且$$\\left { {{a}_{n}} \\right }$$递减，从而$$\\left { {{a}_{n}} \\right }$$中最大的项是$${{a}_{10}}$$． 故选$$\\text{C}$$． " ]

[ "2010年AMC12竞赛B第16题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{29}{81}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{31}{81}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{11}{27}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{13}{27}$$ " } ] ]

[ "课内体系->知识点->统计与概率->概率->事件与概率->事件的独立性->相互独立事件的概率乘法公式", "美国AMC10/12->Knowledge Point->Number Theory->Aliquot Theory->Divisibility Rules of Certain Numbers (2/3/5/9/11)", "美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Classical Models of Probabilities" ]

[ "首先, 注意到$abc+ab+a=a(bc+b+1)$, 因此当$3\\textbar a$时，有$3\\textbar abc+ab+a$. 然后, 当$3\\nmid a$时, $3\\textbar abc+ab+a\\Leftrightarrow 3\\textbar bc+b+1\\Leftrightarrow b(c+1)\\equiv 2\\pmod 3$. 在模$$3$$的意义下, $(b,c)$共有$$3\\times 3=9$$种可能的取值, 其中只有$(2, 0)$和$(1, 1)$满足$b(c+1)\\equiv 2\\pmod 3$. 因此, 总的概率为$\\frac{1}{3}+\\frac{2}{3}\\times\\frac{2}{9}=\\frac{13}{27}$. " ]

[ "全国高中数学联赛竞赛模拟一试（十五）第1题" ]

[ [ { "aoVal": "A", "content": "$$\\left[ 0,+\\infty \\right)$$ " } ], [ { "aoVal": "B", "content": "$$\\left[ 1,+\\infty \\right)$$ " } ], [ { "aoVal": "C", "content": "$$\\left[ 2,+\\infty \\right)$$ " } ], [ { "aoVal": "D", "content": "以上都不对 " } ] ]

[ "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域", "竞赛->知识点->函数->基本初等函数", "竞赛->知识点->函数->二次函数" ]

[ "注意到， $$f\\left( x \\right)={{\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}} \\right)}^{2}}-\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}} \\right)-2$$ $$={{\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}}-\\frac{1}{2} \\right)}^{2}}-\\frac{9}{4}$$． 由于$${{x}^{2}}+\\frac{1}{{{x}^{2}}}\\geqslant 2$$， 从而，所求函数的值域为$$\\left[ 0,+\\infty \\right)$$． " ]