Datasets:

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math-eval

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TAL-SCQ5K

like 57

[ "全国高中数学联赛竞赛模拟一试（十五）第1题" ]

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

[ "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->二次函数", "竞赛->知识点->函数->基本初等函数" ]

[ "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$$\\left[ 0,+\\infty \\right)$$． " ]

[ "2017~2018学年浙江温州瓯海区浙江省瓯海中学高三上学期期中理科第9题4分", "2007年全国全国高中数学联赛竞赛一试第2题6分" ]

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

[ "课内体系->素养->数学运算", "课内体系->知识点->等式与不等式->不等式->解不等式->含绝对值的不等式", "课内体系->知识点->等式与不等式->不等式->解不等式->一元二次不等式" ]

[ "令$$x=\\frac{2}{3}a$$，则有$$\\textbar a\\textbar\\leqslant \\frac{1}{3}$$，排除$$\\text{B}$$、$$\\text{D}$$，由对称性排除$$\\text{C}$$，从而只有$$\\text{A}$$正确． 一般地，对$$k\\in \\mathbf{R}$$，令$$x=\\frac{1}{2}ka$$，则原不等式为$$\\textbar a\\textbar\\cdot \\textbar k-1\\textbar+\\frac{3}{2}\\textbar a\\textbar\\cdot \\left\\textbar{} k-\\frac{4}{3} \\right\\textbar\\geqslant \\textbar a{{\\textbar}^{2}}$$，由此易知原不等式等价于$$\\textbar a\\textbar\\leqslant \\textbar k-1\\textbar+\\frac{3}{2}\\left\\textbar{} k-\\frac{4}{3} \\right\\textbar$$，对任意的$$k\\in \\mathbf{R}$$成立．由于 $$\\textbar k-1\\textbar+\\frac{3}{2}\\left\\textbar{} k-\\frac{4}{3} \\right\\textbar=\\begin{cases} \\frac{5}{2}k-3， k\\geqslant \\frac{4}{3} 1-\\frac{1}{2}k， 1\\leqslant k ~\\textless{} ~\\frac{4}{3} 3-\\frac{5}{2}k， k ~\\textless{} ~1 \\end{cases}$$， 所以$$_{k\\in \\mathbf{R}}^{\\min }\\left { \\textbar k-1\\textbar+\\frac{3}{2}\\left\\textbar{} k-\\frac{4}{3} \\right\\textbar{} \\right }=\\frac{1}{3}$$，从而上述不等式等价于$$\\textbar a\\textbar\\leqslant \\frac{1}{3}$$． 故选$$\\text{A}$$． " ]

[ "2009年浙江全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{15}{2}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{5}{4}$$ " } ] ]

[ "竞赛->知识点->解析几何->椭圆" ]

[ "因为$$a=5,b=3$$，则$$c=4$$．于是$$P$$到另一个焦点$$(-4,0)$$的距离等于$$2\\times 5-4=6$$．由于直线$$x=-\\frac{25}{4}$$为椭圆的左准线方程，则$$P$$到直线$$x=-\\frac{25}{4}$$的距离为$$d=\\frac{6}{e}=7.5$$． " ]

[ "1992年全国高中数学联赛竞赛一试第6题" ]

[ [ { "aoVal": "A", "content": "偶函数，又是周期函数 " } ], [ { "aoVal": "B", "content": "偶函数，但不是周期函数 " } ], [ { "aoVal": "C", "content": "奇函数，又是周期函数 " } ], [ { "aoVal": "D", "content": "奇函数，但不是周期函数 " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "由所给第一式得 $$f\\left[ 10+\\left( 10-x \\right) \\right]=f\\left[ 10-\\left( 10-x \\right) \\right]$$， ∴~~~~ $$f\\left( x \\right)=f\\left( 20-x \\right)$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ① 又由所给第二式得 $$f\\left( x \\right)=-f\\left( 20-x \\right)$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ② ∴~~~~ $$\\begin{matrix}f\\left( 40+x \\right)=f\\left[ 20+\\left( 20+x \\right) \\right] =-f\\left( 20+x \\right)=f\\left( x \\right) \\end{matrix}$$ 可见$$f(x)$$是周期函数． 由①，②得 $$f\\left( -x \\right)=f\\left( 20+x \\right)=-f\\left( x \\right)$$， ∴$$f\\left( x \\right)$$是奇函数．因此答案是（$$\\text{C}$$）． " ]

[ "2010年河南全国高中数学联赛竞赛初赛第3题5分" ]

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

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "本题从题目上可预知$${{f}_{n}}(x)$$具有周期性，为了寻找周期，采用赋值法容易计算．记$${{a}_{1}}={{f}_{1}}(2), {{a}_{n}}={{f}_{n}}(2)$$，则$${{a}_{1}}=\\frac{1}{3},{{a}_{2}}=\\frac{\\frac{1}{3}-1}{\\frac{1}{3}+1}=-\\frac{2}{4}=-\\frac{1}{2},{{a}_{3}}=\\frac{-\\frac{1}{2}-1}{-\\frac{1}{2}+1}=-3,$$ $${{a}_{4}}=\\frac{-3-1}{-3+1}=2, {{a}_{5}}=\\frac{2-1}{2+1}=\\frac{1}{3},\\cdots $$． 故可知周期$$T=4$$，所以$${{a}_{2010}}={{f}_{2010}}(2)={{f}_{2}}(2)={{a}_{2}}=-\\frac{1}{2}$$，将$$x=2$$代入可得选项B. " ]

[ "2003年AMC10竞赛B第22题" ]

[ [ { "aoVal": "A", "content": "March $$8$$ " } ], [ { "aoVal": "B", "content": "March $$9$$ " } ], [ { "aoVal": "C", "content": "March $$10$$ " } ], [ { "aoVal": "D", "content": "March $$20$$ " } ], [ { "aoVal": "E", "content": "March $$21$$ " } ] ]

[ "课内体系->知识点->等式与不等式->等式->方程组的解集", "美国AMC10/12->Knowledge Point->Combination->Reasoning->Recurrence and Recursion" ]

[ "First, find how many chimes will have already happened before midnight (the beginning of the day) of February $$27$$, $$2003$$. $$13$$ half-hours have passed, and the number of chimes according to the hour is $$1+2+3+\\cdots +12$$. The total number of chimes is $$13+78=91$$. Every day, there will be $$24$$ half-hours and $$2(1+2+3+\\cdots 12)$$ chimes according to the arrow, resulting in $$24+156=180$$ total chimes. On February $$26$$, the number of chimes that still need to occur is $$2003-91=1912$$. $$1912\\div 180=10\\rm R112$$. Rounding up, it is $$11$$ days past February $$26$$, which is March $$9$$. " ]

[ "1997年全国高中数学联赛竞赛一试第4题6分" ]

[ [ { "aoVal": "A", "content": "$$(0,1)$$ " } ], [ { "aoVal": "B", "content": "$$(1,+\\infty )$$ " } ], [ { "aoVal": "C", "content": "$$(0,5)$$ " } ], [ { "aoVal": "D", "content": "$$(5,+\\infty )$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->知识点->圆锥曲线->椭圆->椭圆的定义、标准方程->椭圆的标准方程", "课内体系->知识点->直线和圆的方程->直线与方程->平面中的距离->点到直线的距离公式" ]

[ "看成是轨迹上点到$$(0,-1)$$的距离与到直线$$x-2y+3=0$$的距离的比： $$\\frac{\\sqrt{{{x}^{2}}+{{\\left( y+1 \\right)}^{2}}}}{\\frac{\\left\\textbar{} x-2y+3 \\right\\textbar}{\\sqrt{{{1}^{2}}+{{\\left( -2 \\right)}^{2}}}}}=\\sqrt{\\frac{5}{m}}\\textless{}1\\Rightarrow m\\textgreater5$$． 故选$$\\text{D}$$． " ]

[ "2008年AMC10竞赛B第18题", "2008年AMC12竞赛B第10题" ]

[ [ { "aoVal": "A", "content": "$$500$$ " } ], [ { "aoVal": "B", "content": "$$900$$ " } ], [ { "aoVal": "C", "content": "$$950$$ " } ], [ { "aoVal": "D", "content": "$$1000$$ " } ], [ { "aoVal": "E", "content": "$$1900$$ " } ] ]

[ "课内体系->知识点->等式与不等式->等式->方程组的解集", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems" ]

[ "Let $$x$$ be the number of bricks in the chimney, Using $$d=vt$$, we get $$x=\\left( \\frac{x}{9}+\\frac{x}{10}-10 \\right)\\cdot \\left( 5 \\right)$$. Solving for $$x$$, we get $$900\\Rightarrow \\text{B}$$. " ]

[ "1998年全国高中数学联赛竞赛一试第4题6分" ]

[ [ { "aoVal": "A", "content": "是命题$$P$$的充分必要条件 " } ], [ { "aoVal": "B", "content": "是命题$$P$$的充分条件但不是必要条件 " } ], [ { "aoVal": "C", "content": "是命题$$P$$的必要条件但不是充分条件 " } ], [ { "aoVal": "D", "content": "既不是是命题$$P$$的充分条件也不是命题$$P$$的必要条件 " } ] ]

[ "竞赛->知识点->不等式->不等式的解法" ]

[ "若两个不等式的解集都是$$\\mathbf{R}$$，否定$$\\text{A}$$、$$\\text{C}$$，若比值为$$-1$$，否定$$\\text{A}$$、$$\\text{B}$$． 故选$$\\text{D}$$． " ]

[ "2006年上海复旦大学自主招生千分考第16题", "1981年全国高中数学联赛竞赛一试第2题" ]

[ [ { "aoVal": "A", "content": "甲是乙的充分必要条件 " } ], [ { "aoVal": "B", "content": "甲是乙的必要条件 " } ], [ { "aoVal": "C", "content": "甲是乙的充分条件 " } ], [ { "aoVal": "D", "content": "甲不是乙的必要条件，也不是充分条件 " } ] ]

[ "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" ]

[ "因为$$\\sqrt{1+\\sin \\theta }=\\sqrt{{{\\sin }^{2}}\\frac{\\theta }{2}+2\\sin \\frac{\\theta }{2}\\cos \\frac{\\theta }{2}+{{\\cos }^{2}}\\frac{\\theta }{2}}=\\sqrt{{{\\left( \\sin \\frac{\\theta }{2}+\\cos \\frac{\\theta }{2} \\right)}^{2}}}=\\left\\textbar{} \\sin \\frac{\\theta }{2}+\\cos \\frac{\\theta }{2} \\right\\textbar$$ 所以条件甲等价于：$$\\left\\textbar{} \\sin \\frac{\\theta }{2}+\\cos \\frac{\\theta }{2} \\right\\textbar=a$$． 在条件甲中，易知$$a\\geqslant 0$$，而在条件乙中，$$a$$可能取负值． 所以由条件甲推不出乙，由乙也推不出甲， 即甲不是乙的必要条件，也不是充分条件． " ]

[ "2014年辽宁全国高中数学联赛竞赛初赛第4题5分" ]

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

[ "竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->三角函数->三角函数的图像与性质" ]

[ "因为 $$\\frac{\\sin A+\\sqrt{3}\\cos A}{\\cos A-\\sqrt{3}\\sin A}=\\frac{\\tan A+\\tan \\frac{ \\pi }{3}}{1-\\tan A\\tan \\frac{ \\pi }{3}}=\\tan \\left( A+\\frac{ \\pi }{3} \\right)$$， 故$$A+\\frac{ \\pi }{3}=\\frac{7 \\pi }{12}$$，$$A=\\frac{ \\pi }{4}$$． 由于$$2B+2C=\\frac{3 \\pi }{2}$$，故 $$\\sin 2B+2\\cos C=\\sin \\left( \\frac{3 \\pi }{2}-2C \\right)+2\\cos C$$ $$=-\\cos 2C+2\\cos C$$ $$=1-2{{\\cos }^{2}}C+2\\cos C$$ $$=-2{{\\left( \\cos C-\\frac{1}{2} \\right)}^{2}}+\\frac{3}{2}\\leqslant \\frac{3}{2}$$． 当$$C=\\frac{ \\pi }{3}$$时取等号．故$$\\sin 2B+2\\cos C$$的最大值为$$\\frac{3}{2}$$． " ]

[ "全国高中数学联赛竞赛模拟一试（五）第3题" ]

[ [ { "aoVal": "A", "content": "$$ \\frac{7}{15}$$ " } ], [ { "aoVal": "B", "content": "$$ \\frac{8}{15}$$ " } ], [ { "aoVal": "C", "content": "$$ \\frac{2}{5}$$ " } ], [ { "aoVal": "D", "content": "$$ \\frac{3}{5}$$ " } ], [ { "aoVal": "E", "content": "$$ \\frac{1}{2}$$ " } ] ]

[ "竞赛->知识点->排列组合与概率->两个基本计数原理", "竞赛->知识点->排列组合与概率->概率初步", "竞赛->知识点->排列组合与概率->排列与组合", "课内体系->知识点->计数原理" ]

[ "设抽出的号码$$i$$，$$j$$，$$k$$（$$0\\leqslant i\\textless{}j\\textless{}k\\leqslant 9$$）中任何两个均不相邻， 则$$i$$，$$j-1$$，$$k-2$$互不相邻，且只能从$$ {0,1,\\cdots ,7 }$$中取值， 故所求的概率$$ p=1- \\frac{\\text{C}_{8}^{3}}{\\text{C}_{15}^{3}}= \\frac{8}{15}$$． " ]

[ "2008年山东全国高中数学联赛竞赛初赛第6题6分" ]

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

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "方法一：在原函数图象上取点$$\\left( 1,\\frac{1}{a}-1 \\right)$$， 则点$$\\left( \\frac{1}{a}-1,1 \\right)$$在其反函数图象上， 它关于点$$(-13)$$的对称点为$$\\left( -1-\\frac{1}{a},5 \\right)$$，从而点$$\\left( 5,-1-\\frac{1}{a} \\right)$$在原函数图象上， 所以有$$-1-\\frac{1}{a}=\\frac{5-a}{5-a-1}$$， 解得$$a=2$$．故选$$\\text{A}$$． 方法二：因函数$$y=-\\frac{x-a}{x-a-1}$$与其反函数的图象关于直线$$y=x$$对称， 故它们的对称中心必也关于$$y=x$$对称． 可知点$$\\left( -1,3 \\right)$$关于直线$$y=x$$的对称点为$$(3-1)$$． 又因为函数$$y=-\\frac{x-a}{x-a-1}$$，即$$y=-1+\\frac{1}{x-(a+1)}$$， 图象关于点$$(a+1,-1)$$成中心对称图形，故$$a+1=3$$，得$$a=2$$．故选$$\\text{A}$$． " ]

[ "2010年黑龙江全国高中数学联赛竞赛初赛第9题5分" ]

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

[ "竞赛->知识点->三角函数->三角函数的图像与性质" ]

[ "$$y=\\sqrt{3}\\cos x-\\sin x=2\\cos \\left( x+\\frac{ \\pi }{6} \\right),$$对称轴方程$$x=k \\pi -\\frac{ \\pi }{6}, k\\in Z$$． " ]

[ "2002年全国高中数学联赛竞赛一试第2题10分" ]

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

[ "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->等式与不等式->不等式->基本不等式->基本不等式的概念", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件", "课内体系->素养->数学运算" ]

[ "令$$x+5=14\\cos \\theta $$,$$y-12=14\\sin \\theta $$ ,则$${{x}^{2}}+{{y}^{2}}=196+28\\left( 5\\cos \\theta -12\\sin \\theta \\right)+169=365+364\\sin \\left( \\theta +\\varphi \\right)\\geqslant 1$$. " ]

[ "2017年山西全国高中数学联赛竞赛初赛第8题8分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$65$$ " } ], [ { "aoVal": "C", "content": "$$1008$$ " } ], [ { "aoVal": "D", "content": "$$1677$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->数列的通项与求和", "竞赛->知识点->集合->集合的划分与覆盖" ]

[ "$$S=1+2+\\cdots +2017=\\frac{2018\\cdot 2017}{2}=1009\\cdot 2017=2035153\\in \\left( {{1400}^{2}},{{1500}^{2}} \\right)$$， 而$${{1450}^{2}}={{\\left( 1400+50 \\right)}^{2}}=1960000+140000+2500=2102500\\textgreater S$$， 又$${{1425}^{2}}={{\\left( 1400+25 \\right)}^{2}}=1960000+70000+625=2030625\\textless{}S$$， $${{1426}^{2}}={{\\left( 1425+1 \\right)}^{2}}=2030625+2850+1=2033476$$， 所以$$S={{1426}^{2}}+1677$$． " ]

[ "2016年AMC10竞赛第16题5分" ]

[ [ { "aoVal": "A", "content": "counterclockwise retation about the origin by $$90^{}\\circ $$ " } ], [ { "aoVal": "B", "content": "clockwise rotation about the origin by $$90^{}\\circ $$ " } ], [ { "aoVal": "C", "content": "reflection about the x---axis " } ], [ { "aoVal": "D", "content": "reflection about the liney=x " } ], [ { "aoVal": "E", "content": "reflection about the y---axis. " } ] ]

[]

[ "Consider a point $$(x,y)$$. Reflecting it about the $$x$$-axis will map it to $$(x,-y)$$, and rotating it counterclockwise about the origin by $$90^{}\\circ $$will map it to $$(y,x)$$. The operation that undoes this is a reflection about the $$y=x$$, so the answer is $$\\rm D$$. " ]

[ "2010年黑龙江全国高中数学联赛竞赛初赛第12题5分", "2020~2021学年广东深圳福田区深圳市高级中学高中部高二下学期期中第8题5分" ]

[ [ { "aoVal": "A", "content": "$$6$$个 " } ], [ { "aoVal": "B", "content": "$$10$$个 " } ], [ { "aoVal": "C", "content": "$$12$$个 " } ], [ { "aoVal": "D", "content": "$$16$$个 " } ] ]

[ "课内体系->知识点->计数原理->两个基本计数原理->分类加法计数原理", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->函数的定义", "课内体系->素养->数学运算" ]

[ "由$$\\overrightarrow{DA}+\\overrightarrow{DC}=\\overrightarrow{\\gamma DB}$$（$$\\gamma \\in \\mathbf{R}$$），分析可得$$\\triangle ABC$$是等腰三角形， 且$$BA=BC$$，必有$$f(1)=f(3)$$，$$f(1)\\ne f(2)$$． 点$$A(1,f(1))$$、当$$f(1)=1=f(3)$$时$$f(2)=2$$、$$3$$、$$4$$，三种情况． $$f(1)=f(3)=2$$；$$f(2)=1$$、$$3$$、$$4$$，有三种． $$f(1)=f(3)=3$$；$$f(2)=2$$、$$1$$、$$4$$，有三种． $$f(1)=f(3)=4$$；$$f(2)=2$$、$$3$$、$$1$$，有三种． 因而满足条件的函数$$f(x)$$有$$12$$种． 故选$$\\text{C}$$． " ]

[ "1988年全国高中数学联赛竞赛一试第5题" ]

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

[ "竞赛->知识点->集合->集合的概念与运算", "竞赛->知识点->解析几何->直线与方程" ]

[ "直线$$y=\\frac{1}{2}$$不通过任何整点，所以表达式（2）正确； 直线$$y=\\sqrt{2}x$$无穷多个，恰好通过一个整点$$\\left( 0,0 \\right)$$，所以表达式（$$3$$）正确， 直线$$y=x$$通过多个整点，所以表达式（$$4$$）正确； 我们来证明，若 $$ax+by+c=0$$ 通过两个整点$$\\left( {{x}_{1}},{{y}_{1}} \\right)$$、$$\\left( {{x}_{2}},{{y}_{2}} \\right)$$， 则通过无穷多个整点 $$\\left( {{x}_{2}}+k\\left( {{x}_{2}}-{{x}_{1}} \\right),{{y}_{1}}+k\\left( {{y}_{2}}-{{y}_{1}} \\right) \\right)$$，$$k\\in \\bf Z$$． 事实上 $$a\\left[ {{x}_{1}}+k\\left( {{x}_{2}}-{{x}_{1}} \\right) \\right]+b\\left[ {{y}_{1}}+k\\left( {{y}_{2}}-{{y}_{1}} \\right) \\right]+c$$ $$=a{{x}_{1}}+b{{y}_{1}}+c+k\\left( a{{x}_{2}}+b{{y}_{2}}+c \\right)-k\\left( a{{x}_{1}}+b{{y}_{1}}+c \\right)$$ $$=0$$， 故选$$\\text{D}$$． " ]

[ "1989年全国高中数学联赛竞赛一试第2题" ]

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

[ "知识标签->题型->函数->函数及其表示->函数的值域->用单调性观察法求值域", "知识标签->素养->数学运算", "知识标签->知识点->三角函数->三角函数的图象与性质->反三角函数" ]

[ "$$f(x)$$的定义域是$$[-1,1]$$，此时 $$-\\frac{ \\pi }{4}{\\leqslant }\\text{arctg}x{\\leqslant }\\frac{ \\pi }{4}$$，$$-\\frac{ \\pi }{4}{\\leqslant }\\frac{1}{2}\\arcsin x{\\leqslant }\\frac{ \\pi }{4}$$． 而且$$\\text{arctg}x$$和$$\\arcsin x$$是单调增加的，从而$$f(x)$$的值域是$$\\left[ -\\frac{ \\pi }{2},-\\frac{ \\pi }{2} \\right]$$． 故选$$\\text{D}$$． " ]

[ "2018年AMC10竞赛A第14题" ]

[ [ { "aoVal": "A", "content": "$$255$$ " } ], [ { "aoVal": "B", "content": "$$256$$ " } ], [ { "aoVal": "C", "content": "$$270$$ " } ], [ { "aoVal": "D", "content": "$$264$$ " } ], [ { "aoVal": "E", "content": "$$265$$ " } ] ]

[ "美国AMC10/12->Knowledge Point->Algebra->Calculation->Exponentiation", "课内体系->知识点->集合->集合的基本关系->子集个数的计算" ]

[ "We write $$\\frac{4^{100}+3^{100}}{4^{96}+3^{96}}=\\frac{4^{96}}{4^{96}+3^{96}}\\cdot \\frac{4^{100}}{4^{96}}+\\frac{3^{96}}{4^{96}+3^{96}}\\cdot \\frac{3^{100}}{3^{96}}=\\frac{4^{96}}{4^{96}+3^{96}}\\cdot 256+\\frac{3^{96}}{4^{96}+3^{96}}\\cdot 81$$. Hence we see that our number is less than $256$ but more than $81$. So the answer is $$\\boxed{\\rm (A)\\textasciitilde255}$$. " ]

[ "2018年吉林全国高中数学联赛竞赛初赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$$(0,\\sqrt{3}]$$ " } ], [ { "aoVal": "B", "content": "$$(1,\\sqrt{3}]$$ " } ], [ { "aoVal": "C", "content": "$$\\left( 0,\\frac{4}{3} \\right]$$ " } ], [ { "aoVal": "D", "content": "$$\\left( 1,\\frac{4}{3} \\right]$$ " } ] ]

[ "竞赛->知识点->不等式->换元技巧->代数换元" ]

[ "由已知$$y=\\frac{x}{x-1}$$得，$$z=\\frac{x+y}{xy-1}=\\frac{xy}{xy-1}={{\\left( 1-\\frac{1}{xy} \\right)}^{-1}}$$ $$={{\\left( 1-\\frac{1}{x}+\\frac{1}{{{x}^{2}}} \\right)}^{-1}}$$． 因为$$x\\textgreater1$$， 即$$0\\textless{}\\frac{1}{x}\\textless{}1$$， 所以，$$\\frac{3}{4}\\leqslant 1-\\frac{1}{x}+\\frac{1}{{{x}^{2}}}\\textless{}1$$． 故$$1\\textless{}z\\leqslant \\frac{4}{3}$$． 故选$$\\text{D}$$． " ]

[ "2000年AMC10竞赛第17题" ]

[ [ { "aoVal": "A", "content": "＄$$3.63$$ " } ], [ { "aoVal": "B", "content": "＄$$5.13$$ " } ], [ { "aoVal": "C", "content": "＄$$6.30$$ " } ], [ { "aoVal": "D", "content": "＄$$7.45$$ " } ], [ { "aoVal": "E", "content": "＄$$9.07$$ " } ] ]

[ "课内体系->知识点->函数的应用->函数与方程", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems" ]

[ "Consider what happens each time he puts a coin in. If he puts in a quarter, he gets five nickels back, so the amount of money he has doesn\\textquotesingle t change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesn\\textquotesingle t change. However, if he puts a penny in, he gets five quarters back, increasing the amount of money he has by~~cents. This implies that the only possible values, in cents, he can have are the ones one more than a multiple of . Of the choices given, the only one is D " ]

[ "2019年江西全国高中数学联赛竞赛初赛改编第8题" ]

[ [ { "aoVal": "A", "content": "$3029+\\dfrac{\\sqrt{3}-1}{2}$ " } ], [ { "aoVal": "B", "content": "$3032+\\dfrac{\\sqrt{3}-1}{2}$ " } ], [ { "aoVal": "C", "content": "$3035+\\dfrac{\\sqrt{3}-1}{2}$ " } ], [ { "aoVal": "D", "content": "前三个答案都不对 " } ] ]

[ "知识标签->知识点->数列->数列的概念->数列的表示方法->通项公式" ]

[ "$$1^{\\circ}$$迭代几项尝试一下， $$a_1=\\sqrt{3}=1+(\\sqrt{3}-1)$$， $$\\Rightarrow{a_2}=[a_1]+\\frac{1}{\\left { {{a}_{n}} \\right }}=1+\\frac{1}{\\sqrt{3}-1}=2+\\frac{\\sqrt{3}-1}{2}$$， $$\\Rightarrow{{a}_{3}}=[{{a}_{2}}]+\\frac{1}{\\left { {{a}_{2}} \\right }}=2+\\frac{2}{\\sqrt{3}-1}=4+(\\sqrt{3}-1)$$， $$\\Rightarrow{{a}_{4}}=[{{a}_{3}}]+\\frac{1}{\\left { {{a}_{3}} \\right }}=4+\\frac{1}{\\sqrt{3}-1}=5+\\frac{\\sqrt{3}-1}{2}$$， $$\\Rightarrow{{a}_{5}}=[{{a}_{4}}]+\\frac{1}{\\left { {{a}_{4}} \\right }}=5+\\frac{2}{\\sqrt{3}-1}=7+(\\sqrt{3}-1)$$， $$\\Rightarrow{{a}_{6}}=[{{a}_{5}}]+\\frac{1}{\\left { {{a}_{5}} \\right }}=7+\\frac{1}{\\sqrt{3}-1}=8+\\frac{\\sqrt{3}-1}{2}$$． $$2^{\\circ}$$猜想$$a_{2n}=3n-1+\\frac{\\sqrt{3}-1}{2}$$， 用数学归纳法证明， ①$$n=1$$时显然； ②若$$n=k$$时成立，即$$a_{2k}=3k-1+\\frac{\\sqrt{3}-1}{2}$$， 则$$a_{2k+1}=\\left[a_{2k}\\right]+\\frac{1}{ {a_{2k} }}=3k-1+\\frac{2}{\\sqrt{3}-1}=3k+1+\\left(\\sqrt{3}-1\\right)$$， $$\\Rightarrow a_{2k+2}=\\left[a_{2k+1}\\right]+\\frac{1}{ {a_{2k+1} }}=3k+1+\\frac{1}{\\sqrt{3}-1}=3k+2+\\frac{\\sqrt{3}-1}{2}$$，得证． 故，$${{a}_{2020}}=3\\times 1010-1+\\frac{\\sqrt{3}-1}{2}=3029+\\frac{\\sqrt{3}-1}{2}$$． " ]

[ "2017年天津南开区高三二模文科第1题5分", "2015年黑龙江全国高中数学联赛竞赛初赛第1题5分", "2014年重庆高三零模文科第4题5分", "2017年天津南开区高三二模理科第1题5分", "2014年重庆高三零模理科第2题5分" ]

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

[ "课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->思想->分类讨论思想", "课内体系->知识点->集合->集合的概念与表示方法->集合的含义、元素与集合->元素与集合之间的关系->元素与集合的关系判断" ]

[ "∵集合$$A= {-1,0,2 }$$，集合$$B= {-x\\textbar x\\in A,2-x\\notin A }$$， ∴$$-1\\in A$$，且$$2-\\left( -1 \\right)=3\\notin A$$，故$$1\\in B$$； $$0\\in A$$，但$$2-0=2\\in A$$，不满足题意； $$2\\in A$$，但$$2-2=0\\in A$$，不满足题意； 故$$B= {1 }$$， 故选$$\\text{A}$$． " ]

[ "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", "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", "2015年北京海淀区高三三模第4题", "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学年广东深圳高一上学期期末高中联考联盟第8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", "2018~2019学年江苏盐城高一上学期期末第5题5分" ]

[ [ { "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", "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" ]

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

[ "2015年四川全国高中数学联赛竞赛初赛第3题5分" ]

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

[ "竞赛->知识点->立体几何与空间向量->空间中的角与距离", "竞赛->知识点->解析几何->椭圆" ]

[ "设圆的半径为$$1$$，则椭圆的长轴长为$$2$$，短轴长为$$2\\cos 30{}^{}\\circ =\\sqrt{3}$$，则$$c=\\frac{1}{2}$$． " ]

[ "2016年陕西全国高中数学联赛竞赛初赛第3题6分" ]

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

[ "课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积的运算律", "课内体系->知识点->平面向量->平面向量的运算->数量积->向量的数量积的定义", "课内体系->知识点->平面向量->平面向量的运算->数量积->利用向量数量积求模长", "课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积运算（非坐标）", "课内体系->知识点->平面向量->平面向量的运算->数量积->利用向量数量积求夹角", "课内体系->知识点->平面向量->平面向量的基本概念->向量的概念->向量的夹角的判断", "课内体系->知识点->平面向量->平面向量的基本概念->向量的模", "课内体系->素养->数学运算" ]

[ "$$(\\vec{c}-\\vec{a})\\cdot (\\vec{c}-\\vec{b})=1-\\overrightarrow{c}\\left( \\overrightarrow{a}+\\overrightarrow{b} \\right)=1-\\left\\textbar{} \\overrightarrow{c} \\right\\textbar\\left\\textbar{} \\overrightarrow{a}+\\overrightarrow{b} \\right\\textbar\\text{cos }\\theta \\leqslant 1+\\sqrt{2}$$，当$$\\overrightarrow{c}$$与$$\\overrightarrow{a}+\\overrightarrow{b}$$的夹角为$$180{}^{}\\circ $$时取等号． 故选$$\\text{A}$$． " ]

[ "2019~2020学年4月湖南长沙天心区长郡中学高一下学期月考第12题3分", "1998年全国高中数学联赛竞赛一试第3题6分", "2014~2015学年10月辽宁沈阳和平区东北育才中学高二上学期月考理科第7题5分", "2018~2019学年10月山东济宁市中区济宁市第一中学高二上学期月考第6题5分" ]

[ [ { "aoVal": "A", "content": "$$150$$ " } ], [ { "aoVal": "B", "content": "$$-200$$ " } ], [ { "aoVal": "C", "content": "$$150$$或$$-200$$ " } ], [ { "aoVal": "D", "content": "$$-50$$或$$400$$~ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "首先$$q\\ne 1$$，于是$$\\frac{{{a}_{1}}}{q-1}\\left( {{q}^{10}}-1 \\right)=10$$，$$\\frac{{{a}_{1}}}{q-1}\\left( {{q}^{30}}-1 \\right)=70$$， ∴$$\\textasciitilde{{q}^{20}}+{{q}^{10}}+1=7\\Rightarrow {{q}^{10}}=2$$（$$-3$$舍）． ∴$${{S}_{40}}=10\\left( {{q}^{40}}1 \\right)=150$$． 故选$$\\text{A}$$． " ]

[ "2006年上海复旦大学自主招生千分考第6题", "2010年浙江全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$60{}^{}\\circ $$ " } ], [ { "aoVal": "B", "content": "$$75{}^{}\\circ $$ " } ], [ { "aoVal": "C", "content": "$$90{}^{}\\circ $$ " } ], [ { "aoVal": "D", "content": "$$105{}^{}\\circ $$ " } ] ]

[ "竞赛->知识点->立体几何与空间向量->空间中的角与距离" ]

[ "取$${{B}_{1}}{{A}_{1}}$$的中点$$M$$，易知$${{C}_{1}}M\\bot $$平面$$A{{A}_{1}}{{B}_{1}}B$$，所以$${{C}_{1}}M\\bot A{{B}_{1}}$$， 在矩形$$AB{{B}_{1}}{{A}_{1}}$$中，$$\\tan \\angle {{B}_{1}}BM=\\tan \\angle {{B}_{1}}AB=\\frac{\\sqrt{2}}{2}$$， 于是$$\\angle {{B}_{1}}BM=\\angle {{B}_{1}}AB$$，从而$$MB\\bot A{{B}_{1}}$$， 因此，$$A{{B}_{1}}\\bot $$平面$$BM{{C}_{1}}$$，所以$$A{{B}_{1}}\\bot B{{C}_{1}}$$，即$$A{{B}_{1}}$$与$${{C}_{1}}B$$所成的角的大小为$$90{}^{}\\circ $$． " ]

[ "2010年四川全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$30{}^{}\\circ $$ " } ], [ { "aoVal": "B", "content": "$$45{}^{}\\circ $$ " } ], [ { "aoVal": "C", "content": "$$60{}^{}\\circ $$ " } ], [ { "aoVal": "D", "content": "$$\\arctan \\sqrt{2}$$ " } ] ]

[ "竞赛->知识点->立体几何与空间向量->空间中的角与距离" ]

[ "设顶点$$S$$在底面$$\\triangle ABC$$的射影是$$H$$，则$$H$$为$$\\triangle ABC$$的外心．从而$$AH=\\frac{2}{3}\\times 3\\times \\frac{\\sqrt{3}}{2}=\\sqrt{3}$$，于是可得$$\\angle SAH={{30}^{\\circ }}$$．故选$$\\text{A}$$． " ]

[ "1994年全国高中数学联赛竞赛一试第4题" ]

[ [ { "aoVal": "A", "content": "$$x\\textless{}z\\textless{}y$$ " } ], [ { "aoVal": "B", "content": "$$y\\textless{}z\\textless{}x$$ " } ], [ { "aoVal": "C", "content": "$$z\\textless{}x\\textless{}y$$ " } ], [ { "aoVal": "D", "content": "$$x\\textless{}y\\textless{}z$$ " } ] ]

[ "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->指对幂比较大小", "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->素养->数学运算" ]

[ "∵$$0\\textless{}b\\textless{}1$$， ∴$$f(x)={{\\log }_{b}}x$$是减函数，又 ∵$$0\\textless{}a\\textless{}\\frac{ \\pi }{4}$$， ∴$$0\\textless{}\\sin a\\textless{}\\cos a\\textless{}1$$， ∴$${{\\log }_{b}}\\sin a\\textgreater{{\\log }_{b}}\\cos a\\textgreater0$$， ∴$${{(\\sin a)}^{{{\\log }_{b}}\\sin a}}\\textless{}{{(\\sin a)}^{{{\\log }_{b}}\\cos a}}$$，即$$x\\textgreater z$$； 又$${{(\\sin a)}^{{{\\log }_{b}}\\sin a}}\\textless{}{{(\\cos a)}^{{{\\log }_{b}}\\cos a}}$$，即$$z\\textless{}y$$，故有$$x\\textless{}z\\textless{}y$$． 故选$$\\text{A}$$． " ]

[ "2008年四川全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$38$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$42$$ " } ], [ { "aoVal": "D", "content": "$$44$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->思想->方程思想", "课内体系->知识点->数列->等差数列->等差数列的概念与通项公式->等差数列求通项问题", "课内体系->知识点->数列->等差数列->等差数列的前n项和->等差数列求和问题", "课内体系->知识点->数列->等差数列->等差数列的性质及应用", "课内体系->知识点->数列->数列的概念->数列的前n项和", "课内体系->知识点->数列->数列的概念->数列的定义", "课内体系->知识点->数列->数列的概念->数列的表示方法->通项公式->常见数列的通项公式", "课内体系->方法->方程组法" ]

[ "因为等差数列$$ {{{a}_{n}} }$$的公差$$d=4$$， 则$${{a}_{3}}={{a}_{1}}+8$$，$${{S}_{3}}=3{{a}_{1}}+12$$， 从而$$\\frac{({{a}_{1}}+8)+2}{2}=\\sqrt{2(3{{a}_{1}}+12)}$$， 解得$${{a}_{1}}=2$$．所以$${{a}_{10}}=2+9\\times 4=38$$． 故选$$\\text{A}$$． " ]

[ "2012年吉林全国高中数学联赛竞赛初赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$0\\textless{}{{x}_{1}}{{x}_{2}}\\textless{}1$$ " } ], [ { "aoVal": "B", "content": "$${{x}_{1}}{{x}_{2}}=1$$ " } ], [ { "aoVal": "C", "content": "$$1\\textless{}{{x}_{1}}{{x}_{2}}\\textless{}2$$ " } ], [ { "aoVal": "D", "content": "$${{x}_{1}}{{x}_{2}}\\geqslant 2$$ " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "显然$${{x}_{2}}=\\frac{1}{2}$$．设$$f\\left( x \\right)={{\\log }_{4}}x-{{\\left( \\frac{1}{4} \\right)}^{x}}$$，则$$f\\left( 1 \\right)\\cdot f\\left( 2 \\right)\\textless{}0$$，所以$$1\\textless{}{{x}_{2}}\\textless{}2$$． " ]

[ "2017年湖南全国高中数学联赛竞赛初赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$\\left( y,z,w \\right)\\in S$$且$$\\left( x,y,w \\right)\\notin S$$ " } ], [ { "aoVal": "B", "content": "$$\\left( y,z,w \\right)\\in S$$且$$\\left( x,y,w \\right)\\in S$$ " } ], [ { "aoVal": "C", "content": "$$\\left( y,z,w \\right)\\notin S$$且$$\\left( x,y,w \\right)\\in S$$ " } ], [ { "aoVal": "D", "content": "$$\\left( y,z,w \\right)\\notin S$$且$$\\left( x,y,w \\right)\\notin S$$ " } ] ]

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

[ "特殊值排除法， 取$$x=2$$，$$y=3$$，$$z=4$$，$$w=1$$，显然满足$$(x,y,z)$$和$$(z,w,x)$$都在$$S$$中， 此时$$(y ,z,w)=(3,4,1)\\in S$$， $$(x,y,w)=(2,3,1)\\in S$$，故$$\\text A$$、$$\\text C$$、$$\\text D$$均错误； 只有$$\\text B$$成立，故选$$\\text{B}$$． " ]

[ "竞赛第20题" ]

[ [ { "aoVal": "A", "content": "$x=2\\pi$ " } ], [ { "aoVal": "B", "content": "$x\\geq 2\\pi$ " } ], [ { "aoVal": "C", "content": "$x\\leq 2\\pi$ " } ], [ { "aoVal": "D", "content": "不能确定 " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 首先外角和与内角和之和为$4\\pi$，再根据各角的不等关系可证内角和不大于$2\\pi$，外角和不小于$2\\pi$．\\\\ 【详解】\\\\ 对于空间四边形，外角和与内角和之和为$4\\pi$，因此考虑内角和与$2\\pi$的大小关系．\\\\ 取\\emph{A}，\\emph{B}，\\emph{C}，\\emph{D}为正四面体的四个顶点，则内角和为$\\frac{\\pi }{3}\\times 4=\\frac{4\\pi }{3}\\textless{} 2\\pi$，\\\\ 因此选项AC都错误，\\\\ 接下来证明内角和不大于$2\\pi$．\\\\ 事实上，有$\\left {\\begin{array}{l} \\angle ABD+\\angle CBD\\geq \\angle ABC, \\angle ADB+\\angle CDB\\geq \\angle CDA, \\end{array}\\right.$\\\\ 从而$2x=(\\angle ABD+\\angle BDA+\\angle DAB)+(\\angle DBC+\\angle BDC+\\angle BCD)$\\\\ $\\geq \\angle DAB+\\angle ABC+\\angle BCD+\\angle CDA$，\\\\ 因此内角和不大于$2\\pi$，外角和不小于$2\\pi$．\\\\ 故选：B. " ]

[ "2009年黑龙江全国高中数学联赛竞赛初赛第5题5分" ]

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

[ "竞赛->知识点->排列组合与概率->概率初步", "竞赛->知识点->集合->集合的划分与覆盖", "竞赛->知识点->集合->集合的概念与运算" ]

[ "不难算出平面区域$$\\Omega $$的面积为$$\\frac{1}{2}\\cdot 6\\cdot 6=18$$，$$A$$的面积为$$\\frac{1}{2}\\cdot 4\\cdot 2=4$$， 所以所求概率为$$\\frac{4}{18}=\\frac{2}{9}$$． " ]

[ "2016年天津全国高中数学联赛竞赛初赛第1题6分" ]

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

[ "课内体系->素养->直观想象", "课内体系->素养->数学运算", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦型函数的图象与性质->求正弦型函数的周期", "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->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->函数的应用->函数与方程->函数的图象变换问题->翻折变换问题", "课内体系->思想->函数思想" ]

[ "$$f\\left( x \\right)=\\left\\textbar{} \\sin 2x+\\cos 2x \\right\\textbar$$可以改写为 $$f\\left( x \\right)=\\left\\textbar{} \\sqrt{2}\\sin \\left( 2x+\\frac{\\pi }{4} \\right) \\right\\textbar$$， 由函数图象之间的关系可见$$f\\left( x \\right)$$与$$g\\left( x \\right)=\\left\\textbar{} \\sin 2x \\right\\textbar$$有相同的最小正周期． 因为$$\\left\\textbar{} \\sin x \\right\\textbar$$的最小正周期为$$\\pi $$，所以$$g\\left( x \\right)$$的最小正周期为$$\\frac{\\pi }{2}$$． 故选$$\\text{C}$$． " ]

[ "2022~2023学年3月江苏常州武进区江苏省前黄高级中学高一下学期月考第2题", "2022~2023学年3月江苏常州武进区江苏省前黄高级中学高一下学期月考第2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4题" ]

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

[ "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" ]

[ "\\hfill\\break 【分析】\\\\ 根据题意，结合$\\alpha $所在象限，得到$\\sin \\alpha $和$\\cos \\alpha $的值，再根据公式，求得答案.\\\\ 【详解】\\\\ 由角$\\alpha $终边上一点\\emph{M}的坐标为$\\left( 1,\\sqrt{3} \\right)$，\\\\ 得$\\sin \\alpha =\\frac{\\sqrt{3}}{2}$，$\\cos \\alpha =\\frac{1}{2}$，\\\\ 故$\\sin 2\\alpha =2\\sin \\alpha \\cos \\alpha =\\frac{\\sqrt{3}}{2}$，\\\\ 故选D.\\\\ 【点睛】\\\\ 本题考查已知角的终边求对应的三角函数值，二倍角公式，属于简单题. " ]

[ "2008年河南全国高中数学联赛竞赛初赛第2题5分" ]

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

[ "竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "因为$$B=60{}^{}\\circ $$，$${{\\sin }^{2}}B=\\sin A\\sin C$$， 所以$${{\\sin }^{2}}60{}^{}\\circ =\\sin A\\sin (120{}^{}\\circ -A)$$， 可知$$\\frac{3}{4}=\\sin A\\left( \\frac{\\sqrt{3}}{2}\\cos A+\\frac{1}{2}\\sin A \\right)$$ $$=\\frac{\\sqrt{3}}{2}\\sin A\\cos A+\\frac{1}{2}{{\\sin }^{2}}A$$ $$=\\frac{\\sqrt{3}}{4}\\sin 2A+\\frac{1-\\cos 2A}{4}$$ 因此$$\\sin (2A-30{}^{}\\circ )=1$$，而$$0{}^{}\\circ \\textless{}A\\textless{}120{}^{}\\circ $$， 易知$$2A-30{}^{}\\circ =90{}^{}\\circ $$，即$$A=60{}^{}\\circ $$， 因此$$\\triangle ABC$$是等边三角形．故选$$\\text{D}$$． " ]

[ "2020年贵州全国高中数学联赛竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$-1010-1010\\text{i}$$ " } ], [ { "aoVal": "B", "content": "$$1010+1010\\text{i}$$ " } ], [ { "aoVal": "C", "content": "$$-1010+1010\\text{i}$$ " } ], [ { "aoVal": "D", "content": "$$1010-1010\\text{i}$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->数列的通项与求和", "课内体系->知识点->复数->复数的运算->复数的四则运算综合" ]

[ "设$$S=\\text{i}+2{{\\text{i}}^{2}}+3{{\\text{i}}^{3}}+\\cdots +2019{{\\text{i}}^{2019}}+2020{{\\text{i}}^{2020}}$$， 则$$\\text{i}S={{\\text{i}}^{2}}+2{{\\text{i}}^{3}}+3{{\\text{i}}^{4}}+\\cdots +2019{{\\text{i}}^{2020}}+2020{{\\text{i}}^{2021}}$$， 两式相减，得： $$S-\\text{i}S=\\text{i}+{{\\text{i}}^{2}}+{{\\text{i}}^{3}}+\\cdots +{{\\text{i}}^{2020}}-{{2020}^{2021}}$$ $$=\\dfrac{\\text{i}\\left( 1-{{\\text{i}}^{2020}} \\right)}{1-\\text{i}}-2020{{\\text{i}}^{2021}}$$ $$=-2021\\text{i}$$， 故$$S=-\\dfrac{2020\\text{i}}{1-\\text{i}}=1010-1010\\text{i}$$， 即$$\\sum\\limits_{k=1}^{2020}{\\left( k\\cdot {{\\text{i}}^{k+1}} \\right)}=1010+1010\\text{i}$$． 故选$$\\text{B}$$． " ]

[ "1994年全国高中数学联赛竞赛一试第2题" ]

[ [ { "aoVal": "A", "content": "命题（$$1$$）正确，命题（$$2$$）也正确 " } ], [ { "aoVal": "B", "content": "命题（$$1$$）正确，命题（$$2$$）错误 " } ], [ { "aoVal": "C", "content": "命题（$$1$$）错误，命题（$$2$$）也错误 " } ], [ { "aoVal": "D", "content": "命题（$$1$$）错误，命题（$$2$$）正确 " } ] ]

[ "竞赛->知识点->复数与平面向量->复数的概念与运算" ]

[ "命题（$$1$$）是正确的．$${{a}^{2}}+{{b}^{2}}\\textgreater{{c}^{2}}$$表明$${{a}^{2}}+{{b}^{2}}$$与$${{c}^{2}}$$都是实数，因此，根据移项法则有$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}\\textgreater0$$． 命题（$$2$$）是错误的．$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}\\textgreater0$$仅表明$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}$$是实数，并不能保证$${{a}^{2}}+{{b}^{2}}$$与$${{c}^{2}}$$都是实数，故$${{a}^{2}}+{{b}^{2}}\\textgreater{{c}^{2}}$$不一定成立． 例如，取$$a=2+\\text{i}$$，$$b=\\text{i}$$，$$c=\\sqrt{2}+\\sqrt{\\text{2i}}$$，则有 $${{a}^{2}}+{{b}^{2}}-{{c}^{2}}=(3+4\\text{i})+(-1)-4\\text{i}=2\\textgreater0$$，但并没有$${{a}^{2}}+{{b}^{2}}=2+4\\text{i}\\textgreater4\\text{i}={{c}^{2}}$$． 故选$$\\text{B}$$． " ]

[ "2021年贵州全国高中数学联赛竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$M=N$$ " } ], [ { "aoVal": "B", "content": "$$M\\subseteq N$$ " } ], [ { "aoVal": "C", "content": "$$N\\subseteq M$$ " } ], [ { "aoVal": "D", "content": "前三个都不正确 " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "对于任意$$\\left( x,y \\right)\\in M$$，均有$$\\textbar x\\textbar+\\textbar y\\textbar\\leqslant 1$$，故$$\\begin{cases}\\textbar x\\textbar\\leqslant 1 \\textbar y\\textbar\\leqslant 1 \\end{cases}$$，故有$$\\begin{cases}{{x}^{2}}\\leqslant \\textbar x\\textbar{} {{y}^{2}}\\leqslant \\textbar y\\textbar{} \\end{cases}\\Rightarrow {{x}^{2}}+{{y}^{2}}\\leqslant \\textbar x\\textbar+\\textbar y\\textbar$$， 故$$\\left( x,y \\right)\\in N$$，所以$$M\\subseteq N$$； 又因为$$\\left( 1,1 \\right)\\in N$$，$$\\left( 1,1 \\right)\\notin M$$， 故$$M\\subsetneqq N$$，选$$\\text{B}$$． " ]

[ "2011年黑龙江全国高中数学联赛竞赛初赛第9题5分" ]

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

[ "竞赛->知识点->解析几何->直线与圆锥曲线" ]

[ "由题知，可转换为该椭圆与直线平行并且到直线$$\\frac{x}{4}+\\frac{y}{3}=1$$的距离为$$\\frac{6}{5}$$的直线与椭圆的交点个数，与直线平行并且到直线$$\\frac{x}{4}+\\frac{y}{3}=1$$的距离为$$\\frac{6}{5}$$的直线方程为$$3x+4y-6=0$$或$$3x+4y-18=0$$，由$$\\begin{cases}\\frac{{{x}^{2}}}{16}+\\frac{{{y}^{2}}}{9}=1 3x+4y-18=0 \\end{cases}$$，消去$$y$$可得$$9{{x}^{2}}+{{\\left( 3x-18 \\right)}^{2}}=144$$，化简可得$${{x}^{2}}-6x+10=0$$，由判别式可知该方程无解，所以直线$$3x+4y-6=0$$与椭圆必有$$2$$个交点，故选$$\\text{B}$$. " ]

[ "2010年四川全国高中数学联赛竞赛初赛第8题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{12}{8}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{16}{5}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{18}{5}$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "竞赛->知识点->解析几何->双曲线" ]

[ "设动点$$P\\left( x, -\\frac{x}{3} \\right)$$与$$Q\\left( y,\\frac{3}{y} \\right)$$，则$$F(x, y)={{\\left\\textbar{} PQ \\right\\textbar}^{2}}$$，点$$P$$的轨迹为直线$$y=-\\frac{x}{3}$$，点$$Q$$的轨迹为双曲线$$y=\\frac{3}{x}$$，双曲线上的任一点$$\\left( {{x}_{0}},\\frac{3}{{{x}_{0}}} \\right)$$到直线$$x+3y=0$$的距离: $$d=\\frac{\\left\\textbar{} {{x}_{0}}+3\\cdot \\frac{3}{{{x}_{0}}} \\right\\textbar}{\\sqrt{10}}\\geqslant \\frac{6}{\\sqrt{10}}$$， 当$${{x}_{0}}=\\pm 3$$时等号成立．故$$F(x, y)$$的最小值为$$\\frac{18}{5}$$．故选$$\\text{C}$$． " ]

[ "2010年浙江全国高中数学联赛竞赛初赛第2题5分" ]

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

[ "竞赛->知识点->不等式->不等式的解法" ]

[ "$$p$$成立$$\\Leftrightarrow x\\geqslant -3$$，所以$$p $$成立，推不出$$q$$一定成立． " ]

[ "2009年浙江全国高中数学联赛竞赛初赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$\\overrightarrow{a}\\bot \\overrightarrow{b}$$ " } ], [ { "aoVal": "B", "content": "$$(\\overrightarrow{a}+\\overrightarrow{b})\\bot (\\overrightarrow{a}-\\overrightarrow{b})$$ " } ], [ { "aoVal": "C", "content": "$$(\\overrightarrow{a}+\\overrightarrow{b})//(\\overrightarrow{a}-\\overrightarrow{b})$$ " } ], [ { "aoVal": "D", "content": "$$\\overrightarrow{a}\\bot (\\overrightarrow{a}+\\overrightarrow{b})$$ " } ] ]

[ "课内体系->方法->图象法", "课内体系->知识点->平面向量->向量应用->平面几何中的向量方法", "课内体系->知识点->平面向量->平面向量的运算->数量积->数量积的坐标表达式", "课内体系->知识点->平面向量->平面向量的运算->数量积->向量的数量积的定义", "课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积运算（非坐标）", "课内体系->知识点->平面向量->平面向量的运算->数量积->利用数量积解决向量垂直问题（非坐标运算）", "课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积的运算律", "课内体系->知识点->平面向量->平面向量的运算->数量积->线性运算和数量积综合问题", "课内体系->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量的加法运算及运算规则", "课内体系->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量的减法运算及运算规则", "课内体系->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量线性运算综合（非坐标）", "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->平面向量基本定理及其意义", "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->平面向量的正交分解及坐标表示", "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->平面向量加、减、数乘的坐标运算", "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->坐标表示平面向量的垂直", "课内体系->知识点->平面向量->平面向量的基本概念->单位向量", "课内体系->知识点->平面向量->平面向量的基本概念->向量的概念", "课内体系->素养->直观想象", "课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->思想->数形结合思想" ]

[ "因为$$\\overrightarrow{a},\\overrightarrow{b}$$都是非零单位向量，则以$$\\overrightarrow{a},\\overrightarrow{b}$$为边，$$\\overrightarrow{a}-\\overrightarrow{b},\\overrightarrow{a}+\\overrightarrow{b}$$为对角线构成一个菱形，所以$$(\\overrightarrow{a}+\\overrightarrow{b})\\bot (\\overrightarrow{a}-\\overrightarrow{b})$$． " ]

[ "2009年四川全国高中数学联赛竞赛初赛第4题5分" ]

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

[ "竞赛->知识点->数列与数学归纳法->（模）周期数列" ]

[ "因为 $${{a}_{n+2}}=1-\\frac{1}{{{a}_{n+1}}}=1-\\frac{1}{1-\\frac{1}{{{a}_{n}}}}=\\frac{-1}{{{a}_{n}}-1}$$， 于是 $${{a}_{n+3}}=1-\\frac{1}{{{a}_{n+2}}}=1-\\frac{1}{\\frac{-1}{{{a}_{n}}-1}}={{a}_{n}}$$， 故$$ {{{a}_{n}} }$$是以$$3$$为周期的周期数列． 又$${{a}_{1}}=2,{{a}_{2}}=\\frac{1}{2},{{a}_{3}}=-1$$，从而$${{P}_{3}}=-1$$，所以， $${{P}_{2009}}={{(-1)}^{669}}{{P}_{2}}=-1$$． " ]

[ "2010~2011学年北京海淀区高三上学期期末理科第7题", "2018~2019学年北京西城区北京市第十三中学高二上学期期中第10题3分", "2019~2020学年11月重庆沙坪坝区重庆市南开中学高二上学期周测A卷第6题5分", "2019~2020学年10月北京海淀区北京市第五十七中学高二上学期月考第10题", "2016年吉林全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$kx+y+k=0$$ " } ], [ { "aoVal": "B", "content": "$$kx-y-1=0$$ " } ], [ { "aoVal": "C", "content": "$$kx+y-k=0$$ " } ], [ { "aoVal": "D", "content": "$$kx+y-2=0$$ " } ] ]

[ "课内体系->知识点->圆锥曲线->直线与圆锥曲线问题->弦长求解问题", "课内体系->知识点->圆锥曲线->椭圆->直线和椭圆的位置关系", "课内体系->素养->数学运算" ]

[ "由数形结合可知，当$$l$$过点$$(-1,0)$$时，直线$$l$$和选项A中的直线关于$$x$$对称，被椭圆$$E$$所截得的弦长相同，故不能选A． 当$$l$$过点$$(1,0)$$时，直$$l$$和选项C中的直线关于$$x$$轴对称，被椭圆$$E$$所截得的弦长相同，故不能选C． 当$$k=0$$时，直线$$l$$和选项B中的直线关于$$x$$轴对称，被椭圆$$E$$所截得的弦长相同，故不能选B． 直线$$l$$斜率为$$k$$，在$$y$$轴上的截距为$$1$$；选项D中的直线$$kx+y-2=0$$ 斜率为$$-k$$，在$$y$$轴上的截距为$$2$$，这两直线不关于$$x$$轴、$$y$$轴、原点对称，故被椭圆$$E$$所截得的弦长不可能相等．故选D． " ]

[ "2005年高考真题福建卷理科第12题5分", "2009年黑龙江全国高中数学联赛竞赛初赛第6题5分" ]

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

[ "课内体系->知识点->函数的概念与性质->函数的性质->周期性->函数周期性与奇偶性综合问题", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->抽象函数", "课内体系->知识点->函数的应用->函数的零点->零点的个数问题->求零点个数问题（不含参）", "课内体系->素养->数学抽象", "课内体系->素养->数学运算" ]

[ "$$f(x)$$是定义在$$\\text{R}$$周期为$$3$$的奇函数，∴$$f(0)=0=f(3)$$，$$f(5)=f(2)=-f(-2)=-f(1)=0$$，∴$$f(1)=0=f(4)$$，∴选D. " ]

[ "2011年黑龙江全国高中数学联赛竞赛初赛第7题5分" ]

[ [ { "aoVal": "A", "content": "$$30$$种 " } ], [ { "aoVal": "B", "content": "$$60$$种 " } ], [ { "aoVal": "C", "content": "$$120$$种 " } ], [ { "aoVal": "D", "content": "$$720$$种 " } ] ]

[ "竞赛->知识点->排列组合与概率->两个基本计数原理", "竞赛->知识点->排列组合与概率->排列与组合", "课内体系->知识点->计数原理->两个基本计数原理->分步乘法计数原理" ]

[ "总共$$10$$个球，先考虑选$$5$$个位置放篮球，共有$$\\text{C}_{10}^{5}=252$$种，再从剩下的$$5$$个位置种选$$3$$个位置放排球，总共有$$10$$种方法，剩下的$$2$$个位置放橄榄球，故有$$252\\times 10=2520$$种. " ]

[ "2008年贵州全国高中数学联赛竞赛初赛第10题5分" ]

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

[ "竞赛->知识点->解析几何->双曲线" ]

[ "由已知可得$$M\\left( 3,\\frac{\\sqrt{6}}{2} \\right)$$，则$$M{{F}_{1}}=\\frac{\\sqrt{6}}{2}$$， 故$$M{{F}_{2}}=2\\sqrt{6}+\\frac{\\sqrt{6}}{2}=\\frac{5\\sqrt{6}}{2}$$， 故$${{F}_{1}}$$到直线$${{F}_{2}}M$$的距离为： $$\\frac{\\left\\textbar{} {{F}_{1}}{{F}_{2}} \\right\\textbar\\cdot \\left\\textbar{} M{{F}_{1}}_{2} \\right\\textbar}{\\left\\textbar{} M{{F}_{2}} \\right\\textbar}=\\frac{6\\times \\frac{\\sqrt{6}}{2}}{\\frac{5\\sqrt{6}}{2}}=\\frac{6}{5}$$，故选$$\\text{A}$$. " ]

[ "2009年河北全国高中数学联赛竞赛初赛第1题6分" ]

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

[ "竞赛->知识点->数列与数学归纳法->数列的综合应用", "竞赛->知识点->数列与数学归纳法->数列的通项与求和" ]

[ "前$$n$$组所含项数之和为$$2+{{2}^{2}}+\\cdots +{{2}^{n}}={{2}^{n+1}}-2$$， $$2009$$是数列的第$$1005$$项． $$510={{2}^{9}}-2\\textless{}1005\\textless{}{{2}^{10}}-2=1022$$． 因此，$$2009$$在第$$9$$组，选$$\\text{C}$$． " ]

[ "2008年河北全国高中数学联赛竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$(1,-3)$$ " } ], [ { "aoVal": "B", "content": "$$(-1,3)$$ " } ], [ { "aoVal": "C", "content": "$$(-3,-3)$$ " } ], [ { "aoVal": "D", "content": "$$(-3,3)$$ " } ] ]

[ "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" ]

[ "函数$$y=f(x+2)$$的图象过点$$(-1,3)$$，则$$f(-1+2)=3$$，即$$f(1)=3$$，所以数$$f(x)$$的图象关于$$y$$轴对称的图形一定过点$$f(-1)=3$$．故选$$\\text{B}$$． " ]

[ "2009年AMC10竞赛B第23题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{16}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{8}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{16}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{4}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{5}{16}$$ " } ] ]

[ "课内体系->学科符号->数与式->新定义符号->取最值函数->max{f(x),g(x)}", "美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Geometric Models of Probabilities" ]

[ "After $$10$$ minutes ($$600$$ seconds), Rachel will have completed $$6$$ laps and be $$30$$ seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in $$22.5$$ seconds, she will be in the picture between $$18.75$$ seconds and $$41.25$$ seconds of the tenth minute. After $$10$$ minutes Robert will have completed $$7$$ laps and will be $$40$$ seconds past the starting line. Because Robert runs one-fourth of a lap in $$20$$ seconds, he will be in the picture between $$30$$ and $$50$$ seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between $$30$$ and $$41.25$$ seconds of the tenth minute. So the probability that both runners are in the picture is $$\\frac{41 .25-30}{60}= \\boxed{\\frac{3}{16}}$$. The answer is $$\\rm(C)$$. " ]

[ "2018~2019学年10月四川成都金牛区成都市实验外国语学校高二上学期周测C卷第9题5分", "2017年河南洛阳高三二模文科第9题5分", "2016年广东揭阳高三一模理科第12题5分", "2014年黑龙江全国高中数学联赛竞赛初赛第10题5分", "2016年山东日照高三二模理科第9题5分" ]

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

[ "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->圆的切线的相关问题", "课内体系->素养->数学运算" ]

[ "设$$AB$$中点为$$D$$，则$$OD\\bot AB$$， ∵$$\\left\\textbar{} \\overrightarrow{OA}+\\overrightarrow{OB}\\left\\textbar{} \\geqslant \\frac{\\sqrt{3}}{3} \\right\\textbar\\overrightarrow{AB} \\right\\textbar$$， ∴$$\\left\\textbar{} 2\\overrightarrow{OD}\\left\\textbar{} \\geqslant \\frac{\\sqrt{3}}{3} \\right\\textbar\\overrightarrow{AB} \\right\\textbar$$， ∴$$\\left\\textbar{} \\overrightarrow{AB}\\left\\textbar{} \\leqslant 2\\sqrt{3} \\right\\textbar\\overrightarrow{OD} \\right\\textbar$$， ∵$$\\left\\textbar{} \\overrightarrow{OD}{{\\textbar}^{2}}+\\frac{1}{4} \\right\\textbar\\overrightarrow{AB}{{\\textbar}^{2}}=4$$， ∴$$\\textbar\\overrightarrow{OD}{{\\textbar}^{2}}\\geqslant 1$$， ∵直线$$x+y-k=0\\left( k\\textgreater0 \\right)$$与圆$${{x}^{2}}+{{y}^{2}}=4$$交于不同的两点$$A,B$$， ∴$$\\textbar\\overrightarrow{OD}{{\\textbar}^{2}}\\textless{}4$$ ∴$$4\\textgreater\\textbar\\overrightarrow{OD}{{\\textbar}^{2}}\\geqslant 1$$ ∴$$4\\textgreater{{\\left( \\frac{\\left\\textbar{} -k \\right\\textbar}{\\sqrt{2}} \\right)}^{2}}\\geqslant 1$$ ∵$$k\\textgreater0$$， ∴$$\\sqrt{2}\\leqslant k\\textless{}2\\sqrt{2}$$. 故选：$$\\text{C}$$． " ]

[ "2018年黑龙江全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$208$$ " } ], [ { "aoVal": "B", "content": "$$216$$ " } ], [ { "aoVal": "C", "content": "$$217$$ " } ], [ { "aoVal": "D", "content": "$$218$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->知识点->计数原理->二项式定理->求项的系数或二项式系数", "课内体系->知识点->计数原理->二项式定理->求二项式展开式的特定项" ]

[ "依题意，知$$ab{{c}^{2}}$$的系数为$$\\text{C}_{4}^{2}{{(-3)}^{2}}\\text{C}_{2}^{1}\\times 2\\times 1=216$$． " ]

[ "1989年全国高中数学联赛竞赛一试第3题" ]

[ [ { "aoVal": "A", "content": "关于$$x$$轴对称； " } ], [ { "aoVal": "B", "content": "关于直线$$x=1$$对称 " } ], [ { "aoVal": "C", "content": "关于直线$$x=-1$$对称 " } ], [ { "aoVal": "D", "content": "关于$$y$$轴对称 " } ] ]

[ "课内体系->知识点->函数的概念与性质->函数的性质->对称性->两个函数的互对称问题", "竞赛->知识点->函数->函数的图像与性质" ]

[ "$$f(x)$$和$$f(-x)$$的图象关于直线$$x=0$$对称， $$f(x-1)$$与$$f(-x+1)$$的图象关于直线$$x=1$$对称． 故选$$\\text{B}$$． " ]

[ "2018年天津全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$\\ln \\sqrt{2}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{\\text{e}}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{\\ln \\pi }{\\pi }$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{\\sqrt{10}\\ln 10}{20}$$ " } ] ]

[ "竞赛->知识点->导数模块->导数" ]

[ "考虑函数$$f\\left( x \\right)=\\frac{\\ln x}{x}$$． 则四个选项分别为$$f\\left( 2 \\right)$$、$$f\\left( \\text{e} \\right)$$、$$f\\left( \\pi \\right)$$、$$f\\left( \\sqrt{10} \\right)$$， 又$${f}'\\left( x \\right)=\\frac{1-\\ln x}{{{x}^{2}}}$$，则$$f\\left( x \\right)$$在区间$$\\left( 0,\\text{e} \\right)$$上单调递增，在区间$$\\left( \\text{e},+\\infty \\right)$$上单调递减． 从而，$$f\\left( x \\right)$$在区间$$\\left( 0,+\\infty \\right)$$上的最大值为$$f\\left( \\text{e} \\right)$$． " ]

[ "2022年湖南湘西吉首市竞赛（第一届中小学生教师解题大赛）第2题" ]

[ [ { "aoVal": "A", "content": "$\\text{ } ! !\\pi ! !\\text{ }$ " } ], [ { "aoVal": "B", "content": "$3\\text{ } ! !\\pi ! !\\text{ }$ " } ], [ { "aoVal": "C", "content": "$5\\text{ } ! !\\pi ! !\\text{ }$ " } ], [ { "aoVal": "D", "content": "$6\\text{ } ! !\\pi ! !\\text{ }$ " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 令$z=a+b\\text{i}$且$a,b\\in \\text{R}$，可得$1\\le {{(a-1)}^{2}}+{{(b+1)}^{2}}\\le 4$，然后根据复数模的几何意义结合条件即得.\\\\ 【详解】\\\\ 令$z=a+b\\text{i}$且$a,b\\in \\text{R}$，则$2\\le \\left\\textbar{} (a-1)+(b+1)\\text{i} \\right\\textbar\\le 3$，\\\\ 所以$4\\le {{(a-1)}^{2}}+{{(b+1)}^{2}}\\le 9$，\\\\ 所以复数$z$在复平面内对应的点$Z$所在区域是圆${{(a-1)}^{2}}+{{(b+1)}^{2}}=4$和圆${{(a-1)}^{2}}+{{(b+1)}^{2}}=9$围成的圆环，\\\\ 所以点$Z$所在区域的面积为$9\\text{ } ! !\\pi ! !\\text{ }-4\\text{ } ! !\\pi ! !\\text{ }=5\\text{ } ! !\\pi ! !\\text{ }$.\\\\ 故选：C. " ]

[ "2016年高考真题天津卷", "2016年天津全国高中数学联赛竞赛初赛第3题6分" ]

[ [ { "aoVal": "A", "content": "$$x\\textgreater y$$ " } ], [ { "aoVal": "B", "content": "$$x=y$$ " } ], [ { "aoVal": "C", "content": "$$x\\textless{}y$$ " } ], [ { "aoVal": "D", "content": "既有$$x\\textgreater y$$的情形，也有$$x\\textless{}y$$的情形 " } ] ]

[ "竞赛->知识点->不等式->几个重要的不等式->排序", "竞赛->知识点->不等式->几个重要的不等式->均值", "竞赛->知识点->不等式->不等式的综合应用（数列与不等式）", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件" ]

[ "由题设有$$x=\\frac{2a+b}{3}\\cdot \\frac{a+2b}{3}\\left( a+b \\right)$$，$$y=ab\\left( \\sqrt[3]{{{a}^{2}}b}+\\sqrt[3]{a{{b}^{2}}} \\right)$$， 由均值不等式，$$\\frac{2a+b}{3}\\geqslant \\sqrt[3]{{{a}^{2}}b}$$，$$\\frac{a+2b}{3}\\geqslant \\sqrt[3]{a{{b}^{2}}}$$（等号都不成立）， 由排序不等式，$$a+b=\\sqrt[3]{{{a}^{3}}}+\\sqrt[3]{{{b}^{3}}}\\geqslant \\sqrt[3]{{{a}^{2}}b}+\\sqrt[3]{a{{b}^{2}}}$$（等号不成立）， 所以，$$x\\textgreater y$$． 故选$$\\text{A}$$． " ]

[ "2014年四川全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$32\\sqrt{3}$$ " } ], [ { "aoVal": "B", "content": "$$54\\sqrt{3}$$ " } ], [ { "aoVal": "C", "content": "$$64\\sqrt{3}$$ " } ], [ { "aoVal": "D", "content": "$$72\\sqrt{3}$$ " } ] ]

[ "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", "竞赛->知识点->三角函数->三角恒等变换" ]

[ "设棱锥的高与一条侧棱的夹角为$$\\theta $$，侧棱长$$l=12\\cos \\theta $$，高$$h=12{{\\cos }^{2}}\\theta $$，底面正三角形的外接圆半径$$R=12\\sin \\theta \\cos \\theta $$，底面边长为$$a=12\\sqrt{2}\\sin \\theta \\cos \\theta $$，从而 $$V=432\\sqrt{3}{{\\cos }^{4}}\\theta {{\\sin }^{2}}\\theta $$ $$=216\\sqrt{3}\\cdot {{\\cos }^{2}}\\theta \\cdot {{\\cos }^{2}}\\theta \\cdot 2{{\\sin }^{2}}\\theta $$ $$\\leqslant 216\\sqrt{3}\\cdot {{\\left( \\frac{{{\\cos }^{2}}\\theta +{{\\cos }^{2}}\\theta +2{{\\sin }^{2}}\\theta }{3} \\right)}^{3}}$$ $$=64\\sqrt{3}$$， 当$${{\\tan }^{2}}\\theta =\\frac{1}{2}$$时等号成立． " ]

[ "2009年河北全国高中数学联赛竞赛初赛第3题6分" ]

[ [ { "aoVal": "A", "content": "$$f(x)\\textgreater g(x)$$ " } ], [ { "aoVal": "B", "content": "$$f(x)\\textless{}g(x)$$ " } ], [ { "aoVal": "C", "content": "$$f(x)=g(x)$$ " } ], [ { "aoVal": "D", "content": "不确定 " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->导数模块->导数" ]

[ "$$f(x)$$与$$g(x)$$的图象在$$x$$轴上有唯一公共点$$(1,0)$$， 所以$$g(1)=0$$，即$$a+b=0$$． 因为$${{f}^{\\prime }}(x)=\\frac{1}{x}$$，$${{g}^{\\prime }}(x)=a-\\frac{b}{{{x}^{2}}}$$， 由题意$${{f}^{\\prime }}(1)={{g}^{\\prime }}(1)=1$$，即$$a-b=1$$． 所以$$a=\\frac{1}{2}$$，$$b=-\\frac{1}{2}$$． 令$$F(x)=f(x)-g(x)=\\ln x-\\left( \\frac{1}{2}x-\\frac{1}{2x} \\right)$$， 则$${{F}^{\\prime }}(x)=\\frac{1}{x}-\\frac{1}{2}-\\frac{1}{2{{x}^{2}}}=-\\frac{1}{2}{{\\left( \\frac{1}{x}-1 \\right)}^{2}}\\leqslant 0$$． 所以$$F(x)$$在$$(1,+\\infty )$$内单调递减． " ]

[ "2002年全国高中数学联赛竞赛一试第3题6分" ]

[ [ { "aoVal": "A", "content": "是偶函数但不是奇函数 " } ], [ { "aoVal": "B", "content": "是奇函数但不是偶函数 " } ], [ { "aoVal": "C", "content": "既是奇函数又是偶函数 " } ], [ { "aoVal": "D", "content": "既不是奇函数又不是偶函数 " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "$$f\\left( x \\right)$$定义域为$$(-\\infty ,0)\\cup (0,+\\infty )$$；$$f\\left( x \\right)-f\\left( -x \\right)=\\frac{x}{1-{{2}^{x}}}-\\frac{x}{2}-\\frac{-x}{1-{{2}^{-x}}}+\\frac{-x}{2}=\\frac{x-x{{2}^{x}}}{1-{{2}^{x}}}-x=0$$．即$$f\\left( x \\right)$$是偶函数. " ]

[ "2019年吉林全国高中数学联赛竞赛初赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{4}{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{5}{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{15}{8}$$ " } ], [ { "aoVal": "D", "content": "$$2$$ " } ] ]

[ "竞赛->知识点->复数与平面向量->平面向量的概念与运算", "课内体系->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量的加法运算及运算规则" ]

[ "注意到， $$\\overrightarrow{AC}=\\lambda \\overrightarrow{AM}+\\mu \\overrightarrow{BD}$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\lambda \\left( \\overrightarrow{AB}+\\overrightarrow{BM} \\right)+\\mu \\left( \\overrightarrow{BA}+\\overrightarrow{AD} \\right)$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\lambda \\left( \\overrightarrow{AB}+\\dfrac{1}{2}\\overrightarrow{AD} \\right)+\\mu \\left( -\\overrightarrow{AB}+\\overrightarrow{AD} \\right)$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\left( \\lambda -\\mu \\right)\\overrightarrow{AB}+\\left( \\dfrac{1}{2}\\lambda +\\mu \\right)\\overrightarrow{AD}$$． 又$$\\overrightarrow{AC}=\\overrightarrow{AB}+\\overrightarrow{AD}$$， 则$$\\begin{cases}\\lambda -\\mu =1 \\dfrac{1}{2}\\lambda +\\mu =1 \\end{cases}\\Rightarrow \\begin{cases}\\lambda =\\dfrac{4}{3} \\mu =\\dfrac{1}{3} \\end{cases}$$， 故$$\\lambda +2\\mu =2$$． " ]

[ "高二下学期单元测试《整除与同余》自招", "第二十届全国希望杯高一竞赛复赛邀请赛第6题4分" ]

[ [ { "aoVal": "A", "content": "方程没有整数根 " } ], [ { "aoVal": "B", "content": "方程有两个相等的整数根 " } ], [ { "aoVal": "C", "content": "方程有两个不相等的整数根 " } ], [ { "aoVal": "D", "content": "不能判定方程整数根的情况 " } ] ]

[ "竞赛->知识点->多项式与方程->解方程（组）" ]

[ "若方程有整数根，则根据题意 $$a{{x}^{2}}+bx+c=\\begin{cases}c(\\text{mod}2)\\left. 2 \\right\\textbar x a+b+c(\\text{mod}2)\\left. 2 \\right\\textbar x \\end{cases}\\equiv 1(\\text{mod}2)$$ 矛盾，因此题中方程没有整数根． " ]

[ "2019~2020学年4月山西太原迎泽区太原市第五中学高二下学期周测C卷理科第6题6分", "2012年黑龙江全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$57$$ " } ], [ { "aoVal": "B", "content": "$$56$$ " } ], [ { "aoVal": "C", "content": "$$49$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "$$S\\subseteq A$$，且$$S\\cap B\\ne \\varnothing $$，说明$$S$$是$$A$$的子集，且$$S$$与$$B$$有公共元素； ∴$$A$$的构成情况为：①含一个元素：从$$4$$，$$5$$，$$6$$中选一个元素，个数为$$\\text{C}_{3}^{1}=3$$； ②含两个元素:从$$4$$，$$5$$，$$6$$选两个元素，或从$$1$$，$$2$$，$$3$$选一个，从$$4$$，$$56$$选一个，个数为：$$\\text{C}_{3}^{2}+\\text{C}_{3}^{1}\\text{C}_{3}^{1}=12$$；③含三个元素：从$$4$$，$$5$$，$$6$$选三个，或从$$4$$，$$5$$，$$6$$选两个，从$$1$$，$$2$$，$$3$$选一个或从$$4$$，$$5$$，$$6$$选一个，从$$1$$，$$2$$，$$3$$选两个，个数为：$$\\text{C}_{3}^{3}+\\text{C}_{3}^{2}\\text{C}_{3}^{1}+\\text{C}_{3}^{1}\\text{C}_{3}^{2}=19$$； ④含四个元素：从$$4$$，$$5$$，$$6$$选三个从$$1$$，$$2$$，$$3$$选一个，或从$$4$$，$$5$$，$$6$$选两个，或从$$4$$，$$5$$，$$6$$选一个，从$$1$$，$$2$$，$$3$$选三个个数为：$$\\text{C}_{3}^{3}\\text{C}_{3}^{1}+\\text{C}_{3}^{2}\\text{C}_{3}^{2}+\\text{C}_{3}^{1}\\text{C}_{3}^{3}=15$$； ⑤含五个元素：从$$4$$，$$5$$，$$6$$选三个，从$$1$$，$$2$$，$$3$$选两个，或从$$4$$，$$5$$，$$6$$选两个，从$$1$$，$$2$$，$$3$$选三个，个数为：$$\\text{C}_{3}^{3}\\text{C}_{3}^{2}+\\text{C}_{3}^{2}\\text{C}_{3}^{3}=6$$含\"6\"个元素：从$$4$$，$$5$$，$$6$$选三个，从$$1$$，$$2$$，$$3$$选三个，个数为$$\\text{C}_{3}^{3}\\text{C}_{3}^{3}=1$$； ∴集合$$S$$的个数为：$$2+12+19+15+6+1=56$$． 故选$$\\text{B}$$． " ]

[ "2020~2021学年浙江杭州萧山区杭州第二中学钱江学校高一下学期期中第5题5分", "2012年浙江全国高中数学联赛竞赛初赛第3题5分", "2016年上海闸北区高三二模理科第11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": "充分且必要条件 " } ], [ { "aoVal": "D", "content": "非充分且非必要条件 " } ] ]

[ "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->平面向量->平面向量的运算->数量积->平面向量数量积的运算律", "课内体系->知识点->平面向量->平面向量的基本概念->向量的概念->向量的夹角的判断", "课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与向量结合" ]

[ "若$$\\left\\textbar{} \\overrightarrow{a}-\\overrightarrow{b} \\right\\textbar\\textgreater1$$，则$${{\\overrightarrow{a}}^{2}}-2\\overrightarrow{a}\\cdot \\overrightarrow{b}+{{\\overrightarrow{b}}^{2}}=2-2\\overrightarrow{a}\\cdot \\overrightarrow{b}\\textgreater1$$，即$$\\overrightarrow{a}\\cdot \\overrightarrow{b}\\textless{}\\frac{1}{2}$$， 则$$\\cos \\theta =\\frac{\\overrightarrow{a}\\cdot \\overrightarrow{b}}{\\left\\textbar{} \\overrightarrow{a} \\right\\textbar\\cdot \\left\\textbar{} \\overrightarrow{b} \\right\\textbar}=\\overrightarrow{a}\\cdot \\overrightarrow{b}\\textless{}\\frac{1}{2}$$，∴$$\\theta \\in \\left( \\frac{ \\pi }{3}, \\pi ~\\right]$$，即$$P:\\theta \\in \\left( \\frac{ \\pi }{3}, \\pi ~\\right]$$， ∵$$Q:\\theta \\in \\left[ \\frac{ \\pi }{2},\\frac{5 \\pi }{6} \\right)$$，∴$$P$$是$$Q$$的必要不充分条件， 故选$$\\text{B}$$． " ]

[ "2014年AMC12竞赛A第11题", "2014年AMC10竞赛A第15题" ]

[ [ { "aoVal": "A", "content": "$$140$$ " } ], [ { "aoVal": "B", "content": "$$175$$ " } ], [ { "aoVal": "C", "content": "$$210$$ " } ], [ { "aoVal": "D", "content": "$$245$$ " } ], [ { "aoVal": "E", "content": "$$280$$ " } ] ]

[ "课内体系->知识点->等式与不等式->等式->方程组的解集", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems" ]

[ "Note that he drives at $$50$$ miles per hour after the first hour and continues doing so until he arrives. Let $$d$$ be the distance still needed to travel after the frst $$1$$ hour.~ We have that $$\\dfrac{d}{50}+1.5= \\dfrac{d}{35}$$, ~where the $$1.5$$ comes from $$1$$ hour late decreased to $$0.5$$ hours early. Simplifying gives $$7d+525=10 d$$, or $$d=175$$. Now, we must add an extra $$35$$ miles traveled in the frst hour, giving a total or $$(\\rm C) 210$$ miles. Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices. Quickly checking, we know that neither choice $$(\\rm A)$$ or choice $$(\\rm B)$$ work, but $$(\\rm C)$$ does. We can verify as follows. After l hour at $$35 \\rm mph$$, David has $$175$$ miles left. This then takes him $$3.5$$ hours at $$50 \\rm mph$$. But $$210/35=6$$ hours. Since $$1+3.5=4.5$$ hours is $$1.5$$ hours less than $$6$$ our anwer is $$(\\rm C)210$$. Let us call the total number of miles after David changes speed $$x$$. We realize that if David is driving $$50$$ miles an hour, and he arrives $$30$$ minutes early, $$x≡25 (\\rm mod 50)$$ The only solution that fits the descripion is $$210$$. " ]

[ "2017年陕西全国高中数学联赛竞赛初赛第3题6分" ]

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

[ "竞赛->知识点->立体几何与空间向量->空间向量", "竞赛->知识点->三角函数->三角恒等变换" ]

[ "$$\\cos A=\\frac{\\overrightarrow{AB}\\cdot \\overrightarrow{AC}}{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}=\\frac{\\left( -3,0,4 \\right)\\cdot \\left( 2,-2,-1 \\right)}{\\sqrt{{{\\left( -3 \\right)}^{2}}+{{4}^{2}}}\\cdot \\sqrt{{{2}^{2}}+{{\\left( -2 \\right)}^{2}}+{{\\left( -1 \\right)}^{2}}}}=-\\frac{2}{3}$$， 又$$\\cos A=\\frac{1-{{\\tan }^{2}}\\frac{A}{2}}{1+{{\\tan }^{2}}\\frac{A}{2}}$$，解得$$\\tan \\frac{A}{2}=\\sqrt{5}$$． " ]

[ "1999年全国全国高中数学联赛竞赛一试第5题6分" ]

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

[ "竞赛->知识点->排列组合与概率->排列与组合" ]

[ "解法一： 设这三名选手之间的比赛场数是$$r$$，共$$n$$名选手参赛． 由题意，可得$$\\text{C}_{n-3}^{2}+6-r=50$$， 即$$\\frac{\\left( n-3 \\right)\\left( n-4 \\right)}{2}=44+r$$． 由于$$0\\leqslant r\\leqslant 3$$， 经检验可知，仅当$$r=1$$时，$$n=13$$为正整数． 解法二： $$3$$名选手之间比赛的可能场数为$$0$$、$$1$$、$$2$$、$$3$$，设总人数为$$N$$人． 那么除这$$3$$人外的$$N-3$$人中比赛场数为$$C_{N-3}^{2}=\\frac{(N-3)(N-4)}{2}$$． ①当这$$3$$人之间比赛$$0$$场时，他们每人与另外$$N-3$$人（以下称为``局外人''）要比赛两场，这些比赛没有重合，共计$$6$$场，则有方程：$$\\frac{(N-3)(N-4)}{2}+6=50$$，$$N$$无整数解，故舍去． ②当这$$3$$人之间比赛$$1$$场时，他们有两人与``局内人''分别比赛一场，另一人两场都是和局内人比赛的，所以共计$$5$$场，则有方程：$$\\frac{(N-3)(N-4)}{2}+5=50$$，$$N=13$$，是整数解，满足条件． ③当这$$3$$人之间比赛$$2$$场时，他们有$$1$$人与另两人分别比赛一场，另两人都有一场与局内人的比赛，所以共计$$4$$场，则有方程：$$\\frac{(N-3)(N-4)}{2}+4=50$$，$$N$$无整数解，故舍去． 故选$$\\text{B}$$． " ]

[ "2019~2020学年上海浦东新区华东师范大学第二附属中学高一上学期单元测试第5题", "2003年全国全国高中数学联赛竞赛一试第13题20分" ]

[ [ { "aoVal": "A", "content": "$${}2\\sqrt{17}$$ " } ], [ { "aoVal": "B", "content": "$${}2\\sqrt{19}$$ " } ], [ { "aoVal": "C", "content": "$${}2\\sqrt{21}$$ " } ], [ { "aoVal": "D", "content": "$${}2\\sqrt{23}$$ " } ] ]

[ "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$${{\\left( a+b+c+d \\right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+2\\left( ab+bc+cd+da+ac+bd \\right)$$可得 $$a+b+c+d\\leqslant 2\\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}}$$当且仅当$$a=b=c=d$$时取等号， 由柯西不等式 $${{\\left( 2\\sqrt{x+1}+\\sqrt{2x-3}+\\sqrt{15-3x} \\right)}^{2}}=\\left( \\sqrt{x+1}+\\sqrt{x+1}+\\sqrt{2x-3}+\\sqrt{15-3x} \\right)$$ $$\\leqslant 4\\left( x+1+x+1+2x-3+15-3x \\right)$$ $$=4(x+14)\\leqslant 4\\times 19$$ ∵$$\\sqrt{x+1}$$，$$\\sqrt{2x-3}$$，$$\\sqrt{15-3x}$$不能同时相等， ∴$$2\\sqrt{x+1}+\\sqrt{2x-3}+\\sqrt{15-3x}\\textless{}2\\sqrt{19}$$． " ]

[ "2008年福建全国高中数学联赛竞赛初赛第3题6分" ]

[ [ { "aoVal": "A", "content": "$$(2,+\\infty )$$ " } ], [ { "aoVal": "B", "content": "$$(3,+\\infty )$$ " } ], [ { "aoVal": "C", "content": "$$[4,+\\infty )$$ " } ], [ { "aoVal": "D", "content": "$$[8,+\\infty )$$ " } ] ]

[ "课内体系->知识点->直线和圆的方程->直线与方程->倾斜角和斜率的概念->斜率计算", "课内体系->知识点->三角函数->三角函数的图象与性质->正切函数的图象和性质", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正切", "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->素养->数学运算" ]

[ "由于$$0\\textless{}\\angle QOA$$，$$\\angle POA\\textless{}\\frac{ \\pi }{2}$$， 因此$$\\angle QOA=2\\angle POA$$的充要条件是$$\\tan \\angle QOA=\\tan 2\\angle POA$$． 由$$\\tan \\angle QOA=\\tan 2\\angle POA=\\frac{2\\tan \\angle POA}{1-{{\\tan }^{2}}\\angle POA}$$， 结合$$\\tan \\angle POA=a$$，$$\\tan \\angle QOA=ak$$， 得$$ak=\\frac{2a}{1-{{a}^{2}}}$$，即$$a=\\sqrt{1-\\frac{2}{k}}$$，因此$$k\\textgreater2$$．故选$$\\text{A}$$． " ]

[ "2008年贵州全国高中数学联赛竞赛初赛第2题5分" ]

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

[ "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" ]

[ "$$f\\left( x \\right)={{\\cos }^{2}}2x+\\sqrt{3}\\sin 2x\\cos 2x$$ $$=\\frac{1+\\cos 4x}{2}+\\frac{\\sqrt{3}}{2}\\sin 4x$$ $$=\\sin \\left( 4x+\\frac{ \\pi }{6} \\right)+\\frac{1}{2}$$， 所以最小正周期$$T=\\frac{2 \\pi }{4}=\\frac{ \\pi }{2}$$，故选$$\\text{B}$$. " ]

[ "1988年全国高中数学联赛竞赛一试第2题" ]

[ [ { "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": "$$0\\textbar k\\textbar1$$ " } ] ]

[ "竞赛->知识点->解析几何->椭圆" ]

[ "椭圆外部的点可以离原点很远，它的坐标$$x\\mathsf{}y$$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），所以答$$\\text{D}$$． " ]

[ "2003年AMC10竞赛B第7题" ]

[ [ { "aoVal": "A", "content": "$$35$$ " } ], [ { "aoVal": "B", "content": "$$38$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$42$$ " } ], [ { "aoVal": "E", "content": "$$136$$ " } ] ]

[ "美国AMC10/12->Knowledge Point->Combination->Reasoning->Information Migration (new definition)", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->取整函数（高斯函数）" ]

[ "The first three values in the sum are equal to $$1$$, the next five equal to $$2$$, the next seven equal to $$3$$, and the last one equal to $$4$$, For example, since $$2^{2}=4$$ any square root of a number less than $$4$$ must be less than $$2$$. Sum them all together to get $$3\\times 1+5\\times 2+7\\times 3+1\\times 4=3+10+21+4=38$$. " ]

[ "2008年黑龙江全国高中数学联赛竞赛初赛第3题5分" ]

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

[ "竞赛->知识点->复数与平面向量->平面向量的概念与运算" ]

[ "因为$$\\begin{cases}\\overrightarrow{PB}-\\overrightarrow{PA}=\\overrightarrow{AB} \\overrightarrow{PB}+\\overrightarrow{PA}-2\\overrightarrow{PC}=\\overrightarrow{CB}+\\overrightarrow{CA} \\end{cases}$$， 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$$\\text{C}$$． " ]

[ "竞赛第9题" ]

[ [ { "aoVal": "A", "content": "1 " } ], [ { "aoVal": "B", "content": "2 " } ], [ { "aoVal": "C", "content": "$\\frac{5}{4}$ " } ], [ { "aoVal": "D", "content": "以上答案都不对 " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 利用基本不等式可求最大值.\\\\ 【详解】\\\\ $2x^{2}y^{2}\\leq xy\\left(x^{2}+y^{2}\\right)$，$2y^{2}z^{2}\\leq yz\\left(y^{2}+z^{2}\\right)$，$2z^{2}x^{2}\\leq zx\\left(z^{2}+x^{2}\\right)$，\\\\ 所以$2x^{2}y^{2}+2y^{2}z^{2}+2z^{2}x^{2}+2xyz\\leq xy\\left(x^{2}+y^{2}\\right)+yz\\left(y^{2}+z^{2}\\right)+zx\\left(z^{2}+x^{2}\\right)+2xyz$，\\\\ $\\leq xy\\left(x^{2}+y^{2}\\right)+yz\\left(y^{2}+z^{2}\\right)+zx\\left(z^{2}+x^{2}\\right)+\\left(x+y+z\\right)xyz$\\\\ $=\\left(xy+yz+zx\\right)\\left(x^{2}+y^{2}+z^{2}\\right)$\\\\ $=\\frac{1}{2}\\left(2xy+2yz+2zx\\right)\\left(x^{2}+y^{2}+z^{2}\\right)$\\\\ $\\leq \\frac{1}{2}\\frac{\\left[\\left(2xy+2yz+2zx\\right)+\\left(x^{2}+y^{2}+z^{2}\\right)\\right]^{2}}{2}=2$，\\\\ 因此所求代数式的最大值为1．\\\\ 故选：A. " ]

[ "1986年全国高中数学联赛竞赛一试第4题" ]

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

[ "竞赛->知识点->逻辑->逻辑推理" ]

[ "因四面体每个面都不是等腰三角形，故至少有三条棱的长度互不相等． 另一方面，每两条对棱都相等的四面体可以是仅长度互不相等且每个面都不是等腰三角形的四面体． 故选$$\\text{A}$$． " ]

[ "2003年AMC12竞赛B第4题", "2003年AMC10竞赛B" ]

[ [ { "aoVal": "A", "content": "$$0.75$$ " } ], [ { "aoVal": "B", "content": "$$0.8$$ " } ], [ { "aoVal": "C", "content": "$$1.35$$ " } ], [ { "aoVal": "D", "content": "$$1.5$$ " } ], [ { "aoVal": "E", "content": "$$3$$ " } ] ]

[ "课内体系->知识点->函数的应用->函数的实际应用->一次函数模型", "美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Quadrilateral" ]

[ "Since the swath Moe actually mows is $$24$$ inches, or $$2$$ feet wide, he mows $$10000$$ square feet in one hour. His lawn has an area of $$13500$$, so it will take Moe $$1.35$$ hours to finish mowing the lawn. Thus the answer is $$\\boxed{(\\text{C})1.35}$$． " ]

[ "2010年全国全国高中数学联赛竞赛复赛一试第2题8分", "高二上学期单元测试《导数及其应用》自招第8题" ]

[ [ { "aoVal": "A", "content": "$$-\\frac{3}{2}\\leqslant a\\leqslant 12$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{3}{2}\\leqslant a\\leqslant 12$$ " } ], [ { "aoVal": "C", "content": "$$-12\\leqslant a\\leqslant \\frac{3}{2}$$ " } ], [ { "aoVal": "D", "content": "$$-12\\leqslant a\\leqslant- \\frac{3}{2}$$ " } ] ]

[ "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->求函数的值域->用换元法求值域", "课内体系->知识点->等式与不等式->不等式->解不等式->一元二次不等式", "课内体系->素养->数学运算" ]

[ "令$$\\sin x=t$$，则原函数化为$$g(t)=(-a{{t}^{2}}+a-3)t$$，即$$g(t)=-a{{t}^{3}}+(a-3)t$$， 由$$-a{{t}^{3}}+(a-3)t\\geqslant -3$$，$$-at({{t}^{2}}-1)-3(t-1)\\geqslant 0$$，$$(t-1)(-at(t+1)-3)\\geqslant 0$$及$$t-1\\leqslant 0$$，知$$-at(t+1)-3\\leqslant 0$$即$$a({{t}^{2}}+t)\\geqslant -3$$，（$$1$$） 当$$t=0$$，$$1$$时（$$1$$）总成立； 对$$0\\textless{}t\\leqslant 1$$，$$0\\textless{}{{t}^{2}}+t\\leqslant 2$$，对$$-1\\textless{}t\\textless{}0$$，$$-\\frac{1}{4}\\leqslant {{t}^{2}}+t\\textless{}0$$， 从而可知$$-\\frac{3}{2}\\leqslant a\\leqslant 12$$． 问题即函数$$f\\left( x \\right)=-a{{x}^{3}}+(a-3)x$$在区间$$[-1,1]$$上的最小值为$$-3$$．函数$$f\\left( x \\right)$$的导函数为$${{f}^{\\prime }}\\left( x \\right)=-3a{{x}^{2}}+(a-3)$$． 注意到$$f(1)=-3$$，于是$${{f}^{\\prime }}\\left( 1 \\right)\\leqslant 0$$，可得$$a\\geqslant -\\frac{3}{2}$$． 情形一：$$-\\frac{3}{2}\\leqslant a\\leqslant 3$$， 此时恒有$${{f}^{\\prime }}\\left( x \\right)\\leqslant 0$$，函数$$f(x)$$在区间$$[-1,1]$$上单调递减，符合题意． 情形二：$$a\\textgreater3$$． 此时函数$$f(x)$$在区间$$[-1,1]$$上有极小值点$$x=-\\sqrt{\\frac{a-3}{3a}}$$， 于是根据题意，有$$f\\left( -\\sqrt{\\frac{a-3}{3\\alpha }} \\right)\\geqslant -3$$， 即$$-\\frac{2(a-3)}{3}\\cdot \\sqrt{\\frac{a-3}{3a}}\\geqslant -3$$， 整理得$$(a-12)(4{{a}^{2}}+12a+9)\\leqslant 0$$， 解得$$a\\leqslant 12$$． 综上所述，$$a$$的取值范围是$$\\left[ -\\frac{3}{2},12 \\right]$$． " ]

[ "2011年黑龙江全国高中数学联赛竞赛初赛第7题5分" ]

[ [ { "aoVal": "A", "content": "$$420$$种 " } ], [ { "aoVal": "B", "content": "$$1260$$种 " } ], [ { "aoVal": "C", "content": "$$5040$$种 " } ], [ { "aoVal": "D", "content": "$$2520$$种 " } ] ]

[ "竞赛->知识点->排列组合与概率->两个基本计数原理", "竞赛->知识点->排列组合与概率->排列与组合" ]

[ "总共$$10$$个球，先考虑选$$5$$个位置放篮球，共有$$\\text{C}_{10}^{5}=252$$种，再从剩下的$$5$$个位置种选$$3$$个位置放排球，总共有$$10$$种方法，剩下的$$2$$个位置放橄榄球，故有$$252\\times 10=2520$$种. " ]

[ "1984年全国高中数学联赛竞赛一试第6题" ]

[ [ { "aoVal": "A", "content": "$$F\\left( -2-x \\right)=-2-F\\left( x \\right)$$ " } ], [ { "aoVal": "B", "content": "$$F\\left( -x \\right)=F\\left( \\frac{1+x}{1-x} \\right)$$ " } ], [ { "aoVal": "C", "content": "$$F\\left( {{x}^{-1}} \\right)=F\\left( x \\right)$$ " } ], [ { "aoVal": "D", "content": "$$F\\left( F\\left( x \\right) \\right)=-x$$ " } ] ]

[ "竞赛->知识点->函数->函数的概念", "竞赛->知识点->不等式->换元技巧->代数换元" ]

[ "令 $$u=\\frac{1-x}{1+x}$$，解得 $$x=\\frac{1-u}{1+u}$$． 所以 $$F\\left( u \\right)=\\frac{1-u}{1+u}$$，即 $$F\\left( x \\right)=\\frac{1-x}{1+x}$$ 然后逐一验证 $$F\\left( -2-x \\right)=\\frac{1-\\left( -2-x \\right)}{1+\\left( -2-x \\right)}=-\\frac{3+x}{1+x}$$ $$=-2-\\frac{1-x}{1+x}=-2-F\\left( x \\right)$$ 选择$$A$$． 又本题也可用特殊值验证排除法． 由 $$F\\left( \\frac{1-x}{1+x} \\right)=x$$中． 令 $$x=1$$，得 $$F\\left( 0 \\right)=1$$ 令 $$x=0$$，得 $$F\\left( 1 \\right)=0$$． 而在$$\\text{B}$$的等式中，令$$x=0$$，有$$F\\left( 0 \\right)=F\\left( 1 \\right)$$，所以应排除$$\\text{B}$$，在$$\\text{C}$$中的等式中，$$x=0$$无意义，排除$$\\text{C}$$．在等式$$\\text{D}$$中，左边$$F\\left( F\\left( 1 \\right) \\right)=F\\left( 0 \\right)=1\\ne -1$$，所以也排除$$\\text{D}$$． " ]

[ "2012年黑龙江全国高中数学联赛竞赛初赛第9题5分" ]

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

[ "竞赛->知识点->排列组合与概率->概率初步", "竞赛->知识点->函数->基本初等函数", "竞赛->知识点->函数->函数的图像与性质" ]

[ "$$M=\\left { x\\textbar-1\\textless{}x\\textless{}5 \\right }, {{\\log }_{2}}\\left( x+1 \\right)\\textless{}1$$的解集为$$\\left { x\\textbar-1\\textless{}x\\textless{}1 \\right }$$，所以概率为$$\\frac{1}{3}$$． " ]

[ "2021年新疆全国高中数学联赛竞赛初赛第3题8分" ]

[ [ { "aoVal": "A", "content": "$$10000$$ " } ], [ { "aoVal": "B", "content": "$$13205$$ " } ], [ { "aoVal": "C", "content": "$$13321$$ " } ], [ { "aoVal": "D", "content": "$$14102$$ " } ] ]

[ "课内体系->知识点->计数原理->排列与组合", "竞赛->知识点->排列组合与概率->两个基本计数原理" ]

[ "方法一：所有数码不超过$$5$$的一位正整数有$$5$$个，两位正整数有$$5\\times 6=30$$个，三位正整数有$$5\\times {{6}^{2}}=180$$个，四位正整数有$$5\\times {{6}^{3}}=1080$$个，共有$$1295$$个； 万位数为$$1$$，千位为$$0$$，共$$216$$个； 万位数为$$1$$，千位为$$1$$，共$$216$$个； 万位数为$$1$$，千位为$$2$$，共$$216$$个；共$$1943$$个， 万位数为$$1$$，千位为$$3$$，百位是$$0$$，$$1$$各$$36$$个，共$$72$$个， 一共$$1943+72=2015$$个，还差$$6$$个，百位是$$2$$，个位取$$0$$，$$1$$，$$2$$，$$3$$，$$4$$，$$5$$， 所以第$$2021$$个数是$$13205$$． 方法二：数码不超过$$5$$的数可以与一个六进制数建立一一对应关系，$$2021=1\\times {{6}^{4}}+3\\times {{6}^{3}}+2\\times {{6}^{2}}+0\\times 6+5$$，利用除$$6$$取余法可得， 即$${{\\left( 2021 \\right)}_{10}}={{\\left( 13205 \\right)}_{6}}$$， 所以答案是：$$13205$$． " ]

[ "第二十届全国希望杯高一竞赛复赛邀请赛第3题4分" ]

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

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "注意到$$x\\to -\\infty $$时，$$f(x)\\textgreater0$$；$$f(0)=-1\\textless{}0$$；$$f(5)=3\\textgreater0$$；$$x\\to +\\infty $$时，$$f(x)\\textless{}0$$，于是$$f(x)=0$$有$$3$$个实根． " ]

[ "2005年AMC12竞赛A第21题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ], [ { "aoVal": "E", "content": "$$4$$ " } ] ]

[ "美国AMC10/12->Knowledge Point->Algebra->Calculation->Solve Quadratic Equations", "课内体系->知识点->基本初等函数->对数的概念及其运算->指对化简求值" ]

[ "由$\\log_a{b}=c^{2005}$可知$b=a^{c\\^{2005}}$. 又由$a+b+c=2005$知$a\\textless2005$, 故$c=0$或$c=1$. $c=0$时, $b=a^{0}=1$, $a=2005-b-c=2004$. $c=1$时, $b=a^{1}=a$, 再由$a+b+c=2005$知$(a,b,c)=(1002, 1002, 1)$. 共两种, 选$$\\text{C}$$. " ]

[ "2015年黑龙江全国高中数学联赛竞赛初赛第15题5分" ]

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

[ "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->函数的概念与性质->函数的性质->奇偶性" ]

[ "解析：第二个式子可以化成$${{(2y)}^{3}}+\\sin 2y+2a=0$$，由于$${{x}^{3}}+\\sin x$$是奇函数，且$$2y\\in \\left[ -\\frac{\\pi }{2},\\frac{\\pi }{2} \\right]$$， 所以$$x=-2y$$，$$\\cos (x+2y)=\\cos 0=1$$． 故答案为：$$1$$． " ]

[ "2009年浙江全国高中数学联赛竞赛初赛第7题5分" ]

[ [ { "aoVal": "A", "content": "$$\\text{C}_{2009}^{286}$$ " } ], [ { "aoVal": "B", "content": "$$\\text{C}_{2009}^{287}$$ " } ], [ { "aoVal": "C", "content": "$$-\\text{C}_{2009}^{286}$$ " } ], [ { "aoVal": "D", "content": "$$-\\text{C}_{2009}^{287}$$ " } ] ]

[ "课内体系->知识点->计数原理->二项式定理->求二项式展开式的特定项", "课内体系->知识点->计数原理->二项式定理->二项式定理的展开式", "课内体系->知识点->计数原理->二项式定理->求项的系数或二项式系数", "课内体系->素养->数学运算" ]

[ "由于 $${{\\left( x-\\frac{1}{{{x}^{6}}} \\right)}^{2009}}=\\sum\\limits_{k=0}^{2009}{\\text{C}_{2009}^{k}{{x}^{2009-k}}{{(-{{x}^{-6}})}^{k}}}$$， 则若要出现常数项，须满足$$2009-k-6k=0\\Rightarrow k=287$$． " ]

[ "2014年天津全国高中数学联赛竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "直线 " } ], [ { "aoVal": "B", "content": "椭圆 " } ], [ { "aoVal": "C", "content": "抛物线 " } ], [ { "aoVal": "D", "content": "双曲线 " } ] ]

[ "竞赛->知识点->解析几何->轨迹方程（二试）" ]

[ "设$$m$$在平面$$ \\pi $$上的投影为$${m}'$$，$${m}'$$交$$l$$于$$O$$点．在平面$$ \\pi $$上，以$$O$$为原点，$$l$$为$$y$$轴建立直角坐标系，则可设$${m}'$$的方程为$$y=k \\pi $$．又设$$P$$点的坐标为$$\\left( x, y \\right)$$，则$$P$$到$$l$$的距离为$$\\left\\textbar{} x \\right\\textbar$$；它到$${m}'$$的距离为$$\\frac{\\left\\textbar{} y-kx \\right\\textbar}{\\sqrt{1+{{k}^{2}}}}$$，从而$$P$$到$$m$$的距离平方等于 $$\\frac{{{\\left( y-kx \\right)}^{2}}}{1+{{k}^{2}}}+{{a}^{2}}$$， 其中$$a$$为直线$$m$$到平面$$ \\pi $$的距离．因此，$$P$$点的轨迹方程是 $$\\frac{{{\\left( y-kx \\right)}^{2}}}{1+{{k}^{2}}}+{{a}^{2}}={{x}^{2}}$$． 可见轨迹是双曲线． " ]

[ "1999年全国全国高中数学联赛竞赛一试第3题6分" ]

[ [ { "aoVal": "A", "content": "$$x-y\\geqslant 0$$ " } ], [ { "aoVal": "B", "content": "$$x+y\\geqslant 0$$ " } ], [ { "aoVal": "C", "content": "$$x-y\\leqslant 0$$ " } ], [ { "aoVal": "D", "content": "$$x+y\\leqslant 0$$ " } ] ]

[ "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", "课内体系->素养->数学运算" ]

[ "记$$f\\left( t \\right)={{\\left( {{\\log }_{2}}3 \\right)}^{t}}-{{\\left( {{\\log }_{5}}3 \\right)}^{t}}$$， 则$$f\\left( t \\right)$$在$$\\mathbf{R}$$上是严格增函数． 原不等式即$$f\\left( x \\right)\\geqslant f\\left( -y \\right)$$． 故$$x\\geqslant -y$$， 即$$x+y\\geqslant 0$$． " ]

[ "2011年吉林全国高中数学联赛竞赛初赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$30{}^{}\\circ $$ " } ], [ { "aoVal": "B", "content": "$$45{}^{}\\circ $$ " } ], [ { "aoVal": "C", "content": "$$60{}^{}\\circ $$ " } ], [ { "aoVal": "D", "content": "$$135{}^{}\\circ $$ " } ] ]

[ "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->三角形中的问题->解三角形" ]

[ "设$$\\triangle ABC$$的三边分别为$$a,b,c,$$ 由已知$$6\\overrightarrow{AC}\\cdot \\overrightarrow{AB}=2\\overrightarrow{AB}\\cdot \\overrightarrow{BC}=3\\overrightarrow{BC}\\cdot \\overrightarrow{CA}$$， 可得$$6bc\\cos A=2ac\\cos \\left( ~\\pi -B \\right)=3ab\\cos \\left( ~\\pi -C \\right)$$ 即$$6bc\\cos A=-2ac\\cos B=-3ab\\cos C$$， 再利用余弦定理可得$$6bc\\cdot \\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=-2ac\\cdot \\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}=-3ab\\cdot \\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}$$， 化简可得$${{a}^{2}}=5{{b}^{2}},{{c}^{2}}=2{{b}^{2}},$$ 所以$$\\cos A=\\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=-\\frac{\\sqrt{2}}{2}$$， 故$$A=135{}^{}\\circ $$，故选$$\\text{D}$$. " ]

[ "1993年全国高中数学联赛竞赛一试第4题" ]

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

[ "竞赛->知识点->解析几何->圆与方程", "竞赛->知识点->解析几何->直线与方程" ]

[ "由题设知，曲线$$C$$是以$${{P}_{1}}\\arcsin a,\\arcsin a$$，$${{P}_{2}}\\arccos a-\\arccos a$$两点为直径端点的圆．其圆心的横坐标为 $${{x}_{0}}=\\frac{\\arcsin a+\\arccos a}{2}=\\frac{\\mathsf{\\pi }}{4}$$， 故直线$$x=\\frac{\\pi }{4}$$过圆心，$$d$$就是该圆的直径．而 $${{d}^{2}}=\\left[ {{(\\arcsin a)}^{2}}+(\\arccos a) \\right]\\mathsf{\\geqslant }{{(\\arcsin a+\\arccos a)}^{2}}={{\\frac{ \\pi }{4}}^{2}}$$． ∴$$d\\mathsf{\\geqslant }\\frac{ \\pi }{2}$$，又当$$\\arcsin a=\\arccos a$$时相应的$$d\\mathsf{\\geqslant }\\frac{ \\pi }{2}$$， 故选$$\\text{C}$$． " ]

[ "2018年河北唐山高三二模理科第13题5分", "2017~2018学年陕西西安碑林区西北工业大学附属中学高二上学期期中理科第14题3分", "2018~2019学年山东济南高三上学期期末理科第13题5分", "2020年山东日照高三一模第14题5分", "2020~2021学年11月广东广州海珠区广州市第五中学高三上学期月考第14题5分", "2008年贵州全国高中数学联赛竞赛初赛第13题4分", "2017年湖北武汉高三三模理科第13题5分" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ] ]

[ "课内体系->知识点->计数原理->二项式定理->二项式定理的展开式", "课内体系->知识点->计数原理->二项式定理->求项的系数或二项式系数", "课内体系->素养->数学运算" ]

[ "$${{T}_{r+1}}=\\text{C}_{6}^{r}{{({{x}^{2}})}^{6-r}}{{\\left( -\\frac{1}{x} \\right)}^{2}}$$ $$={{(-1)}^{r}}\\text{C}_{6}^{r}{{x}^{12-2r}}\\cdot {{x}^{-r}}$$ $$={{(-1)}^{r}}\\text{C}_{6}^{r}{{x}^{12-3r}}$$， ∵求常数项， ∴令$$12-3r=0$$，则$$r=4$$， ∴$${{T}_{5}}={{(-1)}^{4}}\\cdot \\text{C}_{6}^{4}{{x}^{0}}=15$$． 故答案为：$$15$$． " ]

[ "2014年AMC10竞赛A第11题" ]

[ [ { "aoVal": "A", "content": "＄$$179.95$$ " } ], [ { "aoVal": "B", "content": "＄$$199.95$$ " } ], [ { "aoVal": "C", "content": "＄$$219.95$$ " } ], [ { "aoVal": "D", "content": "＄$$239.95$$ " } ], [ { "aoVal": "E", "content": "＄$$259.95$$ " } ] ]

[ "课内体系->知识点->函数的应用->函数的实际应用->一次函数模型", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems" ]

[ "Let the listed price be $$x$$. Since all the answer choices are above ＄$$100$$, we can assume $$x\\textgreater{} 100$$. Thus the discounts after the coupons are used will be as follows: Coupon $$1$$: $$x\\times10$$\\%$$=$$$$0.1 x$$. Coupon $$2$$: $$20$$. Coupon $$3$$: $$18$$\\%$$\\times (x-100)=0.18 x-18$$. For coupon $$1$$ to give a greater price reduction than the other coupons, we must have $$0.1 x\\textgreater20 \\Rightarrow x\\textgreater200$$ and $$0.1x\\textgreater0.18 x-18 \\Rightarrow0 .08 x\\textless18 \\Rightarrow x\\textless225$$. From the first inequality, the listed price must be greater than ＄$$200$$, so answer choices $$(\\rm A)$$ and $$(\\rm B)$$ are eliminated. From the second inequality, the listed price must be less than ＄$$225$$, so answer choices $$(\\rm D)$$ and $$(\\rm E)$$ are eliminated. The only answer choice that remains is $$(\\rm C)$$＄$$219.95$$. For coupon $$1$$ to be the most effective, we want $$10$$\\% of the price to be greater than $$20$$. This clearly ~occurs if the value is over $$200$$. For coupon $$1$$ to be more effective than coupon $$3$$, we want to minimize the value over $$200$$, so $$(\\rm C)$$＄$$219.95$$ is the smallest number over $$200$$. " ]

[ "2018年安徽全国高中数学联赛竞赛初赛第3题8分" ]

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

[ "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" ]

[ "由和差化积公式得： $$\\sin 2x+\\sin 4x=2\\cos x\\sin 3x$$， ∴原式$$=\\left\\textbar{} 2\\cos x\\sin 3x+\\sin 3x \\right\\textbar$$$$=\\left\\textbar{} 2\\cos +1 \\right\\textbar\\cdot \\left\\textbar{} \\sin 3x \\right\\textbar$$， 又$$\\left\\textbar{} 2\\cos x+1 \\right\\textbar$$最小正周期为$$2\\pi $$， $$\\left\\textbar{} \\sin 3x \\right\\textbar$$最小正周期为$$\\frac{\\pi }{3}$$， 故$$f\\left( x \\right)$$最小正周期为$$2\\pi $$． " ]

[ "1997年全国高中数学联赛竞赛一试第5题6分" ]

[ [ { "aoVal": "A", "content": "$$f\\left( \\alpha \\right)\\textgreater f\\left( \\beta \\right)\\textgreater f\\left( \\delta \\right)\\textgreater f\\left( \\gamma \\right)$$ " } ], [ { "aoVal": "B", "content": "$$f\\left( \\alpha \\right)\\textgreater f\\left( \\delta \\right)\\textgreater f\\left( \\beta \\right)\\textgreater f\\left( \\gamma \\right)$$ " } ], [ { "aoVal": "C", "content": "$$f\\left( \\delta \\right)\\textgreater f\\left( \\alpha \\right)\\textgreater f\\left( \\beta \\right)\\textgreater f\\left( \\gamma \\right)$$ " } ], [ { "aoVal": "D", "content": "$$f\\left( \\delta \\right)\\textgreater f\\left( \\alpha \\right)\\textgreater f\\left( \\gamma \\right)\\textgreater f\\left( \\beta \\right)$$ " } ] ]

[ "课内体系->知识点->三角函数->反三角函数", "课内体系->知识点->函数的概念与性质->二次函数->二次函数的图象及性质->二次函数相关的恒成立问题", "课内体系->知识点->函数的概念与性质->二次函数->二次函数的图象及性质->含参二次函数的图象及性质", "课内体系->素养->直观想象" ]

[ "$$f(x)$$的对称轴为$$x=\\frac{ \\pi }{2}$$， 易得$$0\\textless{}\\alpha \\textless{}\\frac{ \\pi }{6}\\textless{}\\frac{ \\pi }{4}\\textless{}\\beta \\textless{}\\frac{ \\pi }{3}\\textless{}\\frac{ \\pi }{2}\\textless{}\\gamma \\textless{}\\frac{2 \\pi }{3}\\textless{}\\frac{3 \\pi }{4}\\textless{}\\delta \\textless{}\\frac{5 \\pi }{6}$$． 故选$$\\text{B}$$． " ]

[ "2012年天津全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "正数 " } ], [ { "aoVal": "B", "content": "零 " } ], [ { "aoVal": "C", "content": "负数 " } ], [ { "aoVal": "D", "content": "以上皆有可能 " } ] ]

[ "竞赛->知识点->函数->基本初等函数" ]

[ "令$$m=\\ln x$$，则$${{x}^{\\ln \\ln x}}-{{\\left( \\ln x \\right)}^{\\ln x}}={{\\left( {{\\text{e}}^{m}} \\right)}^{\\ln m}}-{{m}^{m}}=0$$． " ]

[ "第二十届全国希望杯高一竞赛复赛邀请赛第5题4分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "$$M*N= {0,1,2,3,4,6 }$$，故所求的和为$$16$$． " ]

[ "2019年吉林全国高中数学联赛竞赛初赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$-4$$ " } ], [ { "aoVal": "B", "content": "$$-5$$ " } ], [ { "aoVal": "C", "content": "$$-6$$ " } ], [ { "aoVal": "D", "content": "$$-7$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "易知，$$B=\\left { -2,-3 \\right }$$， 则集合$$B$$中所有元素的和为$$-5$$． 故选$$\\text{B}$$． " ]