Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2011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": "$$1$$ " } ] ]

[ "竞赛->知识点->立体几何与空间向量->空间中的平行和垂直" ]

[ "①~~ 错误，$$\\beta $$不一定是经过直线$$m$$的平面； ②~~ 正确； ③错误，例如教室的墙角，不妨设$$\\alpha $$为东墙面，$$\\gamma $$为北墙面，$$\\beta $$为底面，满足$$\\alpha \\bot \\beta ,\\gamma \\bot \\beta $$，但$$\\alpha $$与$$\\gamma $$相交； ④正确，故选$$\\text{A}$$. " ]

[ "2015年吉林全国高中数学联赛竞赛初赛第1题6分" ]

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

[ "竞赛->知识点->函数->函数的概念", "竞赛->知识点->函数->函数方程", "竞赛->知识点->函数->基本初等函数" ]

[ "$$f\\left[ f\\left( -1 \\right) \\right]=f\\left[ {{\\left( \\frac{1}{4} \\right)}^{-1}} \\right]=f\\left( 4 \\right)={{\\log }_{\\frac{1}{2}}}4=-2$$ " ]

[ "2015年广东全国高中数学联赛竞赛初赛第3题8分" ]

[ [ { "aoVal": "A", "content": "$$a\\textless{}1$$ " } ], [ { "aoVal": "B", "content": "$$1\\textless{}a\\textless{}2$$ " } ], [ { "aoVal": "C", "content": "$$a\\textgreater{}2$$ " } ], [ { "aoVal": "D", "content": "以上选项都不对 " } ] ]

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

[ "因为$${{x}^{2}}-ax+1$$无最大值，所以$$a\\textgreater1$$．函数有最小值，则$${{x}^{2}}-ax+1$$有最小值，$$\\Delta \\textless{}0$$． " ]

[ "2004年AMC10竞赛A第17题", "2004年AMC12竞赛A第15题" ]

[ [ { "aoVal": "A", "content": "$$250$$ " } ], [ { "aoVal": "B", "content": "$$300$$ " } ], [ { "aoVal": "C", "content": "$$350$$ " } ], [ { "aoVal": "D", "content": "$$400$$ " } ], [ { "aoVal": "E", "content": "$$500$$ " } ] ]

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

[ "Call the length of the race track $$x$$. When they meet at the first meeting point, Brenda has run $$100$$ meters, while Sally has run $$\\dfrac{x}{2}-100$$ meters. By the second meeting point, Sally has run $$150$$ meters, while Brenda has run $$x-150$$ meters. Since they run at a constant speed, we can set up a proportion: $$\\dfrac{100}{x-150}=\\dfrac{\\dfrac{x}{2}-100}{150}$$. Cross-multiplying, we get that $$x=350\\Rightarrow\\boxed {(\\text{C})350}$$. Sidenote by carlos $$8$$: Since they run at constant speeds, Brenda must\\textquotesingle ve ran $$200$$ meters to get to the second meeting point, therefore we can make an equation $$200=x-150$$, solving for $$x$$, gives us our answer $$350$$. The total distance the girls run between the start and the first meeting is one half of the track length. The total distance they run between the two meetings is the track length. As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting. Thus between the meetings Brenda will run $$2\\times100=200$$ meters. Therefore the length of the track is $$150+200=350$$ meters $$\\Rightarrow \\boxed {(\\text{C})350}$$. " ]

[ "2018年辽宁全国高中数学联赛竞赛初赛第3题6分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{30}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{31}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{32}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{33}$$ " } ] ]

[ "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）", "竞赛->知识点->复数与平面向量" ]

[ "由题设知 $${{S}_{\\triangle BCP}}:{{S}_{\\triangle CAP}}:{{S}_{\\triangle ABP}}=1:2:3$$， $${{S}_{\\triangle BCQ}}:{{S}_{\\triangle CAQ}}:{{S}_{\\triangle ABQ}}=2:3:5$$， 则$${{S}_{\\triangle ABP}}={{S}_{\\triangle ABQ}}=\\frac{1}{2}{{S}_{\\triangle ABC}}$$，于是，$$PQ\\text{//}AB$$， 且$${{S}_{\\triangle BCP}}=\\frac{1}{6}{{S}_{\\triangle ABC}}$$，$${{S}_{\\triangle BCQ}}=\\frac{1}{5}{{S}_{\\triangle ABC}}$$， 故$$\\frac{\\textbar\\overrightarrow{PQ}\\textbar}{\\textbar\\overrightarrow{AB}\\textbar}=\\frac{1}{5}-\\frac{1}{6}=\\frac{1}{30}$$． 故选$$\\text{A}$$． " ]

[ "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3题" ]

[ [ { "aoVal": "A", "content": "$[\\frac{1}{2},+\\infty )$ " } ], [ { "aoVal": "B", "content": "(1+∞) " } ], [ { "aoVal": "C", "content": "$\\text{ } ! !\\textasciitilde ! !\\text{ (-1,}\\frac{\\text{1}}{\\text{2}}\\text{)}\\cup \\text{(1,+}\\infty \\text{) } ! !\\textasciitilde ! !\\text{ }$ " } ], [ { "aoVal": "D", "content": "$[\\frac{\\text{1}}{\\text{2}}\\text{,1)U(1,+}\\infty \\text{)}$ " } ] ]

[]

[ "\\hfill\\break 根据偶次方根下非负，且分母不为零，列式即可得解.\\\\ 【详解】\\\\ 由$f(x)=\\frac{\\sqrt{2x-1}}{{{x}^{2}}-1}$，\\\\ 可得：$\\left { \\begin{align} \\& 2x-1\\ge 0 \\& {{x}^{2}}-1\\ne 0 \\end{align} \\right.$，解得：$x\\ge \\frac{1}{2}$且$x\\ne 1$，\\\\ 故选：D. " ]

[ "2008年安徽全国高中数学联赛竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$g(x)={{f}^{-1}}(-x)$$ " } ], [ { "aoVal": "B", "content": "$$g(x)={{f}^{-1}}(x)$$ " } ], [ { "aoVal": "C", "content": "$$g(x)=-{{f}^{-1}}(-x)$$ " } ], [ { "aoVal": "D", "content": "$$g(x)=-{{f}^{-1}}(x)$$ " } ] ]

[ "竞赛->知识点->平面几何->几何变换（二试）->平移和旋转（二试）" ]

[ "若原函数为$$f\\left( a,b \\right)$$，则其反函数为$$f\\left( b,a \\right)$$， 所以原函数顺时针旋转$$90{}^{}\\circ $$后为$$f\\left( b,-a \\right)$$， 所以$$g\\left( x \\right)$$为$$-{{f}^{-1}}(x)$$，故选$$\\text{D}$$. " ]

[ "2008年黑龙江全国高中数学联赛竞赛初赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$$1372$$ " } ], [ { "aoVal": "B", "content": "$$2024$$ " } ], [ { "aoVal": "C", "content": "$$3136$$ " } ], [ { "aoVal": "D", "content": "$$4495$$ " } ] ]

[ "竞赛->知识点->组合->图论（二试）", "竞赛->知识点->排列组合与概率->两个基本计数原理" ]

[ "首先注意到三角形的三个顶点不在正方形的同一边上，任选正方形的三边，使三个顶点分别在其上，有$$4$$种方法；再在选出的三条边上各选一点，有$${{7}^{3}}$$种方法．这类三角形共有$$4\\times {{7}^{3}}=1372$$（个）．另外，若三角形有两个顶点在正方形的一条边上，第三个顶点在另一条边上，则先取一边使其上有三角形的两个顶点，有$$4$$种方法，再在这条边上任取两点有$$21$$种方法，然后在其余的$$21$$个分点中任取一点作为第三个顶点，这类三角形共有$$4\\times 21\\times 21=1764$$（个）．综上可知，不同三角形的个数为$$1 372+1 764=3 136$$．故选$$\\text{C}$$． " ]

[ "2011年浙江全国高中数学联赛竞赛初赛第8题5分", "2018~2019学年浙江宁波北仑区浙江省北仑中学高一下学期期中A卷第10题4分", "高考真题" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ] ]

[ "课内体系->知识点->等式与不等式->不等式->线性规划->二元一次不等式组的概念", "课内体系->知识点->等式与不等式->不等式->线性规划->含参线性规划相关问题->参数与可行域面积", "课内体系->素养->直观想象" ]

[ "平面区域$$\\left { (x,y)\\textbar\\left\\textbar{} x \\right\\textbar\\leqslant 1,\\left\\textbar{} y \\right\\textbar\\leqslant1 \\right }$$的四个边界点$$(-1,-1)$$，$$(-1,1)$$，$$(1,-1)$$，$$(1,1)$$满足$$ax-2by\\leqslant 2$$，即有$$a+2b\\leqslant 2,a-2b\\leqslant 2,-a-2b\\leqslant 2,-a+2b\\leqslant 2$$，由此计算动点$$P(a,b)$$所形成平面区域的面积为$$4$$．故选$$\\text{A}$$． " ]

[ "2013年天津全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$125$$ " } ], [ { "aoVal": "B", "content": "$$85$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ] ]

[ "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->数列与数学归纳法->数列的通项与求和" ]

[ "由于$${{S}_{25}}=25\\cdot {{a}_{13}}$$，所以由已知条件可得$${{a}_{13}}:{{a}_{23}}=1:5$$，从而$${{a}_{23}}:{{a}_{33}}=5:9$$，$${{a}_{33}}:{{a}_{43}}=9:13$$．现在，$${{S}_{65}}=65{{a}_{33}}$$，所以$${{S}_{65}}:{{a}_{43}}=45:1$$． " ]

[ "高一上学期单元测试《等差、等比数列、递推数列1》竞赛第2题", "2018~2019学年内蒙古呼和浩特回民区呼和浩特市回民中学高二上学期期中（体、艺班）第16题5分", "2017~2018学年浙江杭州上城区杭州第四中学高一下学期期中第14题4分", "2007年高考真题江西卷文科第13题4分" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{7}{2}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{7}{4}$ " } ] ]

[ "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", "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->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->等差数列前n项和的性质", "知识标签->题型->数列->等差数列->等差数列的性质问题->求等差数列的通项公式", "知识标签->题型->数列->等差数列->等差数列的性质问题->等差数列的前n项和计算  " ]

[ "$${{a}_{1}}+{{a}_{4}}+{{a}_{7}}+{{a}_{10}}$$，$${{a}_{2}}+{{a}_{5}}+{{a}_{8}}+{{a}_{11}}$$，$${{a}_{3}}+{{a}_{6}}+{{a}_{9}}+{{a}_{12}}$$构成等差数列， ∴原式$$=21\\div 3=7$$． 因为$${{a}_{1}}+{{a}_{12}}={{a}_{2}}+{{a}_{11}}={{a}_{3}}+{{a}_{10}}={{a}_{4}}+{{a}_{9}}={{a}_{5}}+{{a}_{8}}={{a}_{6}}+{{a}_{7}}$$， 且他们加起来为$${{S}_{12}}=21$$，故$${{a}_{2}}+{{a}_{5}}+{{a}_{8}}+{{a}_{11}}=\\frac{2}{6}\\cdot {{S}_{12}}=7$$． $${{S}_{12}}=6\\left( {{a}_{1}}+{{a}_{12}} \\right)=21$$，$${{a}_{1}}+{{a}_{12}}=\\frac{7}{2}$$，$${{a}_{2}}+{{a}_{5}}+{{a}_{8}}+{{a}_{11}}=2\\left( {{a}_{1}}+{{a}_{12}} \\right)=7$$． 故答案为：$$7$$． " ]

[ "2016年浙江全国高中数学联赛竞赛初赛第8题6分" ]

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

[ "课内体系->素养->数学运算", "课内体系->素养->直观想象", "课内体系->特色题型->新定义", "课内体系->思想->分类讨论思想", "课内体系->知识点->等式与不等式->不等式->线性规划->简单线性规划->画出二元一次不等式（组）所表示的平面区域", "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" ]

[ "当$$\\begin{cases}\\textbar[x+y]\\textbar=1 \\textbar[x-y]\\textbar=0 \\end{cases}$$时，$$\\begin{cases}1\\leqslant x+y ~\\textless{} ~2 0\\leqslant x-y ~\\textless{} ~1 \\end{cases}$$或$$\\begin{cases}-1\\leqslant x+y ~\\textless{} ~0 0\\leqslant x-y ~\\textless{} ~1 \\end{cases}$$ 同理$$\\begin{cases}\\textbar[x+y]\\textbar=0 \\textbar[x-y]\\textbar=1 \\end{cases}$$或$$\\begin{cases}\\textbar[x+y]\\textbar=0 \\textbar[x-y]\\textbar=0 \\end{cases}$$都可以类似表示，画图即可求出面积. 故选$$\\text{A}$$． " ]

[ "2008年天津全国高中数学联赛竞赛初赛第2题6分" ]

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

[ "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->圆锥曲线->双曲线->双曲线的定义、标准方程->双曲线的标准方程", "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->思想->方程思想", "课内体系->方法->换元法" ]

[ "抛物线$$y=a{{x}^{2}}+bx+1$$顶点的坐标为$$\\left( -\\frac{b}{2a},\\frac{4a-{{b}^{2}}}{4a} \\right)$$，设$$x=-\\frac{b}{2a},y=\\frac{4a-{{b}^{2}}}{4a}$$，则有$$\\frac{b}{a}=-2x, y=1-\\frac{{{b}^{2}}}{4a}=1+\\frac{bx}{2}$$．因为$$a\\ne 0$$，所以$$a, b$$满足的条件等价于$$8+4\\frac{b}{a}=b{{\\left( \\frac{b}{a} \\right)}^{2}}$$，于是有$$8+4\\left( -2x \\right)=b{{\\left( -2x \\right)}^{2}}=4x\\left( 2y-2 \\right)$$，即$$xy=1$$．故选$$\\text{B}$$． " ]

[ "2016年湖南全国高中数学联赛竞赛初赛第5题5分", "2010年上海复旦大学自主招生千分考第9题" ]

[ [ { "aoVal": "A", "content": "顺时针旋转$$60{}^{}\\circ $$所得 " } ], [ { "aoVal": "B", "content": "顺时针旋转$$120{}^{}\\circ $$所得 " } ], [ { "aoVal": "C", "content": "逆时针旋转$$60{}^{}\\circ $$所得 " } ], [ { "aoVal": "D", "content": "逆时针旋转$$120{}^{}\\circ $$所得 " } ] ]

[ "竞赛->知识点->复数与平面向量->平面向量的应用" ]

[ "两向量对应的角度分别是$$45{}^{}\\circ $$和$$105{}^{}\\circ $$． " ]

[ "1989年全国高中数学联赛竞赛一试第1题" ]

[ [ { "aoVal": "A", "content": "第一象限 " } ], [ { "aoVal": "B", "content": "第二象限 " } ], [ { "aoVal": "C", "content": "第三象限 " } ], [ { "aoVal": "D", "content": "第四象限 " } ] ]

[ "竞赛->知识点->复数与平面向量->复数的概念与运算" ]

[ "因为$$A+B\\textgreater\\frac{ \\pi }{2}$$，进而$$0\\textless{}\\frac{ \\pi }{2}-A\\textless{}B\\textless{}\\frac{ \\pi }{2}$$， 所以$$\\cos B\\textless{}\\cos \\left( \\frac{ \\pi }{2}-A \\right)=\\sin A$$， $$\\sin B\\textgreater\\sin \\left( \\frac{ \\pi }{2}-A \\right)=\\cos A$$， $$\\cos B-\\sin A\\textless{}0$$，$$\\sin B-\\cos A\\textgreater0$$． 说明$$z$$位于第二象限． 故选$$\\text{B}$$． " ]

[ "2021年吉林全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$a=-1$$ " } ], [ { "aoVal": "B", "content": "$$a=1$$ " } ], [ { "aoVal": "C", "content": "$$a=2$$ " } ], [ { "aoVal": "D", "content": "$$a$$的值不唯一 " } ] ]

[ "竞赛->知识点->函数->函数方程" ]

[ "函数$$f\\left( x \\right)={{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-a\\cos \\left( 1-x \\right)$$的图象关于直线$$x=1$$对称， 又方程只有一个实数解， ∴$$f\\left( 1 \\right)=0$$，得$$a=1$$， 当$$a=1$$时，$$f\\left( x \\right)={{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-\\cos \\left( 1-x \\right)\\geqslant 1-1=0$$， 当且仅当$$x=1$$时取等号， 即方程$${{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-a\\cos \\left( 1-x \\right)=0$$只有一个实数解，符合题设． 故选$$\\text{B}$$． " ]

[ "1990年全国高中数学联赛竞赛一试第1题" ]

[ [ { "aoVal": "A", "content": "$${{\\left( \\cos \\alpha \\right)}^{\\cos \\alpha }} ~\\textless{} ~{{\\left( \\sin \\alpha \\right)}^{\\cos \\alpha }} ~\\textless{} ~{{\\left( \\cos \\alpha \\right)}^{\\sin \\alpha }}$$； " } ], [ { "aoVal": "B", "content": "$${{\\left( \\cos \\alpha \\right)}^{\\cos \\alpha }} ~\\textless{} ~{{\\left( \\cos \\alpha \\right)}^{\\sin \\alpha }} ~\\textless{} ~{{\\left( \\sin \\alpha \\right)}^{\\cos \\alpha }}$$； " } ], [ { "aoVal": "C", "content": "$${{\\left( \\sin \\alpha \\right)}^{\\cos \\alpha }} ~\\textless{} ~{{\\left( \\cos \\alpha \\right)}^{\\cos \\alpha }} ~\\textless{} ~{{\\left( \\cos \\alpha \\right)}^{\\sin \\alpha }}$$； " } ], [ { "aoVal": "D", "content": "$${{\\left( \\cos \\alpha \\right)}^{\\sin \\alpha }} ~\\textless{} ~{{\\left( \\cos \\alpha \\right)}^{\\cos \\alpha }} ~\\textless{} ~{{\\left( \\sin \\alpha \\right)}^{\\cos \\alpha }}$$． " } ] ]

[ "竞赛->知识点->三角函数->三角函数的图像与性质" ]

[ "∵$$\\alpha \\in \\left( \\frac{ \\pi }{4}，\\frac{ \\pi }{2} \\right)$$， ∴ $$0 ~\\textless{} ~\\cos \\alpha ~~\\textless{} ~\\frac{\\sqrt{2}}{2} ~\\textless{} ~\\sin \\alpha ~~\\textless{} ~1$$． 故 $$\\cos \\alpha {{}^{\\sin \\alpha }} ~\\textless{} ~\\cos \\alpha {{}^{\\cos \\alpha }}$$． ∴ $$\\cos \\alpha {{}^{\\cos \\alpha }} ~\\textless{} ~{{(\\sin \\alpha )}^{\\cos \\alpha }}$$． " ]

[ "1989年第7届美国数学邀请赛（AIME）竞赛第10题" ]

[ [ { "aoVal": "A", "content": "$\\dfrac{1}{2021}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{1}{1010}$ " } ], [ { "aoVal": "C", "content": "$1010$ " } ], [ { "aoVal": "D", "content": "$2021$ " } ] ]

[ "知识标签->题型->解三角形->正余弦定理的简单应用->利用正、余弦定理求解边角", "知识标签->知识点->解三角形->正弦定理", "知识标签->知识点->解三角形->余弦定理", "知识标签->素养->数学运算" ]

[ "由余弦定理$${{a}^{2}}+{{b}^{2}}-2ab\\cos \\gamma ={{c}^{2}}$$，代入条件式得 $$2ab\\cos \\gamma =1988{{c}^{2}}$$，或$$\\left( \\frac{a}{b} \\right)\\left( \\frac{b}{c} \\right)\\cos \\gamma =994$$． 由下正弦定理$$\\frac{a}{c}=\\frac{\\sin \\alpha }{\\sin \\gamma }$$，$$\\frac{b}{c}=\\frac{\\sin \\beta }{\\sin \\gamma }$$， 于是$$\\frac{\\sin \\alpha \\sin \\beta \\cos \\gamma }{{{\\sin }^{2}}\\gamma }=994$$． 因为$$\\alpha +\\beta +\\gamma = \\pi $$，所以$$\\alpha +\\beta = \\pi -\\gamma $$．从而$$\\sin \\left( \\alpha +\\beta \\right)=\\sin \\gamma $$． $$\\frac{\\cot \\gamma }{\\cot \\alpha +\\cot \\beta }=\\frac{\\sin \\alpha \\sin \\beta \\cos \\gamma }{\\sin \\left( \\alpha +\\beta \\right)\\sin \\gamma }=994$$． " ]

[ "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. " ]

[ "2009年吉林全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$-\\frac{23}{8}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{11}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{19}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{24}$$ " } ] ]

[ "竞赛->知识点->函数->基本初等函数" ]

[ "因为$$1\\textless{}{{\\log }_{2}}3\\textless{}2$$，所以 $$f\\left( {{\\log }_{2}}3 \\right)=f\\left( {{\\log }_{2}}3+1 \\right)=f\\left( {{\\log }_{2}}3+2 \\right)=f\\left( {{\\log }_{2}}3+3 \\right)={{\\left( \\frac{1}{2} \\right)}^{{{\\log }_{2}}3+3}}=\\frac{1}{3}\\cdot \\frac{1}{8}=\\frac{1}{24}$$． " ]

[ "1982年全国高中数学联赛竞赛第6题" ]

[ [ { "aoVal": "A", "content": "$$19$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$5\\frac{5}{9}$$ " } ], [ { "aoVal": "D", "content": "不存在 " } ] ]

[ "竞赛->知识点->函数->二次函数" ]

[ "由韦达定理得 $$x_{1}^{2}+x_{2}^{2}={{({{x}_{1}}+{{x}_{2}})}^{2}}-2{{x}_{1}}{{x}_{2}}$$ $$={{(k-2)}^{2}}-2({{k}^{2}}+3k+5)$$ $$=-{{(k+5)}^{2}}+19$$． 又因方程有两个实根，所以$$k$$值要受下面条件的约束： $$\\Delta ={{(k-2)}^{2}}-4({{k}^{2}}+3k+5)\\mathsf{\\geqslant }0$$， 即 $$3{{k}^{2}}+16k+16\\mathsf{\\leqslant }0$$， 解得 $$-4\\mathsf{\\leqslant }k\\mathsf{\\leqslant }-\\frac{4}{3}$$ 因此，求$$x_{1}^{2}+x_{2}^{2}$$的最大值问题，就是当$$-4\\mathsf{\\leqslant }k\\mathsf{\\leqslant }-\\frac{4}{3}$$的条件下，求函数 $$y=-{{(k+5)}^{2}}+19$$ 的最大值． 由于对称轴$${{a}_{1}}$$在$$[-4\\mathsf{，}-\\frac{4}{3}]$$的外面，所以在$$[-4\\mathsf{，}-\\frac{4}{3}]$$上函数单调递减，在$$k=-4$$取得最大值$$18$$． $$\\therefore $$ $$x_{1}^{2}+x_{2}^{2}$$的最大值是$$18$$． " ]

[ "2016年浙江全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{\\sqrt{15}}{10}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{4\\sqrt{15}}{15}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{\\sqrt{15}}{15}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2\\sqrt{15}}{15}$$ " } ] ]

[ "课内体系->知识点->立体几何初步->基本立体图形->空间几何体的体积、表面积->组合体求体积、表面积问题", "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->基本立体图形->平面图形、空间几何体的直观图认识", "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->空间中的基本事实与定理->已知线面位置关系判定结论的问题", "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", "课内体系->素养->逻辑推理", "课内体系->素养->直观想象" ]

[ "设截面与棱$$SC$$交于$$D$$点，由已知条件可知，点$$D$$为棱$$SC$$的中点． 取$$AB$$的中点$$E$$，连结$$EC$$、$$DE$$、$$SE$$， 则$$\\angle DEC$$为截面与底面所成二面角的平面角，设为$$\\theta $$． 在$$\\triangle SEC$$中，$$SE=\\frac{\\sqrt{15}}{2}$$，$$EC=\\frac{\\sqrt{3}}{2}$$，$$SC=2$$，所以中线$$DE=\\frac{\\sqrt{5}}{2}$$． 在$$\\triangle DEC$$应用余弦定理得$$\\cos \\theta =\\frac{2\\sqrt{15}}{15}$$．正确答案为$$D$$． " ]

[ "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5题" ]

[ [ { "aoVal": "A", "content": "\\emph{f}(3)\\textless{}\\emph{f}(-2)\\textless{}\\emph{f}(1) " } ], [ { "aoVal": "B", "content": "\\emph{f}(1)\\textless{}\\emph{f}(-2)\\textless{}\\emph{f}(3) " } ], [ { "aoVal": "C", "content": "\\emph{f}(3)\\textless{}\\emph{f}(1)\\textless{}\\emph{f}(-2) " } ], [ { "aoVal": "D", "content": "\\emph{f}(-2)\\textless{}\\emph{f}(1)\\textless{}\\emph{f}(3) " } ] ]

[ "课内体系->知识点->函数的概念与性质->函数的性质->单调性" ]

[ "\\hfill\\break 已知不等关系说明函数是减函数，再由偶函数得$f(-2)=f(2)$，然后由单调性可得大小关系．\\\\ 【详解】\\\\ ∵对任意的${{x}_{1}},{{x}_{2}}\\in [0,+\\infty ),{{x}_{1}}\\ne {{x}_{2}}$，有$({{x}_{2}}-{{x}_{1}})[f({{x}_{2}})-f({{x}_{1}})]\\textless{} 0$，\\\\ ∴$f(x)$在$[0,+\\infty )$上是减函数．\\\\ ∴$f(3) \\textless{} f(2) \\textless{} f(1)$，\\\\ ∵$f(x)$是偶函数，∴$f(-2)=f(2)$\\\\ ∴$f(3)\\textless{} f(-2)\\textless{} f(1)$．\\\\ 故选：A． " ]

[ "高考真题", "2009年第二十届全国希望杯高二竞赛复赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{7}{11}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{7}{13}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{11}{23}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{9}{23}$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->知识点->数列->等差数列->等差数列的前n项和->等差数列前n项和的性质" ]

[ "因为等差数列$$n$$项和公式形如$$A{{n}^{2}}+Bn$$，可设$${{S}_{n}}=kn(2n+4)$$；$${{T}_{n}}=kn(3n+7)$$，则$${{a}_{5}}=22k$$，$${{b}_{7}}=46k$$，所以$$\\frac{{{a}_{5}}}{{{b}_{7}}}=\\frac{11}{23}$$． " ]

[ "2018年吉林全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$\\sqrt{10}$$ " } ], [ { "aoVal": "C", "content": "$$\\sqrt{11}$$ " } ], [ { "aoVal": "D", "content": "$$2\\sqrt{3}$$ " } ] ]

[ "竞赛->知识点->立体几何与空间向量" ]

[ "解：$${{V}_{P-ABC}}={{V}_{A-PBC}}=\\frac{1}{3}\\times \\left( \\frac{1}{2}\\times 3\\times 4 \\right)\\times \\sqrt{{{3}^{2}}-{{\\left( \\frac{5}{2} \\right)}^{2}}}=\\sqrt{11}$$． " ]

[ "1992年全国高中数学联赛竞赛一试第3题" ]

[ [ { "aoVal": "A", "content": "$$2\\textless{}\\lambda \\mathsf{\\leqslant }4$$ " } ], [ { "aoVal": "B", "content": "$$3\\textless{}\\lambda \\textless{}4$$ " } ], [ { "aoVal": "C", "content": "$$2.5\\textless{}\\lambda \\mathsf{\\leqslant }4.5$$ " } ], [ { "aoVal": "D", "content": "$$3.5\\textless{}\\lambda \\textless{}5.5$$ " } ] ]

[ "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" ]

[ "因为 $${{S}_{i}}\\mathsf{\\leqslant }S$$ $$\\left( i=1,2,3,4 \\right)$$ 所以 $$\\sum\\limits_{i=1}^{4}{{{S}_{i}}}\\mathsf{\\leqslant }4S$$． 进而 $$\\lambda =\\frac{\\sum\\limits_{i=1}^{4}{{{S}_{i}}}}{S}\\mathsf{\\leqslant }4$$． 特别地，当四面体为正四面体时，式中的等号成立，从而否定（$$\\text{B}$$）． 考察侧面与底面所成的二面角的平面角均等于$$45{}^{}\\circ $$的任一个正三棱锥，以$${{S}_{4}}$$表示该正三棱锥的底面面积，则 $$S={{S}_{4}}=\\left( {{S}_{1}}+{{S}_{2}}+{{S}_{3}} \\right)\\cos {{45}^{\\circ }}=\\frac{\\sqrt{2}}{2}\\left( {{S}_{1}}+{{S}_{2}}+{{S}_{3}} \\right)$$， 于是 $$\\left( {{S}_{1}}+{{S}_{2}}+{{S}_{3}} \\right)=\\sqrt{2}S$$， $${{S}_{1}}+{{S}_{2}}+{{S}_{3}}+{{S}_{4}}=\\left( 1+\\sqrt{2} \\right)S$$． 此时 $$\\lambda =\\frac{\\sum\\limits_{i=1}^{4}{{{S}_{i}}}}{S}=1+\\sqrt{2}\\textless{}2.5$$， 从而否定（$$\\text{C}$$）与（$$\\text{D}$$）．因此答案是（$$\\text{A}$$）． " ]

[ "2014年江西全国高中数学联赛竞赛初赛第6题8分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{20}{7}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{40}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{64}{9}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{72}{7}$$ " } ] ]

[ "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->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->等差数列->等差数列的前n项和->等差数列前n项和的性质" ]

[ "因为$$ {{{a}_{n}} }$$，$$ {{{b}_{n}} }$$为等差数列，故可设 $${{S}_{n}}=kn(5n-3)$$，$${{T}_{n}}=kn(2n+1)$$， 当$$n\\geqslant 2$$时， $${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}=k(10n-8)$$，$${{b}_{n}}={{T}_{n}}-{{T}_{n-1}}=k(4n-1)$$， 所以$$\\frac{{{a}_{20}}}{{{b}_{7}}}=\\frac{k(200-8)}{k(28-1)}=\\frac{64}{9}$$． 故答案为：$$\\frac{64}{9}$$． " ]

[ "2016~2017学年北京东城区北京市第一七一中学高一下学期期中第8题", "1997年全国高中数学联赛竞赛一试第1题6分" ]

[ [ { "aoVal": "A", "content": "$${{x}_{100}}=-a$$，$${{S}_{100}}=2b-a$$ " } ], [ { "aoVal": "B", "content": "$${{x}_{100}}=-b$$，$${{S}_{100}}=2b-a$$ " } ], [ { "aoVal": "C", "content": "$${{x}_{100}}=-b$$，$${{S}_{100}}=b-a$$ " } ], [ { "aoVal": "D", "content": "$${{x}_{100}}=-a$$，$${{S}_{100}}=b-a$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->数列的通项与求和" ]

[ "已知中$${{x}_{n+1}}={{x}_{n}}-{{x}_{n-1}}$$（$$n\\geqslant 2$$），$${{x}_{1}}=a$$，$${{x}_{2}}=b$$． ∴$${{x}_{3}}={{x}_{2}}-{{x}_{1}}=b-a$$， $${{x}_{4}}={{x}_{3}}-{{x}_{2}}=-a$$， $${{x}_{5}}={{x}_{4}}-{{x}_{3}}=-b$$， $${{x}_{6}}={{x}_{5}}-{{x}_{4}}=a-b$$， $${{x}_{7}}={{x}_{6}}-{{x}_{5}}=a$$， $${{x}_{8}}={{x}_{7}}-{{x}_{6}}=b$$， $$\\vdots $$ 由此可知：$${{x}_{n}}$$每隔六项开始重复， 即$${{x}_{i}}={{x}_{i}}+6(i\\geqslant 1)$$． ∴$${{S}_{100}}=16({{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}+{{x}_{6}})+{{x}_{97}}+{{x}_{98}}+{{x}_{99}}+{{x}_{100}}$$ $$=16(a+b+b-a-a-b+a-b)+a+b+b-a-a$$ $$=2b-a$$． " ]

[ "2011年AMC10竞赛A第10题", "2011年AMC12竞赛A第7题" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$17$$ " } ], [ { "aoVal": "D", "content": "$$23$$ " } ], [ { "aoVal": "E", "content": "$$77$$ " } ] ]

[ "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 $$x\\textgreater15$$ be the number of students at the bookstore， $$p\\textgreater1$$ be the number of pencils each student bought, and $$c\\textgreater p$$ be the price of one pencil. We see that $$xcp=1771=7\\cdot 11\\cdot 23$$. Then we must have $$x=23$$，$$c=11$$，$$p=7$$ by the original conditions and the answer is $$c=11$$． " ]

[ "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": "$$1900$$ " } ] ]

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

[ "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 $$900\\Rightarrow \\text{B}$$. " ]

[ "2015年四川全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{ \\pi }{6}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{ \\pi }{4}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{ \\pi }{3}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{5 \\pi }{12}$$ " } ] ]

[ "竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->三角函数->三角形中的问题->解三角形" ]

[ "结合正弦定理，有$$\\tan A=\\frac{1}{2}\\tan B=\\frac{1}{3}\\tan C$$， 熟知$$\\tan A+\\tan B+\\tan C=\\tan A\\tan B\\tan C$$，代入解得$$\\tan A=1$$． " ]

[ "2018年吉林全国高中数学联赛竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$f(x)\\geqslant\\frac13sinx$$ " } ], [ { "aoVal": "B", "content": "$$\\textbar f(x)\\textbar\\leqslant\\textbar x\\textbar$$ " } ], [ { "aoVal": "C", "content": "$$\\textbar f(x)\\textbar\\leqslant \\frac{\\sqrt3}3$$ " } ], [ { "aoVal": "D", "content": "$$f(\\pi+x)+f(\\pi-x)=0$$ " } ] ]

[ "课内体系->知识点->三角函数->三角函数的图象与性质" ]

[ "$$A$$ " ]

[ "2018年辽宁全国高中数学联赛竞赛初赛第3题6分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{30}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{31}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{32}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{33}$$ " } ] ]

[ "竞赛->知识点->复数与平面向量" ]

[ "由题设知 $${{S}_{\\triangle BCP}}:{{S}_{\\triangle CAP}}:{{S}_{\\triangle ABP}}=1:2:3$$， $${{S}_{\\triangle BCQ}}:{{S}_{\\triangle CAQ}}:{{S}_{\\triangle ABQ}}=2:3:5$$， 则$${{S}_{\\triangle ABP}}={{S}_{\\triangle ABQ}}=\\frac{1}{2}{{S}_{\\triangle ABC}}$$，于是，$$PQ\\text{//}AB$$， 且$${{S}_{\\triangle BCP}}=\\frac{1}{6}{{S}_{\\triangle ABC}}$$，$${{S}_{\\triangle BCQ}}=\\frac{1}{5}{{S}_{\\triangle ABC}}$$， 故$$\\frac{\\textbar\\overrightarrow{PQ}\\textbar}{\\textbar\\overrightarrow{AB}\\textbar}=\\frac{1}{5}-\\frac{1}{6}=\\frac{1}{30}$$． 故选$$\\text{A}$$． " ]

[ "2015年四川全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{6}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{\\sqrt{2}}{8}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{5}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{\\sqrt{3}}{3}$$ " } ] ]

[ "竞赛->知识点->立体几何与空间向量->空间中的角与距离" ]

[ "由题意，$$O$$应该在$$P-ABC$$的底面$$ABC$$上，设底面边长为$$2$$，则外接球半径等于$$\\frac{2}{3}\\sqrt{3}$$，于是侧棱长$$\\frac{2}{3}\\sqrt{6}$$．作$$CH\\bot PB$$于$$H$$，由对称性知$$AH\\bot PB$$，所以$$\\angle AHC$$就是二面角$$A-PB-C$$的平面角，不难算出$$CH=\\frac{\\sqrt{10}}{2}$$． 于是由余弦定理，可得$$\\cos \\angle AHC=\\frac{1}{5}$$． " ]

[ "2021年第31届浙江绍兴上虞区希望杯高一竞赛复赛第1题" ]

[ [ { "aoVal": "A", "content": "$$1$$或$$3$$ " } ], [ { "aoVal": "B", "content": "$$\\left { 1,3 \\right }$$ " } ], [ { "aoVal": "C", "content": "$$\\varnothing $$ " } ], [ { "aoVal": "D", "content": "$$\\left { 0,1,2,3 \\right }$$ " } ] ]

[ "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->集合->集合的基本运算->交集", "课内体系->素养->数学运算" ]

[ "由题意知$$B=\\left { -2,-1,0,1,2,3 \\right }$$所以$$ A \\cap B= \\left { 1,3\\right } $$，故选$$\\text{B}$$． " ]

[ "2002年AMC10竞赛A第17题" ]

[ [ { "aoVal": "A", "content": "$$\\frac 14$$ " } ], [ { "aoVal": "B", "content": "$$\\frac 13$$ " } ], [ { "aoVal": "C", "content": "$$\\frac 38$$ " } ], [ { "aoVal": "D", "content": "$$\\frac 25$$ " } ], [ { "aoVal": "E", "content": "$$\\frac 12$$ " } ] ]

[ "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems", "课内体系->知识点->集合->容斥原理" ]

[ "We will simulate the process in steps. In the beginning, we have: ▪ $$4$$ ounces of coffee in cup $$1$$ ▪ $$4$$ ounces of cream in cup $$2$$ In the first step we pour $$\\frac 42=2$$ ounces of coffee from cup $$1$$ to cup $$2$$; getting: ▪ $$2$$ ounces of coffee in cup $$1$$ ▪ $$2$$ ounces of coffee and $$4$$ ounces of cream in cup $$2$$; In the second step we pour $$\\frac 22=1$$ ounce of coffee and $$\\frac 42=2$$ ounces of cream from cup $$2$$ to cup $$1$$, getting: ▪ $$2+1=3$$ ounces of coffee and $$0+2=2$$ ounces of cream in cup $$1$$ ▪ the rest in cup $$2$$ Hence at the end we have $$3+2=5$$ ounces of liquid in cup $$1$$, and out of these $$2$$ ounces is cream. Thus the answer is $$\\frac 25$$. " ]

[ "2010年河北全国高中数学联赛竞赛初赛第5题6分" ]

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

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "因为$$f(x)={{(x-1)}^{3}}+2$$，所以函数图象的对称中心为$$(1,2)$$． " ]

[ "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5题", "2022~2023学年四川泸州泸县泸县第一中学高一上学期期末第5题", "2022~2023学年甘肃兰州高一上学期期末（兰州市第六十三中学 兰化三中 线上 考试）第10题" ]

[ [ { "aoVal": "A", "content": "1 " } ], [ { "aoVal": "B", "content": "3 " } ], [ { "aoVal": "C", "content": "5 " } ], [ { "aoVal": "D", "content": "7 " } ] ]

[ "课内体系->知识点->函数的应用->函数的实际应用" ]

[ "由题意得：$100\\text{ml}$血液中酒精含量低于$20\\text{mg}$的驾驶员可以驾驶汽车故${{\\left( 1-30 ! \\% ! \\right)}^{x}}\\textless{} 0.2$，即${{0.7}^{x}}\\textless{} 0.2$两边取对数即可得$\\lg {{0.7}^{x}}\\textless{} \\lg 0.2$，即$x\\textgreater\\frac{\\lg 0.2}{\\lg 0.7}\\approx 4.67$那么他至少经过$$5$$个小时才能驾驶汽车故选：$$\\text{C}$$ " ]

[ "2016年湖南全国高中数学联赛竞赛初赛第6题5分" ]

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

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

[ "设选手人数为$$n$$，$$3$$名退出的选手之间的比赛场数为$$x$$，$$0\\leqslant x\\leqslant 3$$，由题意 $$\\mathrm{C}_{n}^{2}+\\left( 2\\times 3-x \\right)=50$$ 则$$x=\\mathrm{C}_{n}^{2}-44$$． 注意到$$\\mathrm{C}_{10}^{2}=45\\mathrm{C}_{11}^{2}=55$$，所以$$x=1$$． 故选$$\\text{B}$$． " ]

[ "2019年全国高中数学联赛竞赛初赛第10题" ]

[ [ { "aoVal": "A", "content": "$\\frac{8}{3}$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$\\frac{16}{3}$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "竞赛->知识点->函数->基本初等函数" ]

[ "记分别记$$\\lg a$$，$$\\lg b$$，$$\\lg c$$为$$x$$，$$y$$，$$z$$，则$$x$$，$$y$$，$$z\\textgreater0$$且$$\\begin{cases}x+\\frac{z}{y}=3 y+\\frac{z}{x}=4 \\end{cases}\\Rightarrow \\begin{cases}xy+z=3y xy+z=4x \\end{cases}$$，设$$x=3t$$，$$y=4t$$，则$$z=12t-12{{t}^{2}}$$， 其中$$t\\in (0,1)$$， 因此$$\\lg a\\cdot \\lg c=xz=3t(12t-12{{t}^{2}})=18\\cdot (t\\cdot t\\cdot (2-2t))\\leqslant 18\\cdot {{\\left( \\frac{2}{3} \\right)}^{3}}=\\frac{16}{3}$$， 等号当$$t=\\frac{2}{3}$$时取得，因此所求最大值为$$\\frac{16}{3}$$． " ]

[ "高二上学期单元测试《不等式专题》自招", "高二上学期单元测试《不等式专题》自招第3题", "高二上学期单元测试《不等式专题（1）》自招第1题", "2009年黑龙江全国高中数学联赛竞赛初赛第10题5分", "2020~2021学年浙江杭州西湖区杭州学军中学高一上学期单元测试《基本不等式》第43题" ]

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

[ "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->转化化归思想", "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->基本不等式的实际应用" ]

[ "由$$a\\textgreater b\\textgreater0$$知$$a-b\\textgreater0$$，故有$$b(a-b)\\leqslant \\dfrac{{{a}^{2}}}{4}$$． ∴$${{a}^{2}}+\\dfrac{1}{b(a-b)}\\geqslant {{a}^{2}}+\\dfrac{4}{{{a}^{2}}}\\geqslant 4$$． 当且仅当$$\\begin{cases}b=a-b {{a}^{2}}=\\dfrac{4}{{{a}^{2}}} \\end{cases}$$，即$$\\begin{cases}a=\\sqrt{2} b=\\dfrac{\\sqrt{2}}{2} \\end{cases}$$时等号成立． ∴$${{a}^{2}}+\\dfrac{1}{b(a-b)}$$的最小值为$$4$$，故选$$\\text{C}$$． " ]

[ "2010年河南全国高中数学联赛竞赛初赛第6题5分", "2018~2019学年上海静安区上海市市北中学高三上学期期中第16题5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": "$$\\left( \\frac{\\sqrt{5}-1}{2}, +\\infty \\right)$$ " } ] ]

[ "竞赛->知识点->不等式->不等式的解法", "竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->三角函数->三角形中的问题->解三角形" ]

[ "设$$a, b, c$$的公比为$$q$$，则$$b=aq, c=a{{q}^{2}}$$，而 $$\\frac{\\sin A+\\cos A\\tan C}{\\sin B+\\cos B\\tan C}$$ $$=\\frac{\\sin A\\cos C+\\cos A\\sin C}{\\sin B\\cos C+\\cos B\\sin C}$$ $$=\\frac{\\sin (A+C)}{\\sin (B+C)}=\\frac{\\sin ( \\pi -B)}{\\sin ( \\pi -A)}$$ $$=\\frac{\\sin B}{\\sin A}=\\frac{b}{a}=q.$$ 因此，只需求$$q$$的取值范围． 因$$a, b, c$$成等比数列，最大边只能是$$a$$或$$c$$，因此$$a, b, c$$要构成三角形的三边，必须且只需$$a+bc$$且$$b+ca$$．即有不等式组$$\\begin{cases}a+aq\\textgreater a{{q}^{2}} aq+a{{q}^{2}}\\textgreater a \\end{cases}$$，即$$\\begin{cases}{{q}^{2}}-q-1\\textless{}0 {{q}^{2}}+q-1\\textgreater0 \\end{cases}$$ 解得$$\\begin{cases}\\frac{1-\\sqrt{5}}{2}\\textless{}q\\textless{}\\frac{\\sqrt{5}+1}{2} q\\textgreater\\frac{\\sqrt{5}-1}{2} \\end{cases}$$或$$\\begin{cases}\\frac{1-\\sqrt{5}}{2}\\textless{}q\\textless{}\\frac{\\sqrt{5}+1}{2} q\\textless{}-\\frac{\\sqrt{5}+1}{2} \\end{cases}$$ 从而$$\\frac{\\sqrt{5}-1}{2}\\textless{}q\\textless{}\\frac{\\sqrt{5}+1}{2},$$ 因此所求的取值范围是$$\\left( \\frac{\\sqrt{5}-1}{2},\\frac{\\sqrt{5}+1}{2} \\right)$$．故选$$\\text{C}$$． " ]

[ "2008年河北全国高中数学联赛竞赛初赛第3题6分", "2019~2020学年重庆沙坪坝区重庆市第七中学高一上学期单元测试《集合与函数》第29题" ]

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

[ "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->复合函数", "课内体系->知识点->基本初等函数->对数函数->对数函数的图象及性质", "课内体系->知识点->基本初等函数->指数函数->利用指数函数性质求最值", "课内体系->素养->数学运算" ]

[ "当$$a\\textgreater1$$时，$$y$$有最小值，则说明$${{x}^{2}}-ax+1$$有最小值，故$${{x}^{2}}-ax+1=0$$中$$\\Delta ~~\\textless{} ~0$$， 即$${{a}^{2}}-4 ~\\textless{} ~0$$，所以$$2\\textgreater a\\textgreater1$$． 当$$0 ~\\textless{} ~a ~\\textless{} ~1$$时，$$y$$有最小值， 则说明$${{x}^{2}}-ax+1$$有最大值，与二次函数性质相互矛盾，舍去． 故选$$\\text{C}$$． " ]

[ "竞赛" ]

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

[ "知识标签->知识点->函数->函数的性质" ]

[ "条件等式变形为$(3x+y)^{5}+(3x+y)=-(x^{5}+x)$，设$f(t)=t^{5}+t$，显然$f(t)$是单调奇函数，又$f(3x+y)=-f(x)=f(-x)$，故$3x+y=-x,\\dfrac{x}{y}=-\\dfrac{1}{4}$ " ]

[ "2008年AMC12竞赛A第17题" ]

[ [ { "aoVal": "A", "content": "$$250$$ " } ], [ { "aoVal": "B", "content": "$$251$$ " } ], [ { "aoVal": "C", "content": "$$501$$ " } ], [ { "aoVal": "D", "content": "$$502$$ " } ], [ { "aoVal": "E", "content": "$$1004$$ " } ] ]

[ "课内体系->知识点->数列->数列的概念->数列的定义", "美国AMC10/12->Knowledge Point->Geometry->Basic Analytic Geometry->Curves and Equations" ]

[ "首先, 若$a_1$是偶数, 则$a_2=\\frac{a_1}{2}\\textless a_1$, 不满足条件. 若$a_1$是奇数, 则$a_2=3a_1+1$, $a_3=\\frac{3}{2}a_1+\\frac{1}{2}$必然都大于$a_1$, 问题出在$a_4$上$$.$$ 我们不知道$a_3$的奇偶, 也就无法计算$a_4$, 更无法讨论它和$a_1$的大小关系$$.$$ 因此, 接下来, 我们先对$a_3$的奇偶性进行讨论. 若$a_3$为偶数, 这等价于$3a_1+1$是$$4$$的倍数, 也就是说$a_1\\equiv 1\\pmod 4$. 此时$a_4=\\frac{a_3}{2}=\\frac{3}{4}a_1+\\frac{1}{4}\\leq a_1$, 不满足条件. 若$a_3$为奇数, 也就是说$a_1\\equiv 3\\pmod 4$时, $a_4=3a_3+1\\textgreater a_1$, 满足条件. 综上可知有且只有模$$4$$余$$3$$的$a_1$满足条件, 这些数在总共$$2008$$个数中占了$\\frac{1}{4}$, 有$\\frac{1}{4}\\times 2008=502$个, 选$$\\text{D}$$. 拓展: 数学上有一个著名的问题, 叫$3n+1$猜想, 讲的是: 在本题中所给的数列中, 无论$a_1$取什么正整数, $1$最终都会出现在数列的某一项$$.$$ 这一猜想至今未得到证明. " ]

[ "2015年天津全国高中数学联赛竞赛初赛第6题6分" ]

[ [ { "aoVal": "A", "content": "$$\\triangle ABC$$必为直角三角形 " } ], [ { "aoVal": "B", "content": "$$\\triangle ABC$$必为锐角三角形 " } ], [ { "aoVal": "C", "content": "$$\\triangle ABC$$必为直角三角形或锐角三角形 " } ], [ { "aoVal": "D", "content": "以上结论都不对 " } ] ]

[ "竞赛->知识点->三角函数->三角形中的问题->解三角形" ]

[ "设三边为$$abc$$，则$$\\tan \\frac{A}{2}=\\frac{1}{p-a}=\\frac{1}{6-a}$$（$$p$$是半周长）． 当$$a\\leqslant 5$$时，$$0\\textless{}\\tan \\frac{A}{2}\\leqslant 1$$，则$$A\\leqslant 90{}^{}\\circ $$； 当$$a\\textgreater5$$时，$$\\tan \\frac{A}{2}\\textgreater1$$，则$$A\\textgreater90{}^{}\\circ $$． 所以若有一条边的长度大于$$5$$，则$$\\triangle ABC$$是钝角三角形，否则$$\\triangle ABC$$是直角或锐角三角形． 设$$a=5+x$$，$$b=c=\\frac{7-x}{2}$$，内切圆半径为$$r$$，由面积法，得 $$\\frac{1}{2}\\cdot \\left( 5+x \\right)\\cdot \\sqrt{{{\\left( \\frac{7-x}{2} \\right)}^{2}}-{{\\left( \\frac{5+x}{2} \\right)}^{2}}}=\\frac{1}{2}\\cdot r\\cdot 12$$ 化简可得$$r=\\frac{\\left( 5+x \\right)\\sqrt{1-x}}{2\\sqrt{6}}$$ 当$$r=1$$时，化简得方程$${{x}^{3}}+9{{x}^{2}}+15x-1=0$$，此方程在$$\\left( 0,1 \\right)$$之间有解$${{x}_{0}}$$，即存在三边长为$$5+{{x}_{0}}\\frac{7-{{x}_{0}}}{2}\\frac{7-{{x}_{0}}}{2}$$，内切圆半径为$$1$$的$$\\triangle ABC$$，是一个钝角三角形． " ]

[ "2009年辽宁全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$${{2}^{10}}+53$$ " } ], [ { "aoVal": "B", "content": "$${{2}^{11}}+53$$ " } ], [ { "aoVal": "C", "content": "$$110\\times \\left( {{2}^{9}}-1 \\right)$$ " } ], [ { "aoVal": "D", "content": "$$110\\times \\left( {{2}^{10}}-1 \\right)$$~ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->数列的通项与求和" ]

[ "$${{a}_{n}}={{2}^{n}}$$，$${{b}_{n}}=n$$，$${{A}_{n}}={{2}^{n+1}}-2$$，$${{B}_{n}}=\\frac{n\\left( n+1 \\right)}{2}$$，于是$${{c}_{n}}=\\frac{n\\left( n+1 \\right)}{2}{{2}^{n}}+n\\cdot {{2}^{n}}-2n$$， 不妨设$${{P}_{n}}=\\sum\\limits_{i=1}^{n}{i\\cdot {{2}^{i}}}$$，$${{Q}_{n}}=\\sum\\limits_{i=1}^{n}{\\frac{i\\left( i+1 \\right)}{2}\\cdot {{2}^{i}}}$$，于是$$\\sum\\limits_{i=1}^{n}{{{c}_{i}}={{P}_{n}}+{{Q}_{n}}-n\\left( n+1 \\right)}$$， $${{Q}_{n}}=1\\times 2+3\\times {{2}^{2}}+\\cdots +\\frac{n\\left( n+1 \\right)}{2}\\cdot {{2}^{n}}$$① $$2{{Q}_{n}}=1\\times {{2}^{2}}+3\\times {{2}^{3}}+\\cdots +\\frac{n\\left( n-1 \\right)}{2}\\cdot {{2}^{n}}+\\frac{n\\left( n+1 \\right)}{2}\\cdot {{2}^{n+1}}$$② ①$$-$$②得$$-{{Q}_{n}}=1\\times {{2}^{1}}+2\\times {{2}^{2}}+\\cdots +n\\cdot {{2}^{n}}-\\frac{n\\left( n+1 \\right)}{2}\\cdot {{2}^{n+1}}$$ 即$${{Q}_{n}}=\\frac{n\\left( n+1 \\right)}{2}\\cdot {{2}^{n+1}}-{{P}_{n}}$$，即$${{Q}_{n}}+{{P}_{n}}=\\frac{n\\left( n+1 \\right)}{2}\\cdot {{2}^{n+1}}=n\\left( n+1 \\right)\\cdot {{2}^{n}}$$ ∴$$\\sum\\limits_{i=1}^{n}{{{c}_{i}}={{P}_{n}}+{{Q}_{n}}-n\\left( n+1 \\right)=n\\left( n+1 \\right)\\left( {{2}^{n}}-1 \\right)}$$，将$$n=10$$代入即可． 故选$$\\text{D}$$． " ]

[ "2008年江苏全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$1$$，$$9$$ " } ], [ { "aoVal": "B", "content": "$$-1$$，$$ 9 $$，$$2-\\sqrt{13}$$ " } ], [ { "aoVal": "C", "content": "$$1 $$，$$9$$， $$2+\\sqrt{13}$$ " } ], [ { "aoVal": "D", "content": "$$1$$，$$ 9$$，$$ 2-\\sqrt{13}$$ " } ] ]

[ "竞赛->知识点->复数与平面向量->复数的概念与运算" ]

[ "将方程写为$${{(x-a)}^{2}}=4a$$．当$$a\\geqslant 0$$时，此时方程有实根， 因为该实根模为$$3$$，故方程有一根为$$3$$或$$-3$$．代入， 由$${{(a\\pm 3)}^{2}}=4a$$，得$$a=1$$或$$9$$； 当$$a\\textless{}0$$时，得$$x=a\\pm 2\\sqrt{\\textbar a\\textbar}\\text{i}$$， 故$$\\textbar x{{\\textbar}^{2}}={{a}^{2}}-4a=9$$，得$$a=2-\\sqrt{13}$$，故选$$\\text{D}$$． " ]

[ "2019年贵州全国高中数学联赛竞赛初赛第5题5分" ]

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

[ "竞赛->知识点->数论模块->同余->平方剩余", "竞赛->知识点->排列组合与概率->概率初步" ]

[ "注意到，素数$$x\\leqslant y\\leqslant z$$． 由$$3x^{2}\\leqslant 2019\\Rightarrow 2\\leqslant x\\leqslant 25$$， 故$$x$$、$$y$$的所有可能取值为：$$2$$、$$3$$、$$5$$、$$7$$、$$11$$、$$13$$、$$17$$、$$19$$、$$23$$． 由$$3z^{2}\\geqslant 2019$$，$$z^{2}\\textless{}2019\\Rightarrow 26\\leqslant z\\textless{}45$$， 故$$z$$的所有可能取值为：$$29$$、$$31$$、$$37$$、$$41$$、$$43$$． 当$$z=29$$时， $$x^{2}+y^{2}=2019-29^{2}=1178$$，无解； 当$$z=31$$时， $$x^{2}+y^{2}=2019-31^{2}=1058=23^{2}+23^{2}$$； 当$$z=37$$时， $$x^{2}+y^{2}=2019-37^{2}=650=11^{2}+23^{2}=17^{2}+19^{2}$$； 当$$z=41$$时， $$x^{2}+y^{2}=2019-41^{2}=338=7^{2}+17^{2}=13^{2}+13^{2}$$； 当$$z=43$$时， $$x^{2}+y^{2}=2019-43^{2}=170$$，无解． 综上，原不定方程共有五个解： $$(x,y,z)=(23,23,31)$$，$$(11,23,37)$$，$$(17,19,37)$$，$$(7,17,41)$$，$$(13,13,41)$$． 故选$$\\text{C}$$． " ]

[ "2021年AMC10竞赛" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ], [ { "aoVal": "E", "content": "$$36$$ " } ] ]

[ "美国AMC10/12->Knowledge Point->Number Theory" ]

[ "N\\A " ]

[ "2021~2022学年福建莆田城厢区莆田第一中学高一下学期开学考试（学科素养能力竞赛）第7~7题" ]

[ [ { "aoVal": "A", "content": "2021年 " } ], [ { "aoVal": "B", "content": "2022年 " } ], [ { "aoVal": "C", "content": "2023年 " } ], [ { "aoVal": "D", "content": "2024年 " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 由题意确定函数模型，再根据指数与对数运算解不等式.\\\\ 【详解】\\\\ 设产量首次超过800万辆的年份为$$x$$，则$$260\\times {{\\left( 1+25\\% \\right)}^{x-2017}}\\textgreater800$$，\\\\ 即$$\\lg 260+\\left( x-2017 \\right)\\lg 1.25\\textgreater\\lg 800$$，\\\\ $$x-2017\\textgreater\\frac{\\lg \\frac{800}{260}}{\\lg 1.25}=\\frac{\\lg 4-\\lg 1.3}{\\lg 1.25}\\approx 5.05$$，\\\\ 即$$x\\textgreater2017+5.05=2022.05$$，\\\\ 所以$$x=2023$$，\\\\ 故选：C. " ]

[ "2008年江西全国高中数学联赛竞赛初赛第2题6分" ]

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

[ "竞赛->知识点->解析几何->直线与圆锥曲线" ]

[ "将$$y=\\frac{2-ax}{b}$$代入椭圆方程并整理得$$\\left( 3{{a}^{2}}+{{b}^{2}} \\right){{x}^{2}}-12ax+12-6{{b}^{2}}=0$$， 因直线和椭圆有公共点，则判别式$${{\\left( 12a \\right)}^{2}}-4\\left( 3{{a}^{2}}+{{b}^{2}} \\right)\\left( 12-6{{b}^{2}} \\right)\\geqslant 0$$， 利用$${{a}^{2}}+{{b}^{2}}=1$$，化简得$${{a}^{2}}\\geqslant {{b}^{2}}$$， 所以$$\\left\\textbar{} \\frac{a}{b} \\right\\textbar\\geqslant 1$$．即$$\\frac{a}{b}\\in \\left( -\\infty ,-1 \\right]\\cup \\left[ 1,+\\infty \\right)$$．故选$$\\text{C}$$． " ]

[ "2015年黑龙江全国高中数学联赛竞赛初赛第9题5分", "2017~2018学年北京东城区北京市第五中学高二下学期期中理科第6题6分", "2015年黑龙江全国高中数学联赛竞赛初赛第9题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{9}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{7}{18}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{4}{9}$$ " } ] ]

[ "课内体系->素养->数学运算", "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（不涉及计数原理）" ]

[ "若``心有灵犀''，则当$$a=1$$时，$$b=1$$或$$b=2$$； 当$$a=2$$时，$$b=1$$或$$b=2$$或$$b=3$$；当$$a=3$$时，$$b=2$$或$$b=3$$或$$b=4$$； 当$$a=4$$时，$$b=3$$或$$b=4$$或$$b=5$$；当$$a=5$$时，$$b=4$$或$$b=5$$或$$b=6$$； 当$$a=6$$时，$$b=5$$或$$b=6$$，即有$$16$$种满足题意的结果，∴$$P=\\frac{16}{36}=\\frac{4}{9}$$． " ]

[ "2012年江西全国高中数学联赛竞赛初赛第4题10分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "以上都不对 " } ] ]

[ "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" ]

[ "提示：据柯西不等式， $$3\\left( a+b+c \\right)=\\left( {{1}^{2}}+{{1}^{2}}+{{1}^{2}} \\right)\\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\right)\\geqslant {{\\left( a+b+c \\right)}^{2}}$$，所以$$a+b+c\\leqslant 3$$，当$$a=b=c=1$$时取得等号． " ]

[ "全国全国高中数学联赛竞赛一试" ]

[ [ { "aoVal": "A", "content": "甲是乙的充分条件但不必要 " } ], [ { "aoVal": "B", "content": "甲是乙的必要条件但不充分 " } ], [ { "aoVal": "C", "content": "甲是乙的充分必要条件 " } ], [ { "aoVal": "D", "content": "$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$都不对 " } ] ]

[ "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", "课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与立体几何结合", "课内体系->素养->逻辑推理" ]

[ "设$$a$$，$$b$$，$$c$$交于一点， 由所成三面角内部一点引三条射线分别垂直于$$\\alpha $$、$$\\beta $$、$$\\lambda $$， 其中每两条射线所成的角都是$$ \\pi -\\varphi $$， $$\\varphi $$为三面角中两两相等的二面角的平面角， 总和$$3( \\pi -\\varphi )\\textless{}2 \\pi $$， ∴$$\\varphi \\textgreater\\frac{1}{3} \\pi $$． 当$$\\varphi \\textgreater\\frac{2}{3} \\pi $$时，$$\\varphi \\textgreater\\frac{1}{3} \\pi $$． 若$$a$$，$$b$$，$$c$$不交于一点，则互相平行， 这时$$\\varphi \\textgreater\\frac{1}{3} \\pi $$． 故选$$\\text{A}$$． （注：认定两平面间夹角范围是$$0\\textless{}\\theta \\mathsf{\\leqslant }\\frac{1}{2} \\pi $$） " ]

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

[ [ { "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->均值", "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->不等式的综合应用（数列与不等式）" ]

[ "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$$\\text{A}$$． " ]

[ "2001年全国高中数学联赛竞赛一试第3题6分" ]

[ [ { "aoVal": "A", "content": "$$y=\\sin \\left\\textbar{} x \\right\\textbar$$ " } ], [ { "aoVal": "B", "content": "$$y=\\cos \\left\\textbar{} x \\right\\textbar$$ " } ], [ { "aoVal": "C", "content": "$$y=\\left\\textbar{} \\text{cot}x \\right\\textbar$$ " } ], [ { "aoVal": "D", "content": "$$y=\\lg \\left\\textbar{} \\sin x \\right\\textbar$$ " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->基本初等函数" ]

[ "可考虑用排除法．$$y=\\sin \\left\\textbar{} x \\right\\textbar$$不是周期函数（可通过作图判断），故$$\\text{A}$$错误； $$y=\\cos \\left\\textbar{} x \\right\\textbar$$的最小正周期为$$2 \\pi $$，且在$$\\left( 0,\\frac{ \\pi }{2} \\right)$$上是减函数，故$$\\text{B}$$错误； $$y=\\left\\textbar{} \\text{cot}x \\right\\textbar$$在$$\\left( 0,\\frac{ \\pi }{2} \\right)$$上是减函数，故$$\\text{C}$$错误． 故选$$\\text{D}$$． " ]

[ "2016年陕西全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{5}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{10}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{19}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{3}{38}$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->素养->数据分析", "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->等差数列的性质及应用", "课内体系->知识点->统计与概率->概率->事件与概率->古典概型", "课内体系->知识点->计数原理->排列与组合->组合" ]

[ "设等差数列的三个数为$${{a}_{1}}$$，$${{a}_{1}}+d$$，$${{a}_{1}}+2d$$，其中$${{a}_{1}}$$，$$d$$都是正整数． 由$${{a}_{1}}+2d\\leqslant 20$$，得$${{a}_{1}}\\leqslant 20-2d$$． 等差数列的个数为$$\\sum\\limits_{d=1}^{9}{\\left( 20-2d \\right)}=180-9\\times 10=90$$． 故所求概率为$$\\frac{90}{\\mathrm{C}_{20}^{3}}=\\frac{3}{38}$$． 故选$$\\text{D}$$． " ]

[ "2008年山东全国高中数学联赛竞赛初赛第4题6分" ]

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

[ "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->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", "课内体系->知识点->圆锥曲线->双曲线->双曲线的几何性质->双曲线的渐近线", "课内体系->知识点->圆锥曲线->双曲线->双曲线的几何性质->双曲线的离心率->求双曲线的离心率", "课内体系->素养->数学运算" ]

[ "易知$$a=2$$，$$b=\\sqrt{m}$$，$$c=4+m$$，$$e=\\frac{\\sqrt{4+m}}{2}$$． 渐近线方程：$$\\frac{x}{2}\\pm \\frac{y}{\\sqrt{m}}=0$$，即$$\\sqrt{m}x\\pm 2y=0$$． 右焦点$$\\left( \\sqrt{4+m},0 \\right)$$到渐近线的距离$$d=b=\\sqrt{m}$$． 从而得方程$$\\sqrt{m}=\\frac{\\sqrt{4+m}}{2}$$，解得$$m=\\frac{4}{3}$$．故选$$\\text{B}$$． " ]

[ "2020年贵州全国高中数学联赛竞赛初赛第3题6分" ]

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

[ "竞赛->知识点->复数与平面向量->平面向量的概念与运算" ]

[ "由已知，可设$$A\\left( -1,0 \\right)$$，$$B\\left( \\frac{1}{2},\\frac{\\sqrt{3}}{2} \\right)$$，$$C\\left( \\cos \\theta , \\sin \\theta \\right)\\left( 0\\leqslant \\theta \\leqslant 2\\pi \\right)$$，则$$\\overrightarrow{AB}\\cdot \\overrightarrow{AC}=\\left( \\frac{3}{2},\\frac{\\sqrt{3}}{2} \\right)\\cdot \\left( \\cos \\theta +1,\\sin \\theta \\right)=\\frac{3}{2}\\cos \\theta +\\frac{\\sqrt{3}}{2}\\sin \\theta +\\frac{3}{2}$$ $$=\\sqrt{3}\\sin \\left( \\theta +\\frac{\\pi }{3} \\right)+\\frac{3}{2}\\leqslant \\sqrt{3}+\\frac{3}{2}$$， 所以，$$\\overrightarrow{AB}\\cdot \\overrightarrow{AC}$$的最大值为$$\\sqrt{3}+\\frac{3}{2}$$． 故选$$\\text{A}$$． " ]

[ "2021~2022学年广东肇庆德庆县广东省德庆县香山中学高一下学期期中第4题", "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1~1题", "2021~2022学年广东肇庆德庆县广东省德庆县香山中学高一下学期期中第4题" ]

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

[]

[ "\\hfill\\break 【分析】\\\\ 直接利用余弦定理即可求得.\\\\ 【详解】\\\\ 在$$\\vartriangle ABC$$中，已知$$B=60{}^{}\\circ ,AC=\\sqrt{3},AB=1$$，即为$$B=60{}^{}\\circ ,b=\\sqrt{3},c=1$$，\\\\ 由余弦定理$${{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\\cos B$$得：$$3={{a}^{2}}+1-2a\\times 1\\times \\frac{1}{2}$$,解得：$$a=2$$（边长大于0，所以$$a=-1$$舍去）\\\\ 即$$BC=2$$.\\\\ 故选：C " ]

[ "2012年辽宁全国高中数学联赛竞赛初赛第1题6分", "2012~2013学年北京东城区高二下学期期末理科第4题3分" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

[ "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->排列->特殊元素优先法" ]

[ "最高位只能填$$1$$、$$2$$、$$3$$，有$$3$$种方法，其余的位置没有限制， 任意排有$$A_{3}^{3}=6$$种方法， 这样的四位数的个数为$$3\\times 6=18$$种． 故选B． " ]

[ "2021年第31届浙江绍兴上虞区希望杯高一竞赛复赛第6题" ]

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

[ "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" ]

[ "由于是$$9$$个点，故$$4$$个为上平面的顶点，下平面$$4$$个顶点，正方体中心为$$1$$个点， 故只需要$$\\begin{cases} a\\geqslant 1 \\sqrt{3}a\\geqslant 2 \\end{cases}$$，解得$$a\\geqslant \\frac{2\\sqrt{3}}{3}$$，故当$$a=\\frac{2\\sqrt{3}}{3}$$时， 一定存在$$9$$个点使得两两距离至少为$$1$$． " ]

[ "2009年河北全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$AB$$与$$CD$$垂直 " } ], [ { "aoVal": "B", "content": "$$AB$$与$$CD$$平行 " } ], [ { "aoVal": "C", "content": "$$AB$$与$$CD$$垂直且异面 " } ], [ { "aoVal": "D", "content": "不能确定 " } ] ]

[ "竞赛->知识点->立体几何与空间向量->空间中的平行和垂直" ]

[ "若$$AE=EF\\ne AF$$，则$$AB$$与$$CD$$不垂直，排除选项$$\\text{A}$$和$$\\text{C}$$； 若$$AB$$与$$CD$$平行，则$$A$$、$$E$$、$$F$$共线，构不成三角形，排除$$\\text{B}$$． 故选$$\\text{D}$$． " ]

[ "2004年AMC10竞赛A第6题", "2004年AMC12竞赛A第4题" ]

[ [ { "aoVal": "A", "content": "$$22$$ " } ], [ { "aoVal": "B", "content": "$$23$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ], [ { "aoVal": "E", "content": "$$26$$ " } ] ]

[ "课内体系->知识点->计数原理->排列与组合->排除法", "美国AMC10/12->Knowledge Point->Combination->Reasoning->Simple Logical Reasoning" ]

[ "Since Bertha has $$6$$ daughters, she has $$30-6=24$$ granddaughters, of which none have daughters. Of Bertha\\textquotesingle s daughters, $$\\dfrac{24}{6}=4$$ have daughters, so $$6-4=2$$ do not have daughters. Therefore, of Bertha\\textquotesingle s daughters and granddaughters, $$24+2=26$$ do not have daughters. $$\\boxed{ (\\text{E}) 26}$$. ▪ ▪ Alcumus** Bertha has $$30-6=24$$ granddaughters, none of whom have any daughters. The granddaughters are the children of $$\\dfrac{24}{6}=4$$ of Bertha\\textquotesingle s daughters, so the number of women having no daughters is $$30-4=\\boxed {26}$$. Draw a tree diagram and see that the answer can be found in the sum of $$6+6$$ granddaughters, $$5+5$$ daughters, and $$4$$ more daughters. Adding them together gives the answer of $$\\boxed{ (\\text{E}) 26}$$. " ]

[ "2008年吉林全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$ {x\\textbar-1\\leqslant x\\textless{}0$$或$$1\\textless{}x\\leqslant 2 }$$ " } ], [ { "aoVal": "B", "content": "$$ {x\\textbar-1\\textless{}x\\textless{}0$$或$$1\\textless{}x\\textless{}2 }$$ " } ], [ { "aoVal": "C", "content": "$$ {x\\textbar0\\leqslant x\\leqslant 1 }$$ " } ], [ { "aoVal": "D", "content": "$$ {x\\textbar-1\\textless{}x\\textless{}2 }$$ " } ] ]

[ "竞赛->知识点->函数->函数综合" ]

[ "设$$\\frac{ \\pi }{2}\\textless{}\\alpha \\textless{} \\pi $$，则$$\\sin \\alpha -\\sqrt{3}\\cos \\alpha =2\\sin \\left( \\alpha -\\frac{ \\pi }{3} \\right)\\in \\left( 1,2 \\right]$$， 可知$$1\\textless{}{{\\log }_{2}}({{x}^{2}}-x+2)\\leqslant 2$$， 即$$2\\textless{}{{x}^{2}}-x+2\\leqslant 4$$， 解得$$-1\\leqslant x\\textless{}0$$或$$1\\textless{}x\\leqslant 2$$．故选$$\\text{A}$$． " ]

[ "2012年浙江全国高中数学联赛竞赛初赛第4题5分" ]

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

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "由$$P\\cap M=P\\Rightarrow P\\subseteq M\\Rightarrow 2-a\\leqslant 1$$，$$1+a\\geqslant 2\\Rightarrow a\\geqslant 1$$． " ]

[ "第二十届全国希望杯高一竞赛复赛邀请赛第2题4分" ]

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

[ "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->基本初等函数" ]

[ "画图可知，$$y={{a}^{-x}}$$与$$y={{\\log }_{a}}x$$的图象只有$$1$$个交点． " ]

[ "2018年天津全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$-34$$ " } ], [ { "aoVal": "C", "content": "$$17$$ " } ], [ { "aoVal": "D", "content": "$$34$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "设公差为$$d$$，则 $${{a}_{2}}+{{a}_{11}}+{{a}_{14}}=3{{a}_{1}}+24d=3{{a}_{9}}$$， 结合已知条件得$${{a}_{9}}=-2$$， 从而，前$$17$$项的和为$$17{{a}_{9}}=-34$$． 故选$$\\text{B}$$． " ]

[ "1994年全国高中数学联赛竞赛一试第6题" ]

[ [ { "aoVal": "A", "content": "三角形 " } ], [ { "aoVal": "B", "content": "正方形 " } ], [ { "aoVal": "C", "content": "非正方形的长方形 " } ], [ { "aoVal": "D", "content": "非正方形的菱形 " } ] ]

[ "竞赛->知识点->解析几何->直线与方程" ]

[ "将直角坐标系$$XOY$$绕原点逆时针旋转$$45{}^{}\\circ $$，得到新坐标系$${{X}^{\\prime }}O{{Y}^{\\prime }}$$，点$$P$$在坐标系$${{X}^{\\prime }}O{{Y}^{\\prime }}$$中的坐标为（$${{x}^{\\prime }}$$，$${{y}^{\\prime }}$$）在坐标系$$XOY$$中的坐标为（$$x$$，$$y$$），则 $$\\begin{cases}{{x}^{\\prime }}=\\frac{1}{\\sqrt{2}}(x+y) {{y}^{\\prime }}=\\frac{1}{\\sqrt{2}}(-x+y) \\end{cases}$$ 题中方程$$\\frac{\\left\\textbar{} x+y \\right\\textbar}{2a}+\\frac{\\left\\textbar{} x-y \\right\\textbar}{2b}=1$$① 化成$$\\frac{\\left\\textbar{} {{x}^{\\prime }} \\right\\textbar}{a}+\\frac{\\left\\textbar{} {{y}^{\\prime }} \\right\\textbar}{b}=\\sqrt{2}$$② 显然，②代表的曲线关于$${{X}^{\\prime }}$$轴、$${{Y}^{\\prime }}$$轴对称，在$${{X}^{\\prime }}O{{Y}^{\\prime }}$$的第Ⅰ象限内，②成为$$\\frac{{{x}^{\\prime }}}{a}+\\frac{{{y}^{\\prime }}}{b}=\\sqrt{2}$$，即为线段$$AB$$，其中$$A$$（$$\\sqrt{2}a$$，$$0$$），$$B$$（$$0$$，$$\\sqrt{2}b$$）． 据对称性，在第Ⅱ象限内方程②是线段$$BC$$，其中$$C$$（$$-\\sqrt{2}a$$，$$0$$）； 在第Ⅲ象限内方程②是线段$$CD$$，其中$$D$$（$$0$$，$$\\sqrt{2}b$$）； 在第Ⅳ象限内方程②是线段$$AD$$． 由对称性知，$$AB=BC=CD=AD$$，又由于$$a\\ne b$$，故$$AC\\ne BD$$，所以$$ABCD$$是非正方形的菱形． 故选$$\\text{D}$$． " ]

[ "2002年全国全国高中数学联赛竞赛一试第8题9分" ]

[ [ { "aoVal": "A", "content": "$2$ " } ], [ { "aoVal": "B", "content": "$3$ " } ], [ { "aoVal": "C", "content": "$4$ " } ], [ { "aoVal": "D", "content": "以上答案均不对 " } ] ]

[ "课内体系->知识点->计数原理->二项式定理", "竞赛->知识点->排列组合与概率->二项式定理及其应用" ]

[ "不难求出前三项的系数分别是$$1,\\frac{1}{2}n,\\frac{1}{8}n(n-1)$$， ∵$$2\\cdot \\frac{1}{2}n=1+\\frac{1}{8}n(n-1)$$， ∴当$$n=8$$时，$${{T}_{r+1}}=C_{n}^{r}{{\\left( \\frac{1}{2} \\right)}^{r}}{{x}^{\\frac{16-3r}{4}}}$$ （$$r=0$$，$$1$$，$$2$$，$$\\cdots $$，$$8$$） ∴$$r=0,4,8$$，即有$$3$$个． " ]

[ "2014年黑龙江全国高中数学联赛竞赛初赛第11题5分", "2013~2014学年北京朝阳区高三上学期期末理科第8题" ]

[ [ { "aoVal": "A", "content": "①② " } ], [ { "aoVal": "B", "content": "②④ " } ], [ { "aoVal": "C", "content": "③④ " } ], [ { "aoVal": "D", "content": "②③ " } ] ]

[ "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", "课内体系->知识点->函数的概念与性质->函数的性质->单调性", "课内体系->思想->函数思想", "课内体系->素养->数学运算", "课内体系->特色题型->解答压轴" ]

[ "①当$$k=\\frac{1}{2}$$时，$${{a}_{n}}=n\\cdot {{\\left( \\frac{1}{2} \\right)}^{n}}(n\\in {{N}^{*}})$$，$$\\because {{a}_{1}}=\\frac{1}{2},{{a}_{2}}=2\\times \\frac{1}{4}=\\frac{1}{2}$$，$$\\therefore {{a}_{1}}={{a}_{2}}$$， 即数列$$\\left { {{a}_{n}} \\right }$$不是递减数列，$$\\therefore $$①错误； ②当$$\\frac{1}{2}\\textless{}k\\textless{}1$$时，$$\\frac{{{a}_{n+1}}}{{{a}_{n}}}=\\frac{(n+1)\\cdot {{k}^{n+1}}}{n\\cdot {{k}^{n}}}=\\frac{(n+1)k}{n}$$， $$\\therefore k\\textless\\frac{k(n+1)}{n}\\textless2k$$，因此数列$$\\left { {{a}_{n}} \\right }$$可有最大项，因此错误； ③当$$0\\textless{}k\\textless{}\\frac{1}{2}$$时，$$\\frac{{{a}_{n+1}}}{{{a}_{n}}}=\\frac{(n+1)\\cdot {{k}^{n+1}}}{n\\cdot {{k}^{n}}}=\\frac{(n+1)k}{n}\\textless{}\\frac{n+1}{2n}\\leqslant 1$$， $$\\therefore {{a}_{n+1}}\\textless{{a}_{n}}$$，故数列$$\\left { {{a}_{n}} \\right }$$为递减数列； ④当$$\\frac{k}{1-k}$$为正整数时，$$\\frac{{{a}_{n+1}}}{{{a}_{n}}}=\\frac{(n+1)\\cdot {{k}^{n+1}}}{n\\cdot {{k}^{n}}}=\\frac{(n+1)k}{n}$$， $$\\therefore $$当$$\\frac{k}{1-k}$$为正整数时，$$1\\textgreater k\\geqslant \\begin{matrix} 1 2 \\end{matrix}$$， 即当$$k=\\frac{1}{2}$$时，$${{a}_{1}}={{a}_{2}}\\textgreater{{a}_{3}}\\textgreater{{a}_{4}}\\textgreater\\cdots $$ 当$$1\\textgreater k\\textgreater\\frac{1}{2}$$时，令$$\\frac{k}{1-k}=m\\in {{\\mathbf{N}}^{*}}$$，解得$$k=\\frac{m}{1+m}$$，则$$\\frac{{{a}_{n+1}}}{{{a}_{n}}}=\\frac{(n+1)m}{n(1+m)}$$， 综上数列$$\\left { {{a}_{n}} \\right }$$必有两项相等的最大项． " ]

[ "2016年AMC10竞赛第12题5分" ]

[ [ { "aoVal": "A", "content": "$$ p\\textless{} \\frac{1}{8} $$ " } ], [ { "aoVal": "B", "content": "$$ p= \\frac{1}{8} $$ " } ], [ { "aoVal": "C", "content": "$$ \\frac{1}{8} \\frac{1}{3} $$ " } ], [ { "aoVal": "D", "content": "$$p=\\frac 13$$ " } ], [ { "aoVal": "E", "content": "$$p\\textgreater\\frac 13$$ " } ] ]

[]

[ "For the product to be odd, all three factors have to be odd. The probability of this is $$\\frac {1008}{2016}\\cdot \\frac {1007}{2015}\\cdot \\frac {1006}{2014}$$. $$\\frac {1008}{2016}=\\frac 12$$, but $$\\frac {1007}{2015}$$ and $$\\frac {1006}{2014}$$ are slightly less than $$\\frac 12$$. Thus, the whole product is slightly less than $$\\frac 12\\cdot \\frac 12\\cdot \\frac 12=\\frac 18$$, so $$p\\textless\\frac 18$$. For the product to be odd, all three factors have to be odd. There are a total of $$\\left(\\frac {2016}3\\right)$$ ways to choose $$3$$ numbers at random, and there are $$\\left(\\frac {1008}3\\right)$$ to choose $$3$$ odd numbers. Therefore, the probability of choosing $$3$$ odd numbers is $$\\frac {(\\frac {1008}3)}{(\\frac {2016}3)}$$. Simplifying this, we obtain $$\\frac {1008\\cdot 1007\\cdot 1006}{2016\\cdot 2015\\cdot 2014}$$, which is slightly less than $$\\frac 18$$, so our answer is $$p\\textless\\frac 18$$. " ]

[ "2003年全国全国高中数学联赛竞赛一试第1题6分" ]

[ [ { "aoVal": "A", "content": "$$2046$$ " } ], [ { "aoVal": "B", "content": "$$2047$$ " } ], [ { "aoVal": "C", "content": "$$2048$$ " } ], [ { "aoVal": "D", "content": "$$2049$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->数列的通项与求和" ]

[ "注意到$${{45}^{2}}=2025$$，$${{46}^{2}}=2116$$，∴$$2026={{a}_{2026-45}}={{a}_{1981}}$$，$$2115={{a}_{2115-45}}={{a}_{2070}}$$， 而且在从第$$1981$$项到第$$2070$$项之间的$$90$$项中没有完全平方数， 又$$1981+22=2003$$，∴$${{a}_{2003}}={{a}_{1981}}+22=2026+22=2048$$． 故选$$\\text{C}$$． " ]

[ "2013年AMC10竞赛B第15题" ]

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

[ "课内体系->知识点->解三角形->三角形面积公式", "美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Regular Polygon" ]

[ "Using the area formulas for an equilateral triangle $$\\left(\\frac{s^{2}\\sqrt{3}}{4}\\right)$$ and regular hexagon $$\\left(\\frac{3s^{2}\\sqrt{3}}{2}\\right)$$ , with side length $$s$$ and plugging $$\\frac a3$$ and $$\\frac b6$$ into each equation, we find that $$\\frac{a^{2}\\sqrt{3}}{36}= \\frac{b^{2}\\sqrt{3}}{24}$$. Simplifying this, we get $$\\frac ab=\\boxed{\\rm(B)\\textasciitilde\\frac{\\sqrt{6}}{2}}$$. The regular hexagon can be broken into $$6$$ small equilateral triangles, each of which is similar to the big equilateral triangle. The big triangle\\textquotesingle s area is $$6$$ times the area of one of the little triangles. Therefore each side of the big triangle is $$\\sqrt{6}$$ times the side of the small triangle. The desired ratio is $$\\frac{3 \\sqrt{6}}{6}= \\frac{\\sqrt{6}}{2}\\Rightarrow (B)$$. " ]

[ "2008年辽宁全国高中数学联赛竞赛初赛第5题5分" ]

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

[ "竞赛->知识点->解析几何->直线与圆锥曲线" ]

[ "切线方程为$$y=kx-\\frac{p}{2}$$，代入抛物线方程得$$\\frac{{{x}^{2}}}{2p}=kx-\\frac{p}{2}$$， 即$${{x}^{2}}-2pkx+{{p}^{2}}=0$$有一个实根， 故$$4{{p}^{2}}{{k}^{2}}-4{{p}^{2}}=0$$， 解得$$k=\\pm 1$$，$$x=\\pm p$$， 所以$$AB$$的长度为$$2p$$．故选$$\\text{C}$$． " ]

[ "2012年黑龙江全国高中数学联赛竞赛初赛第12题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{2010}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{2011}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{2012}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{2013}$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "$${{a}_{n}}\\left( {{a}_{n-1}}+{{a}_{n+1}} \\right)=2{{a}_{n+1}}{{a}_{n-1}}\\Rightarrow \\frac{1}{{{a}_{n+1}}}+\\frac{1}{{{a}_{n-1}}}=2\\frac{1}{{{a}_{n}}}\\left( n\\geqslant 2 \\right)$$． 所以$$\\left { \\frac{1}{{{a}_{n}}} \\right }$$是等差数列，且$$\\frac{1}{{{a}_{1}}}=1, d=1$$，则数列的通项公式$${{a}_{n}}=\\frac{1}{n}$$，故第$$2012$$项为$$\\frac{1}{2012}$$． " ]

[ "2008年AMC12竞赛B第23题" ]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "$$15$$ " } ] ]

[ "美国AMC10/12->Knowledge Point->Algebra->Calculation->Exponentiation", "美国AMC10/12->Knowledge Point->Combination->Reasoning->Simple Logical Reasoning", "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" ]

[ "$$792=\\sum_{i=0}^{}n\\sum_{j=0}^{}n\\left(i\\log{2}+j\\log{5}\\right)=\\sum_{i=0}^{}n\\left((n+1)i\\log{2}+\\frac{n(n+1)}{2}\\log{5}\\right)=\\frac{n(n+1)^{2}}{2}(\\log{2}+\\log{5})=\\frac{n(n+1)^{2}}{2}$$, 其中$\\log$指$\\log_{10}$. 于是有$n(n+1)^{2}=1584=2^{4}\\times 3^{2}\\times 11$. 可以看出$11\\textbar n$, 然后如果再有$2^{2}\\textbar n$或$3^{2}\\textbar n$, $(n+1)^{2}$就太小了, 于是只有$n=11$, 选$$\\text{A}$$. " ]

[ "2010年山东全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "$$-1\\textless{}a\\textless{}2$$ " } ], [ { "aoVal": "B", "content": "$$a\\textgreater2$$ " } ], [ { "aoVal": "C", "content": "$$a\\textless{}-1$$ " } ], [ { "aoVal": "D", "content": "$$a\\textgreater2$$或$$a\\textless{}-1$$ " } ] ]

[ "竞赛->知识点->导数模块->导数" ]

[ "由$$f(x)={{x}^{3}}+(a+1){{x}^{2}}+(a+1)x+a,$$ 得$$f'(x)=3{{x}^{2}}+2(a+1)x+(a+1)$$． 已知$$f(x)$$在其定义域内既有极大值又有极小值， 所以$$f\\prime (x)=0$$一定有两个不相等的实数根． 从而$$\\Delta =4{{(a+1)}^{2}}-12(a+1)\\textgreater0$$， 解得$$a\\textgreater2$$或$$a\\textless{}-1$$． " ]

[ "1991年全国高中数学联赛竞赛一试第3题" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

[ "竞赛->知识点->数论模块->同余->同余的概念与性质" ]

[ "考虑使得$$24\\textbar{{a}^{3}}-1$$成立的$$a$$． $${{a}^{3}}-1=\\left( a-1 \\right)\\left( {{a}^{2}}+a+1 \\right)=\\left[ \\left( a-1 \\right)a\\left( a+1 \\right)+1 \\right]$$． 因$$a\\left( a+1 \\right)+1$$是奇数， 若要$$24\\textbar\\left( {{a}^{3}}-1 \\right)$$，必有$${{2}^{3}}\\left( a-1 \\right)$$． 若$$a-1$$不能被$$3$$整除，则$$3\\textbar a\\left( a+1 \\right)$$， 从而$$a\\left( a+1 \\right)+1$$不能被$$3$$整除． 因此，若要$$24\\textbar\\left( {{a}^{3}}-1 \\right)$$，必有$$3\\left( a-1 \\right)$$．这样就有 $$24\\textbar\\left( a-1 \\right)$$，即 $$a=24k+1$$， $$\\left( k\\in Z \\right)$$． 由$$24k+1\\textless{}100$$，$$k$$可能取的一切值为$$0,1,2,3,4$$，也就是这样的$$a$$有$$1,25,49,73,97$$五个． " ]

[ "2009年黑龙江全国高中数学联赛竞赛初赛第3题5分" ]

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

[ "竞赛->知识点->三角函数->三角不等式", "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" ]

[ "$$y=4\\sin 2x$$向左平移$$\\frac{\\text{ }\\pi }{3}$$个单位，得到$$y=4\\sin 2\\left( x+\\frac{\\text{ }\\pi }{3} \\right)=4\\sin \\left( 2x+\\frac{2}{3}\\text{ }\\pi ~\\right)$$，⑴错； 由$$4\\cos \\left( 2\\cdot \\frac{\\text{ }\\pi }{6}+\\varphi \\right)=0$$，得$$\\frac{\\text{ }\\pi }{3}+\\varphi =k\\text{ }\\pi +\\frac{\\text{ }\\pi }{2}$$，即$$\\varphi =k\\text{ }\\pi +\\frac{\\text{ }\\pi }{6}$$，$$k\\in \\mathbf{Z}$$，⑵对； $$y=\\frac{4\\tan x}{1-{{\\tan }^{2}}x}=2\\tan 2x$$，但$$x\\ne k\\text{ }\\pi \\pm \\frac{\\text{ }\\pi }{4},k\\text{ }\\pi +\\frac{\\text{ }\\pi }{2}$$，$$y$$不是周期函数，⑶错； $$\\sqrt{1+\\sin 2}-\\sqrt{1-\\sin 2}=\\sqrt{{{\\left( \\sin 1+\\cos 1 \\right)}^{2}}}-\\sqrt{{{\\left( \\sin 1-\\cos 1 \\right)}^{2}}}=2\\cos 1$$，⑷错． 正确的命题个数只有$$1$$个． " ]

[ "2018年四川全国高中数学联赛竞赛初赛第6题5分" ]

[ [ { "aoVal": "A", "content": "①② " } ], [ { "aoVal": "B", "content": "①③ " } ], [ { "aoVal": "C", "content": "②③ " } ], [ { "aoVal": "D", "content": "①②③ " } ] ]

[ "竞赛->知识点->数论模块->整除->质数（算数基本定理）" ]

[ "注意到，$$1+2+ \\cdots +m= \\frac{m(m+1)}{2}$$， 在命题①中，由$$p\\left\\textbar{} \\frac{m(m+1)}{2} \\right.$$，知$$2p\\textbar m(m+1)$$， 又$$p$$为奇素数，则$$p\\textbar m$$或$$p\\textbar(m+1)$$， 故$$m$$的最小值为$$p-1$$， 因此，命题①正确． 在命题②中，由$$2^{a}\\left\\textbar{} \\frac{m(m+1)}{2} \\right.$$，知$$2^{a+1}\\textbar m(m+1)$$， 注意到，$$(m,m+1)=1$$， 则$$2^{a+1}\\textbar m$$或$$2^{a+1}\\textbar(m+1)$$， 故$$Z(2^{}a)=2^{a+1}-1\\textgreater2^{}a$$， 因此，命题②正确． 在命题③中，由$$3^{a}\\left\\textbar\\frac{m(m+1)}{2} \\right.$$，知$$2\\times 3^{a}\\textbar m(m+1)$$， 注意到，$$2\\textbar m(m+1)$$，$$(2,3)=1$$， 则$$2\\times3^{}a\\textbar m(m+1)\\Leftrightarrow 3^{}a\\textbar m(m+1)$$， 又$$(m,m+1)=1$$， 于是，$$3^{}a\\textbar m$$或$$3^{}a\\textbar(m+1)$$， 故$$Z(3^{}a)=3^{}a-1$$， 因此，命题③正确． " ]

[ "2017~2018学年4月北京西城区北京市第八中学高三下学期月考理科第8题5分", "2009年竞赛珠海市第12题5分" ]

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

[ "竞赛->知识点->三角函数->三角形中的问题->解三角形" ]

[ "注意到$$C$$点坐标的大小介于$$A,B$$点之间，易知①正确； 取点$$C$$为原点，$$A,B$$点分别在$$x,y$$轴上， 则$$\\textbar\\textbar AB\\textbar{{\\textbar}^{2}}=\\textbar AB{{\\textbar}^{2}}+2\\textbar AC\\textbar\\textbar BC\\textbar\\textgreater\\textbar AC{{\\textbar}^{2}}+\\textbar BC{{\\textbar}^{2}}=\\textbar\\textbar AC\\textbar{{\\textbar}^{2}}+\\textbar\\textbar CB\\textbar{{\\textbar}^{2}}$$，②不正确； 当$$AC$$、$$CB$$分别与$$x$$、$$y$$轴平行时，$$\\textbar\\textbar AC\\textbar\\textbar+\\textbar\\textbar CB\\textbar\\textbar=\\textbar\\textbar AB\\textbar\\textbar$$． " ]

[ "2012年浙江全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{13}{4}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{\\sqrt{13}}{4}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{\\sqrt{13}}{2}$$ " } ], [ { "aoVal": "D", "content": "$$\\sqrt{13}$$ " } ] ]

[ "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" ]

[ "$$y=\\frac{\\sqrt{3}}{2}\\sin \\left( x+\\frac{ \\pi }{2} \\right)+\\cos \\left( \\frac{ \\pi }{6}-x \\right)=\\sqrt{3}\\cos x+\\frac{1}{2}\\sin x=\\frac{\\sqrt{13}}{2}\\sin \\left( x+\\varphi \\right)$$，其中$$\\sin \\varphi =\\frac{2\\sqrt{39}}{13}$$，$$\\cos \\varphi =\\frac{\\sqrt{13}}{13}$$． " ]

[ "2009年吉林全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "竞赛->知识点->解析几何->直线与圆综合" ]

[ "次整点有$$\\left( \\pm 3,0 \\right)$$，$$\\left( \\pm 2,\\sqrt{5} \\right)$$，$$\\left( \\pm 1,2\\sqrt{2} \\right)$$，$$\\left( 0,3 \\right)$$，斜率为负的直线都满足要求，共有$$5+3+1=9$$条，斜率为正的只有$$5$$条满足要求，总共$$9+5=14$$条． " ]

[ "2008年江苏全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "$$\\left[ -\\frac{\\sqrt{3}}{3},\\frac{\\sqrt{3}}{3} \\right]$$ " } ], [ { "aoVal": "B", "content": "$$\\left[ 0,\\frac{\\sqrt{3}}{3} \\right]$$ " } ], [ { "aoVal": "C", "content": "$$\\left[ -\\frac{\\sqrt{3}}{3},0 \\right]$$ " } ], [ { "aoVal": "D", "content": "$$\\left[ \\frac{\\sqrt{3}}{3},+\\infty \\right)$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "集合$$M$$的图形为以$$\\left( 1,0 \\right)$$为圆心、$$1$$为半径的圆的上半圆， 集合$$N$$的图形为过$$\\left( -1,0 \\right)$$的直线． 若直线与圆有公共点，则易得其倾斜角在$$\\left[ 0,\\frac{ \\pi }{6} \\right]$$内， 即$$k\\in \\left[ 0,\\frac{\\sqrt{3}}{3} \\right]$$．故选$$\\text{B}$$． " ]

[ "2021~2022学年福建莆田城厢区莆田第一中学高一下学期开学考试（学科素养能力竞赛）第1~1题" ]

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

[]

[ "\\hfill\\break 【分析】\\\\ 求出集合$$N$$，利用交集的定义可得结果.\\\\ 【详解】\\\\ $$N=\\left { y\\left\\textbar{} y={{x}^{2}} \\right. \\right }=\\left { y\\left\\textbar{} y\\ge 0 \\right. \\right }$$，故$$M\\bigcap N=\\left { 0,1 \\right }$$.\\\\ 故选：C. " ]

[ "2018~2019学年浙江宁波北仑区浙江省北仑中学高一下学期期中A卷第1题4分", "2014年浙江全国高中数学联赛竞赛初赛第5题5分" ]

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

[ "竞赛->知识点->解析几何->直线与圆综合" ]

[ "直线$$l$$即为两圆心$$\\left( -2, 1 \\right)$$、$$\\left( 0, -1 \\right)$$的垂直平分线，即为 $$x-y+1=0$$． " ]

[ "2016年吉林全国高中数学联赛竞赛初赛第6题5分" ]

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

[ "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->转化化归思想", "课内体系->思想->函数思想", "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", "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->等差数列->等差数列的性质及应用" ]

[ "如果有$$6$$项，则 $$0 ~\\textless{} ~\\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}} \\right)+\\left( {{a}_{2}}+{{a}_{3}}+{{a}_{4}}+{{a}_{5}} \\right)+\\left( {{a}_{3}}+{{a}_{4}}+{{a}_{5}}+{{a}_{6}} \\right)$$ $$=\\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}} \\right)+\\left( {{a}_{2}}+{{a}_{3}}+{{a}_{4}} \\right)+\\left( {{a}_{3}}+{{a}_{4}}+{{a}_{5}} \\right)+\\left( {{a}_{4}}+{{a}_{5}}+{{a}_{6}} \\right)$$ $$ ~\\textless{} ~0$$． 矛盾，所以最多$$5$$项． 数列$$2$$，$$2$$，$$-5$$，$$2$$，$$2$$满足要求，所以数列项数最大值为$$5$$． 故选$$\\text{C}$$． " ]

[ "2016~2017学年广东深圳福田区深圳市红岭中学高二下学期期中理科第8题5分", "2014年天津和平区高三二模文科第6题5分", "2008年全国高中数学联赛竞赛一试第2题6分", "2014年天津和平区高三二模理科第6题5分" ]

[ [ { "aoVal": "A", "content": "$$\\left[ -1,2 \\right)$$ " } ], [ { "aoVal": "B", "content": "$$[-1,2]$$ " } ], [ { "aoVal": "C", "content": "$$[0,3]$$ " } ], [ { "aoVal": "D", "content": "$$[0,3)$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->素养->数学抽象", "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->知识点->集合->集合的基本关系->子集", "课内体系->知识点->集合->集合的基本关系->集合关系中的含参问题" ]

[ "∵$$B\\subseteq A$$， ∴①$$B=\\varnothing $$，$$\\Delta \\textless{}0$$，无解， ②$$B\\ne \\varnothing $$，$${{x}^{2}}-ax-4=0$$有两根$${{x}_{1}}$$，$${{x}_{2}}\\in [-2,4)$$， 设$$f(x)=x^{2}-ax-4$$， ∴$$\\begin{cases}f(-2)\\geqslant 0 f(4)\\textgreater0 -2\\textless\\dfrac{a}{2}\\textless{}4 \\end{cases}\\Rightarrow 0\\leqslant a\\textless{}3$$． 故选$$\\text{D}$$． ", "<p>因$${{x}^{2}}-2ax-4=0$$有两个实根$${{x}_{1}}=\\frac{a}{2}-\\sqrt{4+\\frac{{{a}^{2}}}{4}}$$，$${{x}_{2}}=\\frac{a}{2}+\\sqrt{4+\\frac{{{a}^{2}}}{4}}$$，故$$B\\subseteq A$$等价于$${{x}_{1}}\\mathsf{\\geqslant }-2$$且$${{x}_{2}}&lt;{}4$$，即$$\\frac{a}{2}-\\sqrt{4+\\frac{{{a}^{2}}}{4}}\\mathsf{\\geqslant }-2$$且$$\\frac{a}{2}+\\sqrt{4+\\frac{{{a}^{2}}}{4}}&lt;{}4$$，解之得$$0\\mathsf{\\leqslant }a&lt;{}3$$．</p>\n<p>故选$$\\text{D}$$．</p>\n" ]

[ "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$$\\text{A}$$． " ]

[ "2016~2017学年12月北京顺义区顺义区第一中学高一上学期月考第7题5分", "1984年全国高中数学联赛竞赛一试第4题" ]

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

[ "竞赛->知识点->三角函数->三角函数的图像与性质", "竞赛->知识点->函数->基本初等函数" ]

[ "方程$$\\sin x=\\lg x$$的实根个数即函数$$y=\\sin x$$与函数$$y=\\lg x$$的交点的个数， 由函数$$y=\\sin x$$与函数$$y=\\lg x$$的图像可得图像交点为$$3$$个， 即方程$$\\sin x=\\lg x$$的实根的个数是$$3$$．~~~~~~~ " ]

[ "2010年山东全国高中数学联赛竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$31$$ " } ], [ { "aoVal": "D", "content": "$$33$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "设$$\\alpha ,\\beta $$为方程$${{x}^{2}}+mx-36=0$$的两根，则$$\\alpha \\cdot \\beta =-36$$．因此， $$\\left\\textbar{} \\alpha \\right\\textbar=1,\\left\\textbar{} \\beta \\right\\textbar=36$$，有$$m=\\pm 35$$； $$\\left\\textbar{} \\alpha \\right\\textbar=2,\\left\\textbar{} \\beta \\right\\textbar=18$$，有$$m=\\pm 16$$； $$\\left\\textbar{} \\alpha \\right\\textbar=3,\\left\\textbar{} \\beta \\right\\textbar=12$$，有$$m=\\pm 9$$； $$\\left\\textbar{} \\alpha \\right\\textbar=4,\\left\\textbar{} \\beta \\right\\textbar=9$$，有$$m=\\pm 5$$； $$\\left\\textbar{} \\alpha \\right\\textbar=6,\\left\\textbar{} \\beta \\right\\textbar=6$$，有$$m=0$$． $$M=\\left { 0 \\right }\\cup \\left { -5, 5 \\right }\\cup \\left { -9, 9 \\right }\\cup \\left { -16, 16 \\right }\\cup \\left { -35, 35 \\right }$$． 由条件①知$$A\\ne \\varnothing $$．由条件②知$$A$$是由一些成对的相反数所成之集，所以$$M$$的$$5$$对相反数共能组成$${{2}^{5}}-1$$个不同的非空集合$$A$$． " ]

[ "2011年山东全国高中数学联赛竞赛初赛第4题6分" ]

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

[ "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" ]

[ "由对数函数的性质知，$${{x}^{2}}+x-2\\textgreater0$$，则$$x\\textgreater1$$或$$x\\textless{}-2$$．当$$x\\textless{}-2$$时，$$f(x)$$为增函数；当$$x\\textgreater1$$时，$$f(x)$$为减函数． " ]

[ "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6题" ]

[ [ { "aoVal": "A", "content": "$ {x\\textbar0\\textless{} x\\textless{} \\frac{5}{2} , }$ " } ], [ { "aoVal": "B", "content": "$ {x\\textbar-\\frac{3}{2}\\textless{} x\\le 0 , }$ " } ], [ { "aoVal": "C", "content": "$ {x\\textbar-\\frac{3}{2}\\textless{} x\\textless{} 0或0\\le x\\textless{} \\frac{5}{2} , }$ " } ], [ { "aoVal": "D", "content": "$ {x\\textbar x\\textless{} -\\frac{3}{2}或0\\le x\\textless{} \\frac{5}{2} , }$ " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 根据$y=f(x)$是定义在上$\\text{R}$的奇函数，求得函数解析式，然后再由分段函数的定义域，分$x\\textless{} 0$，$x\\textgreater0,x=0$三种情况讨论求解不等式即可.\\\\ 【详解】\\\\ 因为$y=f(x)$是定义在$\\text{R}$上的奇函数，所以$f(0)=0$，\\\\ 当$x\\textless{} 0$时，则$-x\\textgreater0$，所以$f(-x)=-x-2=-f\\left( x \\right)$，所以$f(x)=x+2$，\\\\ 综上所述，$f(x)=\\text{ } ! ! { ! !\\text{ }\\begin{array}{*{35}{l}} x+2,x\\textless{} 0 0,x=0 x-2,x\\textgreater0 \\end{array}\\text{ }$，\\\\ 不等式$f(x)\\textless{} \\frac{1}{2}$等价于$\\text{ } ! ! { ! !\\text{ }\\begin{array}{*{35}{l}} x\\textless{} 0 x+2\\textless{} \\frac{1}{2} \\end{array}\\text{ }$或$\\text{ } ! ! { ! !\\text{ }\\begin{array}{*{35}{l}} x=0 0\\textless{} \\frac{1}{2} \\end{array}\\text{ }$或$\\text{ } ! ! { ! !\\text{ }\\begin{array}{*{35}{l}} x\\textgreater0 x-2\\textless{} \\frac{1}{2} \\end{array}\\text{ }$，\\\\ 解得$x\\textless{} -\\frac{3}{2}$或$x=0$或$0\\textless{} x\\textless{} \\frac{5}{2}$，\\\\ 综上：不等式$f(x)\\textless{} \\frac{1}{2}$的解集是$\\text{ } ! ! { ! !\\text{ }x\\textbar\\text{ } ! !\\textbar ! !\\text{ }x\\textless{} -\\frac{3}{2}\\text{ }$或$\\text{ }0\\le x\\textless{} \\frac{5}{2}\\text{ } ! ! } ! !\\text{ }$.\\\\ 故选：D. " ]

[ "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 " } ] ]

[ "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$$\\text{B}$$． " ]

[ "竞赛第7题5分" ]

[ [ { "aoVal": "A", "content": "$$75$$ " } ], [ { "aoVal": "B", "content": "$$90$$ " } ], [ { "aoVal": "C", "content": "$$135$$ " } ], [ { "aoVal": "D", "content": "$$150$$ " } ], [ { "aoVal": "E", "content": "$$270$$ " } ] ]

[]

[ "We can set up a system of equations where $$x$$ and $$y$$ are the two acute angles. WLOG, assume that $$x\\textless y$$ in order for the complement of $$x$$ to be greater than the complement of $$y$$. Therefore, $$5x=4y$$ and $$\\begin{array}{l} {90-x=2\\left( 90-y\\right)} {90-x=2\\left( 90-1.25x\\right)} {1.5x=90} {x=60} \\end{array}$$ Solving for $$y$$ in the first equation and substituting into the second equation yields Substituting this $$x$$ value back into the first equation yields $$y=75$$, leaving $$x+y$$ equal to $$\\left(\\text{C}\\right) 135$$. " ]

[ "1984年全国高中数学联赛竞赛一试第5题" ]

[ [ { "aoVal": "A", "content": "奇函数 " } ], [ { "aoVal": "B", "content": "偶函数 " } ], [ { "aoVal": "C", "content": "不是奇函数也不是偶函数 " } ], [ { "aoVal": "D", "content": "奇偶性与$$a$$的具体数值有关 " } ] ]

[ "竞赛->知识点->函数->函数综合" ]

[ "∵$$F\\left( x \\right)$$是奇函数，$$F\\left( -x \\right)=-F\\left( x \\right)$$ 令 $$g\\left( x \\right)=\\frac{1}{{{a}^{x}}-1}+\\frac{1}{2}=\\frac{{{a}^{x}}+1}{2\\left( {{a}^{x}}-1 \\right)}$$， 则 $$g\\left( -x \\right)=\\frac{{{a}^{-x}}+1}{2\\left( {{a}^{-x}}-1 \\right)}=\\frac{\\frac{1}{{{a}^{x}}}+1}{2\\left( \\frac{1}{{{a}^{x}}}+1 \\right)}$$， $$=\\frac{{{a}^{x}}+1}{2(1-{{a}^{x}})}=-g\\left( x \\right)$$， 即$$g\\left( x \\right)$$也是奇函数． 当$$a\\textgreater0$$，$$a\\ne 1$$时，$$G\\left( x \\right)=F\\left( x \\right)g\\left( x \\right)$$在$$R$$上有意义． 而 $$G\\left( -x \\right)=F\\left( -x \\right)g\\left( -x \\right)=\\left[ -F\\left( -x \\right) \\right]\\left[ -g\\left( -x \\right) \\right]$$ $$=F\\left( x \\right)g\\left( x \\right)$$， ∴$$G\\left( x \\right)$$是偶函数． " ]

[ "2008年河北全国高中数学联赛竞赛初赛第5题6分" ]

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

[ "竞赛->知识点->不等式->换元技巧->代数换元", "竞赛->知识点->三角函数->三角恒等变换" ]

[ "设$$\\sin x-\\sin y=t$$，易得 $$\\cos x\\cos y-\\sin x\\sin y=\\frac{{{t}^{2}}-1}{2}$$， 即$$\\cos (x+y)=\\frac{{{t}^{2}}-1}{2}$$． 由于$$-1\\leqslant \\cos (x+y)\\leqslant 1$$， 所以$$-1\\leqslant \\frac{{{t}^{2}}-1}{2}\\leqslant 1$$， 解得$$-\\sqrt{3}\\leqslant t\\leqslant \\sqrt{3}$$．故选$$\\text{D}$$． " ]

[ "2019年吉林全国高中数学联赛竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$-7$$ " } ], [ { "aoVal": "B", "content": "$$-5$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "课内体系->知识点->等式与不等式->不等式->线性规划" ]

[ "略 " ]