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12题5分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{33}{8}$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "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->单调性", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域", "课内体系->知识点->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->知识点->等式与不等式->不等式->解不等式->不等式中的恒成立与能成立问题", "课内体系->方法->定义法" ]

[ "易知$$x=2$$时，$$\\varphi \\left( x \\right)$$取得最小值$$4$$，所以$${{x}_{0}}=2$$． 在$$\\left[ 1,\\frac{9}{4} \\right]$$上，$$f\\left( x \\right)$$在$$x=2$$取得最小值，所以$$-\\frac{3}{4}p=2$$，解得$$p=-\\frac{8}{3}$$． 又$$f\\left( 2 \\right)=\\varphi \\left( 2 \\right)=4$$，$$8+6p+2q=4$$，解得$$q=6$$． $$f\\left( x \\right)=2{{x}^{2}}-8x+12=2{{\\left( x-2 \\right)}^{2}}+4$$，在$$M$$上最大值为$$f\\left( 1 \\right)=6$$． " ]

[ "1992年全国高中数学联赛竞赛一试第4题" ]

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

[ "竞赛->知识点->函数->基本初等函数", "竞赛->知识点->三角函数->三角形中的问题->解三角形" ]

[ "由已知方程得 $${{\\log }_{b}}{{x}^{2}}={{\\log }_{b}}\\left( 4x-4 \\right)$$， 即 $${{x}^{2}}-4x+4=0$$， 求得的根 $${{x}_{1}}={{x}_{2}}=2$$． 故 $$C=2A$$ 以及 $$\\sin B=2\\sin A$$， 因 $$A+B+C=180{}^{}\\circ $$，所以$$3A+B={{180}^{\\circ }}$$， 因此 $$\\sin B=\\sin 3A$$， 所以 $$2\\sin A-4{{\\sin }^{3}}A=2\\sin A$$， 即 $$\\sin A\\left( 1-4{{\\sin }^{2}}A \\right)=0$$，但 $$\\sin A{\\ne }0$$， 所以 $${{\\sin }^{2}}A=\\frac{1}{4}$$， 而 $$\\sin A\\textgreater0$$， 所以 $$\\sin A=\\frac{1}{2}$$， 从而 $$A=30{}^{}\\circ $$，$$C=60{}^{}\\circ $$，$$B=90{}^{}\\circ $$． 因此答案是（$$\\text{B}$$）． " ]

[ "第二十届全国希望杯高二竞赛初赛邀请赛第2题4分" ]

[ [ { "aoVal": "A", "content": "直角三角形或钝角三角形 " } ], [ { "aoVal": "B", "content": "直角三角形或锐角三角形 " } ], [ { "aoVal": "C", "content": "钝角三角形 " } ], [ { "aoVal": "D", "content": "直角三角形 " } ] ]

[ "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->三角函数->三角恒等变换->利用三角恒等变换判断三角形形状" ]

[ "$$\\sin x+\\cos x={{\\sin }^{3}}x+{{\\cos }^{3}}x\\Leftrightarrow \\sin x(1-{{\\sin }^{2}}x)=\\cos x({{\\cos }^{2}}x-1)$$$$\\Leftrightarrow \\sin x{{\\cos }^{2}}x=-\\cos x{{\\sin }^{2}}x\\Leftrightarrow \\sin x\\cos x(\\cos x+\\sin x)=0$$． ∵$$x$$表示三角形一个内角，则$$x\\in (0, \\pi )$$． ∴$$x$$为直角或$$\\cos x+\\sin x=0$$． 当$$\\cos x+\\sin x=0$$时，则$$\\cos x=-\\sin x ~\\textless{} ~0$$，（$$x\\in (0, \\pi )$$） ， ∴$$x$$为钝角，此时$$x=\\frac{3}{4} \\pi $$． 故三角形为直角三角形或钝角三角形． 故选$$\\rm A$$． " ]

[ "1983年全国高中数学联赛竞赛一试第2题" ]

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

[ "竞赛->知识点->函数->基本初等函数" ]

[ "$$x=\\frac{1}{{{\\log }_{\\frac{1}{2}}}\\frac{1}{3}}+\\frac{1}{{{\\log }_{\\frac{1}{5}}}\\frac{1}{3}}=\\frac{1}{{{\\log }_{{{2}^{-1}}}}{{3}^{-1}}}+\\frac{1}{{{\\log }_{{{5}^{-1}}}}{{3}^{-1}}}$$ $$=\\frac{1}{{{\\log }_{2}}3}+\\frac{1}{{{\\log }_{5}}3}={{\\log }_{3}}2+{{\\log }_{3}}5={{\\log }_{3}}10$$， ∵$${{3}^{2}} ~\\textless{} ~10 ~\\textless{} ~{{3}^{3}}$$， ∴$$x\\in \\left( 2,3 \\right)$$． " ]

[ "2019~2020学年4月山西太原迎泽区太原市第五中学校高二下学期周测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": "$$8$$ " } ] ]

[ "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", "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$$\\text{B}$$． " ]

[ "2017年四川全国高中数学联赛竞赛初赛第4题5分" ]

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

[ "竞赛->知识点->解析几何->椭圆" ]

[ "因为$$\\left\\textbar{} \\overrightarrow{{{F}_{1}}A} \\right\\textbar=a+e{{x}_{A}}\\leqslant a+e\\cdot a=a+c$$，$$\\left\\textbar{} \\overrightarrow{{{F}_{2}}B} \\right\\textbar=a-e{{x}_{B}}\\geqslant a-e\\cdot a=a-c$$， 所以$$a+c\\geqslant 3\\left( a-c \\right)$$，取等号时$$AB$$重合，不成立，因此$$a+c\\textgreater3\\left( a-c \\right)$$， 化简得$$\\frac{c}{a}\\textgreater\\frac{1}{2}$$ " ]

[ "2003年全国全国高中数学联赛竞赛一试第4题6分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{12\\sqrt{2}}{5}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{11\\sqrt{2}}{6}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{11\\sqrt{3}}{6}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{12\\sqrt{3}}{5}$$ " } ] ]

[ "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" ]

[ "$$y=\\tan \\left( x+\\frac{2 \\pi }{3} \\right)+\\cot \\left( x+\\frac{2 \\pi }{3} \\right)+\\cos \\left( x+\\frac{ \\pi }{6} \\right)=\\frac{1}{\\cos \\left( x+\\frac{2 \\pi }{3} \\right)\\sin \\left( x+\\frac{2 \\pi }{3} \\right)}+\\cos \\left( x+\\frac{ \\pi }{6} \\right)$$ $$=\\frac{2}{\\sin \\left( 2x+\\frac{4 \\pi }{3} \\right)}+\\cos \\left( x+\\frac{ \\pi }{6} \\right)$$ 因为$$x\\in \\left[ -\\frac{5 \\pi }{12},-\\frac{ \\pi }{3} \\right]$$，∴$$2x+\\frac{4 \\pi }{3}\\in \\left[ \\frac{ \\pi }{2},\\frac{2 \\pi }{3} \\right]$$，$$x+\\frac{ \\pi }{6}\\in \\left[ -\\frac{ \\pi }{4},-\\frac{ \\pi }{6} \\right]$$， 因此$$\\frac{2}{\\sin \\left( 2x+\\frac{4 \\pi }{3} \\right)}$$与$$\\cos \\left( x+\\frac{ \\pi }{6} \\right)$$在$$\\left[ -\\frac{5 \\pi }{12},-\\frac{ \\pi }{3} \\right]$$上同为增函数， 故当$$x=-\\frac{ \\pi }{3}$$时，$$y$$取最大值$$\\frac{11\\sqrt{3}}{6}$$． 故选$$\\text{C}$$． " ]

[ "2014年辽宁全国高中数学联赛竞赛初赛第3题5分" ]

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

[ "课内体系->知识点->三角函数->三角恒等变换->倍角、和差角公式综合", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->辅助角公式", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦函数的图象和性质", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦型函数的图象与性质", "课内体系->知识点->数列->等比数列->等比数列的性质及应用", "课内体系->知识点->解三角形->正弦定理", "课内体系->素养->数学运算" ]

[ "因为$$a$$，$$b$$，$$c$$成等比数列，所以$${{b}^{2}}=ac$$． 则$${{b}^{2}}=ac={{a}^{2}}+{{c}^{2}}-2ac\\cos B{\\geqslant }2ac-2ac\\cos B$$， $$\\cos B{\\geqslant }\\frac{1}{2}$$． 于是，$$0 \\textless{} B{\\leqslant }60{}^{}\\circ $$． 由$$\\frac{1}{2}{\\leqslant }\\cos B \\textless{} 1$$及$$0 \\textless{} \\sin B{\\leqslant }\\frac{\\sqrt{3}}{2}$$，得 $$\\frac{1}{2} \\textless{} \\cos B+\\sin B \\textless{} \\frac{\\sqrt{3}}{2}+1$$． ① 另一方面$$\\sin B+\\cos B=\\sqrt{2}\\sin (B+45{}^{}\\circ )$$． 而$$45{}^{}\\circ \\textless{} \\angle B+45{}^{}\\circ {\\leqslant }105{}^{}\\circ $$． 故$$1 \\textless{} \\sin B+\\cos B{\\leqslant }\\sqrt{2}$$． ② 综合①、②，有$$1 \\textless{} \\sin B+\\cos B{\\leqslant }\\sqrt{2}$$． 故选$$\\text{B}$$. " ]

[ "2021年吉林全国高中数学联赛竞赛初赛第5题5分" ]

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

[ "竞赛->知识点->多项式与方程->解方程（组）" ]

[ "由题设得：$$\\begin{cases}x-1=0 {{x}^{2}}+{{y}^{2}}+2x-4=0 \\end{cases}$$或$$\\begin{cases}x-1\\textgreater0 x+y-1=0 {{x}^{2}}+{{y}^{2}}+2x-4=0 \\end{cases}$$， 有三组解$$\\left( 1,1 \\right)$$，$$\\left( 1,-1 \\right)$$，$$\\left( \\frac{\\sqrt{6}}{2},1-\\frac{\\sqrt{6}}{2} \\right)$$． 故选$$\\text{C}$$． " ]

[ "2012年浙江全国高中数学联赛竞赛初赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$\\text{i}$$ " } ], [ { "aoVal": "B", "content": "$$-\\text{i}$$ " } ], [ { "aoVal": "C", "content": "$$-\\frac{4}{5}-\\frac{3}{5}\\text{i}$$ " } ], [ { "aoVal": "D", "content": "$$-\\frac{4}{5}+\\frac{3}{5}\\text{i}$$ " } ] ]

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

[ "$$\\frac{1+2\\text{i}}{\\text{i}-2}=\\frac{\\left( 1+2\\text{i} \\right)\\left( \\text{i}+2 \\right)}{\\left( \\text{i}-2 \\right)\\left( \\text{i}+2 \\right)}=-\\text{i}$$． " ]

[ "2016年天津河北区高三一模理科第13题5分", "2017~2018学年浙江杭州西湖区杭州学军中学高二上学期期中理科第14题4分", "2016年天津河北区高三二模文科第14题5分", "2015年甘肃全国高中数学联赛竞赛初赛第9题7分", "2016年天津河北区高三二模理科第14题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{4}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{4}{15}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{3}$$ " } ] ]

[ "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->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" ]

[ "设$$x+2=s$$，$$y+1=t$$，则$$s+t=x+y+3=4$$， 所以$$\\frac{{{x}^{2}}}{x+2}+\\frac{{{y}^{2}}}{y+1}$$ $$=\\frac{{{(s-2)}^{2}}}{s}+\\frac{{{(t-1)}^{2}}}{t}$$ $$=\\left( s-4+\\frac{4}{s} \\right)+\\left( t-2+\\frac{1}{t} \\right)$$ $$=(s+t)+\\left( \\frac{4}{s}+\\frac{1}{t} \\right)-6$$ $$=\\left( \\frac{4}{s}+\\frac{1}{t} \\right)-2$$． 因为$$\\frac{4}{s}+\\frac{1}{t}=\\frac{1}{4}\\left( \\frac{4}{s}+\\frac{1}{t} \\right)(s+t)$$ $$=\\frac{1}{4}\\left( \\frac{4t}{s}+\\frac{s}{t}+5 \\right)\\geqslant \\frac{9}{4}$$， 当且仅当$\\left {\\begin{align}\\&\\frac{4t}{s}=\\frac{s}{t} \\&s+t=4\\end{align}\\right.$即$\\left {\\begin{align}\\&s=\\frac83 \\&t=\\frac43\\end{align}\\right.$也即$\\left {\\begin{align}\\&x=\\frac23 \\&y=\\frac13\\end{align}\\right.$时取等， 故当$x=\\frac23,y=\\frac13$时，$\\frac{{{x}^{2}}}{x+2}+\\frac{{{y}^{2}}}{y+1}$取得最小值$\\frac{1}{4}$． " ]

[ "2020~2021学年河北石家庄新华区石家庄市第二中学高二上学期期中（竞赛班）第11题3分" ]

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

[ "课内体系->知识点->圆锥曲线->椭圆->椭圆的定义、标准方程->椭圆的定义", "课内体系->知识点->解三角形->正弦定理->利用正弦定理求解边角->边角互化（利用正弦定理）", "课内体系->素养->数学抽象", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理" ]

[ "由题设知：椭圆的长半轴长$$a=4$$，半焦距$$c=3$$， $$B$$、$$C$$为椭圆的左右焦点， ∵顶点$$A$$在椭圆上， ∴$$\\left\\textbar{} AB \\right\\textbar+\\left\\textbar{} AC \\right\\textbar=2a=8$$， $$\\textbar BC\\textbar=2c=6$$， 由正弦定理可得： $$\\frac{\\sin B+\\sin C}{\\sin A}=\\frac{\\left\\textbar{} AC \\right\\textbar+\\left\\textbar{} AB \\right\\textbar}{\\left\\textbar{} BC \\right\\textbar}=\\frac{8}{6}=\\frac{4}{3}$$． 故选$$\\text{D}$$． " ]

[ "2015年天津全国高中数学联赛竞赛初赛第3题6分" ]

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

[ "竞赛->知识点->数论模块->取整函数->与[x]有关的方程和不等式" ]

[ "$$x-1\\textless{}{{x}^{2}}-2=[x]\\leqslant x$$，解得$$\\frac{1+\\sqrt{5}}{2}\\textless{}x\\leqslant 2$$或者$$-1\\leqslant x\\textless{}\\frac{1-\\sqrt{5}}{2}$$，所以$$[x]=2$$、$$1$$或$$-1$$．代入原方程，解得$$x=2$$、$$\\sqrt{3}$$或$$-1$$． " ]

[ "2014年浙江全国高中数学联赛竞赛初赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$a=3$$，$$b=1$$ " } ], [ { "aoVal": "B", "content": "$$a=3$$，$$b=-1$$ " } ], [ { "aoVal": "C", "content": "$$a=-3$$，$$b=1$$ " } ], [ { "aoVal": "D", "content": "$$a=-3$$，$$b=-1$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "由$${{\\complement }_{U}}\\left { P\\cup Q \\right }=\\left { 6 \\right }\\Rightarrow {{a}^{2}}+{{b}^{2}}+a+b=6$$，显然只有答案$$D$$符合． " ]

[ "2015年福建全国高中数学联赛竞赛初赛第8题6分" ]

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

[ "知识标签->素养->数学运算", "知识标签->知识点->函数->函数的应用->函数的零点->函数零点的概念", "知识标签->知识点->导数->导数的应用->导数与极值", "知识标签->知识点->导数->导数的应用->导数与单调性", "知识标签->题型->导数->导数的应用->利用导数研究函数的极值问题->已知极值情况求参数的取值范围" ]

[ "$${{f}^{\\prime }}(x)={{\\text{e}}^{x}}(x-a{{\\text{e}}^{x}})+{{\\text{e}}^{x}}(1-a{{\\text{e}}^{x}})={{\\text{e}}^{x}}[-2a{{\\text{e}}^{x}}+x+1]=0$$有两个相异的实根， 即$$2a=\\frac{x+1}{{{\\text{e}}^{x}}}=g(x)$$有两个解， $${{g}^{\\prime }}(x)=\\frac{-x}{{{\\text{e}}^{x}}}$$， $$g(x)$$在$$(-\\infty ,0)$$单调递增；在$$(0,+\\infty )$$单调递减， $$x\\textless{}0$$，$$g(x)\\textless{}1$$，$$x\\textgreater0$$时，$$0\\textless{}g(x)\\textless{}1$$， 因此$$2a\\in (0,1)$$，即$$a\\in \\left( 0,\\frac{1}{2} \\right)$$． " ]

[ "2013年浙江全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$3\\sqrt[9]{81}$$ " } ], [ { "aoVal": "B", "content": "$$3\\sqrt[7]{81}$$ " } ], [ { "aoVal": "C", "content": "$$\\sqrt[3]{9}$$ " } ], [ { "aoVal": "D", "content": "$$3\\sqrt{3}$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "计算得$$q={{3}^{\\frac{2}{7}}},{{a}_{3}}=3\\sqrt[7]{81}$$． " ]

[ "2010年AMC10竞赛A第10题" ]

[ [ { "aoVal": "A", "content": "$$2011$$ " } ], [ { "aoVal": "B", "content": "$$2012$$ " } ], [ { "aoVal": "C", "content": "$$2013$$ " } ], [ { "aoVal": "D", "content": "$$2015$$ " } ], [ { "aoVal": "E", "content": "$$2017$$ " } ] ]

[ "美国AMC10/12->Knowledge Point->Combination->Reasoning->Recurrence and Recursion", "课内体系->知识点->推理与证明->逻辑推理题" ]

[ "$$\\boxed{\\rm(E)}2017$$. There are $$365$$ days in a non-leap year. There are $$7$$ days in a week. Since $$365=52\\cdot7+1$$(or $$365$$ is congruent to $$1\\textasciitilde\\rm mod\\textasciitilde7$$), the same date (after February) moves \"forward\" one day in the subsequent year, if that year is not a leap year. For example: $$5/27/08$$ Tue $$5/27/09$$ Wed However, a leap year has $$366$$ days, and $$366=52\\cdot7+2$$. So the same date (after February) moves \"forward\" two days in the subsequent year, if that year is a leap year. For example: $$5/27/11$$ Fri $$5/27/12$$ Sun You can keep count forward to find that the first time this date falls on a Saturday is in $$2017$$: $$5/27/13$$ Mon $$5/27/14$$ Tue $$5/27/15$$ Wed $$5/27/16$$ Fri $$5/27/17$$ Sat " ]

[ "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3~3题" ]

[ [ { "aoVal": "A", "content": "$$\\left( 2,3 \\right)$$ " } ], [ { "aoVal": "B", "content": "$$\\left( 3,4 \\right)$$ " } ], [ { "aoVal": "C", "content": "$$\\left( 5,6 \\right)$$ " } ], [ { "aoVal": "D", "content": "$$\\left( 6,7 \\right)$$ " } ] ]

[ "课内体系->知识点->函数的应用->函数的零点->求零点所在区间的问题" ]

[ "\\hfill\\break 设$$f(x)={{\\log }_{3}}x-x+4$$，证明$$f(5)\\cdot f(6)\\textless{} 0$$即得解.\\\\ 【详解】\\\\ 因为$${{\\log }_{3}}x=x-4$$，所以$${{\\log }_{3}}x-x+4=0$$.\\\\ 设$$f(x)={{\\log }_{3}}x-x+4$$，\\\\ 所以$$f(5)={{\\log }_{3}}5-5+4={{\\log }_{3}}5-1\\textgreater0$$，\\\\ $$f(6)={{\\log }_{3}}6-6+4={{\\log }_{3}}6-2={{\\log }_{3}}\\frac{6}{9}\\text{=}{{\\log }_{3}}\\frac{2}{3}\\textless{} 0$$，\\\\ 所以$$f(5)\\cdot f(6)\\textless{} 0$$.\\\\ 故选：C\\\\ 【点睛】\\\\ 本题主要考查零点问题，考查零点区间的确定，意在考查学生对这些知识的理解掌握水平. " ]

[ "2015年辽宁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": "$$\\frac{ \\pi }{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{ \\pi }{2}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2 \\pi }{3}$$ " } ] ]

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

[ "$$\\sin A\\sin B+\\sin B\\sin C=1-\\cos 2B=2{{\\sin }^{2}}B\\Rightarrow a+c=2b$$，又$$ac={{b}^{2}}$$， 则$${{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\\cos B={{\\left( a+c \\right)}^{2}}-2ac\\left( 1+\\cos B \\right)=4{{b}^{2}}-2{{b}^{2}}\\left( 1+\\cos B \\right)$$， 解得$$B=\\frac{ \\pi }{3}$$． " ]

[ "2008年福建全国高中数学联赛竞赛初赛第6题6分" ]

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

[ "竞赛->知识点->不等式->不等式的解法" ]

[ "令$$f(x)={{x}^{2}}+2ax+3a$$，由$$f\\left( -\\frac{3}{2} \\right)=\\frac{9}{4}$$知， 函数$$f(x)$$的图象经过点$$\\left( -\\frac{3}{2},\\frac{9}{4} \\right)$$． 欲使得不等式$$\\left\\textbar{} {{x}^{2}}+2ax+3a \\right\\textbar\\leqslant 2$$只有一个解， 则抛物线$$f(x)={{x}^{2}}+2ax+3a$$的图像必须与直线$$y=2$$相切， 即$${{x}^{2}}+2ax+3a-2=0$$的差别式$$\\Delta =4{{a}^{2}}-4(3a-2)=0$$． 解得$$a=1$$，$$2$$．故选$$\\text{B}$$． " ]

[ "2011年黑龙江全国高中数学联赛竞赛初赛第10题5分" ]

[ [ { "aoVal": "A", "content": "$$2x-y-3=0$$ " } ], [ { "aoVal": "B", "content": "$$x-2y-3=0$$ " } ], [ { "aoVal": "C", "content": "$$x+2y-4=0$$ " } ], [ { "aoVal": "D", "content": "$$2x+y-3=0$$ " } ] ]

[ "竞赛->知识点->解析几何->直线与方程" ]

[ "$${{k}_{BC}}=\\frac{3+1}{1+1}=2$$，可得$$BC:y+1=2\\left( x+1 \\right)$$，即$$2x-y+1=0$$， 设$$P\\left( m,2m+1 \\right)$$为直线$$BC$$上一点，$$Q\\left( x,y \\right)$$， 由$$\\overrightarrow{PQ}=\\overrightarrow{PA}+\\overrightarrow{PB}+\\overrightarrow{PC}$$，可得$$\\left( x-m,y-2m-1 \\right)$$ $$=\\left( 2-m,1-2m-2 \\right)+\\left( -1-m,-1-2m-1 \\right)+\\left( 1-m,3-2m-2 \\right)$$ $$=\\left( 2-3m,-6m \\right)$$， 所以有$$\\left( x-m,y-2m-1 \\right)=\\left( 2-3m,-6m \\right)$$， 即$$\\begin{cases}x-m=2-3m y-2m-1=-6m \\end{cases}$$，即$$\\begin{cases}2m=2-x 4m=1-y \\end{cases}$$， 所以$$1-y=2\\left( 2-x \\right)$$，即$$2x-y-3=0$$，故选$$\\text{A}$$. " ]

[ "1995年全国高中数学联赛竞赛一试第3题" ]

[ [ { "aoVal": "A", "content": "$$1$$个 " } ], [ { "aoVal": "B", "content": "$$2$$个 " } ], [ { "aoVal": "C", "content": "$$50$$个 " } ], [ { "aoVal": "D", "content": "$$100$$个 " } ] ]

[ "竞赛->知识点->逻辑->逻辑推理" ]

[ "取$$100$$个小伙子为这样一种特殊情况，他们的身高与体重互不相等，并且最高者同时也就是最轻者，次高者同时也就是次轻者，$$\\ldots $$，第$$k$$高者同时也就是第$$k$$轻者（$$k=1$$，$$2$$，$$\\ldots $$，$$100$$），显然这$$100$$个小伙子都是棒小伙子． 故选$$\\text{D}$$． " ]

[ "2008年黑龙江全国高中数学联赛竞赛初赛第11题5分" ]

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

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "由于$$\\lg x=3-x$$，$${{10}^{x}}=3-x$$，令$$y=3-x$$， 则$${{y}_{1}}=\\lg x$$，$${{y}_{2}}={{10}^{x}}$$互为反函数． 所以$${{y}_{1}}=\\lg x$$与$${{y}_{2}}={{10}^{x}}$$和直线$$y=3-x$$的交点$$A$$、$$B$$关于直线$$y=x$$对称， 由$$y=x$$与$$y=3-x$$联立解得线段$$AB$$的中点$$C\\left( \\frac{3}{2},\\frac{3}{2} \\right)$$， 因此$${{x}_{1}}+{{x}_{2}}=3$$．故选$$\\text{A}$$． " ]

[ "2016年广东全国高中数学联赛高三竞赛初赛第7题8分" ]

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

[ "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的实际应用", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件", "课内体系->知识点->等式与不等式->不等式->不等式的性质", "课内体系->知识点->等式与不等式->不等式->柯西不等式", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式的化简和求值", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->和差角公式化简求值综合运用", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->两角和与差的余弦", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理" ]

[ "∵$$\\frac{{{\\sin }^{4}}\\alpha }{{{\\cos }^{2}}\\beta }+\\frac{{{\\cos }^{4}}\\alpha }{{{\\sin }^{2}}\\beta }=1$$， 即$$\\frac{{{\\sin }^{4}}\\alpha }{{{\\cos }^{2}}\\beta }+\\frac{{{\\cos }^{4}}\\alpha }{{{\\sin }^{2}}\\beta }-1={{\\left( \\frac{{{\\sin }^{2}}\\alpha }{\\cos \\beta }-\\cos \\beta \\right)}^{2}}+{{\\left( \\frac{{{\\cos }^{2}}\\alpha }{\\sin \\beta }-\\sin \\beta \\right)}^{2}}=0$$， ∴$$\\frac{{{\\sin }^{2}}\\alpha }{\\cos \\beta }-\\cos \\beta =0$$，$$\\frac{{{\\cos }^{2}}\\alpha }{\\sin \\beta }-\\sin \\beta =0$$， ∴$${{\\sin }^{2}}\\alpha ={{\\cos }^{2}}\\beta $$，$${{\\cos }^{2}}\\alpha ={{\\sin }^{2}}\\beta $$， 根据$$\\alpha $$、$$\\beta $$为锐角，可得$$\\alpha +\\beta =90{}^{}\\circ $$． 故答案为：$$90{}^{}\\circ $$． 由柯西不等式，$$\\left( {{\\cos }^{2}}\\beta +{{\\sin }^{2}}\\beta \\right)\\left( \\frac{{{\\sin }^{4}}\\alpha }{{{\\cos }^{2}}\\beta }+\\frac{{{\\cos }^{4}}\\alpha }{{{\\sin }^{2}}\\beta } \\right)\\geqslant {{\\left( {{\\sin }^{2}}\\alpha +{{\\cos }^{2}}\\alpha \\right)}^{2}}=1$$， 因此，$$\\frac{{{\\cos }^{4}}\\beta }{{{\\sin }^{4}}\\alpha }=\\frac{{{\\sin }^{4}}\\beta }{{{\\cos }^{4}}\\alpha }$$，即$$\\cos \\alpha \\cos \\beta =\\sin \\alpha \\sin \\beta $$，$$\\cos \\left( \\alpha +\\beta \\right)=0$$，故$$\\alpha +\\beta =\\frac{ \\pi }{2}$$． 故答案为：$$\\frac{ \\pi }{2}$$． " ]

[ "2012年湖南全国高中数学联赛竞赛初赛第3题6分", "2011年北京清华大学自主招生夏令营第8题" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$120$$ " } ], [ { "aoVal": "C", "content": "$$140$$ " } ], [ { "aoVal": "D", "content": "$$160$$ " } ] ]

[ "竞赛->知识点->排列组合与概率->排列与组合" ]

[ "设$${{b}_{k+1}}={{a}_{k+1}}-{{a}_{k}}$$，则$${{b}_{k+1}}=\\pm 1$$，且$${{b}_{2}}+{{b}_{3}}+\\cdots+{{b}_{11}}=4$$，只需$${{b}_{2}}$$，$${{b}_{3}}$$，$$\\cdots$$，$${{b}_{11}}$$中有$$7$$个$$1$$和$$3$$个$$-1$$即可，所求的方法数为$$\\text{C}_{10}^{3}=120$$． " ]

[ "2011年高考真题湖北卷理科第1题", "2012年黑龙江全国高中数学联赛竞赛初赛第1题5分" ]

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

[ "知识标签->题型->复数->复数的运算->复数中的周期问题", "知识标签->素养->数学运算", "知识标签->知识点->复数->复数的运算->复数的乘法和除法" ]

[ "因为$$\\frac{1+\\text{i}}{1-\\text{i}}=\\frac{{{\\left( 1+\\text{i} \\right)}^{2}}}{1-{{\\text{i}}^{2}}}=\\text{i}$$，所以$${{\\left(\\frac{1+\\text{i}}{1-\\text{i}} \\right)}^{2011}}={{\\text{i}}^{2011}}={{\\text{i}}^{4\\times502+3}}={{\\text{i}}^{3}}=-\\text{i}$$， 故选$$\\text{A}$$． " ]

[ "全国高中数学联赛竞赛模拟一试（十八）第3题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{53}{30}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{59}{30}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{61}{30}$$ " } ], [ { "aoVal": "D", "content": "前三个答案都不对 " } ] ]

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

[ "令$$\\frac{y}{x+y}=m$$，$$\\frac{z}{y+z}=n$$，$$\\frac{x}{z+x}=l$$． 则$$\\frac{14}{15}=\\frac{x-y}{x+y}\\cdot\\frac{y-z}{y+z}\\cdot\\frac{z-x}{z+x}=(1-2m)(1-2n)(1-2l )$$． 即$$1-2(m+n+l)+4(mn+nl+lm)-8mnl=\\frac{14}{15}$$． 又$$\\left( {\\frac{1}{m}-1} \\right)\\left( {\\frac{1}{n}-1} \\right)\\left( {\\frac{1}{l}-1} \\right)=\\frac{x}{y}\\cdot\\frac{y}{z}\\cdot\\frac{z}{x}=1$$ $$\\Rightarrow 1-\\left( m+n+l \\right)+\\left( mn+nl+lm \\right)-mnl=mnl$$， 故$$1-2(m+n+l)-4+4(m+n+l)=\\frac{14}{15}$$ $$\\Rightarrow m+n+l=\\frac{59}{30}$$ $$\\Rightarrow \\frac{y}{x+y}+\\frac{z}{y+z}+\\frac{x}{z+x}=\\frac{59}{30}$$． " ]

[ "2008年浙江全国高中数学联赛竞赛初赛第5题6分" ]

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

[ "竞赛->知识点->函数->二次函数", "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->三角函数->三角恒等变换" ]

[ "由已知条件可知$${{a}^{2}}+{{b}^{2}}-1=0$$， 函数图象与$$y$$轴交点的纵坐标为$${{a}^{2}}+2ab-{{b}^{2}}$$． 令$$a=\\cos \\theta $$，$$ b=\\sin \\theta $$， 则$${{a}^{2}}+2ab-{{b}^{2}}={{\\cos }^{2}}\\theta +2\\sin \\theta \\cos \\theta -{{\\sin }^{2}}\\theta $$ $$=\\cos 2\\theta +\\sin 2\\theta \\leqslant \\sqrt{2}$$， 故选$$\\text{A}$$. " ]

[ "2008年河南全国高中数学联赛竞赛初赛第4题5分" ]

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

[ "竞赛->知识点->解析几何->直线与圆锥曲线" ]

[ "设$$AB$$在准线$$l$$上的射影分别为$${A}'$$、$${B}'$$， 则$$\\textbar M{M}'\\textbar=\\frac{\\textbar A{A}'\\textbar+\\textbar B{B}'\\textbar}{2}$$． 又因为$$\\textbar A{A}'\\textbar=\\textbar AF\\textbar$$，$$\\textbar B{B}'\\textbar=\\textbar BF\\textbar$$， 所以$$\\textbar M{M}'\\textbar=\\frac{\\textbar AF\\textbar+\\textbar BF\\textbar}{2}$$． 因为$$\\angle AFB=\\frac{2}{3} \\pi $$， 所以$$\\textbar AB\\textbar=\\textbar FA{{\\textbar}^{2}}+\\textbar FB{{\\textbar}^{2}}-2\\textbar FA\\textbar\\cdot \\textbar FB\\textbar\\cos \\angle AFB$$ $$=\\textbar FA{{\\textbar}^{2}}+\\textbar FB{{\\textbar}^{2}}+\\textbar FA\\textbar\\cdot \\textbar FB\\textbar$$ 因此$$\\textbar AB{{\\textbar}^{2}}+\\textbar M{M}'{{\\textbar}^{2}}=\\textbar AB{{\\textbar}^{2}}+{{\\left( \\frac{\\textbar AF\\textbar+\\textbar BF\\textbar}{2} \\right)}^{2}}$$ $$\\geqslant \\textbar AB{{\\textbar}^{2}}+\\textbar FA\\textbar\\cdot \\textbar FB\\textbar$$ $$={{(\\textbar FA\\textbar+\\textbar FB\\textbar)}^{2}}=4\\textbar M{M}'{{\\textbar}^{2}}$$， 则$$\\textbar AB{{\\textbar}^{2}}\\geqslant 3\\textbar M{M}'{{\\textbar}^{2}}$$，即$$\\frac{\\textbar M{M}'\\textbar}{\\textbar AB\\textbar}\\leqslant \\frac{\\sqrt{3}}{3}$$．故选$$\\text{B}$$． " ]

[ "1985年全国高中数学联赛竞赛一试第1题" ]

[ [ { "aoVal": "A", "content": "甲是乙的充分而不必要条件 " } ], [ { "aoVal": "B", "content": "甲是乙的必要而不充分条件 " } ], [ { "aoVal": "C", "content": "甲是乙的充分必要条件 " } ], [ { "aoVal": "D", "content": "甲既不是乙的充分条件也不是乙的必要条件． " } ] ]

[ "竞赛->知识点->逻辑->常用逻辑用语" ]

[ "由于``甲''成立时，``乙''不一定成立，（例如$$a=3$$，$$b=2$$时甲成立，但乙不成立），因此可知``甲''不是``乙''的充分条件， 接着看``乙''在什么情况下成立，很明显，当且仅当$$a\\textgreater0$$且$$b\\textless{}0$$时``乙''才能成立，由此可知，``甲''是``乙''成立的必要条件，综上所述，``甲''是``乙''的必要而不充分条件． " ]

[ "2008年陕西全国高中数学联赛竞赛初赛第7题5分" ]

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

[ "竞赛->知识点->三角函数->三角函数的图像与性质" ]

[ "要使$$y$$最大，应有$$\\tan x\\textgreater0$$． 不妨设$$0\\textless{}x\\textless{}\\frac{ \\pi }{2}$$，则$$y=\\tan x-\\frac{2}{\\cos x}=\\frac{\\sin x-2}{\\cos x}$$， 即$$\\sin x-y\\cos x=2$$．所以$$\\sin (x-\\theta )=\\frac{2}{\\sqrt{1+{{y}^{2}}}}$$，其中$$\\tan \\theta =y$$． 由$$\\sin (x-\\theta )\\leqslant 1$$，得$$\\frac{2}{\\sqrt{1+{{y}^{2}}}}\\leqslant 1$$， 解得$$y\\leqslant -\\sqrt{3}$$或$$y\\geqslant 3$$． 易知$$y\\textless{}0$$，所以$$y\\leqslant -\\sqrt{3}$$． 将$$y=-\\sqrt{3}$$代入$$\\sin x-y\\cos x=2$$，得$$x=\\frac{ \\pi }{6}$$． 故$${{y}_{\\max }}=-\\sqrt{3}$$．故选$$\\text{C}$$． " ]

[ "2016年四川全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$-2$$ " } ], [ { "aoVal": "B", "content": "$$0$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{4}$$ " } ], [ { "aoVal": "D", "content": "$$1$$ " } ] ]

[ "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->函数的概念与性质->函数的概念及其表示->函数的概念->分段函数", "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->函数求值问题", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域", "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->利用单调性求函数最值", "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", "课内体系->知识点->函数的概念与性质->二次函数->二次函数的概念" ]

[ "$$f\\left( x \\right)={{\\left( x-t \\right)}^{2}}+t-{{t}^{2}}$$，$$m=\\begin{cases}t-{{t}^{2}},-1\\leqslant t\\leqslant 1 1-t,t\\textgreater1 1+3t,t\\textless{}-1 \\end{cases}$$ 因此，$${{m}_{\\max }}=t-{{t}^{2}}=-{{\\left( t-\\frac{1}{2} \\right)}^{2}}+\\frac{1}{4}\\leqslant \\frac{1}{4}$$． 故选$$\\text{C}$$． " ]

[ "2008年全国高中数学联赛竞赛一试第4题6分" ]

[ [ { "aoVal": "A", "content": "$$764\\text{c}{{\\text{m}}^{3}}$$或$$586\\text{c}{{\\text{m}}^{3}}$$ " } ], [ { "aoVal": "B", "content": "$$764\\text{c}{{\\text{m}}^{3}}$$ " } ], [ { "aoVal": "C", "content": "$$586\\text{c}{{\\text{m}}^{3}}$$或$$564\\text{c}{{\\text{m}}^{3}}$$ " } ], [ { "aoVal": "D", "content": "$$586\\text{c}{{\\text{m}}^{3}}$$ " } ] ]

[ "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" ]

[ "设这三个正方体的棱长分别为$$a,b,c$$，则有$$6({{a}^{2}}+{{b}^{2}}+{{c}^{2}})=564$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=94$$， 不妨设$$1\\leqslant a\\leqslant b\\leqslant c\\textless{}10$$，从而$$3{{c}^{2}}\\geqslant {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=94$$，$${{c}^{2}}\\textgreater31$$， 故$$6\\leqslant c\\leqslant 10$$．$$c$$只能取$$9$$，$$8$$，$$7$$，$$6$$， 若$$c=9$$，则$${{a}^{2}}+{{b}^{2}}=94-{{9}^{2}}=13$$，易知$$a=2$$，$$b=3$$，得一组解$$(a,b,c)=(2,3,9)$$， 若$$c=8$$，则$${{a}^{2}}+{{b}^{2}}=94-64=30$$，$$b\\leqslant 5$$，但$$2{{b}^{2}}\\geqslant 30$$，$$b\\geqslant 4$$，从而$$b=4$$或$$5$$， 若$$b=5$$，则$${{a}^{2}}=5$$无解，若$$b=4$$，则$${{a}^{2}}=14$$无解，此时无解， 若$$c=7$$，则$${{a}^{2}}+{{b}^{2}}=94-49=45$$，有唯一解$$a=3$$，$$b=6$$， 若$$c=6$$，则$${{a}^{2}}+{{b}^{2}}=94-36=58$$，此时$$2{{b}^{2}}\\geqslant {{a}^{2}}+{{b}^{2}}=58$$，$${{b}^{2}}\\geqslant 29$$，故$$b\\geqslant 6$$，但$$b\\leqslant c=6$$，故$$b=6$$，此时$${{a}^{2}}=58-36=22$$无解， 综上，共有两组解$$\\begin{cases}a=2 b=3 c=9 \\end{cases}$$或$$\\begin{cases}a=3 b=6 c=7 \\end{cases}$$， 体积为$${{V}_{1}}={{2}^{3}}+{{3}^{3}}+{{9}^{3}}=764\\text{c}{{\\text{m}}^{3}}$$或$${{V}_{2}}={{3}^{3}}+{{6}^{3}}+{{7}^{3}}=586\\text{c}{{\\text{m}}^{3}}$$． 故选$$\\text{A}$$． " ]

[ "2016年浙江全国高中数学联赛竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$a=0$$ " } ], [ { "aoVal": "B", "content": "$$a=1$$ " } ], [ { "aoVal": "C", "content": "$$a=-1$$ " } ], [ { "aoVal": "D", "content": "$$a\\in \\mathbf{R}$$~ " } ] ]

[ "课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与函数结合", "课内体系->知识点->函数的应用->函数与方程", "课内体系->知识点->直线和圆的方程->直线与方程->直线方程的五种形式->直线过定点问题", "课内体系->知识点->直线和圆的方程->直线与方程->直线方程的五种形式->直线的一般式方程", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理" ]

[ "三条直线过原点. " ]

[ "2011年福建全国高中数学联赛竞赛初赛第5题6分" ]

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

[ "课内体系->知识点->函数的概念与性质", "竞赛->知识点->函数->函数的图像与性质" ]

[ "在条件等式（$$3$$）中，令$$x=\\frac{2}{3}, y=\\frac{1}{3},$$得 $$f(1)-f\\left( \\frac{1}{3} \\right)=2f\\left( \\frac{1}{3} \\right)f\\left( \\frac{1}{3} \\right),$$ 结合$$f(1)=1, f\\left( \\frac{1}{3} \\right)\\textgreater0$$，解得$$f\\left( \\frac{1}{3} \\right)=\\frac{1}{2}$$ " ]

[ "高考真题", "2008年贵州全国高中数学联赛竞赛初赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "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->等差数列的概念与通项公式->等差数列求通项问题" ]

[ "$${{a}_{1}}+2{{a}_{8}}+{{a}_{15}}=4{{a}_{8}}=96$$，$${{a}_{8}}=24$$． $$2{{a}_{9}}-{{a}_{10}}={{a}_{8}}=24$$． " ]

[ "2015年黑龙江全国高中数学联赛竞赛初赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{6}+\\frac{ \\pi }{2}$$ " } ], [ { "aoVal": "B", "content": "$$0$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$ \\pi $$ " } ] ]

[ "竞赛->知识点->导数模块->积分", "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->三角函数->三角函数的图像与性质" ]

[ "因为$$\\sin 5x$$是奇函数，关于原点对称，且积分限关于原点对称，积分就是面积， 所以$$x\\textless{}0$$和$$x\\textgreater0$$时积分的值一正一负，且面积相等， 所以正负抵消，即$$\\int_{-\\frac{ \\pi }{2}}^{\\frac{ \\pi }{2}}{\\sin^{5}x\\text{d}x}=0$$， 所以$$\\int_{-\\frac{ \\pi }{2}}^{\\frac{ \\pi }{2}}{f(x)\\text{d}x}=\\int_{-\\frac{ \\pi }{2}}^{\\frac{ \\pi }{2}}{\\sin^{5}x\\text{d}x}+\\int_{-\\frac{ \\pi }{2}}^{\\frac{ \\pi }{2}}{\\text{1d}x}=x\\textbar_{-\\frac{ \\pi }{2}}^{\\frac{ \\pi }{2}}=\\frac{ \\pi }{2}-\\left( -\\frac{ \\pi }{2} \\right)= \\pi $$. " ]

[ "2014年湖南全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$p$$是$$q$$的充分不必要条件 " } ], [ { "aoVal": "B", "content": "$$p$$是$$q$$的充要不充分条件 " } ], [ { "aoVal": "C", "content": "$$p$$是$$q$$的充要条件 " } ], [ { "aoVal": "D", "content": "$$p$$既不是$$q$$的充分条件又不是$$q$$的必要条件 " } ] ]

[ "竞赛->知识点->逻辑->常用逻辑用语" ]

[ "由$$q:\\begin{cases}0\\textless{}m\\textless{}1 2\\textless{}n\\textless{}3 \\end{cases}\\Rightarrow p:\\begin{cases}2\\textless{}m+n\\textless{}4 0\\textless{}mn\\textless{}3 \\end{cases}$$ 则$$p$$是$$q$$的必要条件；注意到当$$\\begin{cases}m=1 n=2 \\end{cases}$$时，满足$$p:\\begin{cases}2\\textless{}m+n\\textless{}4 0\\textless{}mn\\textless{}3 \\end{cases}$$而不满足$$q:\\begin{cases}0\\textless{}m\\textless{}1 2\\textless{}n\\textless{}3 \\end{cases}$$则$$p$$不是$$q$$的充分条件．所以$$p$$是$$q$$的必要不充分条件． " ]

[ "2015年四川全国高中数学联赛竞赛初赛第1题5分", "2017年北京西城区高三三模理科第2题" ]

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

[ "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->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" ]

[ "$${{T}_{r+1}}=C_{n}^{r}{{({{x}^{2}})}^{n-r}}{{\\left( \\frac{1}{{{x}^{3}}} \\right)}^{r}}=C_{n}^{r}{{x}^{2}}^{n-5r}$$．令$$2n-5r=7$$，所以$$r=1$$时，$$n$$取得最小值$$6$$．选$$\\text{C}$$． " ]

[ "2006年AMC10竞赛B第21题" ]

[ [ { "aoVal": "A", "content": "$$\\dfrac{4}{63}$$ " } ], [ { "aoVal": "B", "content": "$$\\dfrac{1}{8}$$ " } ], [ { "aoVal": "C", "content": "$$\\dfrac{8}{63}$$ " } ], [ { "aoVal": "D", "content": "$$\\dfrac{1}{6}$$ " } ], [ { "aoVal": "E", "content": "$$\\dfrac{2}{7}$$ " } ] ]

[ "课内体系->知识点->统计与概率->概率->事件与概率->互斥事件->互斥事件的概率加法公式", "美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Classical Models of Probabilities" ]

[ "Let $$x$$ be the \\uline{probability} of rolling a $$1$$. The probabilities of rolling a $$2$$, $$3$$, $$4$$, $$5$$, and $$6$$ are $$2x$$, $$3x$$, $$4x$$, $$5x$$, and $$6x$$, respectively. The sum of the probabilties of rolling each number must equal $$1$$, so $$x+2x+3x+4x+5x+6x=1$$, $$21x=1$$, $$x= \\dfrac{1}{21}$$, So the probabilities of rolling a $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, and $$6$$ are respectively $$\\dfrac{1}{21}$$,~ $$\\dfrac{2}{21}$$, $$\\dfrac{3}{21}$$, $$\\dfrac{4}{21}$$, $$\\dfrac{5}{21}$$, and, $$\\dfrac{6}{21}$$. The possible combinations of two rolls that total $$7$$ are: $$(1,6)$$; $$(2,5)$$; $$(3,4)$$; $$(4,3)$$; $$(5,2)$$; $$(6,1)$$. The probability of rolling a total of $$7$$ on the two dice is equal to the sum of the probabilities of rolling each combination. $$P= \\dfrac{1}{21}\\cdot \\dfrac{6}{21}+ \\dfrac{2}{21}\\cdot \\dfrac{5}{21}+ \\dfrac{3}{21}\\cdot \\dfrac{4}{21}+ \\dfrac{4}{21}\\cdot \\dfrac{3}{21}+ \\dfrac{5}{21}\\cdot \\dfrac{2}{21}+ \\dfrac{6}{21}\\cdot \\dfrac{1}{21}= \\dfrac{8}{63}\\Rightarrow C$$ " ]

[ "2018年天津全国高中数学联赛竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$256$$ " } ], [ { "aoVal": "B", "content": "$$959$$ " } ], [ { "aoVal": "C", "content": "$$960$$ " } ], [ { "aoVal": "D", "content": "$$961$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "由满足$$C\\cap B=\\varnothing $$的子集$$C$$有$${{2}^{6}}$$个， 知满足$$C\\cap B\\ne \\varnothing $$的子集$$C$$有$${{2}^{10}}-{{2}^{6}}=960$$个． 故选$$\\text{C}$$． " ]

[ "2016年陕西全国高中数学联赛竞赛初赛第8题6分" ]

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

[ "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->等式与不等式->不等式->不等式的性质", "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->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->基本不等式成立的条件" ]

[ "不妨设$$a\\geqslant b\\geqslant c$$ $$\\left( a+b+c \\right)\\left( \\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca} \\right)=\\left( a+b \\right)\\sqrt{ab}+\\left( b+c \\right)\\sqrt{bc}+\\left( c+a \\right)\\sqrt{ca}+\\left( c\\sqrt{ab}+a\\sqrt{bc}+b\\sqrt{ca} \\right)$$ $$\\geqslant \\left( a+b \\right)\\sqrt{ab}+\\left( b+c \\right)\\sqrt{bc}+\\left( c+a \\right)\\sqrt{ca}$$ $$\\geqslant 2\\sqrt{ab}\\sqrt{ab}+2\\sqrt{bc}\\sqrt{bc}+2\\sqrt{ca}\\sqrt{ca}$$ $$=2\\left( ab+bc+ca \\right)$$ 所以，$$\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca}\\geqslant 2$$，当且仅当$$c=0$$，$$a=b$$时等号成立． 由$$c=0$$，$$a=b$$，解得$$a=b=2$$，$$c=0$$． 所以，$$\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca}$$的最小值为$$2$$． 故选$$\\text{A}$$． " ]

[ "2008年河北全国高中数学联赛竞赛初赛第4题6分", "高二上学期单元测试《代数变形1》自招第7题", "2019~2020学年天津滨海新区塘沽第一中学高一上学期期中第7题5分" ]

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

[ "竞赛->知识点->不等式->换元技巧->代数换元" ]

[ "记$$a-b=t$$，则$$t\\textgreater0$$， $$\\frac{{{a}^{2}}+{{b}^{2}}}{a-b}=\\frac{{{t}^{2}}+2}{t}=t+\\frac{2}{t}\\geqslant 2\\sqrt{2}$$， 当且仅当$$a=\\frac{\\sqrt{6}+\\sqrt{2}}{2}$$，$$b=\\frac{\\sqrt{6}-\\sqrt{2}}{2}$$时取等号．故选$$\\text{A}$$． " ]

[ "2009年湖南全国高中数学联赛竞赛初赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$72$$ " } ], [ { "aoVal": "B", "content": "$$73$$ " } ], [ { "aoVal": "C", "content": "$$144$$ " } ], [ { "aoVal": "D", "content": "$$146$$ " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "用$$(1-x)$$代替条件等式中的$$x$$，得 $$f({{x}^{2}}-3x+2)+2f({{x}^{2}}+x)=9{{x}^{2}}-3x-6$$． 由上式及原式消去$$f({{x}^{2}}-3x+2)$$，得 $$f({{x}^{2}}+x)=3{{x}^{2}}+3x-4=3({{x}^{4}}+x)-4$$． 故$$f(50)=150-4=146$$，故选$$\\text{D}$$． " ]

[ "2011年AMC10竞赛A第18题" ]

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

[ "美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Circle", "课内体系->知识点->三角函数->三角函数的概念->任意角与弧度制->弧长公式与扇形面积" ]

[ "Not specific: Draw a rectangle with vertices at the centers of $$A$$ and $$B$$ and the intersection of $$A$$, $$C$$ and $$B$$, $$C$$. Then, we can compute the shaded area as the area of half of $$C$$ plus the area of the rectangle minus the area of the two sectors created by $$A$$ and $$B$$. This is $$\\dfrac{\\pi(1)^{2}}{2}+(2)(1)-2 \\cdot \\dfrac{\\pi(1)^{2}}{4}= (\\rm C)2$$. " ]

[ "2018~2019学年北京东城区北京市第一六六中学高二上学期期中第12题5分", "2006年全国高中数学联赛竞赛一试第4题6分" ]

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

[ "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->空间向量与立体几何->空间向量的应用->向量法求空间距离", "课内体系->思想->数形结合思想" ]

[ "建立直角坐标系，以$$A$$为坐标原点，$$AB$$为$$x$$轴，$$AC$$为$$y$$轴，$$A{{A}_{{}}}$$为$$z$$轴，则$$F({{t}_{1}},0,0)$$（$$0\\textless{}{{t}_{1}}\\textless{}1$$），$$E\\left( 0,1,\\frac{1}{2} \\right)$$，$$G\\left( \\frac{1}{2},0,1 \\right)$$，$$D\\left( 0,{{t}_{2}},0 \\right)$$（$$0\\textless{}{{t}_{2}}\\textless{}1$$）．所以$$\\overrightarrow{EF}=\\left( {{t}_{1}},-1,-\\frac{1}{2} \\right)$$，$$\\overrightarrow{GD}=\\left( -\\frac{1}{2},{{t}_{2}},-1 \\right)$$．因为$$GD\\bot EF$$，所以$${{t}_{1}}+2{{t}_{2}}=1$$，由此推出$$0\\textless{}{{t}_{2}}\\textless{}\\frac{1}{2}$$．又$$\\overrightarrow{DF}=({{t}_{1}},-{{t}_{2}},0)$$，$$\\left\\textbar{} \\overrightarrow{DF} \\right\\textbar=\\sqrt{{{t}_{1}}^{2}+{{t}_{2}}^{2}}=\\sqrt{5{{t}_{2}}^{2}-4{{t}_{2}}+1}=\\sqrt{5{{({{t}_{2}}-\\frac{2}{5})}^{2}}+\\frac{1}{5}}$$，从而有$$\\sqrt{\\frac{1}{5}}\\leqslant \\left\\textbar{} \\overrightarrow{DF} \\right\\textbar\\textless{}1$$． 故选$$\\text{A}$$． " ]

[ "1997年全国高中数学联赛竞赛一试第3题6分" ]

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

[ "竞赛->知识点->数论模块->不定方程->因式分解与恒等变形" ]

[ "设首项为$$a$$，公差为$$d$$，项数为$$n$$，则$$na+\\frac{1}{2}n\\left( n-1 \\right)d={{97}^{2}}$$，$$n\\left[ 2a+\\left( n-1 \\right)d \\right]=2\\times 9{{7}^{2}}$$，即$$n$$为$$2\\times 9{{7}^{2}}$$的大于$$3$$的约数． ∴①$$n=9{{7}^{2}}$$，$$2a+\\left( 9{{7}^{2}}-1 \\right)d=2$$，$$d=0$$，$$a=1$$；$$d\\geqslant 1$$时$$a\\textless{}0$$．有$$1$$解； ②$$n=97$$，$$2a+96d=194$$，$$d=0$$，$$a=97$$；$$d=1$$，$$a=49$$；$$d=2$$，$$a=1$$．有$$3$$解； ③$$n=2\\times 97$$，$$n=2\\times 9{{7}^{2}}$$，无解．$$n=1$$，$$2$$时$$n\\textless{}3$$． 所以符合要求的数列共有$$1+3=4$$个． 故选$$\\text{C}$$． " ]

[ "第二十届全国希望杯高二竞赛初赛邀请赛第6题4分" ]

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

[ "竞赛->知识点->不等式->不等式的解法", "竞赛->知识点->不等式->换元技巧->三角换元" ]

[ "由$$xy$$的对称性，显然大小不能确定，即$$C$$、$$\\text{D}$$错误； 设$$x=\\tan \\alpha ,y=\\tan \\beta ,\\alpha ,\\beta \\in \\left( -\\frac{ \\pi }{2},\\frac{ \\pi }{2} \\right)$$，于是原不等式可化简为$$\\frac{1-\\sin \\alpha }{\\cos \\alpha }\\cdot \\frac{1-\\sin \\beta }{\\cos \\beta }\\leqslant 1$$，整理可得$$\\sin \\alpha +\\sin \\beta \\geqslant 1-\\cos (\\alpha +\\beta )\\geqslant 0$$，即$$2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2}\\geqslant 0$$，又$$\\cos \\frac{\\alpha -\\beta }{2}\\textgreater0$$，故$$\\sin \\frac{\\alpha +\\beta }{2}\\geqslant 0$$．另外，易知$$\\cos \\frac{\\alpha +\\beta }{2}\\textgreater0$$，从而$$\\sin (\\alpha +\\beta )=2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha +\\beta }{2}\\geqslant 0$$，于是$$x+y=\\tan \\alpha +\\tan \\beta =\\frac{\\sin (\\alpha +\\beta )}{\\cos \\alpha \\cos \\beta }\\geqslant 0$$． " ]

[ "2015年天津全国高中数学联赛竞赛初赛第12题9分" ]

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

[ "待定系数，由柯西不等式， $${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+{{\\left( a+b+c+d \\right)}^{2}}\\geqslant \\frac{{{\\left[ \\left( {{x}_{1}}+{{x}_{5}} \\right)a+\\left( {{x}_{2}}+{{x}_{5}} \\right)b+\\left( {{x}_{3}}+{{x}_{5}} \\right)c+\\left( {{x}_{4}}+{{x}_{5}} \\right)d \\right]}^{2}}}{{{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}+{{x}_{4}}^{2}+{{x}_{5}}^{2}}$$令$${{x}_{1}}+{{x}_{5}}=1$$，$${{x}_{2}}+{{x}_{5}}=2$$，$${{x}_{3}}+{{x}_{5}}=3$$，$${{x}_{4}}+{{x}_{5}}=4$$， 则$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+{{\\left( a+b+c+d \\right)}^{2}}\\geqslant \\frac{10}{{{\\left( 1-{{x}_{5}} \\right)}^{2}}+{{\\left( 2-{{x}_{5}} \\right)}^{2}}+{{\\left( 3-{{x}_{5}} \\right)}^{2}}+{{\\left( 4-{{x}_{5}} \\right)}^{2}}+{{x}_{5}}^{2}}$$ $$=\\frac{10}{5{{x}_{5}}^{2}-20{{x}_{5}}+30}=\\frac{10}{5{{\\left( {{x}_{5}}-2 \\right)}^{2}}+10}$$ 取$${{x}_{5}}=2$$，得$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+{{\\left( a+b+c+d \\right)}^{2}}\\geqslant 1$$． " ]

[ "2018年黑龙江全国高中数学联赛竞赛初赛第2题5分" ]

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

[ "课内体系->思想->转化化归思想", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->知识点->直线和圆的方程->直线与方程->直线的位置关系->判定两条直线的位置关系" ]

[ "由题意知$$x+{{a}^{2}}y+6=0$$与$$\\left( a-2 \\right)x+3ay+2a=0$$均为直线方程． 若$$A\\cap B=\\varnothing $$，则这两条直线斜率相同且两条直线不重合． 由斜率相同得$${{a}^{2}}=\\frac{3a}{a-2}$$． 从而，$${{a}^{3}}-2{{a}^{2}}-3a=0$$．① （1）当$$a=0$$时，经验证成立． （2）当$$a\\ne 0$$时，由式①得， $${{a}^{2}}-2a-3=0\\Rightarrow a=-1$$，$$3$$． 红验证$$a=-1$$满足条件； 当$$a=3$$时，两直线重合，舍去． 综上，$$a=0$$或$$-1$$． " ]

[ "2016年四川全国高中数学联赛竞赛初赛第6题5分" ]

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

[ "课内体系->知识点->圆锥曲线->椭圆->椭圆的简单几何性质->椭圆的离心率->求椭圆的离心率", "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->知识点->解三角形->余弦定理", "课内体系->知识点->解三角形->正余弦定理的综合应用->正余弦定理综合求解边角", "课内体系->素养->直观想象", "课内体系->素养->数学运算" ]

[ "设$$\\left\\textbar{} P{{F}_{1}} \\right\\textbar=x$$，$$\\left\\textbar{} P{{F}_{2}} \\right\\textbar=y$$，$$\\left\\textbar{} {{F}_{1}}{{F}_{2}} \\right\\textbar=2c$$，则$${{e}_{1}}{{e}_{2}}=\\frac{2c}{x+y}\\cdot \\frac{2c}{x-y}$$（不妨设$$x\\textgreater y$$）． 在$$\\triangle P{{F}_{1}}{{F}_{2}}$$中，由余弦定理，$$4{{c}^{2}}={{x}^{2}}+{{y}^{2}}-xy$$． 于是，$${{e}_{1}}{{e}_{2}}=\\frac{4{{c}^{2}}}{{{x}^{2}}-{{y}^{2}}}=\\frac{{{x}^{2}}+{{y}^{2}}-xy}{{{x}^{2}}-{{y}^{2}}}$$，去分母，移项得$$\\left( 1-{{e}_{1}}{{e}_{2}} \\right){{x}^{2}}-xy+\\left( 1+{{e}_{1}}{{e}_{2}} \\right){{y}^{2}}=0$$， 判别式$$\\Delta =1-4\\left( 1-{{e}_{1}}{{e}_{2}} \\right)\\left( 1+{{e}_{1}}{{e}_{2}} \\right)\\geqslant 0$$，解得$${{e}_{1}}{{e}_{2}}\\geqslant \\frac{\\sqrt{3}}{2}$$． 故选$$\\text{B}$$． " ]

[ "2009年AMC10竞赛B第14题" ]

[ [ { "aoVal": "A", "content": "Tuesday " } ], [ { "aoVal": "B", "content": "Wednesday " } ], [ { "aoVal": "C", "content": "Thursday " } ], [ { "aoVal": "D", "content": "Friday " } ], [ { "aoVal": "E", "content": "Saturday " } ] ]

[ "美国AMC10/12->Knowledge Point->Algebra->Sequence->Sum of the First n Terms for Geometric Sequences", "课内体系->知识点->数列->数列的实际应用" ]

[ "On Monday, day $$1$$, the birds find $$\\frac14$$ quart of millet in the feeder. On Tuesday they find $$\\frac{1}{4}+ \\frac{3}{4}\\cdot \\frac{1}{4}$$ quarts of millet. On Wednesday, day $$3$$, they find $$\\frac{1}{4}+ \\frac{3}{4}\\cdot \\frac{1}{4}+\\left(\\frac{3}{4}\\right)^{2}\\cdot \\frac{1}{4}$$ quarts of millet. The number of quarts of millet they find on day $$n$$ is $$\\frac{1}{4}+ \\frac{3}{4}\\cdot \\frac{1}{4}+\\left(\\frac{3}{4}\\right)^{2}\\cdot \\frac{1}{4}+ \\cdots +\\left(\\dfrac{3}{4}\\right)^{n-1}\\cdot \\frac{1}{4}= \\dfrac{\\left(\\dfrac{1}{4}\\right)\\left(1-\\left(\\dfrac{3}{4}\\right)^{n}\\right)}{1- \\dfrac{3}{4}}=1-\\left(\\frac{3}{4}\\right)^{n}$$. The birds always find $$\\frac34$$ quart of other seeds, so more than half the seeds are millet if $$1-\\left(\\frac{3}{4}\\right)^{n}\\textgreater\\frac34$$ , that is, when $$\\left(\\frac{3}{4}\\right)^{n}\\textless{} \\frac{1}{4}$$. Because $$\\left(\\frac{3}{4}\\right)^{4}= \\frac{81}{256}\\textgreater{} \\frac{1}{4}$$ and $$\\left(\\frac{3}{4}\\right)^{5}= \\frac{243}{1024}\\textless{} \\frac{1}{4}$$, this will first occur on day $$5$$ which is $$\\boxed{\\rm Friday}$$. The answer is $$\\rm(D)$$. " ]

[ "2013年辽宁全国高中数学联赛竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$2\\textless{}m\\textless{}4$$ " } ], [ { "aoVal": "B", "content": "$$m\\textless{}2$$或$$m\\textgreater4$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{2}\\textless{}m\\textless{}4$$ " } ], [ { "aoVal": "D", "content": "$$m\\textless{}-\\frac{1}{2}$$或$$m\\textgreater4$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "$$A=\\left { x\\textbar-2{\\leqslant }x{\\leqslant }5 \\right }$$，$$A{\\cap }B=\\varnothing $$，当$$B=\\varnothing $$时，$$2m-1\\textless{}m+1$$，即$$m\\textless{}2$$;当$$B\\ne \\varnothing $$时，$$m{\\geqslant }2$$，$$m+1\\textgreater5$$，即$$m\\textgreater4$$． " ]

[ "2008年贵州全国高中数学联赛竞赛初赛第5题5分" ]

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

[ "竞赛->知识点->复数与平面向量->平面向量的概念与运算" ]

[ "因为$$\\overrightarrow{a}=\\left( 1,2 \\right)$$，$$\\overrightarrow{b}=\\left( x,1 \\right)$$， 所以有$$(\\overrightarrow{a}+2\\overrightarrow{b})=\\left( 1+2x,4 \\right)$$， $$(2\\overrightarrow{a}-2\\overrightarrow{b})=\\left( 2-2x,2 \\right)$$， 因为$$(\\overrightarrow{a}+2\\overrightarrow{b})//(2\\overrightarrow{a}-2\\overrightarrow{b})$$， 所以$$\\left( 1+2x \\right)\\times 2=4\\times \\left( 2-2x \\right)$$， 解得$$x=\\frac{1}{2}$$，故选$$\\text{C}$$. " ]

[ "2009年第二十届全国希望杯高二竞赛复赛第1题5分" ]

[ [ { "aoVal": "A", "content": "奇函数 " } ], [ { "aoVal": "B", "content": "偶函数 " } ], [ { "aoVal": "C", "content": "奇函数也是偶函数 " } ], [ { "aoVal": "D", "content": "非奇非偶函数 " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "令$$a=b=0$$，可得$$f(0)=0$$，令$$a=-b$$，可得$$f(a)=-f(-a)$$． " ]

[ "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7题" ]

[ [ { "aoVal": "A", "content": "$\\left( -\\infty ,\\frac{1}{2} \\right)\\cup \\left( 2,+\\infty \\right)$ " } ], [ { "aoVal": "B", "content": "$\\left[ 2,6 \\right)$ " } ], [ { "aoVal": "C", "content": "$\\left( 0,\\frac{1}{2} \\right]\\cup \\left[ 2,6 \\right)$ " } ], [ { "aoVal": "D", "content": "$\\left( 0,6 \\right]$ " } ] ]

[ "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 【分析】\\\\ 根据$f\\left( x \\right)$的单调性，以及定义域，结合一元二次不等式的求解，直接计算即可.\\\\ 【详解】\\\\ 对$f\\left( x \\right)=\\sqrt{x-2}$，且定义域为$\\left[ 2,+\\infty \\right)$，由复合函数单调性可知其在定义域单调递增，\\\\ 故$f\\left( 2{{a}^{2}}-5a+4 \\right)\\textless{} f\\left( {{a}^{2}}+a+4 \\right)$，等价于$2\\le 2{{a}^{2}}-5a+4\\textless{} {{a}^{2}}+a+4$，\\\\ 由$2\\le 2{{a}^{2}}-5a+4$，即$2{{a}^{2}}-5a+2\\ge 0$，$\\left( 2a-1 \\right)\\left( a-2 \\right)\\ge 0$，解得$a\\in \\left( -\\infty ,\\frac{1}{2} \\right]\\cup \\left[ 2,+\\infty \\right)$；\\\\ 由$2{{a}^{2}}-5a+4\\textless{} {{a}^{2}}+a+4$，即${{a}^{2}}-6a\\textless{} 0$，解得$a\\in \\left( 0,6 \\right)$；\\\\ 故实数$a$的取值范围为$\\left( 0,\\frac{1}{2} \\right]\\cup \\left[ 2,6 \\right)$.\\\\ 故选：C. " ]

[ "2005年AMC12竞赛B第23题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{15}{2}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{29}{2}$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{39}{2}$$ " } ], [ { "aoVal": "E", "content": "$$24$$ " } ] ]

[ "美国AMC10/12->Knowledge Point->Algebra->Calculation->Exponentiation", "课内体系->知识点->基本初等函数->实数指数幂运算" ]

[ "这题主要考代数变形. 由题意, $x+y=10^{}z$, $x^{2}+y^{2}=10^{z+1}$. 由此可得$xy=\\frac{(x+y)^{2}-(x^{2}+y^{2})}{2}=\\frac{1}{2}10^{2z}-5\\times 10^{}z$. 代入$x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$, 得 $x^{3}+y^{3}=10^{}z(10\\times 10^{}z-\\frac{1}{2}10^{2z}+5\\times 10^{}z)=-\\frac{1}{2}10^{3z}+15\\times 10^{2z}$. 故$a+b=15-\\frac{1}{2}=\\frac{29}{2}$, 选$$\\text{B}$$. " ]

[ "2000年AMC12竞赛第13题" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "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->Linear Function Word Problems" ]

[ "Let $$c$$ be the total amount of coffee, $$m$$ of milk, and $$p$$ the number of people in the family. Then each person drinks the same total amount of coffee and milk ($$8$$ ounces), so $$\\left(\\frac{c}{6}+\\frac{m}{4} \\right)p=c+m$$. Regrouping, we get $$2c(6-p)=3m(p-4)$$. Since both $$c$$, $$m$$ are positive, it follows that $$6-p$$ and $$p-4$$ are also positive, which is only possible when $$p=5$$ ($$\\text{C}$$). One could notice that (since there are only two components to the mixture) Angela must have more than her \"fair share\" of milk and less then her \"fair share\" of coffee in order to ensure that everyone has $$8$$ ounces. The \"fair share\" is $$\\frac{1}{p}$$. So, $$\\frac{1}{6}\\textless\\frac{1}{p}\\textless\\frac{1}{4}$$ Which requires that $$p$$ be $$p=5$$ ($$\\text{C}$$), since $$p$$ is a whole number. Again, let $$c$$, $$m$$, sand $$p$$ be the total amount of coffee, total amount of milk, and number of people in the family, respectively. Then the total amount that is drunk is $$8p$$, and also $$c+m$$. Thus, $$c+m=8p$$, so $$m=8p-c$$ and $$c=8p-m$$. We also know that the amount Angela drank, which is is $$\\frac{c}{6}+\\frac{m}{4}$$, equal to $$8$$ ounces, thus $$\\frac{c}{6}+\\frac{m}{4}=8$$. Rearranging gives $$24p-c=96$$. Now notice that $$c\\textgreater0$$ (by the problem statement). In addition, $$m\\textgreater0$$ so $$c=8p-m\\textless8p$$. Therefore, $$0\\textless c\\textless8p$$ ,and so We know that so $$24p\\textgreater24p-c\\textgreater16p$$. We know that $$24p-c=96$$, so $$24p\\textgreater96\\textgreater16p$$. From the leftmost inequality, we get $$p\\textgreater4$$, and from the rightmost inequality, we get $$p\\textless6$$. The only possible value of $$p$$ is $$p=5$$ ($$\\text{C}$$). Let $$c$$, $$m$$, and $$p$$ be the total amount of coffee, total amount of milk, and number of people in the family, respectively. $$c$$ and $$m$$ obviously can\\textquotesingle t be $$0$$. We know $$\\frac{c}{6}+\\frac{m}{4}=8$$ or $$3c+2m=96$$ and $$c+m=8p$$ or $$2c+2m=16p$$. Then, $$(2c+2m)+c=16p+c=96$$. Because $$16p$$ and $$96$$ are both divisible by $$16$$, $$c$$ must also be divisible by $$16$$. Let $$c=16k$$. Now, $$3(16k)+2m=2m=48k+2m=96k$$ can\\textquotesingle t be $$0$$, otherwise $$c$$ is $$0$$, and $$k$$ can\\textquotesingle t be $$2$$, otherwise $$m$$ is $$0$$. Therefore $$k$$ must be $$1$$, $$c=16$$ and $$m=24$$. $$c+m=16+24=40=8p$$. Therefore, $$p=5$$ ($$\\text{C}$$). Let $$m$$,$$c$$ be the total amounts of milk and coffee, respectively. In order to know the number of people, we first need to find the total amount of mixture $$t=m+c$$. We are given that $$\\frac{m}{4}+\\frac{c}{6}=8$$. Multiplying the equation by $$4$$ to get $$m+\\frac{2}{3}x=(m+c)-\\frac{1}{3}c=t-\\frac{1}{3}c=32$$. Since $$\\frac{1}{3}m\\textgreater0$$, we have $$t\\textgreater32$$.~ Now multiplying the equation by $$6$$ to get $$\\frac{3}{2}m+c=(m+c)+\\frac{1}{3}m=t+\\frac{1}{3}m=48$$. Since, $$\\frac{1}{3}m\\textgreater0$$, we have $$t\\textless48$$. Thus, $$32\\textless t\\textless48$$. Since $$t$$ is a multiple of $$8$$, the only possible value for in that range is $$40$$. Therefore, there are $$\\frac{40}{8}=5$$ people in Angela\\textquotesingle s family. ($$\\text{C}$$). " ]

[ "2012年浙江全国高中数学联赛竞赛初赛第8题5分" ]

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

[ "竞赛->知识点->复数与平面向量->平面向量的概念与运算" ]

[ "满足$$\\left\\textbar{} \\overrightarrow{a}-\\overrightarrow{i} \\right\\textbar+\\left\\textbar{} \\overrightarrow{a}-\\overrightarrow{2j} \\right\\textbar=\\sqrt{5}$$的向量$$\\overrightarrow{a}$$的终点在线段$$2x+y-2=0\\left( 0\\leqslant x\\leqslant 1 \\right)$$上，所以$$\\left\\textbar{} \\overrightarrow{a}+\\overrightarrow{2i} \\right\\textbar$$的最大值就是点$$\\left( -2, 0 \\right)$$与$$\\left( 1, 0 \\right)$$距离$$3$$，最小值就是点$$\\left( -2, 0 \\right)$$到线段的距离$$\\frac{6\\sqrt{5}}{5}$$． " ]

[ "1989年全国高中数学联赛竞赛一试第4题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

[ "竞赛->知识点->排列组合与概率->排列与组合" ]

[ "以长方体的$$8$$个顶点中的任意$$3$$个为顶点的三角形共有$$\\text{C}_{8}^{3}$$个，且这些三角形只能是直角三角形或锐角三角形．而以长方体的每个顶点为直角顶点的三角形都是$$6$$个，所以在$$\\text{C}_{8}^{3}$$个三角形中共有$$6\\times 8=48$$（个）直角三角形．从而所求锐角三角形的个数为$$\\text{C}_{8}^{3}-48=8$$． 故选$$\\text{C}$$． " ]

[ "2009年高二竞赛广州市第2题6分" ]

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

[ "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->同角三角函数的基本关系式的化简和求值", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->三角函数的定义->计算任意角的三角函数值", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式", "课内体系->知识点->三角函数->三角函数的概念->任意角与弧度制->弧度制", "课内体系->知识点->三角函数->三角函数的概念->任意角与弧度制->角的定义和分类->任意角的表示", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->和差角公式化简求值综合运用", "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", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->两角和与差的余弦", "课内体系->方法->构造法", "课内体系->素养->数学运算", "课内体系->思想->转化化归思想" ]

[ "$$3=\\frac{\\sin (\\alpha +\\beta +\\beta )}{\\sin (\\alpha +\\beta -\\beta )}=\\frac{\\sin (\\alpha +\\beta )\\cos \\beta +\\cos (\\alpha +\\beta )\\sin \\beta }{\\sin (\\alpha +\\beta )\\cos \\beta -\\cos (\\alpha +\\beta )\\sin \\beta }$$， 分子分母同时除以$$\\cos \\beta \\cos (\\alpha +\\beta )$$， 有$$\\frac{\\tan (\\alpha +\\beta )+\\tan \\beta }{\\tan (\\alpha +\\beta )-\\tan \\beta }=3$$，解得$$\\tan (\\alpha +\\beta )=2\\tan \\beta $$． " ]

[ "2004年全国高中数学联赛竞赛一试第1题6分" ]

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

[ "竞赛->知识点->多项式与方程->解方程（组）" ]

[ "由方程有重根，故$$\\frac{1}{4}\\Delta =\\text{4co}{{\\text{s}}^{2}}\\theta -\\cot \\theta =0$$， ∵$$0\\textless{}\\theta \\textless{}\\frac{ \\pi }{2}\\Rightarrow 2\\sin 2\\theta =1$$，$$\\Rightarrow \\theta =\\frac{ \\pi }{12}$$或$$\\frac{5 \\pi }{12}$$． 故选$$\\text{B}$$． " ]

[ "2017年天津全国高中数学联赛竞赛初赛第1题5分" ]

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

[ "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" ]

[ "正四棱锥的高为$$\\sqrt{{{2000}^{2}}-\\frac{{{2017}^{2}}}{2}}$$， 所以，侧棱与底面所成的角的正弦值为$$\\frac{\\sqrt{{{2000}^{2}}-\\frac{{{2017}^{2}}}{2}}}{2000}\\approx \\frac{1400}{2000}=0.7$$，比$$\\frac{\\sqrt{2}}{2}$$稍微小点，也就是角度比$$45{}^{}\\circ $$小点，因此是$$40{}^{}\\circ $$． " ]

[ "2009年山东全国高中数学联赛竞赛初赛第1题6分" ]

[ [ { "aoVal": "A", "content": "$$ {x\\textbar1\\textless{}x\\textless{}3 }$$ " } ], [ { "aoVal": "B", "content": "$$ {x\\textbar1\\textless{}x\\textless{}2 }$$ " } ], [ { "aoVal": "C", "content": "$$ {x\\textbar0\\textless{}x\\textless{}3 }$$ " } ], [ { "aoVal": "D", "content": "$$ {x\\textbar0\\textless{}x\\textless{}2 }$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "$$x\\in M\\Leftrightarrow {{\\log }_{\\frac{1}{2}}}(x-1)\\textgreater-1\\Leftrightarrow \\begin{cases}x-1\\textgreater0 x-1\\textless{}{{\\left( \\frac{1}{2} \\right)}^{-1}}=2 \\end{cases}\\Leftrightarrow 1\\textless{}x\\textless{}3$$；$$x\\in N\\Leftrightarrow 1\\textless{}{{2}^{x}}\\textless{}4\\Leftrightarrow 0\\textless{}x\\textless{}2$$，所以$$M\\cap N= {x\\textbar1\\textless{}x\\textless{}2 }$$．故选$$\\text{B}$$． " ]

[ "2008年甘肃全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "课内体系->知识点->函数的应用->函数的零点->函数零点的概念", "课内体系->知识点->函数的应用->函数的零点->零点的个数问题->利用函数图象研究方程根的分布问题（图象与零点综合）", "课内体系->知识点->函数的应用->函数的零点->零点的个数问题->零点、交点、根的等价转化", "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->素养->数学抽象", "课内体系->素养->直观想象" ]

[ "由正弦函数的性质同时绘出$$y=10\\sin \\left( x+\\frac{ \\pi }{6} \\right)$$和$$y=x$$的图象， 注意到$$3 \\pi -\\frac{ \\pi }{6}\\textless{}10\\textless{}3 \\pi +\\frac{ \\pi }{2}$$， 及$$\\left\\textbar{} -\\frac{ \\pi }{6}-3 \\pi ~\\right\\textbar\\textless{}10\\textless{}\\left\\textbar{} -\\frac{ \\pi }{6}-3 \\pi -\\frac{ \\pi }{2} \\right\\textbar$$， 可知两条曲线有$$7$$个交点．故选$$\\text{B}$$． " ]

[ "2021年吉林全国高中数学联赛竞赛初赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$$\\left( \\frac{1-\\sqrt{17}}{2},\\frac{1+\\sqrt{17}}{2} \\right)$$ " } ], [ { "aoVal": "B", "content": "$$(0,1)$$ " } ], [ { "aoVal": "C", "content": "$$\\left[ \\frac{1-\\sqrt{17}}{2},\\frac{1+\\sqrt{17}}{2} \\right]$$ " } ], [ { "aoVal": "D", "content": "$$[0,1]$$ " } ] ]

[ "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->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->方法->单调性法" ]

[ "函数$$f(x)$$在$$\\mathbf{R}$$上递减，且$$4f(x)=f(2x)$$， 所以$$f(x+t)\\textless{}4f(x)\\Leftrightarrow f(x+t)\\textless{}f(2x)\\Leftrightarrow x+t\\textgreater2x\\Leftrightarrow t\\textgreater x$$， 所以$$t\\geqslant {{t}^{2}}$$，解得$$0\\leqslant t\\leqslant 1$$． 故选$$\\text{D}$$． " ]

[ "2019~2020学年3月湖北武汉洪山区武汉市洪山高级中学（武汉外国语学校光谷分校）高二下学期月考第10题", "2008年全国高中数学联赛竞赛一试第3题6分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{2}{9}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{20}{81}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{7}{27}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{16}{81}$$ " } ] ]

[ "竞赛->知识点->排列组合与概率->概率初步", "课内体系->知识点->统计与概率->概率->事件与概率->事件的独立性->相互独立事件的概率乘法公式", "课内体系->知识点->统计与概率->概率->事件与概率->互斥事件->互斥事件的概率加法公式" ]

[ "解法一：设比赛结束时已打局数为随机变量$$\\xi $$ 依题意知，$$\\xi $$的所有可能值为$$2$$，$$4$$，$$6$$， 设每两局比赛为一轮，则该轮结束时比赛停止的概率为$${{\\left( \\frac{2}{3} \\right)}^{2}}+{{\\left( \\frac{1}{3} \\right)}^{2}}=\\frac{5}{9}$$， 若该轮结束时比赛还将继续，则甲、乙在该轮中必是各得一分，此时，该轮比赛结果对下轮比赛是否停止没有影响，从而有 $$P(\\xi =2)=\\frac{5}{9}$$， $$P(\\xi =4)=\\left( \\frac{4}{9} \\right)\\left( \\frac{5}{9} \\right)=\\frac{20}{81}$$， $$P(\\xi =6)={{\\left( \\frac{4}{9} \\right)}^{2}}=\\frac{16}{81}$$， 故选$$\\text{B}$$ 解法二： 依题意知，$$\\xi $$的所有可能值为$$2$$，$$4$$，$$6$$， 令$${{A}_{k}}$$表示甲在第$$k$$局比赛中获胜，则$${{\\bar{A}}_{k}}$$表示乙在第$$k$$局比赛中获胜， 由独立性与互不相容性得 $$P(\\xi =2)=P({{A}_{1}}{{A}_{2}})+P({{\\bar{A}}_{1}}{{\\bar{A}}_{2}})=\\frac{5}{9}$$， $$P(\\xi =4)=P({{A}_{1}}{{\\bar{A}}_{2}}{{A}_{3}}{{A}_{4}})+P({{A}_{1}}{{\\bar{A}}_{2}}{{\\bar{A}}_{3}}{{\\bar{A}}_{4}})+P({{\\bar{A}}_{1}}{{A}_{2}}{{A}_{3}}{{A}_{4}})+P({{\\bar{A}}_{1}}{{A}_{2}}{{\\bar{A}}_{3}}{{\\bar{A}}_{4}})$$ $$=2\\left[ {{\\left( \\frac{2}{3} \\right)}^{3}}\\left( \\frac{1}{3} \\right)+{{\\left( \\frac{1}{3} \\right)}^{3}}\\left( \\frac{2}{3} \\right) \\right]=\\frac{20}{81}$$， $$P(\\xi =6)=P({{A}_{1}}{{\\bar{A}}_{2}}{{A}_{3}}{{\\bar{A}}_{4}})+P({{A}_{1}}{{\\bar{A}}_{2}}{{\\bar{A}}_{3}}{{A}_{4}})+P({{\\bar{A}}_{1}}{{A}_{2}}{{A}_{3}}{{\\bar{A}}_{4}})+P({{\\bar{A}}_{1}}{{A}_{2}}{{\\bar{A}}_{3}}{{A}_{4}})$$ $$=4{{\\left( \\frac{2}{3} \\right)}^{2}}{{\\left( \\frac{1}{3} \\right)}^{2}}=\\frac{16}{81}$$， 故选$$\\text{B}$$． " ]

[ "2018~2019学年四川成都青羊区成都市树德中学（光华校区）高一下学期期中理科第6题5分", "2008年全国高中数学联赛竞赛一试第1题6分" ]

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

[ "竞赛->知识点->函数->函数的概念", "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->不等式->几个重要的不等式->均值" ]

[ "当$$x\\textless{}2$$时，$$2-x\\textgreater0$$，因此$$f(x)=\\frac{1+(4-4x+{{x}^{2}})}{2-x}=\\frac{1}{2-x}+(2-x)\\geqslant 2\\cdot \\sqrt{\\frac{1}{2-x}\\cdot (2-x)}=2$$， 当且仅当$$\\frac{1}{2-x}=2-x$$时上式取等号， 而此方程有解$$x=1\\in (-\\infty ,2)$$，因此$$f(x)$$在$$(-\\infty ,2)$$上的最小值为$$2$$． 故选$$\\text{C}$$． " ]

[ "竞赛第13题" ]

[ [ { "aoVal": "A", "content": "有两颗骰子之和为7的概率是$\\frac{7}{12}$ " } ], [ { "aoVal": "B", "content": "有两颗骰子之和为8的概率是$\\frac{19}{54}$ " } ], [ { "aoVal": "C", "content": "所有骰子中最小值为2的概率是$\\frac{65}{216}$ " } ], [ { "aoVal": "D", "content": "所有骰子中最小值为3的概率是$\\frac{41}{216}$ " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 利用古典概率的概率公式可得各选项中的概率.\\\\ 【详解】\\\\ 有两颗骰子之和为7的可能为$1+6,2+5,3+4$，对应的基本事件数为$(4\\times 3!+2\\times 3)\\times 3=90$，\\\\ 因此所求概率为$\\frac{90}{6^{3}}=\\frac{5}{12}$．\\\\ 有两颗骰子之和为8的可能为$2+6,3+5,4+4$，对应的基本事件数为$(4\\times 3!+2\\times 3)\\times 2+(5\\times 3+1)=76$，\\\\ 因此所求概率为$\\frac{76}{6^{3}}=\\frac{19}{54}$．\\\\ 所有骰子中最小值为2，则按2出现次数分类，对应的基本事件数为$(6\\times 3!+4\\times 3)+4\\times 3+1=61$，\\\\ 因此所求概率为$\\frac{61}{6^{3}}=\\frac{61}{216}$．\\\\ 所有骰子中最小值为3，则按3出现次数分类，对应的基本事件数为$(3\\times 3!+3\\times 3)+3\\times 3+1=37$，\\\\ 因此所求概率为$\\frac{37}{6^{3}}=\\frac{37}{216}$．\\\\ 故选：B " ]

[ "2002年AMC12竞赛A第10题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{4}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{8}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2}{5}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{1}{2}$$ " } ] ]

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

[ "她将第一杯咖啡的一半倒入第二杯，充分搅拌后，第二杯中有$$2$$盎司咖啡，$$4$$盎司奶油。将第二杯中的一半液体倒回第一杯，第一杯中的奶油有$$2$$盎司。 第一杯里的液体有$$5$$盎司，有2/5是奶油 " ]

[ "1989年全国高中数学联赛竞赛一试第10题" ]

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

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "设该数为$$x$$，则其整数部分为$$[x]$$，小数部分为$$x-[x]$$，由已知得 $$x(x-[x])={{[x]}^{2}}$$， 其中$$[x]\\textgreater0$$，$$0\\textless{}x-[x]\\textless{}1$$，解得 $$x=\\frac{1+\\sqrt{5}}{2}[x]$$． 由$$0\\textless{}x-[x]\\textless{}1$$知， $$0\\textless{}\\frac{\\sqrt{5}-1}{2}[x]\\textless{}1$$，$$0\\textless{}[x]\\textless{}\\frac{1+\\sqrt{5}}{2}\\textless{}2$$， 即$$[x]=1$$，$$x=\\frac{1+\\sqrt{5}}{2}$$． 故答案为：$$\\frac{1+\\sqrt{5}}{2}$$． " ]

[ "全国全国高中数学联赛竞赛一试" ]

[ [ { "aoVal": "A", "content": "甲是乙的充分非必要条件 " } ], [ { "aoVal": "B", "content": "甲是乙的必要非充分条件 " } ], [ { "aoVal": "C", "content": "甲是乙的充分必要条件 " } ], [ { "aoVal": "D", "content": "甲是乙的非充分非必要条件 " } ] ]

[ "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$$0\\textless{}\\theta {\\leqslant }\\frac{1}{2} \\pi $$） " ]

[ "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", "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->集合的概念" ]

[ "易知，$$B=\\left { -2,-3 \\right }$$， 则集合$$B$$中所有元素的积为$$6$$． 故选$$\\text{C}$$． " ]

[ "1999年全国全国高中数学联赛竞赛一试第6题6分" ]

[ [ { "aoVal": "A", "content": "锐角三角形 " } ], [ { "aoVal": "B", "content": "钝角三角形 " } ], [ { "aoVal": "C", "content": "直角三角形 " } ], [ { "aoVal": "D", "content": "不确定 " } ] ]

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

[ "设$$B\\left( {{t}^{2}},2t \\right)$$，$$C\\left( {{s}^{2}},2s \\right)$$，$$s\\ne t$$，$$s\\ne 1$$，$$t\\ne 1$$， 则直线$$BC$$的方程为$$y-2s=\\frac{2s-2t}{{{s}^{2}}-{{t}^{2}}}\\left( x-{{s}^{2}} \\right)$$， 化简得$$2x-\\left( s+t \\right)y+2st=0$$． 由于直线$$BC$$过点$$\\left( 5,-2 \\right)$$， 故$$2\\times 5-\\left( s+t \\right)\\left( -2 \\right)+2st=0$$， 即$$\\left( s+1 \\right)\\left( t+1 \\right)=-4$$． 因此，$${{k}_{AB}}{{k}_{AC}}=\\frac{4}{\\left( t+1 \\right)\\left( s+1 \\right)}=-1$$， 所以，$$\\angle BAC=90{}^{}\\circ $$， 从而$$\\triangle ABC$$是直角三角形． " ]

[ "2008年浙江全国高中数学联赛竞赛初赛第3题6分", "2019~2020学年辽宁高一上学期期中（辽南协作体）第11题5分" ]

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

[ "课内体系->知识点->等式与不等式->不等式->不等式的性质", "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->函数的概念" ]

[ "函数$$f(x-a)+f(x+a)$$的定义域为$$[a,1+a]\\cap [-a,1-a]$$． 当$$a\\geqslant 0$$时，应有$$a\\leqslant 1-a$$，即$$a\\leqslant \\frac{1}{2}$$； 当$$a\\leqslant 0$$时，应有$$-a\\leqslant 1+a$$，即$$a\\geqslant -\\frac{1}{2}$$．故选$$\\text{B}$$． " ]

[ "2009年AMC10竞赛A第20题" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$55$$ " } ], [ { "aoVal": "D", "content": "$$65$$ " } ], [ { "aoVal": "E", "content": "$$80$$ " } ] ]

[ "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 their speeds in kilometers per hour be $$v_A$$ and $$v_L$$. We know that $$vA=3vL$$ and that $$v_A+v_L=60$$. (The second equation follows from the fact that $$1\\rm km/min=60\\rm km/h)$$. This solves to $$v_A=45$$ and $$v_L=15$$. As the distance decreases at a rate of $$1$$ kilometer per minute, after $$5$$ minutes the distance between them will be $$20-5=15$$ kilometers. From this point on, only~ Lauren will be riding her bike. As there are $$15$$ kilometers remaining and $$vL=15$$, she will need exactly an hour to get to Andrea. Therefore the total time in minutes is $$5+60= 65$$. Because the speed of Andrea is $$3$$ times as fast as Lauren and the distance between them is decreasing at a rate of $$1$$ kilometer per minute, Andrea\\textquotesingle s speed is $$\\dfrac{3}{4}\\rm km/min$$,and Lauen\\textquotesingle s $$\\dfrac{1}{4}\\rm km/min$$. Therefore, after $$5$$ minutes, Andrea will have biked $$\\dfrac{3}{4}\\cdot 5= \\dfrac{15}{4}\\rm km$$. In all, Lauren will have to bike $$20- \\dfrac{15}{4}= \\dfrac{80}{4}- \\dfrac{15}{4}= \\dfrac{65}{4}\\rm km$$. Because her speed is$$\\dfrac{1}{4}\\rm km/\\min$$, the time elapsed will be$$\\dfrac{65}{\\dfrac{4}{\\dfrac{1}{4}}}=\\left( \\rm D\\right)65$$. " ]

[ "2008年四川全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$2\\sqrt{2}$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$2\\sqrt{3}$$ " } ], [ { "aoVal": "D", "content": "$$\\sqrt{13}$$ " } ] ]

[ "竞赛->知识点->解析几何->椭圆" ]

[ "由条件知$$a=2$$，$$b=\\sqrt{3}$$，则$$c=1$$， 故$${{A}_{1}}(-2,0)$$、$${{F}_{2}}\\left( 1,0 \\right)$$．设$$P(xy)$$， 则$$\\overrightarrow{P{{A}_{1}}}=(-2-x,-y)$$，$$\\overrightarrow{P{{F}_{2}}}=(1-x,-y)$$． 于是$$\\overrightarrow{P{{A}_{1}}}\\cdot \\overrightarrow{P{{F}_{2}}}=(x+2)(x-1)+{{y}^{2}}$$ $$={{x}^{2}}+x-2+3\\left( 1-\\frac{{{x}^{2}}}{4} \\right)$$ $$=\\frac{1}{4}{{x}^{2}}+x+1=\\frac{1}{4}{{(x+2)}^{2}}\\geqslant 0$$， 当$$x=-2$$时等号成立． 即当$$P$$与$${{A}_{1}}$$重合时，$$\\overrightarrow{P{{A}_{1}}}\\cdot \\overrightarrow{P{{F}_{2}}}$$取最小值， 此时$$\\textbar\\overrightarrow{P{{A}_{1}}}+\\overrightarrow{P{{F}_{2}}}\\textbar=\\textbar\\overrightarrow{{{A}_{1}}{{F}_{2}}}\\textbar=3$$． 故选$$\\text{B}$$． " ]

[ "2014年吉林全国高中数学联赛竞赛初赛第3题6分" ]

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

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "设$$f\\left( x \\right)=\\ln \\frac{2-x}{2+x}$$， 则$$f\\left( -x \\right)=\\ln \\frac{2+x}{2-x}=-\\ln \\frac{2-x}{2+x}=-f\\left( x \\right)$$， 因此，$$f\\left( x \\right)=\\ln \\frac{2-x}{2+x}$$是奇函数． 又$$t=\\frac{2-x}{2+x}=-1+\\frac{4}{2+x}$$为区间$$\\left[ -1, 1 \\right]$$上的单调递减函数，$$y=\\ln t$$为区间$$\\left( 0, +\\infty \\right)$$上的单调递增函数，而$$f\\left( x \\right)=\\ln \\frac{2-x}{2+x}$$为$$y=\\ln t$$与$$t=\\frac{2-x}{2+x}$$的复合函数，因此函数$$f\\left( x \\right)=\\ln \\frac{2-x}{2+x}$$在区间$$\\left[ -1, 1 \\right]$$上单调递减． " ]

[ "2008年浙江全国高中数学联赛竞赛初赛第6题6分" ]

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

[ "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->截面的探索问题", "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->圆柱、圆锥、圆台和球的结构特征", "课内体系->素养->数学运算", "课内体系->素养->直观想象" ]

[ "略． " ]

[ "2008年福建全国高中数学联赛竞赛初赛第2题6分" ]

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

[ "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" ]

[ "设$$ABC-{{A}_{1}}{{B}_{1}}{{C}_{1}}$$是正三棱柱，侧棱的长为$$a$$， 侧面对角线$$A{{B}_{1}}$$与$$B{{C}_{1}}$$互相垂直．则$$\\overrightarrow{{{B}_{1}}A}\\cdot \\overrightarrow{B{{C}_{1}}}=0$$， 即$$(\\overrightarrow{{{B}_{1}}B}+\\overrightarrow{BA})\\cdot (\\overrightarrow{B{{B}_{1}}}+\\overrightarrow{{{B}_{1}}{{C}_{1}}})=0$$， 所以$$\\overrightarrow{{{B}_{1}}B}\\cdot \\overrightarrow{B{{B}_{1}}}+\\overrightarrow{{{B}_{1}}B}\\cdot \\overrightarrow{{{B}_{1}}{{C}_{1}}}+\\overrightarrow{BA}\\cdot \\overrightarrow{B{{B}_{1}}}+\\overrightarrow{BA}\\cdot \\overrightarrow{{{B}_{1}}{{C}_{1}}}=0$$， 即$$-{{a}^{2}}+0+0+\\cos 60{}^{}\\circ =0$$． 可得$$a=\\frac{\\sqrt{2}}{2}$$．故选$$\\text{B}$$． " ]

[ "2008年山东全国高中数学联赛竞赛初赛第9题6分" ]

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

[ "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->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->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->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->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->数量积->向量的数量积的定义" ]

[ "$$a=\\left( 1,-1 \\right)$$，$$b=\\left( -1,2 \\right)$$，$$\\therefore \\left( 2a+b \\right)\\cdot a=\\left( 1,0 \\right)\\cdot \\left( 1,-1 \\right)=1$$． " ]

[ "2020年北京海淀区北京大学自主招生（强基计划）第2题5分", "2020~2021学年北京高三单元测试", "2020~2021学年北京高二单元测试", "竞赛" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$31$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "前三个答案都不对 " } ] ]

[ "竞赛->知识点->数论模块->整除->带余除法（辗转相除法）" ]

[ "将各正因数的幂指数记为数组$$\\left( {{\\alpha }_{1i}},{{\\alpha }_{2i}},{{\\alpha }_{3i}},{{\\alpha }_{4i}},{{\\alpha }_{5i}} \\right)$$ 即任取正因数$${{a}_{i}}$$，有$${{a}_{i}}={{2}^{{{\\alpha }_{1i}}}}\\times {{3}^{{{\\alpha }_{2i}}}}\\times {{5}^{{{\\alpha }_{3i}}}}\\times {{101}^{{{\\alpha }_{4i}}}}\\times {{673}^{{{\\alpha }_{5i}}}}$$（质数分解唯一定理）， 只需考虑幂指数的奇偶性，故设$${{b}_{i}}=\\left( {{\\beta }_{1i}},{{\\beta }_{2i}},{{\\beta }_{3i}},{{\\beta }_{4i}},{{\\beta }_{5i}} \\right)$$，其中$$\\beta \\in \\left { 0,1 \\right }$$且$${{\\beta }_{ji}}\\equiv {{\\alpha }_{ji}}\\left( \\bmod 2 \\right)$$， 若两正因数$${{a}_{i}}$$，$${{a}_{j}}$$为完全平方数， $$\\because {{a}_{i}}\\times {{a}_{j}}={{2}^{{{\\alpha }_{1i}}+\\alpha 1j}}\\times {{3}^{{{\\alpha }_{2i}}\\times {{\\alpha }_{2j}}}}\\times {{5}^{{{\\alpha }_{3i}}+{{\\alpha }_{3j}}}}\\times {{101}^{{{\\alpha }_{4i}}+{{\\alpha }_{4j}}}}\\times {{673}^{{{\\alpha }_{3i}}+{{d}_{5j}}}}$$， $$\\therefore $$$$2\\left\\textbar{} {{\\alpha }_{ki}}+{{\\alpha }_{kj}} \\right.$$，$$k=1$$，$$2$$，$$3$$，$$4$$，$$5$$ $$\\Leftrightarrow 2\\left\\textbar{} {{\\beta }_{ki}}+{{\\beta }_{kj}} \\right.$$，$$k=1$$，$$2$$，$$3$$，$$4$$，$$5$$， $$\\because \\beta \\in \\left { 0,1 \\right }$$， $$\\therefore $$必有$${{\\beta }_{ki}}={{\\beta }_{kj}}$$， $$\\therefore $$当且仅当$${{\\beta }_{ki}}\\pm {{\\beta }_{kj}}\\left( k=1,2,3,4,5 \\right)$$，即任意两正因数，其各质因数的幂指数奇偶性均不同时，才有$${{a}_{i}}\\times {{a}_{j}}$$不是完全平方数， $$\\because {{b}_{i}}$$为$$5$$组数组，$${{\\beta }_{i}}\\in \\left { 0,1 \\right }$$， $$\\therefore $$共有$${{2}^{5}}$$种，选$$\\text{C}$$． " ]

[ "2023年江苏连云港灌南县灌南县中学高三竞赛（下学期3月解题能力竞赛）第8题" ]

[ [ { "aoVal": "A", "content": "$\\left(\\sqrt{5}+1,5\\right]$ " } ], [ { "aoVal": "B", "content": "$\\left[3,5\\right]$ " } ], [ { "aoVal": "C", "content": "$\\left(\\sqrt{5}+1,2\\sqrt{5}\\right]$ " } ], [ { "aoVal": "D", "content": "$\\left[\\sqrt{3},\\sqrt{5}\\right]$ " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 先求出椭圆左焦点$F_{1}$坐标为$(-2,0)$，由题得$\\textbar{} \\left\\textbar{} PA\\right\\textbar{} -\\left\\textbar{} PF_{1}\\right\\textbar{} \\textbar{} =\\textbar{} 8-2a\\textbar{} \\leq \\textbar{} AF_{1}\\textbar{} =2$，解不等式得到$3\\leq a\\leq 5$，再解不等式$\\frac{4}{a^{2}}+\\frac{4}{4-a^{2}}\\textless{} 1$即得解.\\\\ 【详解】\\\\ 点$Q\\left(a,b\\right)$在双曲线$E$：$\\frac{x^{2}}{4}-\\frac{y^{2}}{4}=1$上，所以$a^{2}-b^{2}=4$.\\\\ 所以椭圆左焦点$F_{1}$坐标为$(-2,0)$.\\\\ 因为$\\left\\textbar{} PA\\right\\textbar{} +\\left\\textbar{} PF_{2}\\right\\textbar{} =8$，所以$\\left\\textbar{} PA\\right\\textbar{} +2a-\\left\\textbar{} PF_{1}\\right\\textbar{} =8,\\therefore \\textbar{} \\left\\textbar{} PA\\right\\textbar{} -\\left\\textbar{} PF_{1}\\right\\textbar{} \\textbar{} =\\textbar{} 8-2a\\textbar{} \\leq \\textbar{} AF_{1}\\textbar{} =2$,\\\\ 所以$3\\leq a\\leq 5$.\\\\ 因为$a^{2}-b^{2}=4$，所以$b^{2}=a^{2}-4$.\\\\ 点$A\\left(-2,2\\right)$为椭圆$C$内一点，所以$\\frac{4}{a^{2}}+\\frac{4}{b^{2}}\\textless{} 1,\\therefore \\frac{4}{a^{2}}+\\frac{4}{a^{2}-4}\\textless{} 1$，\\\\ 所以$a^{4}-12a^{2}+16\\textgreater{} 0,\\therefore a\\textgreater{} \\sqrt{5}+1$或$a\\textless{} \\sqrt{5}-1$.\\\\ 综上：$\\sqrt{5}+1\\textless{} a\\leq 5$.\\\\ 故选：A " ]

[ "2010年河南全国高中数学联赛竞赛初赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$${{x}^{2}}+{{y}^{2}}=1$$ " } ], [ { "aoVal": "B", "content": "$${{x}^{2}}+{{y}^{2}}=2$$ " } ], [ { "aoVal": "C", "content": "$${{x}^{2}}+{{y}^{2}}=4$$ " } ], [ { "aoVal": "D", "content": "$${{x}^{2}}+{{y}^{2}}=8$$ " } ] ]

[ "竞赛->知识点->解析几何->曲线与方程", "竞赛->知识点->解析几何->双曲线" ]

[ "延长$${{F}_{1}}M$$交$$P{{F}_{2}}$$（或$$P{{F}_{2}}$$的延长线）于点$$N$$，则$$\\triangle P{{F}_{1}}N$$是等腰三角形，$$M$$是底边$${{F}_{1}}N$$的中点，连结$$OM$$，$$OM=\\frac{1}{2}{{F}_{2}}N=\\frac{1}{2}\\left\\textbar{} P{{F}_{1}}-P{{F}_{2}} \\right\\textbar=2$$，故选$$\\text{C}$$． " ]

[ "1986年全国高中数学联赛竞赛一试第2题" ]

[ [ { "aoVal": "A", "content": "$$M=$${纯虚数} " } ], [ { "aoVal": "B", "content": "$$M=$${实数} " } ], [ { "aoVal": "C", "content": "{实数}$$\\subset M\\subset $${复数} " } ], [ { "aoVal": "D", "content": "$$M=$${复数} " } ] ]

[ "竞赛->知识点->复数与平面向量->复数的概念与运算" ]

[ "∵$${{(Z-1)}^{2}}=\\textbar Z-1{{\\textbar}^{2}}$$， 即$${{(Z-1)}^{2}}=(Z-1)(\\overline{Z}\\mathsf{-1})$$， ∴$$(Z-1)(\\overline{Z}\\mathsf{-1})=\\mathsf{0}$$， 因此$$Z=1$$或$$Z=\\overline{Z}$$， 即$$Z$$为实数． 故选$$\\text{B}$$． " ]

[ "2012年黑龙江全国高中数学联赛竞赛初赛第1题5分", "2011年高考真题湖北卷理科第1题" ]

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

[ "知识标签->知识点->复数->复数的运算->复数的乘法和除法", "知识标签->素养->数学运算", "知识标签->题型->复数->复数的运算->复数中的周期问题" ]

[ "因为$$\\frac{1+\\text{i}}{1-\\text{i}}=\\frac{{{\\left( 1+\\text{i} \\right)}^{2}}}{1-{{\\text{i}}^{2}}}=\\text{i}$$，所以$${{\\left(\\frac{1+\\text{i}}{1-\\text{i}} \\right)}^{2011}}={{\\text{i}}^{2011}}={{\\text{i}}^{4\\times502+3}}={{\\text{i}}^{3}}=-\\text{i}$$， 故选$$\\text{A}$$． " ]

[ "2008年福建全国高中数学联赛竞赛初赛第4题6分" ]

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

[ "竞赛->知识点->多项式与方程->解方程（组）" ]

[ "原方程可以转化为$$\\left\\textbar{} x-1 \\right\\textbar(\\left\\textbar{} x-2 \\right\\textbar+\\left\\textbar{} x+3 \\right\\textbar)=11$$． 当$$x\\leqslant -3$$时，方程转化为$$2{{x}^{2}}-x-12=0$$， 它的两个根都大于$$-3$$，此时无解； 当$$-3\\textless{}x\\leqslant 1$$时，方程转化为$$-5x+5=11$$， 所以$$x=-\\frac{6}{5}$$，满足题意； 当$$1x\\leqslant 2$$时，方程转化为$$5x-5=11$$， 所以$$x=\\frac{16}{5}\\textgreater2$$，此时无解； 当$$x\\textgreater2$$时，方程转化为$$2{{x}^{2}}-x-12=0$$， 其中一个根$$x=\\frac{1+\\sqrt{97}}{4}$$满足题意． 综上所述，满足方程的实数解有$$2$$个．故选$$\\text{C}$$． " ]

[ "2008年山西全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "垂直 " } ], [ { "aoVal": "B", "content": "平行 " } ], [ { "aoVal": "C", "content": "重合 " } ], [ { "aoVal": "D", "content": "其他情况 " } ] ]

[ "竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->解析几何->直线与方程" ]

[ "由于$$A+C=120{}^{}\\circ $$，则 $$\\tan (60{}^{}\\circ -C)=\\tan \\left( \\frac{A+C}{2}-C \\right)=\\tan \\frac{A-C}{2}$$， $$\\sin A-\\sin C=2\\sin \\frac{A-C}{2}\\cos \\frac{A+C}{2}=\\sin \\frac{A-C}{2}$$， $$\\sin A+\\sin C=2\\sin \\frac{A+C}{2}\\cos \\frac{A-C}{2}=\\sqrt{3}\\cos \\frac{A-C}{2}$$， 故直线$${{l}_{1}}$$、$${{l}_{2}}$$的方程皆可化为$$x+y\\cos \\frac{A-C}{2}+1=0$$， 因此两直线重合．故选$$\\text{C}$$． " ]

[ "2005年全国高中数学联赛竞赛一试第5题6分", "2019~2020学年上海浦东新区华东师范大学第二附属中学高一下学期单元测试《任意角的三角比》（实验班）第7题", "2019~2020学年上海浦东新区华东师范大学第二附属中学高二上学期单元测试第3题" ]

[ [ { "aoVal": "A", "content": "焦点在$$x$$轴上的椭圆 " } ], [ { "aoVal": "B", "content": "焦点在$$x$$轴上的双曲线 " } ], [ { "aoVal": "C", "content": "焦点在$$y$$轴上的椭圆 " } ], [ { "aoVal": "D", "content": "焦点在$$y$$轴上的双曲线 " } ] ]

[ "竞赛->知识点->解析几何->轨迹方程（二试）" ]

[ "∵$$\\sqrt{2}+\\sqrt{3}\\textgreater{} \\pi $$，$$\\therefore 0\\textless{}\\frac{ \\pi }{2}-\\sqrt{2}\\textless{}\\sqrt{3}-\\frac{ \\pi }{2}\\textless{}\\frac{ \\pi }{2}$$，$$\\therefore \\cos \\left( \\frac{ \\pi }{2}-\\sqrt{2} \\right)\\textgreater\\cos \\left( \\sqrt{3}-\\frac{ \\pi }{2} \\right)$$ 即$$\\sin \\sqrt{2}\\textgreater\\sin \\sqrt{3}$$． 又$$0\\textless{}\\sqrt{2}\\textless{}\\frac{ \\pi }{2}$$，$$\\frac{ \\pi }{2}\\textless{}\\sqrt{3}\\textless{} \\pi $$， $$\\therefore \\cos \\sqrt{2}\\textgreater0$$，$$\\cos \\sqrt{3}\\textless{}0$$，$$\\therefore \\cos \\sqrt{2}-\\cos \\sqrt{3}\\textgreater0$$， 方程表示的曲线是椭圆． ∵$$(\\sin \\sqrt{2}-\\sin \\sqrt{3})-(\\cos \\sqrt{2}-\\cos \\sqrt{3})$$ $$=2\\sqrt{2}\\sin \\frac{\\sqrt{2}-\\sqrt{3}}{2}\\sin \\left( \\frac{\\sqrt{2}+\\sqrt{3}}{2}+\\frac{ \\pi }{4} \\right)$$\\ldots\\ldots$$(*)$$ $$-\\frac{ \\pi }{2}\\textless{}\\frac{\\sqrt{2}-\\sqrt{3}}{2}\\textless{}0$$，$$\\therefore \\sin \\frac{\\sqrt{2}-\\sqrt{3}}{2}\\textless{}0$$，$$\\frac{ \\pi }{2}\\textless{}\\frac{\\sqrt{2}+\\sqrt{3}}{2}\\textless{}\\frac{3 \\pi }{4}$$，$$\\therefore \\frac{3 \\pi }{4}\\textless{}\\frac{\\sqrt{2}+\\sqrt{3}}{2}+\\frac{ \\pi }{4}\\textless{} \\pi $$． $$\\therefore \\sin \\left( \\frac{\\sqrt{2}+\\sqrt{3}}{2}+\\frac{ \\pi }{4} \\right)\\textgreater0$$，$$\\therefore (*)$$式$$\\textless{}0$$． 即$$\\sin \\sqrt{2}-\\sin \\sqrt{3}\\textless{}\\cos \\sqrt{2}-\\cos \\sqrt{3}$$． ∴曲线表示焦点在$$y$$轴上的椭圆， 故选$$\\text{C}$$． " ]

[ "2011年山东全国高中数学联赛竞赛初赛第3题6分" ]

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

[ "竞赛->知识点->逻辑->常用逻辑用语", "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "充分性显然成立，必要性不成立．例：$$a=1, b=2, c=5, d=10$$. " ]

[ "2008年吉林全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "$$160$$种 " } ], [ { "aoVal": "B", "content": "$$320$$种 " } ], [ { "aoVal": "C", "content": "$$32$$种 " } ], [ { "aoVal": "D", "content": "$$120$$种 " } ] ]

[ "竞赛->知识点->排列组合与概率->排列与组合" ]

[ "$$\\text{C}_{5}^{1}\\cdot \\text{C}_{4}^{1}\\cdot 2\\cdot 2\\cdot 2=160$$．故选$$\\text{A}$$． " ]

[ "2014~2015学年浙江杭州高一下学期期末理科第24题3分", "2011年辽宁全国高中数学联赛竞赛初赛第2题6分" ]

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

[ "课内体系->方法->换元法", "课内体系->素养->逻辑推理", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域->求函数的值域->用换元法求值域" ]

[ "对于$$f(x)$$，有$$3\\leqslant x\\leqslant 4$$，则$$0\\leqslant x-3\\leqslant 1$$， 令$$x-3={{\\sin }^{2}}\\theta $$，$$0\\leqslant \\theta \\leqslant \\frac{ \\pi }{2}$$， 则$$f\\left( x \\right)=\\sqrt{x-3}+\\sqrt{3\\left( 4-x \\right)}=\\sin \\theta +\\sqrt{3\\left( 1-{{\\sin }^{2}}\\theta \\right)}=\\sin \\theta +\\sqrt{3}\\cos \\theta =2\\sin \\left( \\theta +\\frac{ \\pi }{3} \\right)$$， ∵$$\\frac{ \\pi }{3}\\leqslant \\theta +\\frac{ \\pi }{3}\\leqslant \\frac{5 \\pi }{6}$$， ∴$$\\frac{1}{2}\\leqslant \\sin \\left( \\theta +\\frac{ \\pi }{3} \\right)\\leqslant 1$$，$$1\\leqslant 2\\sin \\left( \\theta +\\frac{ \\pi }{3} \\right)\\leqslant 2$$． 函数$$f\\left( x \\right)=\\sqrt{x-3}+\\sqrt{12-3x}$$的值域为$$[1,2]$$． 故选$$\\text{D}$$． " ]

[ "2019年福建全国高中数学联赛竞赛初赛第4题6分" ]

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

[ "课内体系->知识点->复数->复数的概念及几何意义", "竞赛->知识点->复数与平面向量->复数的概念与运算" ]

[ "先求复数$$-2-2\\sqrt{3}\\text{i}$$的平方根． 设$${{(x+y\\text{i)}}^{2}}=-2-2\\sqrt{3}\\text{i(}x,y\\in \\mathbf{R})$$． 则$$({{x}^{2}}-{{y}^{2}})+2xy\\text{i=}-2-2\\sqrt{3}\\text{i}$$ $$\\Rightarrow \\begin{cases}{{x}^{2}}-{{y}^{2}}=-2 2xy=-2\\sqrt{3} \\end{cases}$$， $$\\Rightarrow \\begin{cases}{{x}_{1}}=1 {{y}_{1}}=-\\sqrt{3} \\end{cases}$$，$$\\begin{cases}{{x}_{2}}=-1 {{y}_{2}}=\\sqrt{3} \\end{cases}$$． 由$$z_{1}^{2}=z_{2}^{2}=-2-2\\sqrt{3}\\text{i}$$，$${{z}_{1}}\\ne {{z}_{2}}$$，知$${{z}_{1}}$$、$${{z}_{2}}$$为复数$$-2-2\\sqrt{3}\\text{i}$$的两个平方根． 由对称性，不妨设$${{z}_{1}}=1-\\sqrt{3}\\text{i}$$，$${{z}_{2}}=-1+\\sqrt{3}\\text{i}$$． 于是，$$\\textbar{{z}_{1}}-{{z}_{2}}\\textbar=\\textbar z-{{z}_{1}}\\textbar=\\textbar z-{{z}_{2}}\\textbar=4$$． 从而，复数$$z$$、$${{z}_{1}}$$、$${{z}_{2}}$$对应的点$$Z$$、$${{Z}_{1}}$$、$${{Z}_{2}}$$构成边长为$$4$$的正三角形． 又复数$${{z}_{1}}$$、$${{z}_{2}}$$对应的点$${{Z}_{1}}$$、$${{Z}_{2}}$$关于原点$$0$$对称，从而，$$OZ$$为$$\\triangle Z{{Z}_{1}}{{Z}_{2}}$$的高． 故$$\\textbar z\\textbar=\\textbar OZ\\textbar=2\\sqrt{3}$$． " ]

[ "2015年湖南全国高中数学联赛竞赛初赛第5题5分", "2005年全国高中数学联赛竞赛一试第1题6分" ]

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

[ "竞赛->知识点->不等式->不等式的解法" ]

[ "令$$y=\\sqrt{x-3}+\\sqrt{6-x}$$，$$3\\leqslant x\\leqslant 6$$，则 $${{y}^{2}}=(x-3)+(6-x)+2\\sqrt{(x-3)(6-x)}\\leqslant 2[(x-3)+(6-x)]=6$$． ∴$$0\\textless{}y\\leqslant \\sqrt{6}$$，实数$$k$$的最大值为$$\\sqrt{6}$$． 故选$$\\text{D}$$． " ]

[ "2010年山东全国高中数学联赛竞赛初赛第1题6分" ]

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

[ "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->逻辑->常用逻辑用语" ]

[ "事实上， $$\\left. \\begin{matrix}{{S}_{m}}=\\frac{m}{2}({{a}_{1}}+{{a}_{m}})\\textgreater0\\Leftrightarrow {{a}_{1}}+{{a}_{m}}\\textgreater0 {{S}_{m+1}}=\\frac{m+1}{2}({{a}_{1}}+{{a}_{m+1}})\\textless{}0\\Leftrightarrow {{a}_{1}}+{{a}_{m+1}}\\textless{}0 \\end{matrix} \\right }$$ $$\\Leftrightarrow {{a}_{1}}+{{a}_{m+1}}\\textless{}0\\textless{}{{a}_{1}}+{{a}_{m}}\\Leftrightarrow -{{a}_{m}}\\textless{}{{a}_{1}}\\textless{}-{{a}_{m+1}}.$$ 所以是充分必要条件． " ]

[ "2012年黑龙江全国高中数学联赛竞赛初赛第3题5分" ]

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

[ "课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与数列结合", "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" ]

[ "因为$$\\lg x$$，$$\\lg y$$，$$\\lg z$$成等差数列， 所以$$2\\lg y=\\lg x+\\lg z$$，所以$${{y}^{2}}=xz$$； 若$${{y}^{2}}=xz$$，当$$x\\textless{}0$$，$$z\\textless{}0$$时， $$\\lg x$$，$$\\lg z$$无意义． 所以``$$\\lg x$$，$$\\lg y$$，$$\\lg z$$成等差数列''是``$${{y}^{2}}=xz$$''成立的充分不必要条件． 故选$$\\text{A}$$． " ]

[ "2010年辽宁全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "充分条件而非必要条件 " } ], [ { "aoVal": "B", "content": "必要条件而非充分条件 " } ], [ { "aoVal": "C", "content": "充分必要条件 " } ], [ { "aoVal": "D", "content": "即非充分条件亦非必要条件 " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "若$$f(x)$$在$$\\left[ a,b \\right]$$上单调，则$$f(a)$$和$$f(b)$$一个是最大值，另一个是最小值，故是充分条件；如，$$\\sin x$$在$$\\left[ 0, \\pi ~\\right]$$上有最大值$$1$$和最小值$$0$$，但不单调，故不是必要条件． " ]

[ "2016年河北全国高中数学联赛高三竞赛初赛第5题8分" ]

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

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

[ "由柯西不等式，$${{\\left( \\sqrt{{{x}^{2}}\\left( 1-{{x}^{2}} \\right)}+\\sqrt{\\left( 2-{{x}^{2}} \\right){{x}^{2}}} \\right)}^{2}}\\leqslant \\left( {{x}^{2}}+2-{{x}^{2}} \\right)\\left( 1-{{x}^{2}}+{{x}^{2}} \\right)=2$$，当$$x=\\pm \\frac{\\sqrt{6}}{3}$$时取等号． " ]

[ "2010年河北全国高中数学联赛竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$a\\textgreater b\\textgreater1$$ " } ], [ { "aoVal": "B", "content": "$$b\\textgreater a\\textgreater1$$ " } ], [ { "aoVal": "C", "content": "$$a\\textgreater1$$，$$0 \\textless{} b \\textless{} 1$$ " } ], [ { "aoVal": "D", "content": "$$0 \\textless{} a \\textless{} 1$$，$$b\\textgreater1$$ " } ] ]

[ "课内体系->知识点->等式与不等式->不等式->证明不等式的基本方法->比较法", "课内体系->知识点->等式与不等式->不等式->不等式的性质", "课内体系->知识点->基本初等函数->指数函数->指数函数的图象及性质", "课内体系->素养->逻辑推理", "课内体系->素养->数学运算" ]

[ "首先，$$a\\textgreater1$$，$$b\\textgreater1$$．否则， 若$$0 \\textless{} a \\textless{} 1$$，则$${{a}^{n}}=a+1\\textgreater1$$，从而$$a\\textgreater1$$，矛盾； 若$$0 \\textless{} b \\textless{} 1$$，$${{b}^{2n}}=b+3a\\textgreater3a\\textgreater3$$，从而$$b\\textgreater1$$，矛盾． 故$$a$$、$$b$$均大于1． 一方面，$${{a}^{2n}}-{{b}^{2n}}={{\\left( a+1 \\right)}^{2}}-\\left( b+3a \\right)={{a}^{2}}-a-b+1$$． 另一方面，$${{a}^{2n}}-{{b}^{2n}}=\\left( a-b \\right)\\left( {{a}^{2n-1}}+{{a}^{2n-2}}b+\\cdots +{{b}^{2n-1}} \\right)$$． 故$${{a}^{2}}-a-b+1$$ $$=\\left( a-b \\right)\\left( {{a}^{2n-1}}+{{a}^{2n-2}}b+\\cdots +{{b}^{2n-1}} \\right)$$， 即$$\\frac{{{a}^{2}}-a-b+1}{a-b}$$ $$={{a}^{2n-1}}+{{a}^{2n-2}}b+\\cdots +{{b}^{2n-1}}\\textgreater1$$． $$\\therefore \\frac{{{a}^{2}}-2a+1}{a-b}\\textgreater0$$，即$$\\frac{{{\\left( a-1 \\right)}^{2}}}{a-b}\\textgreater0$$．故$$a\\textgreater b$$． 综上所述，有$$a\\textgreater b\\textgreater1$$． " ]

[ "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2~2题" ]

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

[ "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的定义域" ]

[ "\\hfill\\break 根据函数解析式，只需解析式有意义，即$$\\left { \\begin{array}{*{35}{l}} 2-x\\textgreater0 x+2\\ne 0 \\end{array} \\right.$$，解不等式即可求解.\\\\ 【详解】\\\\ 由$$f(x)=\\frac{1}{\\sqrt{2-x}}+{{(x+2)}^{0}}$$，\\\\ 则$$\\left { \\begin{array}{*{35}{l}} 2-x\\textgreater0 x+2\\ne 0 \\end{array} \\right.$$，解得$$x\\textless{} 2$$且$$x\\ne -2$$，\\\\ 所以函数的定义域为$$\\left( -\\infty ,-2 \\right)\\cup \\left( -2,2 \\right)$$.\\\\ 故选：B " ]