Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2009年黑龙江全国高中数学联赛竞赛初赛第9题5分" ]

[ [ { "aoVal": "A", "content": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$42$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$54$$ " } ] ]

[ "竞赛->知识点->排列组合与概率->排列与组合" ]

[ "将座位从左至右编为$$1,2,\\cdots ,9$$，考虑从左至右第一个有人坐的座位号： ①如果是$$1$$，则另两个人的座位号只能是$$4$$和$$7$$，只有$$1$$种方法； ②如果是$$2$$，则下一个人可以坐$$4$$或$$5$$．若是$$4$$，则第三个人只能坐$$7$$号；若是$$5$$，则第三个人可以坐$$7$$或$$8$$．共有$$3$$种方法； ③如果是$$3$$，则下一个人可以坐$$5$$或$$6$$．若是$$5$$，则第三个人可以坐$$7$$或$$8$$；若是$$6$$，则第三个人可以坐$$8$$或$$9$$．共有$$2+2=4$$种方法． 所以，不同的坐法种数为$$\\left( 1+3+4 \\right)\\text{A}_{3}^{3}=48$$． " ]

[ "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8~8题" ]

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

[ "课内体系->知识点->解三角形" ]

[ "\\hfill\\break 【分析】\\\\ 根据条件求出$$A=\\frac{\\text{ } ! !\\pi ! !\\text{ }}{3}$$，利用三角形面积公式得到$${{S}_{\\vartriangle ABC}}=\\frac{1}{2}bc\\sin A=\\frac{\\sqrt{3}}{2}c$$，采用极端值方法求出$$c$$的最值，进而得到$$c$$的范围，求出面积的取值范围.\\\\ 【详解】\\\\ $$\\cos A=\\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=\\frac{1}{2}$$，因为$$\\vartriangle ABC$$为锐角三角形，故$$A=\\frac{\\text{ } ! !\\pi ! !\\text{ }}{3}$$，\\\\ $${{S}_{\\vartriangle ABC}}=\\frac{1}{2}bc\\sin A=\\frac{\\sqrt{3}}{2}c$$，当\\emph{BC}⊥\\emph{AB}时，$$c=b\\cos A=1$$，当\\emph{CB}⊥\\emph{AC}时，$$c=\\frac b{\\cos A}=4$$，故$$c\\in \\left( 1,4 \\right)$$，所以$${{S}_{\\vartriangle ABC}}=\\frac{\\sqrt{3}}{2}c\\in \\left( \\frac{\\sqrt{3}}{2},2\\sqrt{3} \\right)$$.\\\\ 故选：C " ]

[ "2011年山东全国高中数学联赛竞赛初赛第2题6分" ]

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

[ "竞赛->知识点->复数与平面向量->复数的概念与运算" ]

[ "$$z={{(\\sqrt{3}-3\\text i)}^{n}}={{(-2\\sqrt{3})}^{n}}{{(-\\frac{1}{2}+\\frac{\\sqrt{3}}{2}\\text i)}^{n}}$$，$$n=3$$是使$$z$$为实数的最小的正整数． " ]

[ "2015年吉林全国高中数学联赛竞赛初赛第5题6分" ]

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

[ "竞赛->知识点->函数->函数的图像与性质" ]

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

[ "2015年黑龙江全国高中数学联赛竞赛初赛第8题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{2}{\\frac{1}{a}+\\frac{1}{b}}\\geqslant \\sqrt{ab}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{a}+\\frac{1}{b}\\geqslant \\frac{4}{a+b}$$ " } ], [ { "aoVal": "C", "content": "$$\\sqrt{\\textbar a-b\\textbar}\\geqslant \\sqrt{a}-\\sqrt{b}$$ " } ], [ { "aoVal": "D", "content": "$${{a}^{2}}+{{b}^{2}}+1\\textgreater a+b$$ " } ] ]

[ "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->针对不等式变形判断正误", "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->基本不等式->利用基本不等式证明其它不等式", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的实际应用" ]

[ "$$\\frac{1}{a}+\\frac{1}{b}\\geqslant 2\\sqrt{\\frac{1}{ab}}$$，则$$\\sqrt{ab}\\left( \\frac{1}{a}+\\frac{1}{b} \\right)\\geqslant 2$$，即$$\\frac{2}{\\frac{1}{a}+\\frac{1}{b}}\\leqslant \\sqrt{ab}$$，当且仅当$$a=b$$时取等号，故不等式不恒成立. " ]

[ "2021年贵州全国高中数学联赛竞赛初赛第7题" ]

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

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

[ "因为$$O$$是$$\\triangle ABC$$的外心，故$$O$$在$$AB$$边上的投影为$$AB$$的中点， 故$$\\overrightarrow{AB}\\cdot \\overrightarrow{AO}=\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar\\left\\textbar{} \\overrightarrow{AO} \\right\\textbar\\cdot \\cos \\angle BAO=\\frac{1}{2}{{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}^{2}}$$， 同理可得，$$\\overrightarrow{AC}\\cdot \\overrightarrow{AO}=\\frac{1}{2}{{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}^{2}}$$， 所以$$\\overrightarrow{AG}\\cdot \\overrightarrow{AO}=\\frac{1}{3}(\\overrightarrow{AB}+\\overrightarrow{AC})\\cdot \\overrightarrow{AO}=\\frac{1}{6}\\left( {{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}^{2}}+{{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}^{2}} \\right)=\\frac{1}{6}(3+9)=2$$． " ]

[ "2012年吉林全国高中数学联赛竞赛初赛第3题5分" ]

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

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

[ "$$1\\times 3\\times 3=9$$． " ]

[ "2013年辽宁全国高中数学联赛竞赛初赛第2题6分" ]

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

[ "竞赛->知识点->解析几何->双曲线" ]

[ "设点$$P$$的坐标为$$\\left( x,y \\right)$$，由对称性$$Q$$的坐标为$$\\left( -x,-y \\right)$$，分别过$$P,Q$$作$$x$$轴的垂线，垂足分别为$${{P}_{1}}\\left( x,0 \\right){ }{{Q}_{1}}\\left( -x,0 \\right)$$，折成直二面角后， $${{\\left\\textbar{} PQ \\right\\textbar}^{2}}={{\\left\\textbar{} P{{P}_{1}} \\right\\textbar}^{2}}+{{\\left\\textbar{} {{P}_{1}}{{Q}_{1}} \\right\\textbar}^{2}}+{{\\left\\textbar{} {{Q}_{1}}Q \\right\\textbar}^{2}}$$ $$={{y}^{2}}+{{\\left( 2x \\right)}^{2}}+{{\\left( -y \\right)}^{2}}=4{{x}^{2}}+2{{\\left( \\frac{-2\\sqrt{2}}{x} \\right)}^{2}}$$ $$=4{{x}^{2}}+\\frac{16}{{{x}^{2}}}{\\geqslant }2\\sqrt{64}=16$$， 即$$\\left\\textbar{} PQ \\right\\textbar{\\geqslant }4$$；当$$4{{x}^{2}}=\\frac{16}{{{x}^{2}}}$$即$$x=\\sqrt{2}$$时，$$\\left\\textbar{} PQ \\right\\textbar=4$$． " ]

[ "竞赛第19题" ]

[ [ { "aoVal": "A", "content": "1125 " } ], [ { "aoVal": "B", "content": "1278 " } ], [ { "aoVal": "C", "content": "1350 " } ], [ { "aoVal": "D", "content": "1542 " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 先考虑任取4个点，求出不共面的情形的个数，从而可求异面直线的对数.\\\\ 【详解】\\\\ 12个顶点构成的4点组有${\\mathrm{C}}_{12}^{4}=495$个，其中共面的4点组可以按在上下底面的个数分类讨论．\\\\ \\textbf{情形一}~~~全部在上底面或全部在下底面．共面4点组数为$2{\\mathrm{C}}_{6}^{4}=30$．\\\\ \\textbf{情形二}~~~两个底面各2个，共面4点组数按在上底面的2个点间的位置关系分类（此时上下底面的点分别连线，则连线平行），为$6\\times 3+6\\times 2+3\\times 3=39$．\\\\ 因此共面的4点组共有69个，进而不共面的4点组有426个，每个不共面的4点组贡献3对异面直线，且不同的4点组贡献的异面直线不同，\\\\ 因此所求异面直线的对数为$426\\times 3=1278$．\\\\ 故选：B. " ]

[ "2008年湖南全国高中数学联赛竞赛初赛第7题5分" ]

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

[ "竞赛->知识点->三角函数->三角函数的图像与性质" ]

[ "因为$$2008{}^{}\\circ =5\\times 360{}^{}\\circ +180{}^{}\\circ +28{}^{}\\circ $$，所以， $$a=\\sin (-\\sin 28{}^{}\\circ )=-\\sin (\\sin 28{}^{}\\circ )\\textless{}0$$； $$b=\\sin (-\\cos 28{}^{}\\circ )=-\\sin (\\cos 28{}^{}\\circ )\\textless{}0$$； $$c=\\cos (-\\sin 28{}^{}\\circ )=\\cos (\\sin 28{}^{}\\circ )\\textgreater0$$； $$d=\\cos (-\\cos 28{}^{}\\circ )=\\cos (\\cos 28{}^{}\\circ )\\textgreater0$$． 又$$\\sin 28{}^{}\\circ \\textless{}\\cos 28{}^{}\\circ $$，故$$b\\textless{}a\\textless{}d\\textless{}c.$$故选$$\\text{B}$$． " ]

[ "2008年天津全国高中数学联赛竞赛初赛第6题6分" ]

[ [ { "aoVal": "A", "content": "$$k=0$$ " } ], [ { "aoVal": "B", "content": "$$1\\leqslant k\\leqslant 2008$$ " } ], [ { "aoVal": "C", "content": "$$k\\textgreater2008$$，但$$k$$是有限的数 " } ], [ { "aoVal": "D", "content": "$$k$$是无穷大 " } ] ]

[ "竞赛->知识点->函数->韦达定理" ]

[ "由题意知$$\\left( {{x}_{0}}, {{y}_{0}}, {{z}_{0}} \\right)=\\left( 3, 4, 5 \\right)$$是原不定方程的一个特解．对于原不定方程的任意一个正整数解$$\\left( {{x}_{1}}, y, z \\right)$$，假设$${{x}_{1}}\\leqslant y\\leqslant z$$，且$${{x}_{1}}z$$．设关于$$x$$的二次方程$${{x}^{2}}-yz\\cdot x+{{y}^{2}}+{{z}^{2}}+10=0$$的两个根为$${{x}_{1}}, {{x}_{2}}$$，由韦达定理，$${{x}_{1}}+{{x}_{2}}=yz, {{x}_{1}}{{x}_{2}}={{y}^{2}}+{{z}^{2}}+10{{z}^{2}}$$，因此$${{x}_{2}}$$是正整数，且大于$$z$$，于是$$\\left( {{x}_{2}}, y, z \\right)$$也是原不定方程的一个解，由于原不定方程是轮换对称的，所以$$\\left( y, z, {{x}_{2}} \\right)$$也是它的解，并且它是由小到大排列的． 如此反复利用上面的结论，可以由一个特解得到无穷多个解，因此满足$$x, y, z$$均大于$$2008$$的解有无穷多个．故选$$\\text{D}$$． " ]

[ "2018年湖北全国高中数学联赛竞赛初赛第1题10分" ]

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

[ "知识标签->知识点->三角函数->三角函数的图象与性质->正弦型函数->正弦型函数的图象与性质", "知识标签->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正弦", "知识标签->题型->三角函数->三角函数的图象与性质->正弦型函数->正弦型三角函数图像与性质问题", "知识标签->题型->三角函数->三角恒等变换->倍角、半角公式问题", "知识标签->素养->数学运算" ]

[ "设$$x=\\sin \\theta +\\cos \\theta =\\sqrt{2}\\sin \\left( \\theta +\\frac{\\pi }{4} \\right)$$，则$$2\\sin \\theta \\cdot \\cos \\theta ={{x}^{2}}-1$$， 当$$\\theta \\in \\left[ 0,\\frac{\\pi }{2} \\right]$$时，得$$1\\leqslant x\\leqslant \\sqrt{2}$$， 故原不等式变为：$${{x}^{2}}-ax+3\\leqslant 0\\Rightarrow a\\geqslant x+\\frac{3}{x}$$， 此时，函数$$f\\left( x \\right)=x+\\frac{3}{x}$$单调递减， 从而，实数$$a$$的最小值为$$f\\left( 1 \\right)=4$$． " ]

[ "竞赛第12题" ]

[ [ { "aoVal": "A", "content": "$x^{2}+2y^{2}=25$ " } ], [ { "aoVal": "B", "content": "$x^{2}+3y^{2}=50$ " } ], [ { "aoVal": "C", "content": "$2x^{2}+5y^{2}=50$ " } ], [ { "aoVal": "D", "content": "$2x^{2}+6y^{2}=75$ " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 根据平面截圆锥所得曲线形状的判断法则可得截得的曲线为椭圆且离心率为$e=\\frac{\\cos 45{^{\\circ}}}{\\cos 30{^{\\circ}}}$，求出其长轴后可得曲线方程.\\\\ 【详解】\\\\ 根据平面截圆锥所得曲线形状的判断法则，截得的曲线为椭圆，离心率$e=\\frac{\\cos 45{^{\\circ}}}{\\cos 30{^{\\circ}}}=\\frac{\\sqrt{6}}{3}$，\\\\ 且根据正弦定理，可得椭圆的长轴长$2a=SC\\cdot \\sin 30{^{\\circ}}\\left(\\frac{1}{\\sin 75{^{\\circ}}}+\\frac{1}{\\sin 15{^{\\circ}}}\\right)=5\\sqrt{6}$，\\\\ 因此椭圆的半焦距$c=5$，进而所求曲线方程为$\\frac{x^{2}}{\\frac{75}{2}}+\\frac{y^{2}}{\\frac{25}{2}}=1\\Rightarrow 2x^{2}+6y^{2}=75$．\\\\ 故选：D. " ]

[ "2008年山西全国高中数学联赛竞赛初赛第6题6分" ]

[ [ { "aoVal": "A", "content": "$$200808$$ " } ], [ { "aoVal": "B", "content": "$$179101$$ " } ], [ { "aoVal": "C", "content": "$$153201$$ " } ], [ { "aoVal": "D", "content": "$$116808$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->数列的通项与求和" ]

[ "当且仅当$$n=3k-2(k=1,2,\\cdots )$$时能被$$3$$整除， 故$${{b}_{k}}=\\frac{(3k-2)(3k-1)}{2}=1+\\frac{9k(k-1)}{2}$$， 所以 $${{b}_{200}}=1+\\frac{9\\times 200\\times 199}{2}=179101$$， 故选$$\\text{B}$$． " ]

[ "2010年辽宁全国高中数学联赛竞赛初赛第3题6分" ]

[ [ { "aoVal": "A", "content": "$$36$$种 " } ], [ { "aoVal": "B", "content": "$$39$$种 " } ], [ { "aoVal": "C", "content": "$$60$$种 " } ], [ { "aoVal": "D", "content": "$$78$$种 " } ] ]

[ "竞赛->知识点->排列组合与概率->排列与组合" ]

[ "不加限制的排法共有$$\\frac{\\text{A}_{5}^{5}}{2}=60$$种；生物排在第一节，物理排在最后一节各有$$\\frac{\\text{A}_{4}^{4}}{2}=12$$种，其中生物排在第一节且物理排在最后一节有$$\\frac{\\text{A}_{3}^{3}}{2}=3$$种．因此符合要求的排法共有$$60-12\\times 2+3=39$$种． " ]

[ "2008年山东全国高中数学联赛竞赛初赛第10题6分" ]

[ [ { "aoVal": "A", "content": "$$17$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->数列的通项与求和" ]

[ "由$${{S}_{n}}=\\frac{1}{2}\\left( {{a}_{n}}+\\frac{1}{{{a}_{n}}} \\right)=\\frac{1}{2}\\left( {{S}_{n}}-{{S}_{n-1}}+\\frac{1}{{{S}_{n}}-{{S}_{n-1}}} \\right)$$， 得$${{S}_{n}}+{{S}_{n-1}}=\\frac{1}{{{S}_{n}}-{{S}_{n-1}}}$$， 即$$S_{n}^{2}=S_{n-1}^{2}+1$$． 因为$${{S}_{1}}={{a}_{1}}=1$$，所以$$S_{n}^{2}=n{{S}_{n}}=\\sqrt{n}$$． 因为$$\\sqrt{n}+\\sqrt{n-1}\\textless{}2\\sqrt{n}\\textless{}\\sqrt{n+1}+\\sqrt{n}$$， 所以$$\\frac{1}{\\sqrt{n+1}+\\sqrt{n}}=\\sqrt{n+1}-\\sqrt{n}\\textless{}\\frac{1}{2\\sqrt{n}}$$ $$\\textless{}\\frac{1}{\\sqrt{n}+\\sqrt{n-1}}=\\sqrt{n}-\\sqrt{n-1}$$， 令$$S=\\frac{1}{{{S}_{1}}}+\\frac{1}{{{S}_{2}}}+\\ldots +\\frac{1}{{{S}_{100}}}$$， 则$$\\frac{S}{2}\\textgreater\\sqrt{101}-1\\textgreater9$$， 解得$$S\\textgreater18$$． 又因为$${{S}_{1}}={{a}_{1}}=1$$，所以$$\\frac{S}{2}-\\frac{1}{2{{S}_{1}}}=\\frac{1}{2{{S}_{2}}}+\\frac{1}{2{{S}_{3}}}+\\ldots +\\frac{1}{2{{S}_{100}}}\\textless{}\\sqrt{100}-1=9$$， 即$$S\\textless{}2\\left( 9+\\frac{1}{2} \\right)=19$$，从而得$$[S]=18$$．故选$$\\text{B}$$． " ]

[ "2008年江西全国高中数学联赛竞赛初赛第6题6分" ]

[ [ { "aoVal": "A", "content": "甲对乙错 " } ], [ { "aoVal": "B", "content": "甲错乙对 " } ], [ { "aoVal": "C", "content": "甲、乙都对 " } ], [ { "aoVal": "D", "content": "甲、乙都不一定对 " } ] ]

[ "竞赛->知识点->数论模块->整除->质数（算数基本定理）" ]

[ "设$$3n+1={{a}^{2}}$$，$$5n-1={{b}^{2}}$$，$$a$$，$$b$$为正整数， 则$$7n+13=9\\left( 3n+1 \\right)-4\\left( 5n-1 \\right)={{\\left( 3a \\right)}^{2}}-{{\\left( 2b \\right)}^{2}}$$ $$=\\left( 3a-2b \\right)\\left( 3a+2b \\right)$$① 由此知，$$3a-2b$$为正整数，且$$3a-2b\\ne 1$$． 因为若$$3a-2b=1$$， 则$$27n+9={{\\left( 3a \\right)}^{2}}={{\\left( 2b+1 \\right)}^{2}}=4{{b}^{2}}+4b+1$$， 即$$27n=4\\left( {{b}^{2}}+b-2 \\right)$$，则$$4\\left\\textbar{} n \\right.$$，记$$n=4k$$， 得$$5n-1=20k-1$$不为平方数，矛盾！ 所以$$3a-2b\\geqslant 2$$，故由①式得，$$7n+13$$为合数； 又因为$$8\\left( 17{{n}^{2}}+3n \\right)=\\left[ \\left( 3n+1 \\right)+\\left( 5n-1 \\right) \\right]\\left[ 4\\left( 3n+1 \\right)+\\left( 5n-1 \\right) \\right]$$ $$=\\left( {{a}^{2}}+{{b}^{2}} \\right)\\left[ {{\\left( 2a \\right)}^{2}}+{{b}^{2}} \\right]={{\\left( 2{{a}^{2}}+{{b}^{2}} \\right)}^{2}}+{{\\left( ab \\right)}^{2}}$$， 故选$$\\text{C}$$．（例如$$65$$即可为上述$$n$$的一个取值） " ]

[ "2008年黑龙江全国高中数学联赛竞赛初赛第1题5分" ]

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

[ "竞赛->知识点->集合->集合的概念与运算", "竞赛->知识点->函数->基本初等函数" ]

[ "$$P\\cap Q$$只有一个子集即$$P\\cap Q$$为空集， 数形结合可知$$k\\leqslant 1$$时，直线$$y=k$$与曲线$$y={{a}^{x}}+1$$无交点．故选$$\\text{B}$$． " ]

[ "第二十届全国希望杯高一竞赛复赛邀请赛第1题4分" ]

[ [ { "aoVal": "A", "content": "$$h(x)$$的最大值是$$M+N$$ " } ], [ { "aoVal": "B", "content": "$$h(x)$$的最小值是$$m+n$$ " } ], [ { "aoVal": "C", "content": "$$h(x)$$的值域为$$ {x\\textbar m+n\\leqslant x\\leqslant M+N }$$ " } ], [ { "aoVal": "D", "content": "$$h(x)$$的值域为$$ {x\\textbar m+n\\leqslant x\\leqslant M+N }$$的一个子集 " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "取$$f(x)=\\sin x,g(x)=\\cos x$$，$$D=\\mathbf{R}$$，即可知$$\\text A\\text B\\text C$$都不正确． " ]

[ "1999年全国全国高中数学联赛竞赛一试第1题6分" ]

[ [ { "aoVal": "A", "content": "是等差数列 " } ], [ { "aoVal": "B", "content": "是公比为$$q$$的等比数列 " } ], [ { "aoVal": "C", "content": "是公比为$${{q}^{3}}$$的等比数列 " } ], [ { "aoVal": "D", "content": "既非等差数列也非等比数列 " } ] ]

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "由题设，$${{a}_{n}}={{a}_{1}}{{q}^{n-1}}$$， $$\\frac{{{b}_{n+1}}}{{{b}_{n}}}=\\frac{{{a}_{3n+1}}+{{a}_{3n+2}}+{{a}_{3n+3}}}{{{a}_{3n-2}}+{{a}_{3n-1}}+{{a}_{3n}}}=\\frac{{{a}_{1}}{{q}^{3n}}+{{a}_{1}}{{q}^{3n+1}}+{{a}_{1}}{{q}^{3n+2}}}{{{a}_{1}}{{q}^{3n-3}}+{{a}_{1}}{{q}^{3n-2}}+{{a}_{1}}{{q}^{3n-1}}}$$， $$=\\frac{{{a}_{1}}{{q}^{3n}}\\left( 1+q+{{q}^{2}} \\right)}{{{a}_{1}}{{q}^{3n-3}}\\left( 1+q+{{q}^{2}} \\right)}={{q}^{3}}$$， 因此，$$\\left { {{b}_{n}} \\right }$$是公比为$${{q}^{3}}$$的等比数列． " ]

[ "2008年陕西全国高中数学联赛竞赛初赛第8题5分" ]

[ [ { "aoVal": "A", "content": "恒大于零 " } ], [ { "aoVal": "B", "content": "恒等于零 " } ], [ { "aoVal": "C", "content": "恒小于零 " } ], [ { "aoVal": "D", "content": "符号不确定 " } ] ]

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "因为$$f(-x)+f(x)$$ $$=-{{x}^{3}}-{{\\log }_{2}}(\\sqrt{{{x}^{2}}+1}+x)+{{x}^{3}}-{{\\log }_{2}}(\\sqrt{{{x}^{2}}+1}-x)$$ $$=-{{\\log }_{2}}[(\\sqrt{{{x}^{2}}+1}+x)(\\sqrt{{{x}^{2}}+1}-x)]=0$$， 所以$$f(-x)=-f(x)$$，即$$f(x)$$为奇函数． 又因为$$f(x)={{x}^{3}}-{{\\log }_{2}}(\\sqrt{{{x}^{2}}+1}-x)={{x}^{3}}+{{\\log }_{2}}(\\sqrt{{{x}^{2}}+1}+x)$$， 在$$(0,+\\infty )$$上单调递增，所以，$$f(x)$$在$$\\mathbf{R}$$上是增函数． 注意到$$f(a)-f(-b)$$与$$a-(-b)$$同号， 所以$$\\frac{f(a)+f(b)}{a+b}\\textgreater0$$． 又因为$${{a}^{3}}+{{b}^{3}}$$与$$a+b$$同号， 故$$\\frac{f(a)+f(b)}{{{a}^{3}}+{{b}^{3}}}\\textgreater0$$．故选$$\\text{A}$$． " ]

[ "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": "$$\\frac{241}{81}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{266}{81}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{274}{81}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{670}{243}$$ " } ] ]

[ "竞赛->知识点->排列组合与概率->概率初步" ]

[ "解法一： 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$$E\\xi =2\\times \\frac{5}{9}+4\\times \\frac{20}{81}+6\\times \\frac{16}{81}=\\frac{266}{81}$$； 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$$E\\xi =2\\times \\frac{5}{9}+4\\times \\frac{20}{81}+6\\times \\frac{16}{81}=\\frac{266}{81}$$． 故选$$\\text{B}$$． " ]

[ "2013年黑龙江全国高中数学联赛竞赛初赛第9题5分", "高二上学期单元测试《三角恒等变形》自招第17题" ]

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

[ "竞赛->知识点->三角函数->三角恒等变换" ]

[ "$$\\frac{\\sin 4\\alpha }{4{{\\sin }^{2}}\\left( \\frac{ \\pi }{4}+\\alpha \\right)\\tan \\left( \\frac{ \\pi }{4}-\\alpha \\right)}=\\frac{\\sin 4\\alpha }{4\\sin \\left( \\frac{ \\pi }{4}+\\alpha \\right)\\sin \\left( \\frac{ \\pi }{4}-\\alpha \\right)}=\\frac{\\sin 4\\alpha }{2\\sin \\left( \\frac{ \\pi }{2}+2\\alpha \\right)}=\\sin 2\\alpha $$ " ]

[ "2017年四川全国高中数学联赛竞赛初赛第12题5分" ]

[ [ { "aoVal": "A", "content": "$$143$$ " } ], [ { "aoVal": "B", "content": "$$224$$ " } ], [ { "aoVal": "C", "content": "$$243$$ " } ], [ { "aoVal": "D", "content": "$$268$$ " } ] ]

[ "竞赛->知识点->数论模块->同余->同余的概念与性质" ]

[ "因为$${{1}^{3}}\\equiv {{4}^{3}}\\equiv {{7}^{3}}\\equiv {{10}^{3}}\\equiv 1\\left( \\bmod 9 \\right)$$，$${{2}^{3}}\\equiv {{5}^{3}}\\equiv {{8}^{3}}\\equiv -1\\left( \\bmod 9 \\right)$$，$${{3}^{3}}\\equiv {{6}^{3}}\\equiv {{9}^{3}}\\equiv 0\\left( \\bmod 9 \\right)$$， 所以若$$9\\left\\textbar{} {{x}^{3}}+{{y}^{3}}+{{z}^{3}} \\right.$$，则$$x,y,z$$都是$$3$$的倍数或者$$3$$类数一样一个． 因此，$$A$$中元素的个数为$${{3}^{3}}+4\\times 3\\times 3\\times 3!=243$$． " ]

[ "2022年江苏徐州贾汪区江苏省贾汪中学高一竞赛（下学期春季）第6题" ]

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

[]

[ "\\hfill\\break 【详解】\\\\ 由$f\\left( x+3 \\right)=-f\\left( x \\right)$，知$f\\left( x+6 \\right)=f\\left( x \\right)$.所以，①正确；\\\\ 将$f\\left( x-\\frac{3}{2} \\right)$的图像向左平移$\\frac{3}{2}$ 个单位，即为$f\\left( x \\right)$的图像，而$f\\left( x-\\frac{3}{2} \\right)$的图像关于原点对称，所以，②正确；\\\\ 由②知，$f\\left( -x \\right)=-f\\left( -3+x \\right)=f\\left( x \\right)$ ，则$f\\left( x \\right)$为偶函数，所以，③正确. " ]

[ "2010年山东全国高中数学联赛竞赛初赛第3题6分" ]

[ [ { "aoVal": "A", "content": "$$c\\leqslant -\\frac{44}{9}$$ " } ], [ { "aoVal": "B", "content": "$$c\\leqslant -\\frac{55}{9}$$ " } ], [ { "aoVal": "C", "content": "$$c\\leqslant -\\frac{66}{9}$$ " } ], [ { "aoVal": "D", "content": "$$c\\leqslant -\\frac{77}{9}$$ " } ] ]

[ "竞赛->知识点->不等式->不等式的解法" ]

[ "因$$x\\ne 5$$，有$$\\frac{\\left\\textbar{} 3-x \\right\\textbar}{\\left\\textbar{} 5-x \\right\\textbar}\\leqslant \\frac{1}{2}\\Leftrightarrow 2\\left\\textbar{} 3-x \\right\\textbar\\leqslant \\left\\textbar{} 5-x \\right\\textbar$$． 当$$x\\leqslant 3$$时，$$2(3-x)\\leqslant (5-x)$$，解得$$x\\geqslant 1$$； 当$$3\\textless{}x\\textless{}5$$时，$$2(x-3)\\leqslant (5-x)$$，解得$$x\\leqslant \\frac{11}{3}$$； 当$$x\\textgreater5$$时，$$2(x-3)\\leqslant x-5$$，无解． 所以$$M=\\left { x\\textbar1\\leqslant x\\leqslant \\frac{11}{3} \\right }$$． 又因$${{x}^{2}}-2x+c=0,x=1\\pm \\sqrt{1-c}$$， 所以$$N=\\left { x\\textbar1-\\sqrt{1-c}\\leqslant x\\leqslant 1+\\sqrt{1-c} \\right }$$． 显然，$$1-\\sqrt{1-c}\\leqslant 1$$， 所以只要$$1+\\sqrt{1-c}\\geqslant \\frac{11}{3}$$，即$$c\\leqslant -\\frac{55}{9}$$，即有$$M\\cap N=M$$． " ]

[ "2022~2023学年陕西西安高一上学期期末（高新唐南中学）第7题", "2021~2022学年重庆高一上学期期末第6题", "2022~2023学年山东滨州高一上学期期末（北镇中学）第6题", "2022~2023学年甘肃兰州高一上学期期末（第六十中学）第6题", "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5~5题", "2022~2023学年甘肃兰州高一上学期期末（第六十中学）第6题", "2022~2023学年山东滨州高一上学期期末（北镇中学）第6题", "2022~2023学年甘肃兰州高一上学期期末（第六十中学）第6题", "2022~2023学年陕西西安高一上学期期末（高新唐南中学）第7题", "2022~2023学年陕西西安高一上学期期末（高新唐南中学）第7题", "2022~2023学年陕西西安高一上学期期末（高新唐南中学）第7题", "2022~2023学年广东广州荔湾区广州市协和中学高二上学期期末第4题" ]

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

[ "课内体系->知识点->基本初等函数" ]

[ "\\hfill\\break 【分析】\\\\ 比较大小，可先与常见的常数$$0,1$$进行比较，然后根据函数的单调性进行比较大小\\\\ 【详解】\\\\ $$c={{\\log }_{0.2}}0.3\\textless{} {{\\log }_{0.2}}0.2=1$$\\\\ $$a={{2}^{0.3}}\\textgreater1$$\\\\ $$b={{3}^{0.4}}\\textgreater1$$\\\\ 则有：$$a\\textgreater c,b\\textgreater c$$\\\\ $$a={{2}^{0.3}}\\textless{} {{3}^{0.3}}\\textless{} {{3}^{0.4}}$$\\\\ 故有：$$b\\textgreater a\\textgreater c$$\\\\ 故选：D " ]

[ "2008年黑龙江全国高中数学联赛竞赛初赛第8题5分", "2017~2018学年江西赣州寻乌县寻乌县第二中学高三上学期期中理科第11题5分", "2005年高考真题福建卷理科第10题5分", "2021年陕西宝鸡渭滨区高三二模理科第6题5分", "2013~2014学年12月辽宁沈阳沈河区沈阳市第二中学高二上学期月考理科第10题5分", "2016~2017学年12月北京海淀区中央民族大学附属中学高二上学期月考理科第6题5分" ]

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

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

[ "$$\\triangle M{{F}_{1}}{{F}_{2}}$$是正三角形，且边$$M{{F}_{1}}$$的中点在双曲线上， 设边$$M{{F}_{1}}$$的中点为$$P$$，则有$$\\angle {{F}_{1}}P{{F}_{2}}=90{}^{}\\circ $$， 从而$$\\left\\textbar{} P{{F}_{2}} \\right\\textbar=\\sqrt{3}c$$，$$\\left\\textbar{} P{{F}_{1}} \\right\\textbar=c$$． 根据双曲线的定义可知$$2a=\\left\\textbar{} P{{F}_{2}} \\right\\textbar-\\left\\textbar{} P{{F}_{1}} \\right\\textbar=\\left( \\sqrt{3}-1 \\right)c$$， 解得$$e=\\frac{c}{a}=\\frac{2}{\\sqrt{3}-1}=\\sqrt{3}+1$$． " ]

[ "1985年全国高中数学联赛竞赛一试第5题" ]

[ [ { "aoVal": "A", "content": "只有（1）和（2）是正确的 " } ], [ { "aoVal": "B", "content": "只有（1）和（3）是正确的 " } ], [ { "aoVal": "C", "content": "只有（1）和（4）是正确的 " } ], [ { "aoVal": "D", "content": "以上$$A$$、$$B$$、$$C$$都不正确 " } ] ]

[ "竞赛->知识点->复数与平面向量->复数的概念与运算" ]

[ "由题中给出的四个结论，可知本题需根据解方程的情况作出选择，于是考虑在方程 $$\\overline{Z}-\\lambda Z=W$$ 的两端同取共轭，得 $$Z-\\overline{\\lambda Z}=\\overline{W}$$， 以$$\\lambda $$乘两端，得 $$\\lambda Z-{{\\left\\textbar{} \\lambda \\right\\textbar}^{2}}\\overline{Z}=\\lambda \\overline{W}$$ 与原方程两端分别相加，得 $$\\overline{Z}(1-{{\\left\\textbar{} \\lambda \\right\\textbar}^{2}})=\\lambda \\overline{W}+W$$ 两端在取共轭，得 $$\\overline{Z}(1-{{\\left\\textbar{} \\lambda \\right\\textbar}^{2}})=\\overline{\\lambda }W+\\overline{W}$$ ∵$$ {{\\left\\textbar{} \\lambda \\right\\textbar}^{2}}\\ne 1$$，∴$$Z=\\frac{\\overline{\\lambda }W+\\overline{W}}{1-{{\\left\\textbar{} \\lambda \\right\\textbar}^{2}}}$$． " ]

[ "2012年山东全国高中数学联赛竞赛初赛第8题5分" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$35$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ] ]

[ "竞赛->知识点->数论模块->整除->最大公约数和最小公倍数" ]

[ "由$$100={{2}^{2}}{{5}^{2}}$$，令 $$A=\\left { n\\left\\textbar{} 2 \\right\\textbar n, n\\leqslant 100, n\\in {{\\mathbf{N}}^{*}} \\right }$$， $$B=\\left { n\\left\\textbar{} 5 \\right\\textbar n, n\\leqslant 100, n\\in {{\\mathbf{N}}^{*}} \\right }$$， 则$$\\left\\textbar{} A\\cup B \\right\\textbar=\\left\\textbar{} A \\right\\textbar+\\left\\textbar{} B \\right\\textbar-\\left\\textbar{} A\\cap B \\right\\textbar=50+20-10=60$$． 因此符合条件的真分数共有$$40$$个． 又若正数$$i$$和$$100$$互质，则$$100-i$$和$$100$$也互质，因此若$$\\frac{i}{100}$$是一个正的既约真分数，则$$\\frac{100-i}{100}$$也为一正的既约真分数，则$$\\frac{100-i}{100}+\\frac{i}{100}=1$$．因此所有真分数之和为$$20$$． " ]

[ "2017年四川全国高中数学联赛竞赛初赛第1题5分" ]

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

[ "竞赛->知识点->导数模块->导数" ]

[ "$${f}'\\left( x \\right)=\\frac{a}{x}+2x$$，由题设$${f}'\\left( 1 \\right)=0$$，解得$$a=-2$$ " ]

[ "1986年全国高中数学联赛竞赛一试第6题" ]

[ [ { "aoVal": "A", "content": "$$s\\textgreater t$$ " } ], [ { "aoVal": "B", "content": "$$s=t$$ " } ], [ { "aoVal": "C", "content": "$$s\\textless{}t$$ " } ], [ { "aoVal": "D", "content": "不确定 " } ] ]

[ "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->不等式->几个重要的不等式->均值" ]

[ "∵$$c=2R\\sin C=2\\sin C$$，又$$\\frac{1}{2}ab\\sin C=\\frac{1}{4}$$， ∴$$abc=1$$，于是 $$t=\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}$$ $$=\\frac{1}{2}\\left( \\frac{1}{a}+\\frac{1}{b} \\right)+\\frac{1}{2}\\left( \\frac{1}{b}+\\frac{1}{c} \\right)+\\frac{1}{2}\\left( \\frac{1}{c}+\\frac{1}{a} \\right)$$ $$\\mathsf{\\geqslant }\\sqrt{\\frac{\\mathsf{1}}{ab}}+\\sqrt{\\frac{1}{bc}}+\\sqrt{\\frac{1}{ca}}$$ $$=\\frac{\\sqrt{c}+\\sqrt{a}+\\sqrt{b}}{\\sqrt{abc}}$$ $$=\\sqrt{a}+\\sqrt{b}+\\sqrt{c}$$ $$=s$$． 且其中等号不可能取到，否则$$a=b=c=R=1$$，这显然是不可能的． 故选$$\\text{C}$$． " ]

[ "2015年河南全国高中数学联赛高二竞赛初赛第5题8分", "2016~2017学年江苏无锡高三上学期期中第13题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{13+2\\sqrt{3}}{14}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{13+3\\sqrt{3}}{14}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{13+4\\sqrt{3}}{14}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{13+5\\sqrt{3}}{14}$$ " } ] ]

[ "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，$$\\left[ 1+a+3\\left( 2+b \\right) \\right]\\left( \\frac{1}{1+a}+\\frac{4}{2+b} \\right)\\geqslant {{\\left( 1+2\\sqrt{3} \\right)}^{2}}$$， 所以$$\\frac{1}{1+a}+\\frac{4}{2+b}\\geqslant \\frac{13+4\\sqrt{3}}{14}$$． $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\frac{1}{1+a}+\\frac{4}{2+b}$$ $$=\\frac{1}{14}\\left[ \\left( a+1 \\right)+3\\left( 2+b \\right) \\right]\\cdot \\left( \\frac{1}{1+a}+\\frac{4}{2+b} \\right)$$ $$=\\frac{1}{14}\\left[ 13+\\frac{3\\left( 2+b \\right)}{a+1}+\\frac{4\\left( a+1 \\right)}{2+b} \\right]$$ $$\\geqslant \\frac{1}{14}\\left[ 13+2\\sqrt{\\frac{3\\left( 2+b \\right)}{a+1}\\cdot \\frac{4\\left( a+1 \\right)}{2+b}} \\right]$$ $$=\\frac{13+4\\sqrt{3}}{14}$$， 当且仅当$$\\frac{3\\left( 2+b \\right)}{a+1}=\\frac{4\\left( a+1 \\right)}{2+b}$$， 即$$a=\\frac{\\sqrt{3}}{2}b+\\sqrt{3-1}$$时等号成立， 故应填$$\\frac{13+4\\sqrt{3}}{14}$$． " ]

[ "2011年全国高中数学联赛竞赛复赛一试第8题8分" ]

[ [ { "aoVal": "A", "content": "$10$ " } ], [ { "aoVal": "B", "content": "$15$ " } ], [ { "aoVal": "C", "content": "$20$ " } ], [ { "aoVal": "D", "content": "$25$ " } ] ]

[ "竞赛->知识点->排列组合与概率->二项式定理及其应用" ]

[ "$${{a}_{n}}=\\text{C}_{200}^{n}\\cdot {{3}^{\\frac{200-n}{3}}}\\cdot {{2}^{\\frac{400-5 \\pi }{6}}}$$， 要使$${{a}_{n}}(1\\leqslant n\\leqslant 95)$$为整数，必有$$\\frac{200-n}{3}$$，$$\\frac{400-5n}{6}$$均为整数，从而$$6\\textbar n+4$$． 当$$n=2,8,14,20,26,32,38,44,50,56,62,68,74,80$$时，$$\\frac{200-n}{3}$$和$$\\frac{400-5n}{6}$$均为非负整数，所以$${{a}_{n}}$$为整数，共有$$14$$个． 当$$n=86$$时，$${{a}_{n}}=\\text{C}_{200}^{86}\\cdot {{3}^{38}}\\cdot 2_{{}}^{-5}$$，在$$\\text{C}_{200}^{86}=\\frac{200!}{86!\\cdot 114!}$$中， $$200$$中因数$$2$$的个数为$$\\left[ \\frac{200}{2} \\right]+\\left[ \\frac{200}{{{2}^{2}}} \\right]+\\left[ \\frac{200}{{{2}^{3}}} \\right]+\\left[ \\frac{200}{{{2}^{4}}} \\right]+\\left[ \\frac{200}{{{2}^{5}}} \\right]+\\left[ \\frac{200}{{{2}^{6}}} \\right]+\\left[ \\frac{200}{{{2}^{7}}} \\right]=197$$， 同理可计算得$$86$$中因数$$2$$的个数为$$82$$，$$114$$中因数$$2$$的个数为$$110$$， 所以$$\\text{C}_{200}^{86}$$中因数$$2$$的个数为$$197-82-110=5$$，故$${{a}_{86}}$$是整数． 当$$n=92$$时，$${{a}_{92}}=\\text{C}_{200}^{92}\\cdot {{3}^{36}}\\cdot {{2}^{-10}} $$在$$\\text{C}_{200}^{92}=\\frac{200!}{92!108!}$$中，同样可求得$$92$$中因数$$2$$的个数为$$88$$，$$108$$中因数$$2$$的个数为$$105$$，故$$\\text{C}_{200}^{86}$$中因数$$2$$的个数为$$197-88-105=4$$，故不是整数． 因此，整数项的个数为$$14+1=15$$． " ]

[ "2021年福建全国高中数学联赛竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "$$31$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$33$$ " } ], [ { "aoVal": "D", "content": "$$34$$ " } ], [ { "aoVal": "E", "content": "$$35$$ " } ] ]

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

[ "设$$f\\left( x \\right)={{x}^{2}}+ax+b$$，$$g\\left( x \\right)={{x}^{2}}+cx+d$$， 则由题设条件可得 $$20\\left( 1-a+b \\right)=21\\left( 1-c+d \\right)\\cdots \\cdots $$①﹐$$20\\left( 1+a+b \\right)=21\\left( 1+c+d \\right)\\cdots \\cdots $$②． ①、②两式左右两边分别相加，得$$40+40b=42+42d$$，$$20b=1+21d$$﹔ ①、②两式左右两边分别相减，得$$-40a=-42c$$，$$20a=21c$$． 另由$$g\\left( 6 \\right)=35$$，得$$36+6c+d=35$$． 所以，$$36+6\\times \\frac{20}{21}a+\\frac{20b-1}{21}=35$$，$$6a+b=-1$$， 所以，$$f\\left( 6 \\right)=36+6a+b=35$$． 设$$h\\left( x \\right)=21g\\left( x \\right)-20f\\left( x \\right)$$， 则由条件知$$h\\left( x \\right)$$是二次项系数为$$1$$的二次函数． 又$$h\\left( -1 \\right)=21g\\left( -1 \\right)-20f\\left( -1 \\right)=0$$，$$h\\left( 1 \\right)=21g\\left( 1 \\right)-20f\\left( 1 \\right)=0$$， 所以，$$h(x)=(x+1)(x-1)={{x}^{2}}-1$$． 因此，$$h\\left( 6 \\right)=21g\\left( 6 \\right)-20f\\left( 6 \\right)={{6}^{2}}-1=35$$． 所以，$$21\\times 35-20f\\left( 6 \\right)=35$$，$$f\\left( 6 \\right)=35$$． " ]

[ "全国高中数学联赛竞赛模拟一试（十三）第6题" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "竞赛->知识点->不等式->多元函数极值（一试）", "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", "课内体系->方法->均值不等式法" ]

[ "由均值不等式及题设条件知 $$2cd\\leqslant {{\\left( \\frac{c+2d}{2} \\right)}^{2}}=\\frac{1}{4}\\Rightarrow cd\\leqslant \\frac{1}{8}$$①． 当且仅当$$c=\\frac{1}{2}$$，$$d=\\frac{1}{4}$$时，以上两式的等号成立． 此时，由式①得$$\\frac{1}{a}+\\frac{1}{bcd}\\geqslant \\frac{1}{a}+\\frac{8}{b}$$． 又由条件及柯西不等式得 $$\\frac{1}{a}+\\frac{8}{b}=\\left( \\frac{1}{a}+\\frac{8}{b} \\right)\\left( a+2b \\right)\\geqslant {{\\left( 1+4 \\right)}^{2}}=25$$， 当且仅当$$b=2a$$时，上式等号成立． 结合条件，知此时$$a=\\frac{1}{5}$$，$$b=\\frac{2}{5}$$． 于是，$$\\frac{1}{a}+\\frac{1}{bcd}\\geqslant 25$$． 当 $$a=\\frac{1}{5}$$，$$b=\\frac{2}{5}$$，$$c=\\frac{1}{2}$$，$$d=\\frac{1}{4}$$时， $$\\frac{1}{a}+\\frac{1}{bcd}$$取得最小值$$25$$． " ]

[ "2008年全国高中数学联赛竞赛一试第10题9分", "2010年河南全国高中数学联赛竞赛初赛第12题5分" ]

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

[ "竞赛->知识点->数列与数学归纳法->数列的通项与求和", "课内体系->知识点->数列" ]

[ "$${{a}_{n+1}}={{S}_{n+1}}-{{S}_{n}}=\\frac{n}{(n+1)(n+2)}-{{a}_{n+1}}-\\frac{n-1}{n(n+1)}+{{a}_{n}}$$， 即$$2{{a}_{n+1}}=\\frac{n+2-2}{(n+1)(n+2)}-\\frac{1}{n+1}+\\frac{1}{n(n+1)}+{{a}_{\\begin{smallmatrix} n \\end{smallmatrix}}}$$ $$=\\frac{-2}{(n+1)(n+2)}+{{a}_{n}}+\\frac{1}{n(n+1)}$$， 由此得$$2\\left( {{a}_{n+1}}+\\frac{1}{(n+1)(n+2)} \\right)={{a}_{n}}+\\frac{1}{n(n+1)}$$， 令$${{b}_{n}}={{a}_{n}}+\\frac{1}{n(n+1)}$$，$${{b}_{1}}={{a}_{1}}+\\frac{1}{2}=\\frac{1}{2}({{a}_{1}}=0)$$， 有$${{b}_{n+1}}=\\frac{1}{2}{{b}_{n}}$$，故$${{b}_{n}}=\\frac{1}{{{2}^{n}}}$$，所以$${{a}_{n}}=\\frac{1}{{{2}^{n}}}-\\frac{1}{n(n+1)}$$． " ]

[ "2019年上海全国高中数学联赛高三竞赛初赛第4题7分", "2019年上海全国高中数学联赛竞赛初赛第4题7分" ]

[ [ { "aoVal": "A", "content": "$$4n-\\frac{3}{4}$$ " } ], [ { "aoVal": "B", "content": "$$4n-\\frac{5}{4}$$ " } ], [ { "aoVal": "C", "content": "$$4n-\\frac{7}{4}$$ " } ], [ { "aoVal": "D", "content": "$$4n-\\frac{9}{4}$$ " } ] ]

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

[ "由$$\\sqrt{8{{S}_{n+1}}+2\\left( n+1 \\right)}-\\sqrt{8{{S}_{n}}+2n}=d\\left( n\\geqslant 1 \\right)$$ $$\\Rightarrow 8{{S}_{n+1}}-8{{S}_{n}}+2=d+2d\\sqrt{8{{S}_{n}}+2n}$$ $$\\Rightarrow 8{{a}_{n+1}}+2=d+2d\\sqrt{8{{S}_{n}}+2n}$$， 类似地， $$8{{a}_{n+2}}+2=d+2d\\sqrt{8{{S}_{n+1}}+2\\left( n+1 \\right)}$$， 将以上两式相减得 $$8\\left( {{a}_{n+2}}-{{a}_{n+1}} \\right)=2d\\left( \\sqrt{8{{S}_{n+1}}+2\\left( n+1 \\right)}-\\sqrt{8{{S}_{n}}+2n} \\right)$$ $$\\Rightarrow 8d=2{{d}^{2}}$$， 而$$d\\ne 0$$，故$$d=4$$， 又$$\\sqrt{8{{S}_{2}}+4}-\\sqrt{8{{S}_{1}}+2}=\\sqrt{16{{a}_{1}}+36}-\\sqrt{8{{a}_{1}}+2}=4$$， 由此解得$${{a}_{1}}=\\frac{7}{4}$$， 从而，$${{a}_{n}}=\\frac{7}{4}+4\\left( n-1 \\right)=4n-\\frac{9}{4}$$． 故答案为：$$4n-\\frac{9}{4}$$． " ]

[ "2010年AMC10竞赛B第11题" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$60$$ " } ], [ { "aoVal": "C", "content": "$$75$$ " } ], [ { "aoVal": "D", "content": "$$80$$ " } ], [ { "aoVal": "E", "content": "$$100$$ " } ] ]

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

[ "Let the listed price be $$(100+p)$$, where $$p\\textgreater0$$. Coupon $$A$$ saves us: $$0.15(100+p)=(0.15 p+15)$$. Coupon $$B$$ saves us: $$30$$. Coupon $$C$$ saves us: $$0.25p$$. Now, the condition is that $$A$$ has to be greater than or equal to either $$B$$ or $$C$$ which give us the following inequalities: $$A≥B⇒0.15 p+15≥30⇒p≥100$$, $$A≥C⇒0.15 p+15≥0.25p⇒p≤150$$. We see here that the greatest possible value for $$p$$ is $$150$$,~ thus $$y= 100+150=250$$ and the smallest value for $$p$$ is $$100$$ so $$x= 100+100=200$$. The difference between $$y$$ and $$x$$ is $$y-x=250-200=(\\rm A) 50$$. " ]

[ "2017年天津全国高中数学联赛竞赛初赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$$z$$可能为纯虚数 " } ], [ { "aoVal": "B", "content": "$$z$$可能为实数 " } ], [ { "aoVal": "C", "content": "$$z$$的实部小于$$2$$ " } ], [ { "aoVal": "D", "content": "$$z$$的辐角可能为$$\\frac{ \\pi }{4}$$ " } ] ]

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

[ "令$$z=b\\text{i}$$，不难算出$$\\left\\textbar{} z-\\left\\textbar{} z+1 \\right\\textbar{} \\right\\textbar=\\sqrt{{{b}^{2}}+{{b}^{2}}+1}=\\left\\textbar{} z+\\left\\textbar{} z-1 \\right\\textbar{} \\right\\textbar$$，$$A$$对； 令$$z=0$$，显然$$\\left\\textbar{} z-\\left\\textbar{} z+1 \\right\\textbar{} \\right\\textbar=\\left\\textbar{} z+\\left\\textbar{} z-1 \\right\\textbar{} \\right\\textbar=1$$，$$B$$对； 设$$z=a+b\\text{i}$$，由题意，$${{\\left( a-\\sqrt{{{\\left( a+1 \\right)}^{2}}+{{b}^{2}}} \\right)}^{2}}={{\\left( a+\\sqrt{{{\\left( a-1 \\right)}^{2}}+{{b}^{2}}} \\right)}^{2}}$$， 若$$a\\geqslant 2$$，则$${{\\left( a-\\sqrt{{{\\left( a+1 \\right)}^{2}}+{{b}^{2}}} \\right)}^{2}}\\textless{}{{\\left( a+1 \\right)}^{2}}+{{b}^{2}}$$， 但$${{\\left( a+\\sqrt{{{\\left( a-1 \\right)}^{2}}+{{b}^{2}}} \\right)}^{2}}\\geqslant {{\\left( a+\\sqrt{1+{{b}^{2}}} \\right)}^{2}}={{a}^{2}}+1+{{b}^{2}}+2a\\sqrt{1+{{b}^{2}}}\\geqslant {{a}^{2}}+1+{{b}^{2}}+2a={{\\left( a+1 \\right)}^{2}}+{{b}^{2}}$$， 矛盾，所以$$a\\textless{}2$$，$$C$$对； " ]

[ "2008年甘肃全国高中数学联赛竞赛初赛第6题6分" ]

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

[ "竞赛->知识点->不等式->换元技巧->三角换元" ]

[ "设$$\\begin{cases}x=3+2\\cos \\theta y=1+\\sin \\theta \\end{cases}$$， 则$$t=\\frac{x+y-3}{x-y+1}=\\frac{2\\cos \\theta +\\sin \\theta +1}{2\\cos \\theta -\\sin \\theta +3}$$， 化为$$(2t-2)\\cos \\theta -(t+1)\\sin \\theta +3t-1=0$$， 由万能公式得$$(t+1){{\\tan }^{2}}\\frac{\\theta }{2}-2(t+1)\\tan \\frac{\\theta }{2}+5t-3=0$$． 当$$t\\ne 1$$时，$$\\frac{1}{4}\\Delta ={{(t+1)}^{2}}-(t+1)(5t-3)\\geqslant 0$$， 所以$$-1\\textless{}t\\leqslant 1$$； 当$$t=-1$$时，解得$$\\cos \\theta =-1$$有意义．故选$$\\text{C}$$． " ]

[ "2012年AMC10竞赛A第16题" ]

[ [ { "aoVal": "A", "content": "$$1000$$ " } ], [ { "aoVal": "B", "content": "$$1250$$ " } ], [ { "aoVal": "C", "content": "$$2500$$ " } ], [ { "aoVal": "D", "content": "$$5000$$ " } ], [ { "aoVal": "E", "content": "$$10000$$ " } ] ]

[ "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" ]

[ "First consider the first two runners. The faster runner will lap the slower runner exactly once, or run $$500$$ meters farther. Let $$x$$ be the time these runners run in seconds. Because $$4.8x-4.4x=500\\Rightarrow x=1250$$ is a multiple of $$500$$, it turns out they just meet back at the start line. Now we must find a time that is a multiple of $$1250$$ and results in the $$5.0\\rm m/s$$ runner to end up on the start line. Every $$1250$$ seconds, that fastest runner goes $$5.0(1250)=6250$$ meters. In $$2(1250)=2500$$ seconds, he goes $$5.0(2500)=12500$$ meters. Therefore the runners run $$\\boxed{\\rm(C)\\textasciitilde2500} $$seconds. Working backwards from the answers starting with the smallest answer, if they had run $$1000$$ seconds, they would have run $$4400$$, $$4800$$, $$5000$$ meters, respectively. The first two runners have a difference of $$400$$ meters, which is not a multiple of $$500$$ (one lap), so they are not in the same place. If they had run $$1250$$ seconds, the runners would have run $$5500$$, $$6000$$, $$6250$$ meters, respectively. The last two runners have a difference of $$250$$ meters, which is not a multiple of $$500$$. If they had run $$2500$$ seconds, the runners would have run $$11000$$, $$12000$$, $$12500$$ meters, respectively. The distance separating each pair of runners is a multiple of $$500$$, so the answer is $$\\boxed{\\rm(C)\\textasciitilde2500}$$ seconds. Let $$t$$ be the time run in seconds, then the difference in meters run between the three runners is $$0.2t$$, $$0.4t$$, $$0.6t$$. For them to be at the same location all of them need to be multiples of $$500$$. It is now easy to see that $$0.2t=500$$, $$0.4t=1000$$, $$0.6t=1500$$, so $$t=\\boxed{\\rm(C)\\textasciitilde2500}$$. After $$t$$ seconds, respectively the runners would\\textquotesingle ve ran $$4.4t$$, $$4.8t$$ and $$5t$$ meters. Their current positions on the track are these values $$(\\rm mod\\textasciitilde500)$$. We\\textquotesingle re trying to find the value of $$t$$ such that $$4.4t\\equiv 4.8t\\equiv 5t\\textasciitilde{} \\rm (mod \\textasciitilde500)$$ Subtracting $$4.4t$$ on all sides, we get $$0\\equiv 0.4t\\equiv 0.6t\\textasciitilde{} \\rm (mod \\textasciitilde500)$$ Now, we must find a value for $$t$$ such that both $$0.6t$$ and $$0.4t$$ are simultaneously multiples of $$500$$. Plugging in $$500$$ for $$0.4t$$ we get $$t=1250$$, but this does not work for $$0.6t$$ ( $$750$$ isn\\textquotesingle t a multiple of $$500$$). Plugging in $$0.4t=1000$$, we get $$t=2500$$, and this does work for $$0.6t$$. Therefore, $$t=2500$$ and the answer is $$\\rm(C)\\textasciitilde2500$$. $$\\cdot$$ Note: Modular Arithmetic works only for integral values, so my usage of decimals is technically incorrect but the intuition leads to the right answer. Similar to the solution above, but is much quicker and does not involve trial and error. This uses decimal mod arithmetic, which can be justified by intuition $$\\cdots$$~~After $$t$$ seconds, respectively the runners would\\textquotesingle ve ran $$4.4t$$, $$4.8t$$, and $$5t$$ meters. These three values are congruent $$\\rm(mod\\textasciitilde500)$$, so $$4.4t\\equiv 4.8t\\equiv 5t\\textasciitilde{} \\rm (mod \\textasciitilde500)$$. Subtract $$4.4t$$ from all three sides to get $$0$$, $$0.4t$$, and $$0.6t$$ are congruent. Now all we need to find is a value of t for which $$0.4t$$ and $$0.6t$$ are congruent mod $$500$$. Subtract $$0.4t$$ from both sides to get $$0.2t$$ and $$0$$ are congruent mod $$500$$, or that $$0.2t=\\frac t5$$ is a multiple of $$500$$. Let $$t=500k$$, so we want $$100k$$ to be a multiple of $$500$$, or $$k$$ to be a multiple of $$5$$. Therefore, thes mallest value of $$t$$ is when $$k=5$$, and when $$t=500k=500(5)=2500\\rm(C)$$. " ]

[ "2009年吉林全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{4}{9}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{5}{9}$$ " } ] ]

[ "竞赛->知识点->复数与平面向量->平面向量的概念与运算", "竞赛->知识点->平面几何->内心相关问题（二试）" ]

[ "在$$\\triangle ABC$$中，$$I$$为内心，连$$AI$$并延长交$$BC$$于$$D$$点， 则$$D$$分$$BC$$的比$$\\lambda =\\frac{AB}{AC}=\\frac{4}{2}=2$$． 故$$\\overrightarrow{AD}=\\frac{1}{3}\\overrightarrow{AB}+\\frac{2}{3}\\overrightarrow{AC}$$．又$$BC=3$$，故$$BD=2, DC=1$$． 又在$$\\triangle ABD$$中，$$I$$分$$AD$$的比$${\\lambda }'=\\frac{AB}{BD}=\\frac{4}{2}=2$$， 即$$\\overrightarrow{AI}=\\frac{2}{3}\\overrightarrow{AD}=\\frac{2}{9}\\overrightarrow{AB}+\\frac{4}{9}\\overrightarrow{AC}$$，所以$$x+y=\\frac{2}{3}$$． " ]

[ "2002年全国全国高中数学联赛竞赛一试第5题6分" ]

[ [ { "aoVal": "A", "content": "$$\\text{C}_{100}^{50}$$ " } ], [ { "aoVal": "B", "content": "$$\\text{C}_{90}^{50}$$ " } ], [ { "aoVal": "C", "content": "$$\\text{C}_{100}^{49}$$ " } ], [ { "aoVal": "D", "content": "$$\\text{C}_{99}^{49}$$ " } ] ]

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

[ "不妨设$${{b}_{1}}\\textless{}{{b}_{2}}\\textless{}\\cdots \\textless{}{{b}_{50}}$$，将$$A$$中元素$${{a}_{1}}$$，$${{a}_{2}}$$，\\ldots，$${{a}_{100}}$$按顺序分为非空的$$50$$组，定义映射$$f:A\\to B$$，使得第$$i$$组的元素在$$f$$之下的象都是$${{b}_{i}}\\left( i=1,2, \\cdots ,50 \\right)$$，易知这样的$$f$$满足题设要求，每个这样的分组都一一对应满足条件的映射，于是满足题设要求的映射$$f$$的个数与$$A$$按足码顺序分为$$50$$组的分法数相等，而$$A$$的分法数为$$\\text{C}_{99}^{49}$$，则这样的映射共有$$\\text{C}_{99}^{49}$$． 故选$$\\text{D}$$． " ]

[ "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1题" ]

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

[]

[ "\\hfill\\break 【分析】\\\\ 先求出集合$B$，再依据并集的定义求并集.\\\\ 【详解】\\\\ $B=\\left {x\\left\\textbar{} x^{2}-2x-3\\leq 0\\right.\\right }=\\left {x\\left\\textbar{} -1\\leq x\\leq 3\\right.\\right }$，又$A=\\left {x\\left\\textbar{} x\\textless{} 2\\right.\\right }$，\\\\ 所以$A\\cup B=\\left(-\\mathrm{\\infty }\\mathrm{，}3\\right]$\\\\ 故选：D " ]

[ "2009年河北全国高中数学联赛竞赛初赛第2题6分" ]

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

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "利用复合函数的单调性求解，令$${{a}^{x}}=t$$，则 $$y=g(t)={{t}^{2}}-(3{{a}^{2}}+1)t$$． ①当$$a\\textgreater1$$时，$$t$$是关于$$x$$的增函数，欲使$$f(x)$$在$$x\\in [0,+\\infty )$$上是增函数，需$$g(t)$$在$$t\\in [1,+\\infty )$$上是增函数，故$$\\frac{3{{a}^{2}}+1}{2}\\leqslant 1$$，即$${{a}^{2}}\\leqslant \\frac{1}{3}$$，矛盾． ②当$$0\\textless{}a\\textless{}1$$时，$$t$$是关于$$x$$的减函数，欲使$$f(x)$$在$$x\\in [0,+\\infty )$$上是增函数，需$$g(t)$$在$$t\\in (0,1]$$上是减函数，故$$\\frac{3{{a}^{2}}+1}{2}\\geqslant 1$$，即$${{a}^{2}}\\geqslant \\frac{1}{3}$$，所以$$\\frac{\\sqrt{3}}{3}\\leqslant a\\textless{}1$$． 综上可知选$$\\text{B}$$． " ]

[ "竞赛第1题" ]

[ [ { "aoVal": "A", "content": "3 " } ], [ { "aoVal": "B", "content": "5 " } ], [ { "aoVal": "C", "content": "7 " } ], [ { "aoVal": "D", "content": "9 " } ] ]

[]

[ "\\hfill\\break 【分析】\\\\ 可证明集合\\emph{A}的正数至多有3个，负数至多有3个，故可判断\\emph{n}的最大值.\\\\ 【详解】\\\\ 不妨设$a_{1}\\textgreater{} a_{2}\\textgreater{} \\cdots \\textgreater{} a_{n}$，若集合\\emph{A}中的正数个数不小于4，取$(i,j,k)=(1,2,3)$，\\\\ 可得$a_{2}+a_{3}=a_{1}$，取$(i,j,k)=(1,2,4)$，可得$a_{2}+a_{4}=a_{1}$，因此$a_{3}=a_{4}$，矛盾．\\\\ 因此集合\\emph{A}的正数至多有3个，同理，集合\\emph{A}中的负数至多有3个．\\\\ 又考虑$A= {3,2,1,0,-1,-2,-3 }$，\\\\ 符合题意，因此\\emph{n}的最大值为7．\\\\ 故选：C. " ]

[ "2005年全国高中数学联赛竞赛一试第2题6分" ]

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

[ "竞赛->知识点->立体几何与空间向量->空间向量" ]

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

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

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

[ "知识标签->知识点->函数->函数的应用->简单的函数方程", "知识标签->知识点->函数->函数及其表示->函数的概念->复合函数", "知识标签->知识点->函数->函数的性质->单调性", "知识标签->知识点->函数->函数的性质->奇偶性", "知识标签->知识点->函数->函数的性质->周期性", "知识标签->素养->数学运算", "知识标签->素养->数据分析", "知识标签->素养->逻辑推理", "知识标签->题型->函数->函数的性质->周期性->函数周期性判断", "知识标签->题型->函数->函数的性质->单调性->用定义法证明函数的单调性", "知识标签->题型->函数->函数的性质->奇偶性->利用定义判断函数奇偶性" ]

[ "若$$f\\left( x \\right)$$是奇函数，则$$f\\left( f\\left( -x \\right) \\right)=f\\left( -f\\left( x \\right) \\right)=-f\\left( f\\left( x \\right) \\right)$$，（$$1$$）对； 若$$f\\left( x+T \\right)=f\\left( x \\right)$$，则$$f\\left( f\\left( x+T \\right) \\right)=f\\left( f\\left( x \\right) \\right)$$，（$$2$$）对； 对任意$${{x}_{1}}\\textless{}{{x}_{2}}$$，有$$f\\left( {{x}_{1}} \\right)\\textgreater f\\left( {{x}_{2}} \\right)$$，则$$f\\left( f\\left( {{x}_{1}} \\right) \\right)\\textless{}f\\left( f\\left( {{x}_{2}} \\right) \\right)$$，（$$3$$）对； 取$$f\\left( x \\right)=\\begin{cases}1,x=0 0,x=1 \\left\\textbar{} x \\right\\textbar+1,其它 \\end{cases}$$，则$$f\\left( f\\left( x \\right) \\right)=\\begin{cases}0,x=0 1,x=1 \\left\\textbar{} x \\right\\textbar+2,其它 \\end{cases}$$，（$$4$$）错． 故选$$\\text{C}$$． " ]

[ "2018年山西全国高中数学联赛竞赛初赛第6题8分" ]

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

[ "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正弦", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->半角公式", "课内体系->素养->数学运算" ]

[ "记$$S=\\cos \\frac{2\\pi }{7}\\cos \\frac{4\\pi }{7}\\cos \\frac{6\\pi }{7}$$，则$$S=-\\cos \\frac{\\pi }{7}\\cos \\frac{2\\pi }{7}\\cos \\frac{4\\pi }{7}$$， $$-S\\cdot 8\\sin \\frac{\\pi }{7}=8\\sin \\frac{\\pi }{7}\\cdot \\cos \\frac{\\pi }{7}\\cos \\frac{2\\pi }{7}\\cos \\frac{4\\pi }{7}=4\\sin \\frac{2\\pi }{7}\\cdot \\cos \\frac{2\\pi }{7}\\cos \\frac{4\\pi }{7}=2\\sin \\frac{4\\pi }{7}\\cos \\frac{4\\pi }{7}$$ $$=\\sin \\frac{8\\pi }{7}=-\\sin \\frac{\\pi }{7}$$，所以，$$S=\\frac{1}{8}$$． " ]

[ "2004年高考真题天津卷文科第9题5分", "2008年贵州全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$y=1+{{\\log }_{3}}x(x\\textgreater0)$$ " } ], [ { "aoVal": "B", "content": "$$y=-1+{{\\log }_{3}}x(x\\textgreater0)$$ " } ], [ { "aoVal": "C", "content": "$$y=1+{{\\log }_{3}}x(1\\leqslant x\\textless{}3)$$ " } ], [ { "aoVal": "D", "content": "$$y=-1+{{\\log }_{3}}x(1\\leqslant x\\textless{}3)$$ " } ] ]

[ "竞赛->知识点->函数->基本初等函数" ]

[ "由$$y={{3}^{x+1}}$$解得，$$x=-1+{{\\log }_{3}}y$$， 因为$$-1\\leqslant x\\textless{}0$$，得$$1\\textless{}y\\textless{}3$$， 所以函数$$y={{3}^{x+1}}(-1\\leqslant x\\textless{}0)$$的反函数是$$y=-1+{{\\log }_{3}}x(1\\leqslant x\\textless{}3)$$， 故选$$\\text{D}$$. " ]

[ "2008年贵州全国高中数学联赛竞赛初赛第8题5分" ]

[ [ { "aoVal": "A", "content": "$$y=\\sin \\left( 4x+\\frac{2 \\pi }{3} \\right)-2$$ " } ], [ { "aoVal": "B", "content": "$$y=\\sin \\left( 4x-\\frac{ \\pi }{6} \\right)$$ " } ], [ { "aoVal": "C", "content": "$$y=\\sin \\left( 2x+\\frac{ \\pi }{6} \\right)$$ " } ], [ { "aoVal": "D", "content": "$$y=\\cos \\left( 4x+\\frac{2 \\pi }{3} \\right)$$ " } ] ]

[ "竞赛->知识点->三角函数->三角函数的图像与性质" ]

[ "由题意，把函数$$y=\\sin \\left( 2x+\\frac{ \\pi }{6} \\right)-1$$的图象按向量$$\\overrightarrow{a}=\\left( \\frac{ \\pi }{6},1 \\right)$$平移， 可得$$y=\\sin \\left[ 2\\left( x-\\frac{ \\pi }{6} \\right)+\\frac{ \\pi }{6} \\right]=\\sin \\left( 2x-\\frac{ \\pi }{6} \\right)$$， 再把所得图象上各点的横坐标缩短为原来的$$\\frac{1}{2}$$， 可得$$y=\\sin \\left( 4x-\\frac{ \\pi }{6} \\right)$$，故选$$\\text{B}$$. " ]

[ "2008年辽宁全国高中数学联赛竞赛初赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{29}{189}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{29}{63}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{34}{63}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{4}{7}$$ " } ] ]

[ "竞赛->知识点->立体几何与空间向量->空间中的角与距离", "竞赛->知识点->排列组合与概率->概率初步" ]

[ "从$$8$$个顶点中任取两点可确定$$\\text{C}_{8}^{2}=28$$（条）直线， 从$$8$$个顶点中任取$$4$$个不共面的点共有$$\\text{C}_{8}^{4}-12$$组， 而其中每一组不共面的$$4$$点可出现$$3$$对异面直线， 所以所求的概率为$$\\frac{3(\\text{C}_{8}^{4}-12)}{\\text{C}_{28}^{2}}=\\frac{29}{63}$$. 故选$$\\text{B}$$． " ]

[ "2011年甘肃全国高中数学联赛竞赛初赛第8题7分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "课内体系->知识点->几何证明选讲", "竞赛->知识点->组合->容斥原理与极端原理（二试）", "竞赛->知识点->数论模块->取整函数->与[x]有关的方程和不等式" ]

[ "在平面上，由满足$${{[x]}^{2}}+{{[y]}^{2}}=50$$的点所形成的图形关于$$x$$轴,y轴对称，可先考察第一象限的情况，即当$$x\\textgreater0, y\\textgreater0$$时，由$${{[x]}^{2}}+{{[y]}^{2}}=50$$得 $$\\begin{cases}{{[x]}^{2}}=49 {{[y]}^{2}}=1 \\end{cases}\\begin{cases}{{[x]}^{2}}=25 {{[y]}^{2}}=25 \\end{cases}\\begin{cases}{{[x]}^{2}}=1 {{\\left[ y \\right]}^{2}}=49 \\end{cases}$$ 所以相应有 $$\\begin{cases}[x]=7 [y]=1 \\end{cases}\\begin{cases}[x]=5 [y]=5 \\end{cases}\\begin{cases}[x]=1 [y]=7 \\end{cases}$$ 因此 $$\\begin{cases}7\\leqslant x\\textless{}8 1\\leqslant y\\textless{}2 \\end{cases}\\begin{cases}5\\leqslant x\\textless{}6 5\\leqslant y\\textless{}6 \\end{cases}\\begin{cases}1\\leqslant x\\textless{}2 7\\leqslant y\\textless{}8 \\end{cases}$$ 所以在第一象限形成的图形的面积由容斥原理计算为$$3$$． 故在平面上，由满足$${{[x]}^{2}}+{{[y]}^{2}}=50$$的点所形成的图形的面积是$$12$$． " ]

[ "1995年全国高中数学联赛竞赛一试第2题" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "竞赛->知识点->复数与平面向量->模、辐角与单位根", "竞赛->知识点->复数与平面向量->复数的概念与运算" ]

[ "设$${{Z}_{1}}=\\cos \\theta +\\text{i}\\sin \\theta $$，则： $${{Z}_{k}}=(\\cos \\theta +\\text{i}\\sin \\theta )(\\cos \\frac{2(k-1)\\mathsf{\\pi }}{20}+\\text{isin }\\frac{2(k-1)\\mathsf{\\pi }}{20})$$，$$1\\mathsf{\\leqslant }k\\mathsf{\\leqslant }20$$由$$1995=20\\times 99+15$$，得 $$\\mathop{Z}_{k}^{1995}=(\\cos 1995\\theta +\\text{isin 1995}\\theta ){{(\\cos \\frac{3\\mathsf{\\pi }}{2}+\\text{isin }\\frac{3\\mathsf{\\pi }}{2})}^{k-1}}$$ $$=(\\cos 1995\\theta +\\text{isin 1995}\\theta ){{(-i)}^{k-1}}$$，$$k=1$$，$$2$$，$$\\ldots $$，$$20$$， 共有四个不同的值． 故选$$\\text{A}$$． " ]

[ "1999年全国全国高中数学联赛竞赛一试第4题6分" ]

[ [ { "aoVal": "A", "content": "命题Ⅰ正确，命题Ⅱ不正确 " } ], [ { "aoVal": "B", "content": "命题Ⅱ正确，命题Ⅰ不正确 " } ], [ { "aoVal": "C", "content": "两个命题都正确 " } ], [ { "aoVal": "D", "content": "两个命题都不正确 " } ] ]

[ "竞赛->知识点->立体几何与空间向量->空间中的平行和垂直" ]

[ "易知命题Ⅰ不正确； 又可以取无穷多个平行平面， 在每个平面上取一条直线， 且使这些直线两两不同向， 则这些直线中的任意两条都是异面直线， 从而命题Ⅱ也不正确． " ]

[ "2012年陕西全国高中数学联赛竞赛初赛第10题8分" ]

[ [ { "aoVal": "A", "content": "$$13000$$ " } ], [ { "aoVal": "B", "content": "$$14000$$ " } ], [ { "aoVal": "C", "content": "$$15000$$ " } ], [ { "aoVal": "D", "content": "以上都不对 " } ] ]

[ "竞赛->知识点->数列与数学归纳法->数列的通项与求和", "课内体系->知识点->数列" ]

[ "先假设要完成$$21$$根电线杆的运栽，每次$$3$$根，设第$$k$$次往返的路程为$${{a}_{k}}\\left( k=1, 2,\\cdots , 7 \\right)$$，则$${{a}_{1}}=2\\times 600=1200$$，且 $${{a}_{k+1}}={{a}_{k}}+2\\times 150\\left( k=1, 2,\\cdots , 6 \\right)$$． 所以$$\\left { {{a}_{n}} \\right }$$是首项为$$1200$$，公差为$$300$$的等差数列． 故$${{S}_{7}}=7\\times 1200+\\frac{7\\times 6}{2}\\times 300=14700$$（m）． 但实际只运栽$$20$$根，那么必有一次运$$2$$根，其余$$6$$次均运$$3$$根． 若将运$$2$$根的安排在第$$k$$次（$$k=1, 2,\\cdots , 7$$），则$${{a}_{1}}, {{a}_{2}},\\cdots , {{a}_{k-1}}$$均不变，$${{a}_{k}}, {{a}_{k+1}},\\cdots , {{a}_{7}}$$各减少$$100\\text{m}$$，所以 $${{S}_{7}}\\left( k \\right)={{S}_{7}}-100\\left( 8-k \\right)=13800+100k$$. 显然，当$$k=1$$，即第$$1$$次运$$2$$根，其余$$6$$次各运$$3$$根时，总的行程最小，最小值为$$14000\\text{m}$$． " ]

[ "2005年全国高中数学联赛竞赛一试第6题6分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{5}{7}+\\frac{5}{{{7}^{2}}}+\\frac{6}{{{7}^{3}}}+\\frac{3}{{{7}^{4}}}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{5}{7}+\\frac{5}{{{7}^{2}}}+\\frac{6}{{{7}^{3}}}+\\frac{2}{{{7}^{4}}}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{7}+\\frac{1}{{{7}^{2}}}+\\frac{0}{{{7}^{3}}}+\\frac{4}{{{7}^{4}}}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{7}+\\frac{1}{{{7}^{2}}}+\\frac{0}{{{7}^{3}}}+\\frac{3}{{{7}^{4}}}$$ " } ] ]

[ "竞赛->知识点->数论模块->整除->质数（算数基本定理）", "竞赛->知识点->集合->集合的概念与运算" ]

[ "用$${{[{{a}_{1}}{{a}_{2}}\\ldots {{a}_{k}}]}_{p}}$$表示$$k$$位$$p$$进制数，将集合$$M$$中的每个数乘以$${{7}^{4}}$$，得 $${{M}^{\\prime }}= {{{a}_{1}}\\cdot {{7}^{3}}+{{a}_{2}}\\cdot {{7}^{2}}+{{a}_{3}}\\cdot 7+{{a}_{4}}\\textbar{{a}_{i}}\\in T,i=1,2,3,4 }= {{{[{{a}_{1}}{{a}_{2}}{{a}_{3}}{{a}_{4}}]}_{7}}\\textbar{{a}_{i}}\\in T,i=1,2,3,4 }$$． $${{M}^{\\prime }}$$中的最大数为$${{[6666]}_{7}}={{[2400]}_{10}}$$． 在十进制数中，从$$2400$$起从大到小顺序排列的第$$2005$$个数是$$2400-2004=396$$． 而$${{[396]}_{10}}={{[1104]}_{7}}$$将此数除以$${{7}^{4}}$$，便得$$M$$中的数$$\\frac{1}{7}+\\frac{1}{{{7}^{2}}}+\\frac{0}{{{7}^{3}}}+\\frac{4}{{{7}^{4}}}$$． 故选$$\\text{C}$$． " ]

[ "2009年山东全国高中数学联赛竞赛初赛第9题6分" ]

[ [ { "aoVal": "A", "content": "$$141$$个 " } ], [ { "aoVal": "B", "content": "$$144$$个 " } ], [ { "aoVal": "C", "content": "$$423$$个 " } ], [ { "aoVal": "D", "content": "$$432$$个 " } ] ]

[ "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" ]

[ "四面体有$$4$$个顶点和$$6$$条棱，每条棱有$$1$$个中点，其有$$10$$个点，从其中任取$$4$$个不共面的点有以下几种情况： ①这$$4$$个点都是顶点的取法只有$$1$$种； ②这$$4$$个点中有$$3$$个顶点的取法共有$$4\\times 3$$=12（种）； ③这$$4$$个点中有$$2$$顶点的取法共有$$6\\times (\\text{C}_{5}^{2}-2)=48$$（种）； ④这$$4$$个点中有$$1$$个顶点的取法共有$$4\\times (\\text{C}_{6}^{3}-3)=68$$（种）； ⑤这$$4$$个点中没有顶点的取法共有$$\\text{C}_{6}^{4}-3=12$$（种）． 综上所述，从这$$10$$个点中取$$4$$个不共面的点的取法共有 $$1+12+48+68+12=141$$（种）． 每不共面的$$4$$个点可以构成$$3$$个不同的空间四边形，则以这$$10$$个点为顶点的空间四边形共有$$141\\times 3=423$$（个）． ", "<p>从这$$10$$个点中任取$$4$$个，共有$$\\text{C}_{10}^{4}=210$$（种）取法，其中取出的$$4$$个点共面的情况共有以下$$3$$种：</p>\n<p>①从每个面上的$$6$$个点中任取$$4$$个点都共面，这样的取法共有$$4\\text{C}_{6}^{4}=60$$（种）；</p>\n<p>②每个棱上的$$3$$个点与相对棱的中点共面，这样的取法共有$$6$$种；</p>\n<p>③去掉$$1$$组相对棱后，其余四棱的中点共面，这样的取法共有$$3$$种．</p>\n<p>综上所述，所取$$4$$点共面的取法共$$60+6+3=69$$（种）．所以，所取$$4$$点不共面的取法共有$$210-69=141$$（种），以下同方法$$1$$．故选$$\\text{C}$$．</p>\n" ]

[ "2012年天津全国高中数学联赛竞赛初赛第4题6分" ]

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

[ "竞赛->知识点->解析几何->椭圆" ]

[ "设椭圆方程为$$\\frac{{{x}^{2}}}{{{a}^{2}}}+\\frac{{{y}^{2}}}{{{b}^{2}}}=1$$，则$$-\\frac{{{b}^{2}}}{{{a}^{2}}}=-\\frac{1}{2}$$，于是$$e=\\sqrt{\\frac{{{c}^{2}}}{{{a}^{2}}}}=\\sqrt{\\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}=\\sqrt{\\frac{1}{2}}$$． " ]

[ "2003年AMC12竞赛B第11题" ]

[ [ { "aoVal": "A", "content": "$$10:22$$ PM and $$24$$ seconds " } ], [ { "aoVal": "B", "content": "$$10:24$$ PM " } ], [ { "aoVal": "C", "content": "$$10:25$$ PM " } ], [ { "aoVal": "D", "content": "$$10:27$$ PM " } ], [ { "aoVal": "E", "content": "$$10:30$$ PM " } ] ]

[ "课内体系->知识点->函数的应用->函数的实际应用->一次函数模型", "美国AMC10/12->Knowledge Point->Combination->Reasoning->Simple Logical Reasoning" ]

[ "$\\dfrac{57.6}{60}=\\dfrac{600}{X}$ $X=625$ " ]

[ "2020~2021学年12月贵州贵阳观山湖区贵阳市第一中学高一上学期月考第12题3分", "2017年天津全国高中数学联赛竞赛初赛第4题5分" ]

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

[ "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" ]

[ "因为$$f\\left( x \\right)={{\\log }_{2}}{{\\log }_{2}}\\left( 2x+2 \\right)+{{2}^{2x+2}}$$在$$\\left[ 0,1 \\right]$$上单调增，且$$f\\left( 0 \\right)=4$$，$$f\\left( 1 \\right)=17$$． 所以，$$f\\left( x \\right)$$可以取到$$4$$至$$17$$之间的所有整数，共$$14$$个． " ]

[ "2008年湖南全国高中数学联赛竞赛初赛第8题5分" ]

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

[ "竞赛->知识点->函数->函数的图像与性质" ]

[ "由$$f(x)={{x}^{3}}+3{{x}^{2}}+6x+14={{\\left( x+1 \\right)}^{3}}+3\\left( x+1 \\right)+10$$， 令$$g(y)={{y}^{3}}+3y$$，则$$g(y)$$为奇函数且单调递增． 而$$f(a)={{\\left( a+1 \\right)}^{3}}+3\\left( a+1 \\right)+10=1$$， $$f(b)={{\\left( b+1 \\right)}^{3}}+3\\left( b+1 \\right)+10=19$$， 所以$$g(a+1)=-9$$，$$g(b+1)=9$$，$$g(-b-1)=-9$$， 从而$$g(a+1)=g(-b-1)$$， 即$$a+1=-b-1$$，故$$a+b=-2$$．选$$\\text{D}$$． " ]

[ "2017年AMC12竞赛B第16题", "2017年AMC10竞赛B第20题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{21}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{19}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{18}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{11}{21}$$ " } ] ]

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

[ "Solution 1 We note that the only thing that affects the parity of the factor are the powers of $$2$$. There are $$10+5+2+1=18$$ factors of $$2$$ in the number. Thus, there are $$18$$ cases in which a factor of $$21!$$ would be even (have a factor of $$2$$ in its prime factorization), and $$1$$ case in which a factor of $$21!$$ would be odd. Therefore, the answer is $$\\frac 1{19}$$. Solution 2 Consider how to construct any divisor $$D$$ of $$21!$$. First by Legendre\\textquotesingle s theorem for the divisors of a factorial , we have that there are a total of $$18$$ factors of $$2$$ in the number. $$D$$ can take up either $$0$$,$$1$$,$$2$$,$$3$$,$$\\cdots$$, or all $$18$$ factors of $$2$$, for a total of $$19$$ possible cases. In order for $$D$$ to be odd, however, it must have $$0$$ factors of $$2$$, meaning that there is a probability of $$1$$ case$$/19$$ cases$$=\\frac 1{19}$$. " ]

[ "1998年全国高中数学联赛竞赛一试第13题20分" ]

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

[ "课内体系->知识点->复数->复数的三角形式", "竞赛->知识点->复数与平面向量->模、辐角与单位根" ]

[ "$$z=1+\\cos \\left( \\frac{ \\pi }{2}+\\theta \\right)+\\text{i}\\sin \\left( \\frac{ \\pi }{2}+\\theta \\right)=2{{\\cos }^{2}}\\frac{\\frac{ \\pi }{2}+\\theta }{2}+2\\text{i}\\sin \\frac{\\frac{ \\pi }{2}+\\theta }{2}\\cos \\frac{\\frac{ \\pi }{2}+\\theta }{2}$$ $$=2\\cos \\frac{\\frac{ \\pi }{2}+\\theta }{2}\\left( \\cos \\frac{\\frac{ \\pi }{2}+\\theta }{2}+\\text{i}\\sin \\frac{\\frac{ \\pi }{2}+\\theta }{2} \\right)$$． 当$$\\frac{ \\pi }{2} ~\\textless{} ~\\theta ~~\\textless{} ~ \\pi $$时，$$\\overline{z}=-2\\cos \\frac{\\frac{ \\pi }{2}+\\theta }{2}\\left( -\\cos \\frac{\\frac{ \\pi }{2}+\\theta }{2}+\\text{i}\\sin \\frac{\\frac{ \\pi }{2}+\\theta }{2} \\right)$$ $$=-2\\cos \\left( \\frac{ \\pi }{4}+\\frac{\\theta }{2} \\right)\\left( \\cos \\left( \\frac{3 \\pi }{4}-\\frac{\\theta }{2} \\right)+\\text{i}\\sin \\left( \\frac{3 \\pi }{4}-\\frac{\\theta }{2} \\right) \\right)$$． ∴辐角主值为$$\\frac{3 \\pi }{4}-\\frac{\\theta }{2}$$． " ]

[ "高二上学期单元测试《代数变形（3）》自招第11题", "1987年全国高中数学联赛竞赛一试第1题" ]

[ [ { "aoVal": "A", "content": "这样的$$a$$有无穷多个 " } ], [ { "aoVal": "B", "content": "这样的$$a$$存在，但只有有限个 " } ], [ { "aoVal": "C", "content": "这样的$$a$$不存在 " } ], [ { "aoVal": "D", "content": "以上$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$的结论都不正确 " } ] ]

[ "竞赛->知识点->数论模块->不定方程->因式分解与恒等变形" ]

[ "对任意自然数$$k$$，取 $$a=3{{n}^{4}}k+9{{n}^{2}}{{k}^{2}}+9{{k}^{2}}$$， 则$${{n}^{6}}+3a={{({{n}^{2}}+3k)}^{3}}$$． 故选$$\\text{A}$$． " ]

[ "1983年全国高中数学联赛竞赛一试第6题" ]

[ [ { "aoVal": "A", "content": "$$P\\mathsf{\\geqslant }Q$$ " } ], [ { "aoVal": "B", "content": "$$P\\mathsf{\\leqslant }Q$$ " } ], [ { "aoVal": "C", "content": "$$P ~\\textless{} ~Q$$ " } ], [ { "aoVal": "D", "content": "$$P，Q$$间的大小关系确定，而与$$m，n$$的大小有关 " } ] ]

[ "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->基本不等式与恒成立问题", "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" ]

[ "$$P=\\sqrt{ab}+\\sqrt{cd}$$，$$Q=\\sqrt{am+nc}\\cdot \\sqrt{\\frac{b}{m}+\\frac{d}{n}}$$． $${{P}^{2}}=ab+cd+2\\sqrt{abcd}$$． $${{Q}^{2}}=(am+nc)(\\frac{b}{m}+\\frac{d}{n})=ab+cd+\\frac{nbc}{m}+\\frac{mad}{n}$$． $$\\therefore \\frac{nbc}{m}+\\frac{mad}{n}\\mathsf{\\geqslant 2}\\sqrt{\\frac{nbc}{m}\\cdot \\frac{mad}{n}}=2\\sqrt{abcd}$$． $$\\because {{Q}^{2}}\\mathsf{\\geqslant }{{P}^{2}}$$， 由于$$P\\mathsf{，}Q$$均为正数． $$\\therefore Q\\mathsf{\\geqslant P}$$． " ]

[ "2019年全国高中数学联赛竞赛初赛第6题" ]

[ [ { "aoVal": "A", "content": "$$31$$ " } ], [ { "aoVal": "B", "content": "$$41$$ " } ], [ { "aoVal": "C", "content": "$$51$$ " } ], [ { "aoVal": "D", "content": "$$61$$ " } ] ]

[ "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" ]

[ "根据二项式定理，通项$${{T}_{r}}=\\text{C}_{n}^{r}{{x}^{n-r}}{{(2\\sqrt{y}-1)}^{r}}$$， 于是根据题意得 $${{T}_{4}}={{T}_{n-1}}\\Rightarrow \\text{C}_{n}^{4}=\\text{C}_{n}^{n-1}\\text{C}_{n-1}^{2}\\cdot 4\\cdot {{(-1)}^{n-3}}$$， 解得$$n=51$$． " ]

[ "2012年湖南全国高中数学联赛竞赛初赛第1题6分" ]

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

[ "竞赛->知识点->函数->二次函数" ]

[ "因为$$a\\textgreater0$$，所以$$a+1\\textgreater1$$，$${{x}^{2}}\\textless{}1+a$$等价于$${{x}^{2}}\\leqslant 1$$，$$-1\\leqslant x\\leqslant 1$$． " ]

[ "2002年AMC10竞赛B第21题", "2002年AMC12竞赛B第17题" ]

[ [ { "aoVal": "A", "content": " Andy " } ], [ { "aoVal": "B", "content": " Beth " } ], [ { "aoVal": "C", "content": " Carlos " } ], [ { "aoVal": "D", "content": "Andy and Carlos tie for first " } ], [ { "aoVal": "E", "content": "All three tie " } ] ]

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

[ "We say Andy\\textquotesingle s lawn has an area of $$x$$. Beth\\textquotesingle s lawn thus has an area of $$\\dfrac{x}{2}$$, and Carlos\\textquotesingle s lawn has an area of $$\\dfrac{x}{3}$$. We say Andy\\textquotesingle s lawn mower cuts at a speed of $$y$$. Carlos\\textquotesingle s cuts at a speed of $$\\dfrac{y}{3}$$, and Beth\\textquotesingle s cuts at a speed $$\\dfrac{2y}{3}$$. Each person\\textquotesingle s lawn is cut at a speed of $$\\dfrac{\\rm area}{\\rm rate}$$, so Andy\\textquotesingle s is cut in $$\\dfrac{x}{y}$$ time, as is Carlos\\textquotesingle s. Beth\\textquotesingle s is cut in $$\\dfrac{3}{4}\\times\\dfrac{x}{y}$$, so the first one to finish is Beth. " ]

[ "1996年全国高中数学联赛竞赛一试第3题" ]

[ [ { "aoVal": "A", "content": "不存在 " } ], [ { "aoVal": "B", "content": "只有一个 " } ], [ { "aoVal": "C", "content": "多于一个，但为有限个 " } ], [ { "aoVal": "D", "content": "有无穷多个 " } ] ]

[ "竞赛->知识点->数论模块->不定方程->因式分解与恒等变形" ]

[ "设$$p$$为任意奇质数，且$$p=2k+1$$，于是$$n={{k}^{2}}$$，便有： $$\\sqrt{p+n}+\\sqrt{n}=\\sqrt{2k+1+{{k}^{2}}}+\\sqrt{{{k}^{2}}}=2k+1$$， 所以，每个奇质数都有题设的性质． 故选$$\\text{D}$$． " ]

[ "2015年吉林全国高中数学联赛竞赛初赛第5题6分" ]

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

[ "竞赛->知识点->函数->函数的图像与性质" ]

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

[ "2011年山东全国高中数学联赛竞赛初赛第7题6分" ]

[ [ { "aoVal": "A", "content": "$$120$$种 " } ], [ { "aoVal": "B", "content": "$$216$$ 种 " } ], [ { "aoVal": "C", "content": "$$384$$ 种 " } ], [ { "aoVal": "D", "content": "$$504$$种 " } ] ]

[ "竞赛->知识点->排列组合与概率->排列与组合" ]

[ "解法一 以$$A$$记甲成绩排名第一的所有可能的排序之集， 以$$B$$记乙成绩排名为最后的所有可能的排序之集，则$$\\left\\textbar{} A \\right\\textbar=\\left\\textbar{} B \\right\\textbar=5!$$，$$\\left\\textbar{} A\\cap B \\right\\textbar=4!$$． 甲排名第一或乙排名最后的所有可能的排序数为 $$\\left\\textbar{} A\\cup B \\right\\textbar=\\left\\textbar{} A \\right\\textbar+\\left\\textbar{} B \\right\\textbar-\\left\\textbar{} A\\cap B \\right\\textbar=216$$． 按照老师所述，这６位同学成绩可能的排序数为$$6!-216=504$$． 解法二 以乙的成绩不在最后为前提，考虑甲的成绩不在第一的所有可能排序． （$$1$$）甲的成绩排在最后的所有可能的排序数为$$A_{5}^{5}=120$$； （$$2$$）甲的成绩不在最后，又不在第一的所有可能排序数为$$C_{4}^{1}\\cdot C_{4}^{1}\\cdot A_{4}^{4}=384$$． 所以甲不在首，乙不在尾的所有可能排序数为$$120+384=504$$． " ]

[ "2008年四川全国高中数学联赛竞赛初赛第4题5分" ]

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

[ "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" ]

[ "由对称性可知，所作的四面体也是正四面体， 设原四面体的棱长、体积分别为$${{a}_{1}}$$、$${{V}_{1}}$$， 所作四面体的棱长、体积分别为$${{a}_{2}}$$、$${{V}_{2}}$$， 则由几何性质得$$\\frac{{{a}_{2}}}{{{a}_{1}}}=\\frac{1}{3}$$，故$$\\frac{{{V}_{2}}}{{{V}_{1}}}={{\\left( \\frac{1}{3} \\right)}^{3}}$$． 所以，所作的四面体的体积$${{V}_{2}}=\\frac{1}{27}{{V}_{1}}=2$$．故选$$\\text{B}$$ " ]

[ "2015年黑龙江全国高中数学联赛竞赛初赛第11题5分" ]

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

[ "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" ]

[ "采用体积法. " ]

[ "2017年陕西全国高中数学联赛竞赛初赛第2题6分" ]

[ [ { "aoVal": "A", "content": "$$a=n+1$$，$$b=2n-1$$ " } ], [ { "aoVal": "B", "content": "$$a=2n-1$$，$$b=5n+2$$ " } ], [ { "aoVal": "C", "content": "$$a=n+1$$，$$b=3n+1$$ " } ], [ { "aoVal": "D", "content": "$$a=3n+1$$，$$b=5n+2$$ " } ] ]

[ "课内体系->素养->数学运算", "课内体系->知识点->函数的应用->函数与方程" ]

[ "$$(3n+1,5n+2)=(3n+1,2n+1)=(n,2n+1)=1$$． 故选$$\\text{D}$$． " ]

[ "2010年山东全国高中数学联赛竞赛初赛第6题6分" ]

[ [ { "aoVal": "A", "content": "$${{(k+1)}^{2}}\\textless{}(m+1)(n+1)$$ " } ], [ { "aoVal": "B", "content": "$${{(k+1)}^{2}}=(m+1)(n+1)$$ " } ], [ { "aoVal": "C", "content": "$${{(k+1)}^{2}}\\textgreater(m+1)(n+1)$$ " } ], [ { "aoVal": "D", "content": "不确定 " } ] ]

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "$$a,k,b$$成等比数列，$${{k}^{2}}=ab$$， 所以$${{(k+1)}^{2}}={{k}^{2}}+2k+1=ab+2\\sqrt{ab}+1$$ $$\\textless{}ab+a+b+1=(a+1)(b+1).$$ $$a,m,n,b$$成等差数列，所以$$a+b=m+n$$，且$$b-a\\textgreater n-m$$， 所以由$$(m+1)+(n+1)=(a+1)+(b+1)$$， 知$$(m+1)(n+1)\\textgreater(a+1)(b+1)$$． 所以$${{(k+1)}^{2}}\\textless{}(a+1)(b+1)\\textgreater(m+1)(n+1)$$． " ]

[ "2016年AMC10竞赛B第12题" ]

[ [ { "aoVal": "A", "content": "$$0.2$$ " } ], [ { "aoVal": "B", "content": "$$0.3$$ " } ], [ { "aoVal": "C", "content": "$$0.5$$ " } ], [ { "aoVal": "D", "content": "$$0.7$$ " } ], [ { "aoVal": "E", "content": "$$0.8$$ " } ] ]

[ "美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Classical Models of Probabilities", "课内体系->知识点->统计与概率->概率" ]

[ "The product will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both numbers are odd is $$\\dfrac{\\left( \\begin{array}{l} {3} {2} \\end{array}\\right)}{\\left( \\begin{array}{l} {5} {2} \\end{array}\\right)}=\\dfrac{3}{10}$$, so the answer is $$1-0.3$$ which is $$\\left(\\text{D}\\right) 0.7$$. " ]

[ "2009年湖南全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$4036080$$ " } ], [ { "aoVal": "B", "content": "$$4036078$$ " } ], [ { "aoVal": "C", "content": "$$4036082$$ " } ], [ { "aoVal": "D", "content": "$$4036099$$ " } ] ]

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "由已知得 $${{a}_{n+1}}+1={{a}_{n}}+1+2\\sqrt{1+{{a}_{n}}}+1={{(\\sqrt{{{a}_{n}}+1}+1)}^{2}}$$． 又因为$${{a}_{n+1}}\\textgreater0$$，所以 $$\\sqrt{{{a}_{n+1}}+1}=\\sqrt{{{a}_{n}}+1}+1$$ 故数列$$ {\\sqrt{{{a}_{n}}+1} }$$是首项为$$1$$，公差为$$1$$的等差数列，$$\\sqrt{{{a}_{n}}+1}=n$$， 即$${{a}_{n}}={{n}^{2}}-1=(n-1)(n+1)$$． 所以$${{a}_{2009}}=2008\\times 2010=4 036 080.$$ 故选$$\\text{A}$$． " ]

[ "2010年黑龙江全国高中数学联赛竞赛初赛第10题5分" ]

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

[ "竞赛->知识点->排列组合与概率->概率初步", "竞赛->知识点->导数模块->导数" ]

[ "$${f}'(x)=2m{{x}^{2}}-n\\geqslant 0$$ 在$$\\left[ 1, +\\infty \\right]$$恒成立，即$$2{{x}^{2}}\\geqslant \\frac{n}{m}$$恒成立， 即$$2m\\geqslant n$$，但符合$$2m\\textless{}n$$条件的有($$1,3$$),($$1,4$$),($$1,5$$),($$1,6$$),($$2,5$$),($$2,6$$)共$$6$$种，故$$1-\\frac{6}{36}=\\frac{5}{6}$$． " ]

[ "2015~2016学年2月湖南长沙开福区长沙市第一中学高三下学期月考理科第8题5分", "2017~2018学年12月北京海淀区北京市育英中学高二上学期月考理科第5题", "2014年黑龙江全国高中数学联赛竞赛初赛第3题5分", "2009年高考真题山东卷理科第9题5分" ]

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

[ "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->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->圆锥曲线->抛物线->直线和抛物线的位置关系", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理" ]

[ "根据双曲线和抛物线的对称性，可设双曲线的渐近线$$y=\\frac{b}{a}x$$与抛物线$$y={{x}^{2}}+1$$只有一个公共点. 将$$y=\\frac{b}{a}x\\left( a\\textgreater0 \\right)$$代入$$y={{x}^{2}}+1$$，得$$a{{x}^{2}}-bx+a=0$$，$$a\\textgreater0$$.令$${{\\left( -b \\right)}^{2}}-4{{a}^{2}}=0$$，得$${{b}^{2}}=4{{a}^{2}}$$，$$a\\textgreater0$$， 所以$${{c}^{2}}={{a}^{2}}+{{b}^{2}}={{a}^{2}}+4{{a}^{2}}=5{{a}^{2}}$$，$$a\\textgreater0$$，$$b\\textgreater0$$，所以$$\\frac{c}{a}=\\sqrt{5}$$. " ]

[ "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7题5分", "2019~2020学年12月福建厦门思明区福建省厦门第二中学高三上学期月考理科第4题5分", "2006年全国高中数学联赛竞赛一试第5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 " } ] ]

[ "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->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->素养->逻辑推理" ]

[ "$$f(x)$$为单调递增函数，且$$f(-x)=-{{x}^{3}}+{{\\log }_{2}}\\left( -x+\\sqrt{{{x}^{2}}+1} \\right)=-{{x}^{3}}+{{\\log }_{2}}\\left( \\frac{1}{x+\\sqrt{{{x}^{2}}+1}} \\right)=-f(x)$$， $$f(x)$$为奇函数，则对任意实数$$a$$，$$b$$，由$$a+b\\geqslant 0$$，得$$a\\geqslant -b$$， ∴$$f(a)\\geqslant f(-b)=-f(b)$$，∴$$f(a)+f(b)\\geqslant 0$$； 由$$f(a)+f(b)\\geqslant 0$$得$$f(a)\\geqslant -f(b)=f(-b)$$， 所以$$a\\geqslant -\\frac{1}{b}$$，∴$$a+b\\geqslant 0$$． 故选$$\\text{A}$$． " ]

[ "2015年福建全国高中数学联赛竞赛初赛第4题6分" ]

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

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

[ "由柯西不等式，$${{\\left( x+2y+3z \\right)}^{2}}\\leqslant \\left( 1+2+3 \\right)\\left( {{x}^{2}}+2{{y}^{2}}+3{{z}^{2}} \\right)=144$$， 所以$$-12\\leqslant x+2y+3z\\leqslant 12$$． " ]

[ "2018~2019学年浙江杭州拱墅区杭州第十四中学康桥校区高一上学期期中第13题3分", "2012年浙江全国高中数学联赛竞赛初赛第10题5分" ]

[ [ { "aoVal": "A", "content": "有四个相异实根 " } ], [ { "aoVal": "B", "content": "有两个相异实根 " } ], [ { "aoVal": "C", "content": "有一个实根 " } ], [ { "aoVal": "D", "content": "无实数根 " } ] ]

[ "竞赛->知识点->函数->二次函数" ]

[ "$$f\\left( x \\right)=x$$无实根，则二次函数图象$$f\\left( x \\right)={{x}^{2}}+bx+c$$在直线$$y=x$$上方，即$$f\\left( x \\right)\\textgreater x$$，所以$$f\\left( f\\left( x \\right) \\right)\\textgreater f\\left( x \\right)\\textgreater x$$． " ]

[ "2016年湖南全国高中数学联赛竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{16} \\pi $$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{8} \\pi $$ " } ], [ { "aoVal": "D", "content": "$$\\frac{13}{16}{{ \\pi }^{2}}$$ " } ] ]

[ "课内体系->知识点->三角函数->三角函数的图象与性质->正弦型函数的图象与性质", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦函数的图象和性质", "课内体系->知识点->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->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->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->素养->数学运算" ]

[ "$$f\\left( {{a}_{1}} \\right)+f\\left( {{a}_{2}} \\right)+\\cdots +f\\left( {{a}_{5}} \\right)=\\sum\\limits_{i=1}^{5}{\\left( 2{{a}_{i}}-\\cos {{a}_{i}} \\right)}$$ $$=10{{a}_{3}}-\\left[ \\cos \\left( {{a}_{3}}-\\frac{ \\pi }{4} \\right)+\\cos \\left( {{a}_{3}}+\\frac{ \\pi }{4} \\right) \\right]-\\left[ \\cos \\left( {{a}_{3}}-\\frac{ \\pi }{8} \\right)+\\cos \\left( {{a}_{3}}+\\frac{ \\pi }{8} \\right) \\right]-\\cos {{a}_{3}}$$ $$=10{{a}_{3}}-2\\cos {{a}_{3}}\\cos \\frac{ \\pi }{4}-2\\cos {{a}_{3}}\\cos \\frac{ \\pi }{8}-\\cos {{a}_{3}}$$ 令$$g\\left( {{a}_{3}} \\right)=10{{a}_{3}}-\\left( \\sqrt{2}+2\\cos \\frac{ \\pi }{8}+1 \\right)\\cos {{a}_{3}}$$，则$${g}'\\left( {{a}_{3}} \\right)=10+\\left( \\sqrt{2}+2\\cos \\frac{ \\pi }{8}+1 \\right)\\sin {{a}_{3}}\\textgreater0$$，$$g\\left( {{a}_{3}} \\right)$$在$$\\mathbf{R}$$上单调增，又$$g\\left( \\frac{ \\pi }{2} \\right)=5 \\pi $$，所以$${{a}_{3}}=\\frac{ \\pi }{2}$$． 从而，$${{a}_{1}}=\\frac{ \\pi }{4}{{a}_{5}}=\\frac{3}{4} \\pi $$，$${{\\left[ f\\left( {{a}_{3}} \\right) \\right]}^{2}}-{{a}_{1}}{{a}_{5}}={{\\left( ~\\pi -\\cos \\frac{ \\pi }{2} \\right)}^{2}}-\\frac{ \\pi }{4}\\cdot \\frac{3}{4} \\pi =\\frac{13}{16}{{ \\pi }^{2}}$$． 故选$$\\text{D}$$． " ]

[ "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10题3分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$0$$ " } ], [ { "aoVal": "C", "content": "$${{2}^{2020}}$$ " } ], [ { "aoVal": "D", "content": "$${{2}^{2021}}$$ " } ] ]

[ "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->计数原理->二项式定理->二项式系数的性质", "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" ]

[ "因为$${{\\left[ x+\\left( 1-x \\right) \\right]}^{2020}}$$ $$={{a}_{0}}{{x}^{2020}}+{{a}_{1}}{{x}^{2019}}\\left( 1-x \\right)+{{a}_{2}}{{x}^{2018}}{{\\left( 1-x \\right)}^{2}}+\\cdots +{{a}_{2020}}{{\\left( 1-x \\right)}^{2020}}$$ $$=1$$． 且由二项式定理得，当$$0\\leqslant k\\leqslant 2020$$， 且$$k\\in \\mathbf{Z}$$时，$${{a}_{k}}=\\text{C}_{2020}^{k}$$， 所以$${{a}_{0}}+{{a}_{1}}+\\cdots +{{a}_{2020}}$$ $$=\\text{C}_{2020}^{0}+\\text{C}_{2020}^{1}+\\text{C}_{2020}^{2}+\\cdots +\\text{C}_{2020}^{2020}$$ $$={{2}^{2020}}$$． 故选$$\\text{C}$$． " ]

[ "2008年黑龙江全国高中数学联赛竞赛初赛第5题5分" ]

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

[ "竞赛->知识点->数列与数学归纳法->（模）周期数列" ]

[ "由于$${{x}_{n+1}}=\\frac{{{x}_{n}}+\\frac{\\sqrt{3}}{3}}{1-\\frac{\\sqrt{3}}{3}{{x}_{n}}}$$，令$${{x}_{n}}=\\tan {{a}_{n}}$$， 因此$${{x}_{n+1}}=\\tan \\left( {{a}_{n}}+\\frac{ \\pi }{6} \\right)$$，$${{x}_{n+6}}={{x}_{n}}$$． 易算得$${{x}_{1}}=1$$，$${{x}_{2}}=2+\\sqrt{3}$$，$${{x}_{3}}=-2-\\sqrt{3}$$， $${{x}_{4}}=-1$$，$${{x}_{5}}=-2+\\sqrt{3}$$，$${{x}_{6}}=2-\\sqrt{3}$$，$${{x}_{7}}=1$$，\\ldots， 所以$$\\sum\\limits_{n=1}^{2008}{{{x}_{n}}={{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=0}$$． 故选$$\\text{A}$$． " ]

[ "2020~2021学年北京高二单元测试", "竞赛", "2020年北京海淀区北京大学自主招生（强基计划）第7题5分", "2020~2021学年北京高三单元测试" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "前三个答案都不对 " } ] ]

[ "竞赛->知识点->多项式与方程->解方程（组）" ]

[ "当$$x$$、$$y$$中有一个为$$0$$时，$$\\left( x,y \\right)$$只能为$$\\left( 0,0 \\right)$$． 当$$x$$、$$y$$均不为$$0$$时，由原式可以得到$$\\begin{cases}19\\frac{x}{y}+93=4x 19+93\\frac{y}{x}=4y \\end{cases}$$，两式左右两端均为整数，设$$k=\\frac{x}{y}$$，则$$19k\\in \\mathbf{Z}$$，$$\\frac{93}{k}\\in \\mathbf{Z}\\Rightarrow k=\\pm \\frac{1}{19}$$，$$\\pm \\frac{3}{19}$$，$$\\pm 1$$，$$\\pm \\frac{31}{19}$$，$$\\pm 3$$，$$\\pm \\frac{93}{19}$$，$$\\pm 31$$，$$\\pm 93$$，共$$8$$组． 由于$$19\\times 93\\equiv 3\\left( \\bmod 4 \\right)$$，$$93\\equiv 1\\left( \\bmod 4 \\right)$$，$$19\\equiv 3\\left( \\bmod 4 \\right)$$，故$$19\\frac{x}{y}\\equiv 3\\left( \\bmod 4 \\right)$$，$$93\\frac{y}{x}\\equiv 1\\left( \\bmod 4 \\right)$$，在$$k$$的$$8$$组取值中，每一组恰有一个取值能同时满足上述两同余方程， 但由于$$x$$、$$y$$均不为$$0$$，须舍去$$k=-\\frac{93}{19}$$，故此时$$\\left( x,y \\right)$$共有$$7$$组． 综上所述，原方程整数解的组数为$$8$$，故选$$\\text{B}$$． " ]

[ "2009年山东全国高中数学联赛竞赛初赛第8题6分" ]

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

[ "竞赛->知识点->复数与平面向量->复数的应用" ]

[ "由已知可得$${{z}_{1}}=t+(1-t)i(0\\leqslant t\\leqslant 1)$$．$${{z}_{2}}=\\cos \\theta +i\\sin \\theta $$，得 $${{z}_{1}}+{{z}_{2}}=t+\\cos \\theta +(1-t+\\sin \\theta )i$$． 设$${{z}_{1}}+{{z}_{2}}=x+yi$$，则$$\\begin{cases}x=t+\\cos \\theta y=1-t+\\sin \\theta \\end{cases}$$．消去$$\\theta $$，得 $${{(x-t)}^{2}}+{{[y-(1-t)]}^{2}}=1$$． 即$${{z}_{1}}+{{z}_{2}}$$对应的点以$$(t,1-t)$$为圆心，半径等于$$1$$的圆上，而因为 （$$0\\leqslant t\\leqslant 1$$）．故$${{z}_{1}}+{{z}_{2}}$$对应点所在区域为图中阴影部分．其面积为 $$2\\cdot \\frac{\\pi }{2}+2\\cdot \\sqrt{2}=2\\sqrt{2}+\\pi $$． 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 " } ] ]

[ "课内体系->知识点->函数的应用->函数的零点->函数零点的概念", "竞赛->知识点->函数->函数方程" ]

[ "函数$$f\\left( x \\right)={{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-a\\cos \\left( 1-x \\right)+1$$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$$a=2$$， 当$$a=1$$时，$$f\\left( x \\right)={{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-2\\cos \\left( 1-x \\right)+1\\geqslant 1-2+1=0$$， 当且仅当$$x=1$$时取等号， 即方程$${{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-a\\cos \\left( 1-x \\right)=-1$$只有一个实数解，符合题设． 故选$$\\text{C}$$． " ]

[ "2015年黑龙江全国高中数学联赛竞赛初赛第12题5分" ]

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

[ "竞赛->知识点->数列与数学归纳法->数列的概念", "竞赛->知识点->数列与数学归纳法->数列的通项与求和", "竞赛->知识点->数列与数学归纳法->数列的综合应用", "竞赛->知识点->三角函数->三角恒等变换" ]

[ "$${{\\sin }^{2}}{{a}_{3}}-{{\\cos }^{2}}{{a}_{3}}+{{\\cos }^{2}}{{a}_{3}}{{\\cos }^{2}}{{a}_{6}}-{{\\sin }^{2}}{{a}_{3}}{{\\sin }^{2}}{{a}_{6}}={{\\sin }^{2}}{{a}_{3}}-{{\\sin }^{2}}{{a}_{6}}$$， 而$${{\\sin }^{2}}{{a}_{3}}-{{\\sin }^{2}}{{a}_{6}}=\\frac{1-\\cos 2{{a}_{3}}}{2}-\\frac{1-\\cos 2{{a}_{6}}}{2}=\\sin \\left( {{a}_{6}}+{{a}_{3}} \\right)\\sin \\left( {{a}_{3}}-{{a}_{6}} \\right)$$， 所以$$\\sin \\left( {{a}_{3}}-{{a}_{6}} \\right)=1$$，$${{a}_{3}}-{{a}_{6}}=-3d=2k \\pi +\\frac{ \\pi }{2}$$，所以$$d=-\\frac{2}{3}k \\pi -\\frac{ \\pi }{6}$$，$$k\\in \\mathbf{Z}$$． 由$$-1\\textless{}d\\textless{}0$$，可得$$-\\frac{1}{4}\\textless{}k\\textless{}-\\frac{1}{4}+\\frac{3}{2 \\pi }$$，所以$$k=0$$，$$d=-\\frac{ \\pi }{6}$$． 由$${{a}_{9}}\\textgreater0$$，$${{a}_{10}}\\textless{}0$$，可得$$\\begin{cases}{{a}_{1}}+8d\\textgreater0 {{a}_{1}}+9d\\textless{}0 \\end{cases}$$，解得$$\\frac{4}{3} \\pi \\textless{}{{a}_{1}}\\textless{}\\frac{3}{2} \\pi $$． " ]

[ "第二十届全国希望杯高一竞赛复赛邀请赛第7题4分" ]

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

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

[ "$$0\\textless{}x\\leqslant \\frac{ \\pi }{3}$$，$$y=\\frac{{{\\cos }^{2}}\\frac{x}{2}-{{\\sin }^{2}}\\frac{x}{2}}{\\cos \\frac{x}{2}-\\sin \\frac{x}{2}}=\\cos \\frac{x}{2}+\\sin \\frac{x}{2}=\\sqrt{2}\\sin \\left( \\frac{x}{2}+\\frac{ \\pi }{4} \\right)$$，即可得$$y$$的值域． " ]

[ "2008年天津全国高中数学联赛竞赛初赛第1题6分" ]

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

[ "课内体系->方法->图象法", "课内体系->方法->换元法", "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->零点、交点、根的等价转化", "课内体系->知识点->函数的应用->函数的实际应用->二次函数模型", "课内体系->素养->数学抽象", "课内体系->素养->逻辑推理", "课内体系->思想->数形结合思想" ]

[ "因为$$f\\left( f\\left( x \\right) \\right)={{\\left( {{x}^{2}}-3x+2 \\right)}^{2}}-3\\left( {{x}^{2}}-3x+2 \\right)+2={{x}^{4}}-6{{x}^{3}}+10{{x}^{2}}-3x$$， 所以有$$x\\left( x-3 \\right)\\left( {{x}^{2}}-3x+1 \\right)=0,{{x}_{1}}=0,{{x}_{2}}=3,{{x}_{34}}=\\frac{3\\pm \\sqrt{5}}{2}$$， 因此原方程有$$4$$个不同实根．故选$$\\text{D}$$． 注 也可以讨论$$f\\left( x \\right)=0$$根的分布情况． 因为当$$x\\leqslant \\frac{3}{2}$$时，函数$$f\\left( x \\right)$$单调递减，当$$x\\textgreater\\frac{3}{2}$$时，函数$$f\\left( x \\right)$$单调递增，且$$f\\left( x \\right)=0$$的两个根为$$1,2$$，所以当$$x\\leqslant \\frac{3}{2}$$时，函数$$f\\left( x \\right)\\in \\left[ -\\frac{1}{4},+\\infty \\right)\\supset \\left[ 1,2 \\right]$$，因此$$f\\left( f\\left( x \\right) \\right)=0$$有两个不同实根；当$$x\\textgreater\\frac{3}{2}$$时，函数$$f\\left( x \\right)\\in \\left( -\\frac{1}{4},+\\infty \\right)\\supset \\left[ 1,2 \\right]$$，因此$$f\\left( f\\left( x \\right) \\right)=0$$也有两个不同实根．综上所述，原方程有$$4$$个不同实根． " ]

[ "2012年吉林全国高中数学联赛竞赛初赛第5题5分" ]

[ [ { "aoVal": "A", "content": "数阵中第一列的数全是$$0$$当且仅当$${{A}_{1}}=\\varnothing $$ " } ], [ { "aoVal": "B", "content": "数阵中第$$n$$列的数全是$$1$$当且仅当$${{A}_{n}}=S$$ " } ], [ { "aoVal": "C", "content": "数阵中第$$j$$行的数字和表明集合$${{A}_{j}}$$含有几个元素 " } ], [ { "aoVal": "D", "content": "数阵中所有的$${{n}^{2}}$$个数字之和不超过$${{n}^{2}}-n+1$$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "当$${{A}_{1}}, {{A}_{2}},\\cdots , {{A}_{n}}$$中一个为$$S$$本身，其余$$n-1$$个子集为$$S$$的互不相同的$$n-1$$元子集时，数阵中所有的$${{n}^{2}}$$个数字之和最大，为$${{n}^{2}}-n+1$$，因此，$$D$$是正确的．数阵中第$$j$$行的数字和表明元素$$j$$属于几个子集，因此，$$C$$是错误的． " ]

[ "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8题", "2022~2023学年10月广东佛山禅城区佛山市第一中学高一上学期月考第7题5分" ]

[ [ { "aoVal": "A", "content": "$$ {m\\textbar m\\textless{} -4$$或$$m\\textgreater2 }$$ " } ], [ { "aoVal": "B", "content": "$$\\left { m\\left\\textbar{} -2 \\right.\\textless{} m\\textless{} 4 \\right }$$ " } ], [ { "aoVal": "C", "content": "$$ {m\\textbar m\\textless{} -2$$或$$m\\textgreater4 }$$ " } ], [ { "aoVal": "D", "content": "$$\\left { m\\left\\textbar{} -4 \\right.\\textless{} m\\textless{} 2 \\right }$$ " } ] ]

[ "课内体系->知识点->等式与不等式->不等式->解不等式", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件" ]

[ "由题意，$$\\dfrac{1}{x}+\\dfrac{2}{y}=\\left(\\frac{1}{x}+\\dfrac{2}{y}\\right)(2x+y)$$ $$=4+\\dfrac{4x}{y}+\\dfrac{y}{x}\\geqslant 4+2\\sqrt{\\dfrac{4x}{y}\\times \\dfrac{y}{x}}=8$$， 当且仅当$$\\dfrac{4x}{y}=\\dfrac{y}{x}$$，即$$y=\\dfrac{1}{2},x=\\dfrac{1}{4}$$时等号成立． 故若存在这样的$$x$$，$$y$$使不等式$$\\dfrac{1}{x}+\\dfrac{2}{y}\\textless{} {{m}^{2}}+2m$$有解． 即$${{m}^{2}}+2m\\textgreater8\\Leftrightarrow (m-2)(m+4)\\textgreater0\\Leftrightarrow m\\textgreater2$$或$$m\\textless{} -4$$． 故选：$$\\text{A}$$． " ]

[ "第二十届全国希望杯高二竞赛初赛邀请赛第8题4分" ]

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

[ "竞赛->知识点->解析几何->直线与圆锥曲线" ]

[ "不难算出与$$x+y=-2$$平行且与$${{y}^{2}}=2x$$相切的直线方程为$$x+y=-\\frac{1}{2}$$，$$PQ$$的最短距离就是$$x+y=-2$$与$$x+y=-\\frac{1}{2}$$之间的距离，易算得为$$\\frac{\\left\\textbar{} -\\frac{1}{2}-(-2) \\right\\textbar}{\\sqrt{2}}=\\frac{3\\sqrt{2}}{4}$$． " ]

[ "2019年全国高中数学联赛竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$[1,4]$$ " } ], [ { "aoVal": "B", "content": "$$[5,8]$$ " } ], [ { "aoVal": "C", "content": "$$[5,20]$$ " } ], [ { "aoVal": "D", "content": "$$[\\sqrt{5},2\\sqrt{5}]$$ " } ] ]

[ "课内体系->知识点->平面向量->平面向量基本定理及其坐标表示", "竞赛->知识点->复数与平面向量->平面向量的概念与运算" ]

[ "设$$\\overrightarrow{e}=\\left( 1,0 \\right)$$，$$\\overrightarrow{a}=\\left( 2,x \\right)$$，则 $$\\forall t\\in \\mathbf{R}$$，$$4+{{x}^{2}}\\leqslant 5\\sqrt{{{\\left( 2+t \\right)}^{2}}+{{x}^{2}}}$$， 即$$4+{{x}^{2}}\\leqslant 5\\textbar x\\textbar\\Leftrightarrow 1\\textless{}\\textbar x\\textbar\\textless{}4$$， 于是$$\\textbar\\overrightarrow{a}\\textbar=\\sqrt{4+{{x}^{2}}}$$的取值范围是$$\\left[ \\sqrt{5},2\\sqrt{5} \\right]$$． 故答案为：$$\\left[ \\sqrt{5},2\\sqrt{5} \\right]$$． " ]

[ "2014年浙江全国高中数学联赛竞赛初赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{2014}{2015}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2013}{2014}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2012}{4028}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2013}{4030}$$ " } ] ]

[ "竞赛->知识点->函数->基本初等函数" ]

[ "设等比数列的公比为$$q$$，则 $$625=5{{q}^{3}}\\Rightarrow q=5\\Rightarrow \\sum\\limits_{k=1}^{2014}{\\frac{1}{{{\\log }_{5}}{{a}_{k}}{{\\log }_{5}}{{a}_{k+1}}}=\\sum\\limits_{k=1}^{2014}{\\frac{1}{k\\left( k+1 \\right)}}=\\frac{2014}{2015}}$$． " ]

[ "2014年天津全国高中数学联赛竞赛初赛第3题6分" ]

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

[ "竞赛->知识点->数列与数学归纳法->等差数列与等比数列" ]

[ "设公比为$$q$$．由于$$q{{S}_{n}}=q\\left( {{a}_{1}}+{{a}_{2}}+\\ldots +{{a}_{n}} \\right)={{a}_{2}}+{{a}_{3}}+\\ldots +{{a}_{n+1}}$$，所以$${{S}_{n+1}}=q{{S}_{n}}+{{a}_{1}}$$．进而$${{S}_{n+2}}=q\\left( q{{S}_{n}}+{{a}_{1}} \\right)+{{a}_{1}}={{q}^{2}}{{S}_{n}}+{{a}_{1}}\\left( q+1 \\right)$$．与已知条件比较可知$${{q}^{2}}=4$$，$${{a}_{1}}\\left( q+1 \\right)=3$$．所以$$q=2$$，$${{a}_{1}}=1$$，或$$q=-2$$，$${{a}_{1}}=-3$$．相应地，$${{a}_{2}}=2$$或$$6$$． " ]

[ "1998年全国高中数学联赛竞赛一试第2题6分", "2009年黑龙江全国高中数学联赛竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$\\left { a\\textbar{} 1\\leqslant a\\leqslant 9 \\right }$$ " } ], [ { "aoVal": "B", "content": "$$\\left { a\\textbar{} 6\\leqslant a\\leqslant 9 \\right }$$ " } ], [ { "aoVal": "C", "content": "$$\\left { a\\textbar a\\leqslant 9 \\right }$$ " } ], [ { "aoVal": "D", "content": "$$\\varnothing $$ " } ] ]

[ "竞赛->知识点->集合->集合的概念与运算" ]

[ "$$A\\subseteq B$$，$$A\\ne \\varnothing $$．$$\\Rightarrow $$$$3\\leqslant 2a+1\\leqslant 3a-5\\leqslant 22$$，$$\\Rightarrow $$$$6\\leqslant a\\leqslant 9$$． 故选$$B$$． " ]