math-eval/TAL-SCQ5K huggingface dataset:
Problem: 设等差数列$$\left { {{a}_{n}} \right }$$与等比数列$$\left { {{b}_{n}} \right }$$满足：$$0\textless{}{{a}_{1}}={{b}_{1}}\textless{}{{a}_{5}}={{b}_{5}}$$，则下述四个结论：  $${}{{a}_{3}}\textless{}{{b}_{3}}$$；~~~~~~~~~~ ②$${{a}_{3}}\textgreater{{b}_{3}}$$；~~~~~~~~~ ③$${{a}_{6}}\textgreater{{b}_{6}}$$；~~~~~~~~~~~~~ ④$${{a}_{6}}\textless{}{{b}_{6}}$$  中正确的有．
Options:
aoVal: A
content: $$0$$个 
aoVal: B
content: $$1$$个 
aoVal: C
content: $$2$$个 
aoVal: D
content: $$3$$个 
Answer: B
Analysis: 因为$${{a}_{3}}=\frac{{{a}_{1}}+{{a}_{5}}}{2}=\frac{{{b}_{1}}+{{b}_{5}}}{2}\textgreater\sqrt{{{b}_{1}}{{b}_{5}}}={{b}_{3}}$$，故②正确，从而①不正确；  令$${{a}_{n}}=1+20\left( n-1 \right)$$，若取$${{b}_{n}}={{3}^{n-1}}$$，可知$${{a}_{6}}\textless{}{{b}_{6}}$$，从而③不正确；  若取$${{b}_{n}}={{\left( -3 \right)}^{n-1}}$$，可知$${{a}_{6}}\textgreater{{b}_{6}}$$，从而④不正确. 
Knowledge points: 竞赛->知识点->数列与数学归纳法->等差数列与等比数列
Source: 2013年四川全国高中数学联赛竞赛初赛第2题5分
Problem: 设$$\alpha \in \left( 0,\frac{ \pi }{2} \right)$$，则$$\frac{{{\sin }^{3}}\alpha }{\cos \alpha }+\frac{{{\cos }^{3}}\alpha }{\sin \alpha }$$的最小值为．
Options:
aoVal: A
content: $$\frac{27}{64}$$ 
aoVal: B
content: $$\frac{3\sqrt{2}}{5}$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$\frac{5\sqrt{3}}{6}$$ 
Answer: C
Analysis: $$\frac{{{\sin }^{3}}\alpha }{\cos \alpha }+\frac{{{\cos }^{3}}\alpha }{\sin \alpha }=\frac{{{\sin }^{4}}\alpha +{{\cos }^{4}}\alpha }{\sin \alpha \cos \alpha }$$  $$=\frac{1-2{{(\sin \alpha \cos \alpha )}^{2}}}{\sin \alpha \cos \alpha }$$，  令$$t=\sin \alpha \cos \alpha =\frac{1}{2}\sin 2\alpha \in \left( 0,\frac{1}{2} \right]$$，  则$$f(t)=\frac{1-2{{t}^{2}}}{t}=\frac{1}{t}-2t$$，  在$$t\in \left( 0,\frac{1}{2} \right]$$是单调递减的．  所以$$f{{(t)}_{\min }}=f\left( \frac{1}{2} \right)=1$$，  即$$\frac{{{\sin }^{3}}\alpha }{\cos \alpha }+\frac{{{\cos }^{3}}\alpha }{\sin \alpha }$$有最小值$$1$$．故选$$\text{C}$$． 
Knowledge points: 竞赛->知识点->三角函数->三角恒等变换
Source: 2008年四川全国高中数学联赛竞赛初赛第6题5分
Problem: 若 $$60$$, $$100$$, $$x$$, $$40$$, $$50$$, $$200$$, $$90$$ 這 $$7$$ 個數的平均数、中位数和众数都是$$x$$，則之値爲何？  The mean, median, and mode of the $$7$$ data values $$60$$, $$100$$, $$x$$, $$40$$, $$50$$, $$200$$, $$90$$ are all equal to $$x$$. What is the value of $$x$$?
Options:
aoVal: A
content: $$50$$ 
aoVal: B
content: $$60$$ 
aoVal: C
content: $$75$$ 
aoVal: D
content: $$90$$ 
aoVal: E
content: $$100$$ 
Answer: D
Analysis: Since $$x$$ is the mean,  $$x=\frac {60+100+x+40+50+200+90}7$$$$=\frac {540+x}7$$.  Therefore, $$7x=540+x$$, so $$x=90$$.  Order the list: $$ {40, 50, 60, 90, 100, 200  }$$. $$x$$ must be either $$60$$ or $$90$$ because it is both the median and the mode of the set. Thus $$90$$ is correct. 
Source: 2016年AMC12竞赛A第4题
2016年AMC10竞赛A第7题
Problem: 设$${{S}_{n}}$$是等差数列$$ {{{a}_{n}} }$$的前$$n$$项和，若$${{S}_{5}}={{S}_{9}}$$，则$${{a}_{3}}:{{a}_{5}}=$$．
Options:
aoVal: A
content: $$9:5$$ 
aoVal: B
content: $$5:9$$ 
aoVal: C
content: $$3:5$$ 
aoVal: D
content: $$5:3$$ 
Answer: A
Analysis: $${{S}_{5}}=\frac{5({{a}_{1}}+{{a}_{5}})}{2}=5{{a}_{3}},  {{S}_{9}}=9{{a}_{5}}$$，由$$5{{a}_{3}}=9{{a}_{5}}$$得$${{a}_{3}}:{{a}_{5}}=9:5$$． 
Knowledge points: 竞赛->知识点->数列与数学归纳法->等差数列与等比数列
Source: 2010年河南全国高中数学联赛竞赛初赛第2题5分
Problem: 设命题$$P$$：关于$$x$$的不等式$${{a}_{1}}{{x}^{2}}+{{b}_{1}}x+c{{}_{1}}\textgreater0$$与$${{a}_{2}}{{x}^{2}}+{{b}_{2}}x+{{c}_{2}}\textgreater0$$的解集相同；命题$$Q:\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}$$．则命题$$Q$$是命题$$P$$的．
Options:
aoVal: A
content: 充要条件 
aoVal: B
content: 充分不必要条件 
aoVal: C
content: 必要不充分条件 
aoVal: D
content: 既不充分也不必要条件 
Answer: D
Analysis: 当两个不等式为$${{x}^{2}}+x+1\textgreater0$$和$${{x}^{2}}+x+2\textgreater0$$时，显然解集相同，但$$Q$$不成立；  反之，当$$Q$$成立，取$$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}=-1$$，显然两个不等式解集不同．  因此，$$Q$$既不是$$P$$的充分条件也不是必要条件． 
Knowledge points: 竞赛->知识点->逻辑->逻辑推理
竞赛->知识点->逻辑->常用逻辑用语
竞赛->知识点->不等式->不等式的证明
竞赛->知识点->不等式->不等式的解法
Source: 2009年黑龙江全国高中数学联赛竞赛初赛第4题5分
Problem: 设$$A,  B,  C,  D$$是以点$$O$$为球心的球面上四点，$$AB,  AC,  AD$$两两互相垂直，且$$AB=3\text{cm}$$，$$AC=4\text{cm}$$，$$AD=\sqrt{11}\text{cm}$$，则球的半径为．
Options:
aoVal: A
content: $$3\text{cm}$$ 
aoVal: B
content: $$4\text{cm}$$ 
aoVal: C
content: $$5\text{cm}$$ 
aoVal: D
content: $$6\text{cm}$$ 
Answer: A
Analysis: 方法一 以点$$A$$为原点，分别以$$AB,  AC,  AD$$所在射线为坐标$$x$$轴、$$y$$轴和$$z$$轴，建立空间直角坐标系，则$$B,  C,  D$$的坐标分别为$$B\left( 3,  0,  0 \right)$$、$$C\left( 0,  4,  0 \right)$$，$$D\left( 0,  0,\sqrt{11} \right)$$．显然球心应为下述三个平面  $$a:x=\frac{3}{2}$$；$$\beta :y=2$$；$$\gamma :z=\frac{\sqrt{11}}{2}$$  的交点，所以球心坐标为$$O\left( \frac{3}{2},  2,\frac{\sqrt{11}}{2} \right)$$，因此球的半径$$r=OA=3\left( \text{cm} \right)$$．  方法二 以点$$A$$为原点，分别以$$AB,  AC,  AD$$所在射线为坐标$$x$$轴、$$y$$轴和$$z$$轴，建立空间直角坐标系，则$$B,  C,  D$$的坐标分别为$$B\left( 3,  0,  0 \right)$$、$$C\left( 0,  4,  0 \right)$$，$$D\left( 0,  0,\sqrt{11} \right)$$．设球面方程为  $${{\left( x-{{x}_{0}} \right)}^{2}}+{{\left( y-{{y}_{0}} \right)}^{2}}+{{\left( z-{{z}_{0}} \right)}^{2}}={{r}^{2}}$$，  其中$$O\left( {{x}_{0}},  {{y}_{0}},  {{z}_{0}} \right)$$为坐标原点，$$r$$为球的半径．因为$$B$$，$$C$$，$$D$$均在球面上，得方程组  $$\begin{cases}x_{0}^{2}+y_{0}^{2}+z_{0}^{2}={{r}^{2}},  {{\left( {{x}_{0}}-3 \right)}^{2}}+y_{0}^{2}+z_{0}^{2}={{r}^{2}}, \x_{0}^{2}+{{\left( {{y}_{0}}-4 \right)}^{2}}+z_{0}^{2}={{r}^{2}}, \x_{0}^{2}+y_{0}^{2}+{{\left( {{z}_{0}}-\sqrt{11} \right)}^{2}}={{r}^{2}}. \end{cases}$$解得$${{x}_{0}}=\frac{3}{2},  {{y}_{0}}=2,  {{z}_{0}}=\frac{\sqrt{11}}{2}$$，$$r=3\left( \text{cm} \right)$$． 
Knowledge points: 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
Source: 2012年山东全国高中数学联赛竞赛初赛第4题5分
Problem: $$2013-AMC10A-17$$  Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next $$365-$$day period will exactly two friends visit her?  她的三个最好的朋友爱丽丝、碧翠丝和克莱尔定期拜访达芙妮。 爱丽丝每三天来一次，碧翠丝每四天来一次，克莱尔每五天来一次。 昨天三位朋友都拜访了达芙妮。 在接下来的 365天期间，正好有两个朋友来拜访她有多少天？
Options:
aoVal: A
content: $$48$$ 
aoVal: B
content: $$54$$ 
aoVal: C
content: $$60$$ 
aoVal: D
content: $$66$$ 
aoVal: E
content: $$72$$ 
Answer: B
Analysis: The $$365-$$day time period can be split up into $$660-$$day time periods, because after $$60$$ days, all three of them visit again $$\left({}\right.$$ Least common multiple of $$3$$, $$4$$, and $$5$$ $$\left.{}\right)$$. You can find how many times each pair of visitors can meet by finding the LCM of their visiting days and dividing that number by $$60$$. Remember to subtract $$1$$, because you do not wish to count the $$60\text{th}$$ day, when all three visit.  A and B visit $$\dfrac{60}{3\cdot4}-1=4$$ times. B and C visit $$\dfrac{60}{4\cdot5}-1=2$$ times. C and A visit $$\dfrac{60}{3\cdot5}-1=3$$ times.  This is a total of $$9$$ visits per $$60$$ day period. Therefore, the total number of $$2-$$person visits is $$9\cdot6=\left( \text{B}\right)54$$.  From the information above, we get that $$A=3x$$, $$B=4x$$, $$C=5x$$.  Now, we want the days in which exactly two of these people meet up The three pairs are $$\left( A,B\right)$$, $$\left( B,C\right)$$, $$\left( A,C\right)$$.  Notice that we are trying to find the LCM of each pair.  Hence, $$LCM\left( A,B\right)=12x$$, $$LCM\left( B,C\right)=20x$$, $$LCM\left( A,C\right)=15x$$.  Notice that we want to eliminate when all these friends meet up. By doing this, we will find the LCM of the three letters.  Hence, $$LCM\left(A,B,C\right)=60x$$.  Now, we add all of the days up $$\left({}\right.$$ including overcount $$\left.{}\right)$$.  We get $$30+18+24=72$$. Now, because $$60\left(6\right)=360$$, we have to subtract $$6$$ days from every pair.  Hence, our answer is $$72-18=\boxed ~{\textasciitilde54\textasciitilde}\Rightarrow\boxed ~{\textasciitilde B\textasciitilde}$$. 
Knowledge points: 课内体系->知识点->集合->容斥原理
课内体系->知识点->计数原理->两个基本计数原理->分类加法计数原理
美国AMC10/12->Knowledge Point->Combination->Reasoning->Simple Logical Reasoning
Source: 2013年AMC10竞赛A第17题
Problem: 已知数列$$\left { {{x}_{n}} \right }$$满足$${{x}_{1}}=20$$，$${{x}_{n+1}}=10{{x}_{n}}-3\times {{5}^{n+1}}$$，则$${{x}_{6}}=$$．
Options:
aoVal: A
content: $${{5}^{7}}$$ 
aoVal: B
content: $${{6}^{7}}$$ 
aoVal: C
content: $$6\times {{5}^{7}}$$ 
aoVal: D
content: $$7\times {{5}^{5}}$$ 
Answer: D
Analysis: 由题得$${{x}_{n+1}}-3\times {{5}^{n+1}}=10\left( {{x}_{n}}-3\times {{5}^{n}} \right)$$，  所以数列 $$\left { {{x}_{n}}-3\times {{5}^{n}} \right }$$ 是以$$5$$为首项，  以$$10$$为公比的等比数列，  所以$${{x}_{6}}-3\times {{5}^{6}}=5\times {{10}^{5}}$$，即$${{x}_{6}}=3\times {{5}^{6}}+5\times {{10}^{5}}=7\times {{5}^{7}}$$．  故选：$$\text{D}$$． 
Knowledge points: 课内体系->素养->数学运算
课内体系->知识点->数列->等比数列->等比数列的概念与通项公式->等比数列求通项问题
Source: 2021年第31届浙江绍兴上虞区希望杯高一竞赛复赛第3题
Problem: 某科室安排国庆节放假期间（共放假$$8$$天）甲、乙、丙、丁四人的值班表．已知甲、乙各值班四天，甲不能在第一天值班且甲、乙不在同一天值班；丙需要值班$$3$$天，且不能连续值班；丁需要值班$$5$$天；规定每天必须两人值班．问符合条件的不同的值班方案共有．
Options:
aoVal: A
content: $$400$$种 
aoVal: B
content: $$700$$种 
aoVal: C
content: $$840$$种 
aoVal: D
content: $$960$$种 
Answer: B
Analysis: 方法一 先考虑甲、乙．甲不在第一天值班，可在后面$$7$$天中任选四天，因此有$$\text{C}_{7}^{4}$$种排法，余下的分派给乙为$$\text{C}_{4}^{4}$$．  再考虑丙、丁．设丙的值班日分别安排在假期的第$${{n}_{1}}$$，第$${{n}_{2}}$$，第$${{n}_{3}}$$天，则由题意知$${{n}_{1}}$$、$${{n}_{2}}$$、$${{n}_{3}}$$互不相邻，且$$1\leqslant {{n}_{1}}\textless{}{{n}_{2}}\textless{}{{n}_{3}}\leqslant 8$$．  令$${{m}_{1}}={{n}_{1}}$$，$${{m}_{2}}={{n}_{2}}-1$$，$${{m}_{3}}={{n}_{3}}-2$$，则有$$1\leqslant {{m}_{1}}\textless{}{{m}_{2}}\textless{}{{m}_{3}}\leqslant 6$$．  显然$${{n}_{1}}$$、$${{n}_{2}}$$、$${{n}_{2}}$$互不相邻等价于$${{m}_{1}}$$、$${{m}_{2}}$$、$${{m}_{3}}$$互不相同，且$$\left { {{n}_{1}},  {{n}_{2}},  {{n}_{3}} \right }$$和$$\left { {{m}_{1}},  {{m}_{2}},  {{m}_{3}} \right }$$的对应是一个一一对应．所以丙的值班安排共有$$\text{C}_{6}^{3}$$个不同方案．  丁的值班方案个数为$$\text{C}_{5}^{5}$$．  由乘法原理知甲、乙、丙、丁的不同值班方案种数为$$\text{C}_{7}^{3}\cdot \text{C}_{4}^{4}\cdot \text{C}_{6}^{3}\cdot \text{C}_{5}^{5}=700$$．  方法二 先考虑甲、乙．甲不在第一天值班，可在后面$$7$$天中任选四天，因此有$$\text{C}_{7}^{4}$$种排法，余下的分派给乙为$$\text{C}_{4}^{4}$$．  再考虑丙、丁．先排丙，如果不计丙的限制，共有$$\text{C}_{8}^{3}$$种排法，如果丙连续三天值班，则有$$\text{C}_{6}^{1}$$种排法，如果连续两天上班，且为前两天或者最后两天值班，则有$$\text{C}_{2}^{1}\text{C}_{5}^{1}$$种排法，如果是中间两天连续值班，则为$$\text{C}_{5}^{1}\text{C}_{4}^{1}$$种排法，因此不符合情况的排法共有$$\text{C}_{6}^{1}+\text{C}_{2}^{1}\text{C}_{5}^{1}+\text{C}_{5}^{1}\text{C}_{4}^{1}=36$$种，因此丙丁的排法共有$$\text{C}_{8}^{3}-\left( \text{C}_{6}^{1}+\text{C}_{2}^{1}\text{C}_{5}^{1}+\text{C}_{5}^{1}\text{C}_{4}^{1} \right)=56-36=20$$种．  由乘法原理知甲、乙、丙、丁的不同值班方案种数为$$\text{C}_{7}^{3}\cdot \text{C}_{4}^{4}\cdot 20=700$$． 
Knowledge points: 竞赛->知识点->排列组合与概率->排列与组合
Source: 2012年山东全国高中数学联赛竞赛初赛第9题5分
Problem: 已知$$\textbar\overrightarrow{a}\textbar$$、$$\textbar\overrightarrow{b}\textbar$$都是整数，且满足$$(\textbar\overrightarrow{a}\textbar+\textbar\overrightarrow{b}\textbar)(\textbar\overrightarrow{a}\textbar+3\textbar\overrightarrow{b}\textbar)=105,(\overrightarrow{a}+\overrightarrow{b})(\overrightarrow{a}+3\overrightarrow{b})=33$$，则$$\overrightarrow{a}$$和$$\overrightarrow{b}$$的夹角为 ．
Options:
aoVal: A
content: $$30{}^{}\circ $$ 
aoVal: B
content: $$60{}^{}\circ $$ 
aoVal: C
content: $$120{}^{}\circ $$ 
aoVal: D
content: $$150{}^{}\circ $$ 
Answer: C
Analysis: 令$$\textbar\overrightarrow{a}\textbar+\textbar\overrightarrow{b}\textbar=m$$，则  $$\textbar m(m+2\textbar\overrightarrow{b}\textbar)=105$$，  由$$\textbar\overrightarrow{b}\textbar\textless{}m\leqslant 10$$，得  $$\textbar\overrightarrow{b}\textbar=\frac{105}{2m}-\frac{m}{2}\textless{}m$$，  解得$$m\textgreater6$$．又$$105=3\times 5\times 7$$，所以$$m=7$$，从而有$$\textbar\overrightarrow{b}\textbar=\frac{105-49}{14}=4$$，$$\textbar\overrightarrow{a}\textbar=7-4=3$$．  $$(\overrightarrow{a}+\overrightarrow{b})(\overrightarrow{a}+3\overrightarrow{b})=\textbar\overrightarrow{a}{{\textbar}^{2}}+3\textbar\overrightarrow{b}{{\textbar}^{2}}+4\overrightarrow{a}\cdot \overrightarrow{b}$$  $$=\textbar\overrightarrow{a}{{\textbar}^{2}}+3\textbar\overrightarrow{b}{{\textbar}^{2}}+4\textbar\overrightarrow{a}\textbar\textbar\overrightarrow{b}\textbar\cos \alpha $$  $$=33$$  这里$$\alpha $$为$$\overrightarrow{a}$$与$$\overrightarrow{b}$$的夹角，从而有  $$\cos \alpha =\frac{1}{4\textbar\overrightarrow{a}\textbar\textbar\overrightarrow{b}\textbar}(33-\textbar\overrightarrow{a}{{\textbar}^{2}}-3\textbar\overrightarrow{b}{{\textbar}^{2}})=-\frac{1}{2}$$．  又$$0{}^{}\circ \leqslant \alpha \textless{}180{}^{}\circ $$，所以$$\alpha =120{}^{}\circ $$． 
<p>令$$|\overrightarrow{a}|+|\overrightarrow{b}|=m,|\overrightarrow{a}|+3|\overrightarrow{b}|=n$$，则$$m&lt;{}n&lt;{}3m$$．</p>
<p>$$105=1\times 105=3\times 35=5\times 21=7\times 15$$，</p>
<p>只有$$7&lt;{}15&lt;{}3\times 7$$．所以$$\begin{cases}|\overrightarrow{a}|+|\overrightarrow{b}|=7 \\ |\overrightarrow{a}|+3|\overrightarrow{b}|=15 \end{cases}$$，解得$$\begin{cases}|\overrightarrow{a}|=3 \\ |\overrightarrow{b}|=4 \end{cases}$$．</p>
<p>下同方法$$1$$．故选$$\text{C}$$．</p>
Knowledge points: 竞赛->知识点->复数与平面向量->平面向量的概念与运算
Source: 2009年山东全国高中数学联赛竞赛初赛第7题6分
Problem: 已知$${{a}^{2}}+{{b}^{2}}=1$$，且$$c\textless{}a+b$$恒成立，则$$c$$的取值范围是．
Options:
aoVal: A
content: $$\left( -\infty ,  -2 \right)$$ 
aoVal: B
content: $$\left( -\infty ,  -\sqrt{2} \right)$$ 
aoVal: C
content: $$\left( -\sqrt{2},\sqrt{2} \right)$$ 
aoVal: D
content: $$\left( -\infty ,\sqrt{2} \right)$$ 
Answer: B
Analysis: 令$$a=\cos \theta ,b=\sin \theta ,\theta \in \mathbf{R}$$，  则$$a+b=\cos \theta +\sin \theta =\sqrt{2}\sin \left( \theta +\frac{ \pi }{4} \right)\geqslant -\sqrt{2}$$，  所以$$c\textless{}-\sqrt{2}$$，故选$$\text{B}$$. 
Knowledge points: 竞赛->知识点->三角函数->三角函数的图像与性质
竞赛->知识点->三角函数->三角恒等变换
Source: 2010年黑龙江全国高中数学联赛竞赛初赛第4题5分
Problem: 已知$${{F}_{1}}$$、$${{F}_{2}}$$分别为椭圆$$\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\left( a\textgreater b\textgreater0 \right)$$的左、右焦点，$$P$$为椭圆上一点，满足$$\angle {{F}_{1}}P{{F}_{2}}=90{}^{}\circ $$，若$$\triangle P{{F}_{1}}{{F}_{2}}$$的面积为$$2$$，则$$b$$的值为．
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$\sqrt{2}$$ 
aoVal: C
content: $$\sqrt{3}$$ 
aoVal: D
content: $$2$$ 
Answer: B
Analysis: 设$$\left\textbar{} P{{F}_{1}} \right\textbar=m$$，$$\left\textbar{} P{{F}_{2}} \right\textbar=n$$．  则$$m+n=2a$$，  $${{m}^{2}}+{{n}^{2}}=4{{c}^{2}}=4\left( {{a}^{2}}-{{b}^{2}} \right)$$．  $$\frac{1}{2}mn=2$$，  由以上三式得$${{b}^{2}}=2$$，即$$b=\sqrt{2}$$．  故选$$\text{B}$$． 
Knowledge points: 竞赛->知识点->解析几何->椭圆
Source: 2018年四川全国高中数学联赛竞赛初赛第2题5分
Problem: 函数$$f\left( x \right)=\frac{\sin x-1}{\sqrt{3-2\cos x-2\sin x}}\left( 0\leqslant x\leqslant 2 \pi ~\right)$$的值域为~\uline{~~~~~~~~~~}~．
Options:
aoVal: A
content: $$[-1,0]$$ 
aoVal: B
content: $$[-1,1]$$ 
aoVal: C
content: $$(-1,0]$$ 
aoVal: D
content: $$(-1,0)$$ 
Answer: A
Analysis: 显然$$\sqrt{3-2\cos x-2\sin x}=\sqrt{3-2\sqrt{2}\sin \left( x+\frac{ \pi }{4} \right)}\textgreater0$$，而$$\sin x-1\leqslant 0$$，  所以，$$f\left( x \right)\leqslant 0$$．  又$${{\left( \sin x-1 \right)}^{2}}=1+{{\sin }^{2}}x-2\sin x=2-{{\cos }^{2}}x-2\sin x\leqslant 3-2\cos x-2\sin x$$，  所以，$$f\left( x \right)\geqslant -1$$．  因此，$$f\left( x \right)$$的值域为$$\left[ -1,0 \right]$$． 
Knowledge points: 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
课内体系->知识点->三角函数->三角函数综合
Source: 2017年山东全国高中数学联赛竞赛初赛第7题8分
Problem: 已知$x$、$y$为正实数，且$\lg {{2}^{x}}+\lg {{8}^{y}}=\lg 4$，则$\frac{1}{x}+\frac{3}{y}$的最小值是（~~~~~）
Options:
aoVal: A
content: 4 
aoVal: B
content: 8 
aoVal: C
content: 12 
aoVal: D
content: 16 
Answer: B
Analysis: \hfill\break 【分析】\\ 根据对数的运算法则可得$x+3y=2$，再利用乘``1''法及基本不等式计算可得．\\ 【详解】\\ 解：$\because x$、$y$为正实数，且$\lg {{2}^{x}}+\lg {{8}^{y}}=\lg 4$，\\ $\therefore \lg ({{2}^{x}}\times {{8}^{y}})=\lg 4$，\\ $\therefore {{2}^{x+3y}}={{2}^{2}}$，\\ $\therefore x+3y=2$．\\ $\therefore $$\frac{1}{x}+\frac{3}{y}=\frac{1}{2}\left( x+3y \right)\left( \frac{1}{x}+\frac{3}{y} \right)=\frac{1}{2}\left( 10+\frac{3y}{x}+\frac{3x}{y} \right)\ge \frac{1}{2}\left( 10+2\sqrt{\frac{3y}{x}\cdot \frac{3x}{y}} \right)=8$，\\ 当且仅当$\frac{3y}{x}=\frac{3x}{y}$，即$x=y=\frac{1}{2}$时取等号．\\ $\therefore $$\frac{1}{x}+\frac{3}{y}$的最小值是$8$．\\ 故选：B． 
Source: 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5题
Problem: 将$$4$$个$$1$$、$$4$$个$$2$$、$$4$$个$$3$$、$$4$$个$$4$$这$$16$$个数分别填入一个$$4\times 4$$的数表中，则使得每行、每列均恰有两个奇数的填法的概率为~\uline{~~~~~~~~~~}~．
Options:
aoVal: A
content: $$\frac{1}{144}$$ 
aoVal: B
content: $$\frac{1}{143}$$ 
aoVal: C
content: $$\frac{1}{169}$$ 
aoVal: D
content: $$\frac{1}{121}$$ 
aoVal: E
content: $$\frac{1}{133}$$ 
Answer: B
Analysis: 首先确定奇数的位置．第一行两个奇数有$$\text{C}_{4}^{2}$$种选法再考虑这两个奇数所在的列，每列还需要再填一个奇数，设为$$a$$，$$b$$．  （$$1$$）若$$a$$，$$b$$位于同行．它们的位置有$$3$$种选择，此时，剩下的四个奇数所填的位置唯一确定；  （$$2$$）若$$a$$，$$b$$位于不同的行．它们的位置有$$6$$种选择，此时，剩下的四个奇数所填的位置有$$2$$种选择．从而，奇数的不同位置数为$$\text{C}_{4}^{2}(3+6\times 2)=90$$．  最后从八个奇数位置中选四个填$$1$$（其余四个填$$3$$)，从其余八个位置中选四个填$$2$$（其余四个填$$4$$)，故总填法数为$$90\text{C}_{8}^{4}\text{C}_{8}^{4}$$，  因此，使得每行、每列均恰有两个奇数的填法的概率为 $$\frac{90\text{C}_{8}^{4}\text{C}_{8}^{4}}{\text{C}_{16}^{2}\text{C}_{12}^{4}\text{C}_{8}^{4}\text{C}_{4}^{4}}=\frac{1}{143}$$． 
Knowledge points: 课内体系->知识点->计数原理
竞赛->知识点->组合->计数问题（二试）
竞赛->知识点->排列组合与概率->排列与组合
竞赛->知识点->排列组合与概率->概率初步
竞赛->知识点->排列组合与概率->两个基本计数原理
Source: 全国高中数学联赛竞赛模拟一试（十四）第7题
Problem: 等差数列$$\left { {{a}_{n}} \right }$$的前$$n$$项和为$${{S}_{n}}$$，已知$${{S}_{10}}=0$$，$${{S}_{15}}=25$$，则$$n{{S}_{n}}$$的最小值为~\uline{~~~~~~~~~~}~．
Options:
aoVal: A
content: $$-47$$ 
aoVal: B
content: $$-48$$ 
aoVal: C
content: $$-49$$ 
aoVal: D
content: $$-50$$ 
Answer: C
Analysis: 设等差数列$$\left { {{a}_{n}} \right }$$的首项为$${{a}_{1}}$$，公差为$$d$$，由等差数列的前$$n$$项和公式可得$$\left {\begin{array}{l}{10 a_{1}+\dfrac{10 \times 9}{2} d=0}    {15 a_{1}+\dfrac{15 \times 14}{2} d=25}\end{array}\right.$$，解得$$\left {\begin{array}{l}{a_{1}=-3}    {d=\frac{2}{3}}\end{array}\right.$$，  ∴$$n S_{n}=n^{2} a_{1}+\frac{n^{2}(n-1)}{2} d=-3 n^{2}+\frac{1}{3}\left(n^{3}-n^{2}\right)=\frac{1}{3} n^{3}-\frac{10 n^{2}}{3}$$，  构造函数$$f(x)=\frac{1}{3} x^{3}-\frac{10}{3} x^{2}$$，则$$f^{\prime}(x)=x^{2}-\frac{20}{3} x$$，  令$$f^{\prime}(x)=0$$，解得$$x=0$$或$$x=\frac{20}{3}$$，  ∴当$$0\textless x\textless\frac{20}{3}$$时，$$f(x)$$单调递减，  当$$x\textgreater\frac{20}{3}$$时，$$f(x)$$单调递增，  ∵$$n \in \mathbf{N}^{*}$$且当$$n=6$$时，$$n S_{n}=\frac{1}{3} \times 6^{3}-\frac{10 \times 6^{2}}{3}=-48$$，当$$n=7$$时，$$n S_{n}=\frac{1}{3} \times 7^{3}-\frac{10 \times 7^{2}}{3}=-49$$，  ∴当$$n=7$$时，$$n S_{n}$$取得最小值$$-49$$．  设等差数列$$\left { {{a}_{n}} \right }$$的首项为$${{a}_{1}}$$，公差为$$d$$，  ∵$${{S}_{10}}=10{{a}_{1}}+45d=0$$，$${{S}_{15}}=15{{a}_{1}}+105d=25$$，  ∴$${{a}_{1}}=-3$$，$$d=\frac{2}{3}$$，  ∴$${{S}_{n}}=n{{a}_{1}}+\frac{n\left( n-1 \right)}{2}d=\frac{1}{3}{{n}^{2}}-\frac{10}{3}n$$，  ∴$$n{{S}_{n}}=\frac{1}{3}{{n}^{3}}-\frac{10}{3}{{n}^{2}}$$，令$$n{{S}_{n}}=f\left( n \right)$$，  ∴$${{f}^{\prime }}\left( n \right)={{n}^{2}}-\frac{20}{3}n$$，  ∴当$$n=\frac{20}{3}$$时，$$f\left( n \right)$$取得极值，当$$n\textless{}\frac{20}{3}$$ 时，$$f\left( n \right)$$递减，  当$$n\textgreater\frac{20}{3}$$时，$$f\left( n \right)$$递增，  因此只需比较$$f\left( 6 \right)$$和$$f\left( 7 \right)$$的大小即可，  $$f\left( 6 \right)=-48$$，$$f\left( 7 \right)=-49$$，  故$$n{{S}_{n}}$$的最小值为$$-49$$．  故答案为：$$-49$$． 
Knowledge points: 课内体系->素养->数学运算
课内体系->知识点->数列->数列的概念->数列的函数特性->数列中最大项与最小项的求解问题
Source: 2018~2019学年北京西城区北京市第十五中学高三下学期开学考试文科第13题5分
2017~2018学年1月重庆沙坪坝区重庆市第八中学高三上学期周测C卷理科第7题5分
2013年高考真题新课标卷II理科第16题
2013年贵州全国高中数学联赛竞赛初赛第5题8分
2016~2017学年北京高一下学期单元测试《数列求和》第33题
Problem: 已知向量$$\overrightarrow{m},  \overrightarrow{n}$$的夹角为$$\frac{ \pi }{6}$$，且$$\left\textbar{} \overrightarrow{m} \right\textbar=\sqrt{3},\left\textbar{} \overrightarrow{n} \right\textbar=2$$．在$$\triangle ABC$$中$$\overrightarrow{AB}=2\overrightarrow{m}+2\overrightarrow{n},\overrightarrow{AC}=$$  $$2\overrightarrow{m}-6\overrightarrow{n}$$，$$D$$为$$BC$$边的中点，则$$\left\textbar{} \overrightarrow{AD} \right\textbar=$$．
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$8$$ 
Answer: A
Analysis: $$\left\textbar{} \overrightarrow{AD} \right\textbar=\left\textbar{} \frac{1}{2}\left( \overrightarrow{AB}+\overrightarrow{AC} \right) \right\textbar=\left\textbar{} 2\overrightarrow{m}-2\overrightarrow{n} \right\textbar$$  $$=\sqrt{4{{\overrightarrow{m}}^{2}}+4{{\overrightarrow{n}}^{2}}-8\overrightarrow{m}\cdot \overrightarrow{n}}=\sqrt{4}=2$$. 
Knowledge points: 竞赛->知识点->复数与平面向量->平面向量的概念与运算
Source: 2011年黑龙江全国高中数学联赛竞赛初赛第4题5分
Problem: 某電影院中有五個朋友坐在五個位置一起的座位上 ,這五個座位由左至右編號從$$1$$到5（此處的左和右是他們坐好時，從觀察者面對他們的觀點而言）．在電影播放时，甲去大厅买爆米花，當她回來時發現「乙」向右挪了兩個位子，「丙」向左移了一个位置;且「丁」與「戊」交換了座位，而將最旁邊的位子留给了甲，試問「甲」在去買爆米花前原來坐在幾號的位子上?  Five friends sat in a movie theater in a row containing o seats, nunbered I to 5 from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had noved two seats to the right, Ceci had moved one seat to the left, and Dee and fdie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: B
Analysis: Bash: we see that the following configuration works.  Bea-Ada-Ceci-Dee-Edie   After moving, it becomes  Ada-Ceci-Bea-Edie-Dee.   Thus, Ada was in seat $$2$$.  Process of elimination of possible configurations.  Let\textquotesingle s say that Ada$$=A$$, Bea$$=B$$, Ceci$$=C$$, Dee$$=D$$, and Edie$$=D$$. Since $$B$$ moved more to the right than $$C$$ did left, this implies that $$B$$ was in a LEFT end seat originally: $$B$$, $$-$$, $$C\rightarrow -$$, $$C$$, $$B$$.  This is affirmed because $$DE\rightarrow ED$$, which there is no new seats uncovered. So $$A$$, $$B$$, $$C$$ are restricted to the same $$1$$, $$2$$, $$3$$ seats. Thus, it must be $$B$$, $$A$$, $$C\rightarrow A$$, $$C$$, $$B$$, and more specifically: $$B,A,C,D,E\rightarrow A,C,B,E,D$$~ So , $$A$$, Ada, was originally in seat $$2$$.  The seats are numbered $$1$$ through $$5$$, so let each letter $$(A,B,C,D,E)$$ correspond to a number. Let a move to the left be subtraction and a move to the right be addition.  We know that $$1+2+3+4+5=A+B+C+D+E=15$$. After everyone moves around, however, our equation looks like $$(A+x)+B+2+C-1+D+E=15$$ because $$D$$ and $$E$$ switched seats, $$B$$ moved two to the right, and $$C$$ moved $$1$$ to the left.  For this equation to be true, $$x$$ has to be$$-1$$, meaning $$A$$ moves $$1$$ left from her original seat. Since $$A$$ is now sitting in a corner seat, the only possible option for the original placement of $$A$$ is in seat number $$2$$. 
Source: 2016年AMC10竞赛第13题5分
Problem: 如果$$\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$的三个内角的余弦值分别等于$$\triangle {{A}_{2}}{{B}_{2}}{{C}_{2}}$$的三个内角的正弦值，则．
Options:
aoVal: A
content: $$\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$是锐角三角形，$$\triangle {{A}_{2}}{{B}_{2}}{{C}_{2}}$$也是锐角三角形 
aoVal: B
content: $$\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$是钝角三角形，$$\triangle {{A}_{2}}{{B}_{2}}{{C}_{2}}$$也是钝角三角形 
aoVal: C
content: $$\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$是锐角三角形，$$\triangle {{A}_{2}}{{B}_{2}}{{C}_{2}}$$也是钝角三角形 
aoVal: D
content: $$\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$是钝角三角形，$$\triangle {{A}_{2}}{{B}_{2}}{{C}_{2}}$$也是锐角三角形 
Answer: C
Analysis: 因为$$\cos {{A}_{1}},\cos {{A}_{2}},\cos {{A}_{3}}$$都大于$$0$$，所以$${{A}_{1}},{{A}_{2}},{{A}_{3}}$$都是锐角，$$\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$是锐角三角形．  不妨设$$\sin {{A}_{2}}=\cos {{A}_{1}}$$，$$\sin {{B}_{2}}=\cos {{B}_{1}}$$，$$\sin {{C}_{2}}=\cos {{C}_{1}}$$，  则$${{A}_{2}}=90{}^{}\circ -{{A}_{1}}$$或$${{A}_{2}}=90{}^{}\circ +{{A}_{1}}$$，$${{B}_{2}}=90{}^{}\circ -{{B}_{1}}$$或$${{B}_{2}}=90{}^{}\circ +{{B}_{1}}$$，$${{C}_{2}}=90{}^{}\circ -{{C}_{1}}$$或$${{C}_{2}}=90{}^{}\circ +{{C}_{1}}$$．  若$${{A}_{2}}=90{}^{}\circ -{{A}_{1}}$$，$${{B}_{2}}=90{}^{}\circ -{{B}_{1}}$$，$${{C}_{2}}=90{}^{}\circ -{{C}_{1}}$$，  则$${{A}_{2}}+{{B}_{2}}+{{C}_{2}}=90{}^{}\circ $$，矛盾．  所以$$\triangle {{A}_{2}}{{B}_{2}}{{C}_{2}}$$存在一个角为钝角． 
Knowledge points: 竞赛->知识点->三角函数->三角形中的问题->解三角形
Source: 2017年湖南全国高中数学联赛竞赛初赛第5题5分
Problem: $$f(x)=a{{x}^{3}}+{{x}^{2}}+x+da,d\in \bf R$$，当$$\textbar x\textbar\leqslant 1$$时，$$\textbar f(x)\textbar\leqslant 1$$，则$$a$$、$$d$$一定属于．
Options:
aoVal: A
content: $$[-2,0]$$ 
aoVal: B
content: $$[0,2]$$ 
aoVal: C
content: $$[-1,0]$$ 
aoVal: D
content: $$[0,1]$$ 
Answer: A
Analysis: 将$$x=1$$，$$-1$$分别代入$$f(x)$$可得  $$\textbar a+1+1+d\textbar\leqslant 1$$和$$\textbar-a+d\textbar\leqslant 1$$，  即$$-1\leqslant a+1+1+d\leqslant 1$$和$$-1\leqslant -a+d\leqslant 1$$．  则$$-3\leqslant a+d\leqslant -1$$①，  $$-1\leqslant -a+d\leqslant 1$$②，  把①和②两个不等式相加得$$-4\leqslant 2d\leqslant 0$$，  即$$-2\leqslant d\leqslant 0$$．类似可得$$-2\leqslant a\leqslant 0$$．故选$$\text{A}$$． 
Knowledge points: 竞赛->知识点->不等式->不等式的解法
Source: 2008年甘肃全国高中数学联赛竞赛初赛第4题6分