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planet/高考数学/2022年北京市高考数学试题.tm

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<TeXmacs|2.1.3>
<style|<tuple|exam|std-latex|chinese>>
<\body>
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<doc-data|<doc-title|2022年数学北京卷>>
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<strong|一、选择题>共10小题每小题4分共40分。在每小题列出的四个选项中选出符合题目要求的一项。
<\enumerate-numeric>
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<item>已知全集<math|U=<around*|{|<around*|\<nobracket\>|x|\|>-3\<less\>x\<less\>3|}>>,集合<math|A=<around*|{|<around*|\<nobracket\>|x|\|>-2\<less\>x\<le\>1|}>>,则<math|C<rsub|U>*A=>
<tabular*|<tformat|<cwith|1|-1|1|-1|cell-valign|c>|<twith|table-width|1par>|<twith|table-hmode|exact>|<cwith|1|1|1|1|cell-halign|l>|<table|<row|<cell|A.
<math|<around*|(|-2,1|]>>>|<cell|B. <math|<around|(|-3,-2|)>\<cup\><around|[|1,3|)>>>|<cell|C.
<math|<around*|[|-2,1|)>>>|<cell|D. <math|<around|(|-3,-2|]>\<cup\><around|(|1,3|)>>>>>>>
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<item>若复数<math|z>满足<math|i\<cdot\>z=3-4*i>,则<math|<around*|\||z|\|>=>
<tabular*|<tformat|<cwith|1|-1|1|-1|cell-valign|c>|<twith|table-width|1par>|<twith|table-hmode|exact>|<cwith|1|1|1|-1|cell-halign|l>|<table|<row|<cell|A.
1>|<cell|B. 5>|<cell|C. 7>|<cell|D. 25>>>>>
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<item>若直线<math|2*x+y-1=0>是圆<math|<around*|(|x-a|)><rsup|2>+y<rsup|2>=1>的一条对称轴,则<math|a=>
<tabular*|<tformat|<cwith|1|-1|1|-1|cell-valign|c>|<twith|table-width|1par>|<twith|table-hmode|exact>|<cwith|1|1|1|-1|cell-halign|l>|<table|<row|<cell|A.
<math|<frac|1|2>>>|<cell|B. <math|-<frac|1|2>>>|<cell|C. 1>|<cell|D.
<math|-1>>>>>>
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<item>己知函数<math|f<around|(|x|)>=<frac|1|1+2<rsup|x>>>,则对任意实数<math|x>,有
<tabular*|<tformat|<cwith|1|-1|1|-1|cell-valign|c>|<cwith|2|2|1|2|cell-valign|c>|<twith|table-width|1par>|<twith|table-hmode|exact>|<cwith|1|-1|1|-1|cell-halign|l>|<table|<row|<cell|A.
<math|f*<around*|(|-x|)>+f<around*|(|x|)>=0>>|<cell|B.
<math|f*<around*|(|-x|)>-f<around*|(|x|)>=0>>>|<row|<cell|C.
<math|f*<around*|(|-x|)>+f<around*|(|x|)>=1>>|<cell|D.
<math|f*<around*|(|-x|)>-f<around*|(|x|)>=<frac|1|3>>>>>>>
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<item>己知函数<math|f<around|(|x|)>=cos<rsup|2> x-sin<rsup|2>
x>,则
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A. <math|f<around*|(|x|)>>在<math|<around*|(|-<frac|\<pi\>|2>,-<frac|\<pi\>|6>|)>>上单调递增
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B. <math|f<around*|(|x|)>>在<math|<around*|(|-<frac|\<pi\>|4>,<frac|\<pi\>|12>|)>>上单调递增
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C. <math|f<around*|(|x|)>>在<math|<around*|(|0,<frac|\<pi\>|3>|)>>上单调递减
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D. <math|f<around*|(|x|)>>在<math|<around*|(|<frac|\<pi\>|4>,<frac|7*\<pi\>|12>|)>>上单调递增
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<item>设<math|<around*|{|a<rsub|n>|}>>是公差不为0的无穷等差数列则``<math|<around*|{|a<rsub|n>|}>>为递增数列''是``存在正整数<math|N<rsub|0>>,当<math|n\<gtr\>N<rsub|0>>时,<math|a<rsub|n>\<gtr\>0>''的
<tabular*|<tformat|<cwith|1|-1|1|-1|cell-valign|c>|<cwith|2|2|1|2|cell-valign|c>|<twith|table-width|1par>|<twith|table-hmode|exact>|<cwith|1|-1|1|-1|cell-halign|l>|<table|<row|<cell|A.
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充分而不必要条件>|<cell|B. 必要而不充分条件>>|<row|<cell|C.
充分必要条件>|<cell|D. 既不充分也不必要条件>>>>>
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<item>在北京冬奥会上,国家速滑馆``冰丝带''使用高效环保的二氧化碳跨临界直冷制冰技术,为实现绿色冬奥作出了贡献,如图描述了一定条件下二氧化碳所处的状态与<math|T>和<math|1*g*P>的关系,其中<math|T>表示温度,单位是<math|K><math|P>表示压强单位是bar下列结论中正确的是
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A. 当<math|T=220><math|P=1026>时,二氧化碳处于液态
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B. 当<math|T=270><math|P=128>时,二氧化碳处于气态
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C. 当<math|T=300><math|P=9987>时,二氧化碳处于超临界状态
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D. 当<math|T=360><math|P=729>时,二氧化碳处于超临界状态
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<item>若<math|<around|(|2*x-1|)><rsup|4>=a<rsub|4>*x<rsup|4>+a<rsub|3>*x<rsup|3>+a<rsub|2>*x<rsup|2>+a<rsub|1>*x+a<rsub|0>>,则<math|a<rsub|0>+a<rsub|2>+a<rsub|4>=>
<tabular*|<tformat|<cwith|1|-1|1|-1|cell-valign|c>|<twith|table-width|1par>|<twith|table-hmode|exact>|<cwith|1|1|1|-1|cell-halign|l>|<table|<row|<cell|A.
40>|<cell|B. 41>|<cell|C. <math|-40>>|<cell|D. <math|-41>>>>>>
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<item>已知正三棱锥<math|P-A*B*C>的六条棱长均为6<math|S>是<with|font-family|rm|\<triangle\>><math|A*B*C>及其内部的点构成的集合,设集合<math|T=<around|{|Q\<in\><around*|\<nobracket\>|S|\|>*P*Q<math-up|}
>5|}>>,则<math|T>表示的区域的面积为
<tabular*|<tformat|<cwith|1|-1|1|-1|cell-valign|c>|<twith|table-width|1par>|<twith|table-hmode|exact>|<cwith|1|1|1|-1|cell-halign|l>|<table|<row|<cell|A.
<math|<frac|3*\<pi\>|4>>>|<cell|B. <math|\<pi\>>>|<cell|C.
<math|2*\<pi\>>>|<cell|D. <math|3*\<pi\>>>>>>>
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<item>在<with|font-family|rm|\<triangle\>><math|A*B*C>中,<math|A*C=3><math|B*C=4><math|\<angle\>*C=90<rsup|\<circ\>>><math|P>为<with|font-family|rm|\<triangle\>><math|A*B*C>所在平面内的动点,且<math|P*C=1>,则<math|<wide|P*A|\<wide-varrightarrow\>>\<cdot\><wide|P*B|\<wide-varrightarrow\>>>的取值范围是
<tabular*|<tformat|<cwith|1|-1|1|-1|cell-valign|c>|<twith|table-width|1par>|<twith|table-hmode|exact>|<cwith|1|1|1|-1|cell-halign|l>|<table|<row|<cell|A.
<math|<around*|[|-5,3|]>>>|<cell|B. <math|<around*|[|-3,5|]>>>|<cell|C.
<math|<around*|[|-6,4|]>>>|<cell|D. <math|<around*|[|-4,6|]>>>>>>>
</enumerate-numeric>
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<strong|二、填空题>共5小题每小题5分共25分。
<\enumerate-numeric>
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<assign|item-nr|10><item>函数<math|f<around*|(|x|)>=<frac|1|x>+<sqrt|1-x>>的定义域是<underline|<space|2cm>>
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<item>已知双曲线<math|y<rsup|2>+<frac|x<rsup|2>|m>=1>的渐近线方程为<math|y=\<pm\><frac|<sqrt|3>|3>*x>,则<math|m=><underline|<space|2cm>>
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<item>若函数<math|f<around|(|x|)>=A*sin x-<sqrt|3>*cos
x>的一个零点为<math|<frac|\<pi\>|3>>,则<math|A=><underline|<space|2cm>><math|f<around*|(|<frac|\<pi\>|12>|)>=><underline|<space|2cm>>
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<item>设函数<math|f<around|(|x|)>=<around*|{|<tabular*|<tformat|<cwith|1|-1|1|1|cell-halign|l>|<cwith|1|-1|1|1|cell-lborder|0ln>|<cwith|1|-1|2|2|cell-halign|l>|<cwith|1|-1|2|2|cell-rborder|0ln>|<table|<row|<cell|-a*x+1,>|<cell|x\<less\>a>>|<row|<cell|<around|(|x-2|)><rsup|2>,>|<cell|x<math-up|\<sim\>>a>>>>>|\<nobracket\>>>,若<math|f<around*|(|x|)>>存在最小值,则<math|a>的一个取值为<underline|<space|2cm>><math|a>的最大值为<underline|<space|2cm>>
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<item>已知数列<math|<around*|{|a<rsub|n>|}>>的各项均为正数,其前<math|n>项和<math|S<rsub|n>>,满足<math|a<rsub|n>\<cdot\>S<rsub|n>=9*<around|(|n=1,2,\<cdots\>|)>>给出下列四个结论:
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①<math|<around*|{|a<rsub|n>|}>>的第2项小于3②<math|<around*|{|a<rsub|n>|}>>为等比数列;
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③<math|<around*|{|a<rsub|n>|}>>为递减数列;④<math|<around*|{|a<rsub|n>|}>>中存在小于<math|<frac|1|100>>的项。
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其中所有正确结论的序号是<underline|<space|2cm>>
</enumerate-numeric>
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<strong|三、解答题>(共6小题共85分。解答应写出文字说明演算步骤或证明过程。)
<\enumerate-numeric>
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<assign|item-nr|15><item>本小题13分
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在<with|font-family|rm|\<triangle\>><math|A*B*C>中,<math|sin
2*C=<sqrt|3>*sin C>
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I求<math|\<angle\>*C>
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II若<math|b=6>,且<with|font-family|rm|\<triangle\>><math|A*B*C>的面积为<math|6*<sqrt|3>>,求<with|font-family|rm|\<triangle\>><math|A*B*C>的周长.<vspace|5cm>
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<item>本小题14分
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如图,在三棱柱<math|A*B*C-A<rsub|1>*B<rsub|1>*C<rsub|1>>中,侧面<math|B*C*C<rsub|1>*B<rsub|1>>为正方形,平面<math|B*C*C<rsub|1>*B<rsub|1>*\<bot\>>平面<math|A*B*B<rsub|1>*A<rsub|1>><math|A*B=B*C=2><math|M,N>分别为<math|A<rsub|1>*B<rsub|1>><math|A*C>的中点.
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I求证<math|M*N><with|font-family|rm|//>平面<math|B*C*C<rsub|1>*B<rsub|1>>
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II再从条件①、条件②这两个条件中选择一个作为已知
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直线<math|A*B>与平面<math|B*M*N>所成角的正弦值。
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条件①:<math|A*B*\<bot\>*M*N>
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条件②:<math|B*M=M*N>
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注:如果选择条件①和条件②分别解答,按第一个解答计分。
<image|<tuple|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
<vspace|5cm>
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<item>本小题13分
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在校运动会上只有甲、乙、丙三名同学参加铅球比赛比赛成绩达到9.50m以上含9.50m的同学将获得优秀奖为预测获得优秀奖的人数及冠军得主收集了甲、乙、丙以往的比赛成绩并整理得到如下数据单位m
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9.80, 9.70, 9.55, 9.54, 9.48, 9.42, 9.40, 9.35, 9.30, 9.25
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9.78, 9.56, 9.51, 9.36, 9.32, 9.23
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9.85, 9.65, 9.20, 9.16
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假设用频率估计概率,且甲、乙、丙的比赛成绩相互独立
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I估计甲在校运动会铅球比赛中获得优秀奖的概率
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II设X是甲、乙、丙在校运动会铅球比赛中获得优秀奖的总人数估计<math|X>的数学期望<math|E*X>
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III在校运动会铅球比赛中甲、乙、丙谁获得冠军的概率估计值最大结论不要求证明<vspace|5cm>
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<item>本小题15分
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已知椭圆<math|E:<frac|x<rsup|2>|a<rsup|2>>+<frac|y<rsup|2>|b<rsup|2>>=1*<around|(|a\<gtr\>b\<gtr\>0|)>>的一个顶点为<math|A*<around*|(|0,1|)>>,焦距为<math|2*<sqrt|3>>
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I求椭圆<math|E>的方程:
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Il过点<math|P*<around*|(|-2,1|)>>作斜率为<math|k>的直线与椭圆<math|E>交于不同的两点<math|B,C>,直线<math|A*B,A*C>分别与<math|x>轴交于点<math|M,N>,当<math|<around*|\||M*N|\|>=2>时,求<math|k>的值。<vspace|5cm>
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<item>本小题15分
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己知函数<math|f<around*|(|x|)>=e<rsup|x><math-up|ln><around*|(|1+x|)>>
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I求曲线<math|y=f<around*|(|x|)>>在点<math|<around*|(|0,f<around*|(|0|)>|)>>处的切线方程;
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I设<math|g<around*|(|x|)>=f<rprime|'><around*|(|x|)>>,讨论函数<math|g<around*|(|x|)>>在<math|<around*|[|0,+\<infty\>|)>>上的单调性;
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III证明对任意的<math|s,t\<in\><around|(|0,+\<infty\>|)>><with|font-family|rm|<space|0.27em>
>,有<math|f*<around|(|s+t|)>\<gtr\>f<around|(|s|)>+f<around|(|t|)>><vspace|5cm>
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<item>本小题15分
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己知<math|Q:a<rsub|1>,a<rsub|2>,\<cdots\>,a<rsub|k>>为有穷整数数列.给定正整数<math|m>,若对任意的<math|n\<in\><around|{|1,2,\<cdots\>,m|}>>,在<math|Q>中存在<math|a<rsub|1>,a<rsub|i+1>,a<rsub|i+2>,\<cdots\>,a<rsub|i+j><around|(|j<math-up|\<sim\>>0|)>>,使得<math|a<rsub|i>+a<rsub|i+1>+a<rsub|i+2>+\<cdots\>+a<rsub|i+j>=n>,则称<math|Q>为<math|m->连续可表数列.
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I判断<math|Q:2,1,4>是否为5-连续可表数列?是否为<math|6->连续可表数列?说明理由;
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II若<math|Q:a<rsub|1>,a<rsub|2>,\<cdots\>,a<rsub|k>>为<math|8->连续可表数列,求证:<math|k>的最小值为4
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III若<math|Q:a<rsub|1>,a<rsub|2>,\<cdots\>,a<rsub|k>>为<math|20->连续可表数列,<math|a<rsub|1>+a<rsub|2>+\<cdots\>+a<rsub|k>\<less\>20>,求证:<math|k\<ge\>7><vspace|5cm>
</enumerate-numeric>
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