planet/高考数学/2022年全国高考乙卷数学理科试题.tm

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  <doc-data|<doc-title|2022年数学全国乙卷（理科）>>

  一、单选题（本大题共12小题，共60分）

  <\enumerate-numeric>
    <item>设全集<math|U=<around|{|1,2,3,4,5|}>>，集合<math|M>满足<math|\<complement\><rsub|U>*M=<around|{|1,3|}>>，则

    <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|2\<in\>M>>|<cell|B. <math|3\<in\>M>>|<cell|C.
    <math|4*\<nin\>M>>|<cell|D. <math|5*\<nin\>M>>>>>>

    <item>已知<math|z=1-2*i>，且<math|z+a*<wide|z|\<bar\>>+b=0>，其中<math|a>，<math|b>为实数，则

    <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>|<cwith|2|2|1|2|cell-valign|c>|<cwith|2|2|1|2|cell-halign|l>|<table|<row|<cell|A.
    <math|a=1>，<math|b=-2>>|<cell|B. <math|a=-1>，<math|b=2>>>|<row|<cell|C.
    <math|a=1>，<math|b=2>>|<cell|D. <math|a=-1>，<math|b=-2>>>>>>

    <item>已知向量<math|a>，<math|b>满足<math|<around|\||<wide|a|\<wide-varrightarrow\>>|\|>=1>，<math|<around|\||<wide|b|\<wide-varrightarrow\>>|\|>=<sqrt|3>>，<math|<around|\||<wide|a|\<wide-varrightarrow\>>-2*<wide|b|\<wide-varrightarrow\>>|\|>=3>，则<math|<wide|a|\<wide-varrightarrow\>>*\<cdot\>*<wide|b|\<wide-varrightarrow\>>=>

    <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|-2>>|<cell|B. <math|-1>>|<cell|C. <math|1>>|<cell|D. <math|2>>>>>>

    <item>嫦娥二号卫星在完成探月任务后，继续进行深空探测，成为我国第一颗环绕太阳飞行的人造行星.为研究嫦娥二号绕日周期与地球绕日周期的比值，用到数列<math|<around|{|b<rsub|n>|}>:b<rsub|1>=1+<frac|1|\<alpha\><rsub|1>>>，<math|b<rsub|2>=1+<frac|1|\<alpha\><rsub|1>+<frac|1|\<alpha\><rsub|2>>>>，<math|b<rsub|3>=1+<frac|1|\<alpha\><rsub|1>+<frac|1|\<alpha\><rsub|2>+<frac|1|\<alpha\><rsub|3>>>>>，\<cdots\>，依此类推，其中<math|\<alpha\><rsub|k>\<in\>N<rsup|\<ast\>>(k=1,2,\<cdots\>)>.则

    <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|b<rsub|1>\<less\>b<rsub|5>>>|<cell|B.
    <math|b<rsub|3>\<less\>b<rsub|s>>>|<cell|C.
    <math|b<rsub|6>\<less\>b<rsub|2>>>|<cell|D.
    <math|b<rsub|4>\<less\>b<rsub|7>>>>>>>

    <item>设<math|F>为抛物线<math|C:y<rsup|2>=4*x>的焦点，点<math|A>在<math|C>上，点<math|B*<around|(|3,0|)>>，若<math|<around|\||A*F|\|>=<around|\||B*F|\|>>，则<math|<around|\||A*B|\|>=>

    <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|2>>|<cell|B. <math|2*<sqrt|2>>>|<cell|C. <math|3>>|<cell|D.
    <math|3*<sqrt|2>>>>>>>

    <item>执行右边的程序框图，输出的<math|n=>

    <image|<tuple|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

    <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|3>>|<cell|B. <math|4>>|<cell|C. <math|5>>|<cell|D. <math|6>>>>>>

    <item>在正方体<math|A*B*C*D-A<rsub|1>*B<rsub|1>*C<rsub|1>*D<rsub|1>>中，<math|E>，<math|F>分别为<math|A*B>，<math|B*C>的中点，则

    <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.
    平面<math|B<rsub|1>*E*F*\<bot\>>平面<math|B*D*D<rsub|1>>>|<cell|B.
    平面<math|B<rsub|1>*E*F*\<bot\>>平面<math|A<rsub|1>*B*D>>>|<row|<cell|C.
    平面<math|B<rsub|1>*E*F//>平面<math|A<rsub|1>*A*C>>|<cell|D.
    平面<math|B<rsub|1>*E*F//>平面<math|A<rsub|1>*C<rsub|1>*D>>>>>>

    <item>已知等比数列<math|<around|{|a<rsub|n>|}>>的前<math|3>项和为<math|168>，<math|a<rsub|2>-a<rsub|5>=42>，则<math|a<rsub|6>=>

    <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|14>>|<cell|B. <math|12>>|<cell|C. <math|6>>|<cell|D. <math|3>>>>>>

    <item>已知球<math|O>的半径为<math|1>，四棱锥的顶点为<math|O>，底面的四个顶点均在球<math|O>的球面上，则当该四棱锥的体积最大时，其高为

    <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|1|3>>>|<cell|B. <math|<frac|1|2>>>|<cell|C.
    <math|<frac|<sqrt|3>|3>>>|<cell|D. <math|<frac|<sqrt|2>|2>>>>>>>

    <item>某棋手与甲、乙、丙三位棋手各比赛一盘，各盘比赛结果相互独立.已知该棋手与甲、乙、丙比赛获胜的概率分别为<math|p<rsub|1>>，<math|p<rsub|2>>，<math|p<rsub|3>>，且<math|p<rsub|3>\<gtr\>p<rsub|2>\<gtr\>p<rsub|1>\<gtr\>0>.记该棋手连胜两盘的概率为<math|p>，则

    A. <math|p>与该棋手和甲、乙、丙的比赛次序无关

    B. 该棋手在第二盘与甲比赛，<math|p>最大

    C. 该棋手在第二盘与乙比赛，<math|p>最大

    D. 该棋手在第二盘与丙比赛，<math|p>最大

    <item>对曲线<math|C>的两个焦点为<math|F<rsub|1>>，<math|F<rsub|2>>，以<math|C>的实轴为直径的圆记为<math|D>，过<math|F<rsub|1>>作<math|D>的切线与<math|C>交于<math|M>，<math|N>两点，且<math|<math-up|cos>\<angle\>*F<rsub|1>*N*F<rsub|2>=<frac|3|5>>，则<math|C>的离心率为

    <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|<sqrt|5>|2>>>|<cell|B. <math|<frac|3|2>>>|<cell|C.
    <math|<frac|<sqrt|13>|2>>>|<cell|D. <math|<frac|<sqrt|17>|2>>>>>>>

    <item>已知函数<math|f<around|(|x|)>>，<math|g<around|(|x|)>>的定义域均为<math|R>，且<math|f<around|(|x|)>+g*<around|(|2-x|)>=5>，
    <math|g<around|(|x|)>-f*<around|(|x-4|)>=7>，
    若<math|y=g<around|(|x|)>>的图像关于直线<math|x=2>对称，<math|g<around|(|2|)>=4>，
    则<math|<big|sum><rsup|22><rsub|k=1>f<around|(|k|)>=>

    <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|-21>>|<cell|B. <math|-22>>|<cell|C. <math|-23>>|<cell|D.
    <math|-24>>>>>>
  </enumerate-numeric>

  二、填空题（本大题共4小题，共20.0分）

  <\enumerate-numeric>
    <assign|item-nr|12><item>从甲、乙等<math|5>名同学中随机选<math|3>名参加社区服务工作，则甲、乙都入选的概率为<underline|<space|2cm>>．

    <item>过四点<math|<around|(|0,0|)>>，<math|<around|(|4,0|)>>，<math|<around|(|-1,1|)>>，<math|<around|(|4,2|)>>中的三点的一个圆的方程为<underline|<space|2cm>>．

    <item>记函数<math|f<around|(|x|)>=<math-up|cos><around|(|\<omega\>*x+\<varphi\>|)>*<around|(|\<omega\>\<gtr\>0,0\<less\>\<varphi\>\<less\>\<pi\>|)>>的最小正周期为<math|T>，若<math|f<around|(|T|)>=<frac|<sqrt|3>|2>>，<math|x=<frac|\<pi\>|9>>为<math|f<around|(|x|)>>的零点，则<math|\<omega\>>的最小值为<underline|<space|2cm>>．

    <item>已知<math|x=x<rsub|1>>和<math|x=x<rsub|2>>分别是函数<math|f<around|(|x|)>=2*a<rsup|x>-e*x<rsup|2>(a\<gtr\>0>且<math|a\<neq\>1)>的极小值点和极大值点，若<math|x<rsub|1>\<less\>x<rsub|2>>，则<math|a>的取值范围是<underline|<space|2cm>>.
  </enumerate-numeric>

  三、解答题（本大题共7小题，共80分）

  \ (一) \ 必考题：共 60 分.

  <\enumerate-numeric>
    <assign|item-nr|16><item>记<math|\<Delta\>*A*B*C>的内角<math|A>、<math|B>、<math|C>的对边分别为<math|a>、<math|b>、<math|c>，已知<math|<math-up|sin>C<math-up|sin><around|(|A-B|)>=<math-up|sin>B<math-up|sin><around|(|C-A|)>>．

    (1)证明：<math|2*a<rsup|2>=b<rsup|2>+c<rsup|2>>;

    (2)若<math|a=5>，<math|<math-up|cos>A=<frac|25|31>>，求<math|\<Delta\>*A*B*C>的周长．<vspace|5cm>

    <item>如图，四面体<math|A*B*C*D>中<math|A*D*\<bot\>*C*D>，<math|A*D=C*D>，<math|\<angle\>*A*D*B=\<angle\>*B*D*C>，<math|E>为<math|A*C>中点．

    (1)证明：平面<math|B*E*D*\<bot\>>平面<math|A*C*D>;

    (2)设<math|A*B=B*D=2>，<math|\<angle\>*A*C*B=60<rsup|\<circ\>>>，点<math|F>在<math|B*D>上，当<math|\<bigtriangleup\>*A*F*C>的面积最小时，求<math|C*F>与平面<math|A*B*D>所成角的正弦值．

    <image|<tuple|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

    <item>某地经过多年的环填治理，已将就山改造成了绿水青山.为估计一林区某种树木的总材积量，随机选取了<math|10>棵这种村木，测量每棵村的根部横截而积(心位：<math|m<rsup|2>>)和材积量<math|<around|(|m<rsup|3>|)>>，得到如下数据：

    <block*|<tformat|<table|<row|<cell|样本号<math|i>>|<cell|1>|<cell|2>|<cell|3>|<cell|4>|<cell|5>|<cell|6>|<cell|7>|<cell|8>|<cell|9>|<cell|10>|<cell|总和>>|<row|<cell|根部横截面积<math|x
    <rsub|i>>>|<cell|0.04>|<cell|0.06>|<cell|0.04>|<cell|0.08>|<cell|0.08>|<cell|0.05>|<cell|0.05>|<cell|0.07>|<cell|0.07>|<cell|0.06>|<cell|0.6>>|<row|<cell|材积量<math|y<rsub|i>>>|<cell|0.25>|<cell|0.40>|<cell|0.22>|<cell|0.54>|<cell|0.51>|<cell|0.34>|<cell|0.36>|<cell|0.46>|<cell|0.42>|<cell|0.40>|<cell|3.9>>>>>

    并计算得<math|<big|sum><rsup|10><rsub|i=1>x<rsub|i><rsup|2>=0.038>，<math|<big|sum><rsup|10><rsub|i=1>y<rsup|2><rsub|i>=1.6158>，<math|<big|sum><rsup|10><rsub|i=1>x<rsub|i>*y<rsub|i>=0.2474>．

    (1)估计该林区这种树木平均一棵的根部横截面积与平均一棵的材积量：

    (2)求该林区这种树木的根部横截面积与材积量的样本相关系数(精确到0.01);

    (3)现测量了该林区所有这种树木的根部横截面积，并得到所有这种树木的根部横截面积总和为<math|186*m<rsup|2>>.已知树木的材积量与其根部横截面积近似成正比.利用以上数据给出该林区这种树木的总材积量的估计值．

    附：相关系数<math|r=<frac|<big|sum><rsup|n><rsub|i=1><around*|(|x<rsub|i>-<wide|x|\<bar\>>|)>*<around*|(|y<rsub|i>-<wide|y|\<bar\>>|)>|<sqrt|<big|sum><rsup|n><rsub|i=1><around*|(|x<rsub|i>-<wide|x|\<bar\>>|)><rsup|2>*<big|sum><rsup|n><rsub|i=1><around*|(|y<rsub|i>-<wide|y|\<bar\>>|)><rsup|2>>>>，<math|<sqrt|1.896>\<approx\>1.377>．<vspace|5cm>

    <item>已知椭圆<math|E>的中心为坐标原点，对称轴为<math|x>轴，<math|y>轴，且过<math|A*<around|(|0,-2|)>>，
    <math|B*<around|(|<frac|3|2>,-1|)>> 两点.

    (1)求<math|E>的方程;

    (2)设过点<math|P<around|(|1,-2|)>>的直线交<math|E>于<math|M>，<math|N>两点，过<math|M>且平行于<math|x>的直线与线段<math|A*B>交于点<math|T>，点<math|H>满足<math|<wide|M*T|\<wide-varrightarrow\>>=<wide|T*H|\<wide-varrightarrow\>>>，证明：直线<math|H*N>过定点．<vspace|5cm>

    <item>已知函数<math|f<around|(|x|)>=<math-up|ln><around|(|1+x|)>+a*x*e<rsup|-x>>．

    (1)当<math|a=1>时，求曲线<math|f<around|(|x|)>>在点<math|<around|(|0,f<around|(|0|)>|)>>处的切线方程：

    (2)若<math|f<around|(|x|)>>在区间<math|<around|(|-1,0|)>>，<math|<around|(|0,+\<infty\>|)>>各恰有一个零点，求<math|a>的取值范围．<vspace|5cm>
  </enumerate-numeric>

  (二) 选考题：共 10 分

  <\enumerate>
    <assign|item-nr|21><item>在直角坐标系<math|x*O*y>中，曲线<math|C>的方程为<math|<around*|{|<tabular*|<tformat|<cwith|1|-1|1|1|cell-halign|l>|<cwith|1|-1|1|1|cell-lborder|0ln>|<cwith|1|-1|1|1|cell-rborder|0ln>|<table|<row|<cell|x=<sqrt|3><math-up|cos>2*t>>|<row|<cell|y=2<math-up|sin>t>>>>>|\<nobracket\>>(t>为参数<math|)>.以坐标原点为极点，<math|x>轴正半轴为极轴建立极坐标系，已知直线<math|l>的极坐标方程为<math|\<rho\><math-up|sin><around|(|\<theta\>+<frac|\<pi\>|3>|)>+m=0>．

    (1)写出<math|l>的直角坐标方程：

    (2)若<math|l>与<math|C>有公共点，求<math|m>的取值范围．<vspace|5cm>

    <item>已知<math|a.*b.*c>为正数，且<math|a<rsup|<frac|3|2>>+b<rsup|<frac|3|2>>+c<rsup|<frac|3|2>>=1>，证明：

    \ <math|<around|(|1|)>*a*b*c\<leq\><frac|1|9>>;

    <math|<around|(|2|)><frac|a|b+c>+<frac|b|a+c>+<frac|c|a+b>\<leq\><frac|1|2*<sqrt|a*b*c>>>.<vspace|5cm>
  </enumerate>

  \;

  \;
</body>