planet/高考数学/2022年全国高考乙卷数学文科试题.tm

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

  一、单选题（本大题共12小题，共60分）

  <\enumerate-numeric>
    <item>集合<math|M=<around|{|2,4,6,8,10|}>>，<math|N=<around|{|x\|-1\<less\>x\<less\>6|}>>，则<math|M*\<cap\>*N=>

    <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|{|2,4|}>>>|<cell|B. <math|<around|{|2,4,6|}>>>|<cell|C.
    <math|<around|{|2,4,6,8|}>>>|<cell|D. <math|<around|{|2,4,6,8,10|}>>>>>>>

    <item>设<math|<around|(|1+2*i|)>*a+b=2*i>，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|<wide|a|\<wide-varrightarrow\>>=<around|(|2,1|)>>，<math|<wide|b|\<wide-varrightarrow\>>=<around|(|-2,4|)>>，则<math|<around|\||<wide|a|\<wide-varrightarrow\>>-<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|3>>|<cell|C. <math|4>>|<cell|D. <math|5>>>>>>

    <item>分别统计了甲、乙两位同学<math|16>周的各周课外体育运动时长(单位：h)，得如下茎叶图：

    <image|<tuple|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

    则下列结论中错误的是

    A. 甲同学周课外体育运动时长的样本中位数为<math|7.4>

    B. 乙同学周课外体育运动时长的样本平均数大于<math|8>

    C. 甲同学周课外体育运动时长大于<math|8>的概率的估计值大于<math|0.4>

    D. 乙同学周课外体育运动时长大于<math|8>的概率的估计值大于<math|0.6>

    <item>若<math|x>，<math|y>满足约束条件<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+y\<geq\>2,>>|<row|<cell|x+2*y\<leq\>4,>>|<row|<cell|y\<geq\>0,>>>>>|\<nobracket\>>>则<math|z=2*x-y>的最大值是

    <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|-2>>|<cell|B. <math|4>>|<cell|C. <math|8>>|<cell|D. <math|12>>>>>>

    <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>右图是下列四个函数中的某个函数在区间<math|<around|[|-3,3|]>>的大致图像，则该函数是

    <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|y=<frac|-x<rsup|3>+3*x|x<rsup|2>+1>>>|<cell|B.
    <math|y=<frac|x<rsup|3>-x|x<rsup|2>+1>>>|<cell|C.
    <math|y=<frac|2*x<math-up|cos>x|x<rsup|2>+1>>>|<cell|D.
    <math|y=<frac|2<math-up|sin>x|x<rsup|2>+1>>>>>>>

    <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>函数<math|f<around|(|x|)>=<math-up|cos>x+<around|(|x+1|)><math-up|sin>x+1>在区间<math|<around|[|0,2*\<pi\>|]>>的最小值，最大值分别为

    <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|\<pi\>|2>>，<math|<frac|\<pi\>|2>>>|<cell|B.
    <math|-<frac|3*\<pi\>|2>>，<math|<frac|\<pi\>|2>>>|<cell|C.
    <math|-<frac|\<pi\>|2>>，<math|<frac|\<pi\>|2>+2>>|<cell|D.
    <math|-<frac|3*\<pi\>|2>>，<math|<frac|\<pi\>|2>+2>>>>>>

    <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>>>>>>>
  </enumerate-numeric>

  二、填空题（本大题共4小题，共20分）

  <\enumerate-numeric>
    <assign|item-nr|12><item>记<math|S<rsub|n>>为等差数列<math|<around|{|a<rsub|n>|}>>的前<math|n>项和.若<math|2*S<rsub|3>=3*S<rsub|2>+6>，则公差<math|d=><underline|<space|2cm>>．

    <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>若<math|f<around|(|x|)>=<math-up|ln>\|a+<frac|1|1-x>\|+b>是奇函数，则<math|a=><underline|<space|2cm>>，<math|b=><underline|<space|2cm>>．
  </enumerate-numeric>

  三、解答题（本大题共7小题，共80.0分）

  \ (一) \ 必考题：共 60 分.

  <\enumerate-numeric>
    <assign|item-nr|16><item>记<math|\<bigtriangleup\>*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)若<math|A=2*B>，求<math|C>:

    (2)证明：<math|2*a<rsup|2>=b<rsup|2>+c<rsup|2>>．<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>的中点．

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

    (1)当<math|a=0>时，求<math|f<around|(|x|)>>的最大值;

    (2)若<math|f<around|(|x|)>>恰有一个零点，求<math|a>的取值范围．<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．
  </enumerate-numeric>

  <vspace|5cm>

  (二) 选考题：共 10 分

  <\enumerate-numeric>
    <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>>>.
  </enumerate-numeric>

  <vspace|5cm>
</body>