planet/高考数学/2022年浙江省高考数学试题.tm

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<style|<tuple|generic|chinese|doc>>

<\body>
  <doc-data|<doc-title|2022年普通高等学校招生全国统一考试>|<doc-subtitle|<with|font|AR
  PL UMing CN|font-base-size|16|数<space|2em>学>>>

  <space|2em>本试题卷分选择题和非选择题两部分。全卷共4页，选择题部分1至3页；非选择题部分3至4页。满分150分，考试时间120分钟。

  \;

  <strong|考生注意：>

  <\enumerate>
    <item>答题前，请务必将自己的姓名、准考证号用黑色字迹的签字笔或钢笔分别填写在试题卷和答题纸规定的位置上。

    <item>答题时，请按照答题纸上\P注意事项\Q的要求，在答题纸相应的位置上规范作答，在本试题卷上的作答一律无效。
  </enumerate>

  \;

  <\with|par-columns|2>
    <strong|参考公式：>

    若事件A,B互斥，则

    <math|P<around*|(|A+B|)>=P<around*|(|A|)>+P<around*|(|B|)>>

    若事件A,B相互独立，则

    <math|P<around*|(|AB|)>=P<around*|(|A|)>*P<around*|(|B|)>>

    若事件A在一次试验中发生的概率是<math|p>，则<math|n>次独立重复试验中事件<math|A>恰好发生<math|k>次的概率

    <math|P<rsub|n><around*|(|k|)>=C<rsup|k><rsub|n>*p<rsup|k><around*|(|1-p|)><rsup|n-k><around*|(|k=0,1,2,\<ldots\>,n|)>>

    台体的体积公式

    <math|V=<frac|1|3><around*|(|S<rsub|1>+<sqrt|S<rsub|1>S<rsub|2>>+S<rsub|2>|)>*h>

    其中<math|S<rsub|1>>,<math|S<rsub|2>>分别表示台体的上、下底面积，<math|h>表示台体的高

    柱体的体积公式

    <math|V=S*h>

    其中<math|S>表示柱体的底面积，h表示柱体的高

    锥体的体积公式

    <math|V=<frac|1|3>S*h>

    其中<math|S>表示锥体的底面积，<math|h>表示锥体的高

    球的表面积公式

    <math|S=4*\<pi\>*R<rsup|2>>

    球的体积公式

    <math|V=<frac|4|3>*\<pi\>*R<rsup|3>>

    其中<math|R>表示球的半径
  </with>

  \;

  \;

  <\with|par-mode|center>
    <strong|<with|font|AR PL UMing CN|font-base-size|14|选择题部分（共40分）>>
  </with>

  \;

  <strong|一、选择题：本大题共10小题，每小题4分，共40分。在每小题给出的四个选项中，只有一项是符合题目要求的。>

  <\enumerate>
    <item>设集合<math|A=<around*|{|1,2|}>,B=<around*|{|2,4,6|}>>，则<math|A\<cup\>B=>

    <\wide-tabular>
      <tformat|<table|<row|<\cell>
        A. <math|2>
      </cell>|<\cell>
        B. <math|<around*|{|1,2|}>>
      </cell>|<\cell>
        C. <math|<around*|{|2,4,6|}>>
      </cell>|<\cell>
        D. <math|<around*|{|1,2,4,6|}>>
      </cell>>>>
    </wide-tabular>

    <item>已知<math|a,b\<in\>\<bbb-R\>,a+3*i=<around*|(|b+i|)>*i>（<math|i>为虚数单位），则

    <\wide-tabular>
      <tformat|<table|<row|<\cell>
        A. <math|a=1,b=-3>
      </cell>|<\cell>
        B. <math|a=-1,b=3>
      </cell>|<\cell>
        C. <math|a=-1,b=-3>
      </cell>|<\cell>
        D. <math|a=1,b=3>
      </cell>>>>
    </wide-tabular>

    <item>若实数<math|x,y>满足约束条件<math|<with|math-display|true|<choice|<tformat|<table|<row|<cell|x-2\<geqslant\>0,>>|<row|<cell|2*x+y-7\<leqslant\>0,>>|<row|<cell|x-y-2\<leqslant\>0,>>>>>>>则<math|z=3*x+4*y>的最大值是

    <\wide-tabular>
      <tformat|<table|<row|<\cell>
        A. <math|20>
      </cell>|<\cell>
        B. <math|18>
      </cell>|<\cell>
        C. <math|13>
      </cell>|<\cell>
        D. <math|6>
      </cell>>>>
    </wide-tabular>

    <item>设<math|x\<in\>\<bbb-R\>>，则“<math|sin x=1>”是“<math|cos
    x=0>”的

    <\wide-tabular>
      <tformat|<table|<row|<\cell>
        A. 充分不必要条件
      </cell>>|<row|<\cell>
        B. 必要不充分条件
      </cell>>|<row|<\cell>
        C. 充分必要条件
      </cell>>|<row|<\cell>
        D. 既不充分也不必要条件
      </cell>>>>
    </wide-tabular>

    <item>某几何体的三视图如图所示（单位：cm），则该几何体的体积（单位：<math|cm<rsup|3>>）是

    <\wide-tabular>
      <tformat|<table|<row|<\cell>
        A. <math|22\<pi\>>
      </cell>|<\cell>
        B. <math|8\<pi\>>
      </cell>>|<row|<\cell>
        C. <math|<frac|22|3>\<pi\>>
      </cell>|<\cell>
        D. <math|<frac|16|3>\<pi\>>
      </cell>>>>
    </wide-tabular>

    <image|<tuple|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

    <item>为了得到函数<math|y=2*sin
    3*x>的图象，只要把函数<math|y=2*sin<around*|(|3*x+<frac|\<pi\>|5>|)>>图象上所有的点

    <\wide-tabular>
      <tformat|<table|<row|<\cell>
        A. 向左平移<math|<frac|\<pi\>|5>>个单位长度
      </cell>|<\cell>
        B. 向右平移<math|<frac|\<pi\>|5>>个单位长度
      </cell>>|<row|<\cell>
        C. 向左平移<math|<frac|\<pi\>|15>>个单位长度
      </cell>|<\cell>
        D. 向右平移<math|<frac|\<pi\>|15>>个单位长度
      </cell>>>>
    </wide-tabular>

    <item>已知<math|2<rsup|a>=5,log<rsub|8>3=b>，则<math|4<rsup|a-3b>=>

    <\wide-tabular>
      <tformat|<table|<row|<\cell>
        A. <math|25>
      </cell>|<\cell>
        B. <math|5>
      </cell>>|<row|<\cell>
        C. <math|<frac|25|9>>
      </cell>|<\cell>
        D. <math|<frac|5|3>>
      </cell>>>>
    </wide-tabular>

    <item>如图，已知正三棱柱<math|<math-it|ABC>-A<rsub|1>B<rsub|1>C<rsub|1>,<math-it|AC>=<math-it|AA<rsub|1>>>，<math|E,F>分别是棱<math|<math-it|BC>>，<math|A<rsub|1>C<rsub|1>>上的点.记<math|<math-it|EF>>与<math|A<math-it|A<rsub|1>>>所成的角为<math|\<alpha\>>，<math|<math-it|EF>>与平面<math|<math-it|ABC>>所成的角为<math|\<beta\>>，二面角<math|F-<math-it|BC>-A>的平面角为<math|\<gamma\>>，则

    <\wide-tabular>
      <tformat|<table|<row|<\cell>
        A. <math|\<alpha\>\<leqslant\>\<beta\>\<leqslant\>\<gamma\>>
      </cell>|<\cell>
        B. <math|\<beta\>\<leqslant\>\<alpha\>\<leqslant\>\<gamma\>>
      </cell>>|<row|<\cell>
        C. <math|\<beta\>\<leqslant\>\<gamma\>\<less\>\<alpha\>>
      </cell>|<\cell>
        D. <math|\<alpha\>\<leqslant\>\<gamma\>\<less\>\<beta\>>
      </cell>>>>
    </wide-tabular>

    <image|<tuple|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

    <item>已知<math|a,b\<in\>\<bbb-R\>>，若对任意<math|x\<in\>\<bbb-R\>>，<math|a*<around*|\||x-b|\|>+<around*|\||x-4|\|>-<around*|\||2*x-5|\|>\<geqslant\>0>，则

    <\wide-tabular>
      <tformat|<table|<row|<\cell>
        A. <math|a\<leqslant\>1,b\<geqslant\>3>
      </cell>|<\cell>
        B. <math|a\<leqslant\>1,b\<leqslant\>3>
      </cell>>|<row|<\cell>
        C. <math|a\<geqslant\>1,b\<geqslant\>3>
      </cell>|<\cell>
        D. <math|a\<geqslant\>1,b\<leqslant\>3>
      </cell>>>>
    </wide-tabular>

    <item>已知数列<math|<around*|{|a<rsub|n>|}>>满足<with|math-display|true|<math|a<rsub|1>=1,a<rsub|n+1>=a<rsub|n>-<frac|1|3>a<rsub|n><rsup|2><around*|(|n\<in\>\<bbb-N\><rsup|\<ast\>>|)>>>，则

    <\wide-tabular>
      <tformat|<table|<row|<\cell>
        A. <math|<with|math-display|true|2\<less\>100*a<rsub|100>\<less\><frac|5|2>>>
      </cell>|<\cell>
        B. <with|math-display|true|<math|<frac|5|2>\<less\>100*a<rsub|100>\<less\>3>>
      </cell>>|<row|<\cell>
        C. <with|math-display|true|<math|3\<less\>100*a<rsub|100>\<less\><frac|7|2>>>
      </cell>|<\cell>
        D. <math|<with|math-display|true|<frac|7|2>\<less\>100*a<rsub|100>\<less\>4>>
      </cell>>>>
    </wide-tabular>
  </enumerate>

  \;

  \;

  \;

  <\with|par-mode|center>
    <strong|<\with|font|AR PL UMing CN|font-base-size|14>
      非选择题部分（共110分）
    </with>>
  </with>

  \;

  <strong|二、填空题：本大题共7小题，单空题每题4分，多空题每空3分，共36分。>

  <\enumerate>
    <assign|item-nr|10><item>我国南宋著名数学叫秦九韶，发现了从三角形三边求面积的公式，他把这种方法称为“三斜求积”，它填补了我国传统数学的一个空白.
    如果把这个方法写成公式，就是<math|<with|math-display|true|S=<sqrt|<frac|1|4><around*|[|c<rsup|2>a<rsup|2>-<around*|(|<frac|c<rsup|2>+a<rsup|2>-b<rsup|2>|2>|)><rsup|2>|]>>>>，其中<math|a,b,c>是三角形的三边，<math|S>是三角形的面积.
    设某三角形的三边<math|a=<sqrt|2>,b=<sqrt|3>,c=2>，则该三角形的面积<math|S=><underline|<space|3em>>.

    <item>已知多项式<math|<with|math-display|true|<around*|(|x+2|)><around*|(|x-1|)><rsup|4>=a<rsub|0>+a<rsub|1>*x+a<rsub|2>*x<rsup|2>+a<rsub|3>*x<rsup|3>+a<rsub|4>*x<rsup|4>+a<rsub|5>*x<rsup|5>>>，则<math|a<rsub|2>=><underline|<space|3em>>,<math|a<rsub|1>+a<rsub|2>+a<rsub|3>+a<rsub|4>+a<rsub|5>=><underline|<space|3em>>.

    <item>若<math|<with|math-display|true|3*sin \<alpha\>-sin
    \<beta\>=<sqrt|10>,\<alpha\>+\<beta\>=<frac|\<pi\>|2>>>，则<math|sin
    \<alpha\>=><underline|<space|3em>>，<math|cos
    2*\<beta\>=><underline|<space|3em>>.

    <item>已知函数<math|<with|math-display|true|f<around*|(|x|)>=<choice|<tformat|<table|<row|<cell|-x<rsup|2>+2,>|<cell|x\<leqslant\>1,>>|<row|<cell|x+<frac|1|x>-1,>|<cell|x\<gtr\>1,>>>>>>>则<math|f<around*|(|f<around*|(|<frac|1|2>|)>|)>=><underline|<space|3em>>；若当<math|x\<in\><around*|[|a,b|]>>时，<math|1\<leqslant\>f<around*|(|x|)>\<leqslant\>3>，则<math|b-a>的最大值是<underline|<space|3em>>.

    <item>现有 7 张卡片，分别写上数字1,2,3,4,5,6.
    从这7张卡片中随机抽取3张，记所抽取卡片上数字的最小值为<math|\<xi\>>，则<math|P<around*|(|\<xi\>=2|)>=><underline|<space|3em>>，<math|E<around*|(|\<xi\>|)>=><underline|<space|3em>>.

    <item>已知双曲线<math|<with|math-display|true|<frac|x<rsup|2>|a<rsup|2>>-<frac|y<rsup|2>|b<rsup|2>>=1<around*|(|a\<gtr\>0,b\<gtr\>0|)>>>的左焦点为<math|F>，过<math|F>且斜率为<math|<with|math-display|true|<frac|b|4a>>>的直线交双曲线于点<math|A<around*|(|x<rsub|1>,y<rsub|1>|)>>，交双曲线的渐近线于点<math|B<around*|(|x<rsub|2>,y<rsub|2>|)>>且<math|x<rsub|1>\<less\>0\<less\>x<rsub|2>>.
    若<math|<around*|\||FB|\|>=3*<around*|\||FA|\|>>，则双曲线的离心率是<underline|<space|3em>>.

    <item>设点<math|P>在单位圆的内接正八边形<math|A<rsub|1>A<rsub|2>\<ldots\>A<rsub|8>>的边<math|A<rsub|1>A<rsub|2>>上，则<math|<wide|PA<rsub|1>|\<vect\>><rsup|2>+<wide|PA<rsub|2>|\<vect\>><rsup|2>+\<cdots\>+<wide|PA<rsub|8>|\<vect\>><rsup|2>>的取值范围是<underline|<space|3em>>.
  </enumerate>

  \;

  \;

  \;

  \;

  \;

  <strong|三、解答题：本大题共5小题，共74分。解答应写出文字说明、证明过程或演算步骤。>

  <\enumerate>
    <assign|item-nr|17><item>（本题满分14分）在<math|\<triangle\><math-it|ABC>>中，角<math|A,B,C>所对的边分别为<math|a,b,c>.
    已知<math|4*a=<sqrt|5>*c,cos C=<frac|3|5>>.

    (I) 求<math|sin A>的值；

    (II) 若<math|b=11>，求<math|\<triangle\><math-it|ABC>>的面积.

    <item>（本题满分15分）如图，已知<math|<math-it|ABCD>>和<math|<math-it|CDEF>>都是直角梯形，<math|<math-it|AB>//<math-it|DC>,<math-it|DC>//<math-it|EF>,<math-it|AB>=5,<math-it|DC>=3,<math-it|EF>=1,\<angle\><math-it|BAD>=\<angle\><math-it|CDE>=60<rsup|\<circ\>>>，二面角<math|F-<math-it|DC>-B>的平面角为<math|60<rsup|\<circ\>>>，设<math|M>，<math|N>分别为<math|<math-it|AE>>，<math|<math-it|BC>>的中点.

    <\with|par-columns|2>
      (I) 证明：<math|<math-it|FN>\<perp\><math-it|AD>>；

      (II) 求直线<math|<math-it|BM>>与平面<math|<math-it|ADE>>所成角的正弦值.

      <image|<tuple|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
    </with>

    <item>（本题满分15分）已知等差数列<math|<around*|{|a<rsub|n>|}>>的首项<math|a<rsub|1>=-1>，公差<math|d\<gtr\>1>.
    记<math|<around*|{|a<rsub|n>|}>>的前<math|n>项和为<math|S<rsub|n><around*|(|n\<in\>\<bbb-N\><rsup|\<ast\>>|)>>.

    (I) 若<math|S<rsub|4>-2a<rsub|2>a<rsub|3>+6>，求<math|S<rsub|n>>；

    (II) 若对于每个<math|n\<in\>\<bbb-N\><rsup|\<ast\>>>，存在实数<math|c<rsub|n>>，使<math|a<rsub|n>+c<rsub|n>,a<rsub|n+1>+4*c<rsub|n>,a<rsub|n+2>+15*c<rsub|n>>成等比数列，求<math|d>的取值范围.

    <item>（本题满分15分）如图，已知椭圆<with|math-display|true|<math|<frac|x<rsup|2>|12>+y<rsup|2>=1>>.
    设<math|A,B>是椭圆上异与<math|P<around*|(|0,1|)>>的两点，且点<math|Q<around*|(|0,<frac|1|2>|)>>在线段<math|<math-it|AB>>上，直线<math|<math-it|PA>,<math-it|PB>>分别交直线<math|<with|math-display|true|y=-<frac|1|2>x+3>>于<math|C,D>两点.

    <\with|par-columns|2>
      (I) 求点<math|P>到椭圆上点的距离的最大值；

      (II) 求<math|<around*|\||<math-it|CD>|\|>>的最小值.

      <image|<tuple|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
    </with>

    <item>（本题满分15分）设函数<math|<with|math-display|true|f<around*|(|x|)>=<frac|\<mathe\>|2*x>+ln
    x<around*|(|x\<gtr\>0|)>>>.

    <\enumerate-Roman>
      <item>求<math|f<around*|(|x|)>>的单调区间；

      <item>已知<math|a,b\<in\>\<bbb-R\>>，曲线<math|y=f<around*|(|x|)>>上不同的三点<math|<around*|(|x<rsub|1>,f<around*|(|x<rsub|1>|)>|)>,<around*|(|x<rsub|2>,f<around*|(|x<rsub|2>|)>|)>,<around*|(|x<rsub|3>,f<around*|(|x<rsub|3>|)>|)>>处的切线都经过点<math|<around*|(|a,b|)>>.
      证明：

      <\enumerate-roman>
        <item>若<math|a\<gtr\>\<mathe\>>，则<math|<with|math-display|true|0\<less\>b-f<around*|(|a|)>\<less\><frac|1|2><around*|(|<frac|a|e>-1|)>>>；

        <item>若<math|0\<less\>a\<less\>\<mathe\>,x<rsub|1>\<less\>x<rsub|2>\<less\>x<rsub|3>>，则<with|math-display|true|<math|<frac|2|\<mathe\>>+<frac|\<mathe\>-a|6*\<mathe\><rsup|2>>\<less\><frac|1|x<rsub|1>>+<frac|1|x<rsub|3>>\<less\><frac|2|a>-<frac|\<mathe\>-a|6e<rsup|2>>>>.
      </enumerate-roman>
    </enumerate-Roman>

    （注：<math|e=2.71828\<ldots\>>是自然对数的底数）
  </enumerate>

  \;

  <\note*>
    本试题由B站UP主<hlink|<strong|沈浪熊猫儿>|https://space.bilibili.com/28058658>于2022年6月10日花费2小时纯手工输入制作。下载墨干编辑器，通过菜单<menu|help|planet>可以得到本试题的可编辑文档。
  </note*>

  \;

  \;
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