planet/高考数学/2021年全国新高考II卷数学试题.tm

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  <doc-data|<doc-title|2021年全国新高考II卷数学试题>>

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    一、单选题
  </with>

  1．复数<math|<frac|2-i|1-3*i>>在复平面内对应的点所在的象限为（<space|1.2em>）

  <tabular|<tformat|<twith|table-width|1par>|<twith|table-hmode|exact>|<table|<row|<cell|A．第一象限>|<cell|B．第二象限>|<cell|C．第三象限>|<cell|D．第四象限>>>>>\ 

  2．设集合<math|U=<around|{|1,2,3,4,5,6|}>,A=<around|{|1,3,6|}>,B=<around|{|2,3,4|}>>，则<math|A\<cap\><around*|(|\<complement\><rsub|U>*B|)>=>（<space|1.2em>）

  <tabular|<tformat|<twith|table-width|1par>|<twith|table-hmode|exact>|<table|<row|<cell|A．<math|<around|{|3|}>>>|<cell|B．<math|<around|{|1,6|}>>>|<cell|C．<math|<around|{|5,6|}>>>|<cell|D．<math|<around|{|1,3|}>>>>>>>\ 

  3．抛物线<math|y<rsup|2>=2*p*x*<around|(|p\<gtr\>0|)>>的焦点到直线<math|y=x+1>的距离为<math|<sqrt|2>>，则<math|p=>（<space|1.2em>）

  <tabular|<tformat|<twith|table-width|1par>|<twith|table-hmode|exact>|<table|<row|<cell|A．1>|<cell|B．2>|<cell|C．<math|2*<sqrt|2>>>|<cell|D．4>>>>>\ 

  4．北斗三号全球卫星导航系统是我国航天事业的重要成果．在卫星导航系统中，地球静止同步卫星的轨道位于地球赤道所在平面，轨道高度为<math|36000>km（轨道高度是指卫星到地球表面的距离）．将地球看作是一个球心为<em|O>，半径<em|r>为<math|6400>km的球，其上点<em|A>的纬度是指<math|O*A>与赤道平面所成角的度数．地球表面上能直接观测到一颗地球静止同步轨道卫星点的纬度最大值为<math|\<alpha\>>，记卫星信号覆盖地球表面的表面积为<math|S=2*\<pi\>*r<rsup|2>*<around|(|1-cos
  \<alpha\>|)>>（单位：km<rsup|<math|2>>），则<em|S>占地球表面积的百分比约为（<space|1.2em>）

  <tabular|<tformat|<twith|table-width|1par>|<twith|table-hmode|exact>|<table|<row|<cell|A．26%>|<cell|B．34%>|<cell|C．42%>|<cell|D．50%>>>>>\ 

  5．正四棱台的上､下底面的边长分别为2，4，侧棱长为2，则其体积为（<space|1.2em>）

  <tabular|<tformat|<twith|table-width|1par>|<twith|table-hmode|exact>|<table|<row|<cell|A．<math|20+12*<sqrt|3>>>|<cell|B．<math|28*<sqrt|2>>>|<cell|C．<math|<frac|56|3>>>|<cell|D．<math|<frac|28*<sqrt|2>|3>>>>>>>\ 

  6．某物理量的测量结果服从正态分布<math|N<around*|(|10,\<sigma\><rsup|2>|)>>，下列结论中不正确的是（<space|1.2em>）

  A．<math|\<sigma\>>越小，该物理量在一次测量中在<math|<around|(|9.9,10.1|)>>的概率越大

  B．<math|\<sigma\>>越小，该物理量在一次测量中大于10的概率为0.5

  C．<math|\<sigma\>>越小，该物理量在一次测量中小于9.99与大于10.01的概率相等

  D．<math|\<sigma\>>越小，该物理量在一次测量中落在<math|<around|(|9.9,10.2|)>>与落在<math|<around|(|10,10.3|)>>的概率相等

  7．已知<math|a=log<rsub|5>2>，<math|b=log<rsub|8>3>，<math|c=<frac|1|2>>，则下列判断正确的是（<space|1.2em>）

  <tabular|<tformat|<twith|table-width|1par>|<twith|table-hmode|exact>|<table|<row|<cell|A．<math|c\<less\>b\<less\>a>>|<cell|B．<math|b\<less\>a\<less\>c>>|<cell|C．<math|a\<less\>c\<less\>b>>|<cell|D．<math|a\<less\>b\<less\>c>>>>>>\ 

  8．已知函数<math|f<around|(|x|)>>的定义域为<math|<math-bf|R>>，<math|f*<around|(|x+2|)>>为偶函数，<math|f*<around|(|2*x+1|)>>为奇函数，则（<space|1.2em>）

  <tabular|<tformat|<twith|table-width|1par>|<twith|table-hmode|exact>|<table|<row|<cell|A．<math|f*<around*|(|-<frac|1|2>|)>=0>>|<cell|B．<math|f*<around|(|-1|)>=0>>|<cell|C．<math|f<around|(|2|)>=0>>|<cell|D．<math|f<around|(|4|)>=0>>>>>>\ 

  <with|font-series|bold|二、多选题>

  9．下列统计量中，能度量样本<math|x<rsub|1>,x<rsub|2>,\<cdots\>,x<rsub|n>>的离散程度的是（<space|1.2em>）

  <tabular|<tformat|<twith|table-width|1par>|<twith|table-hmode|exact>|<table|<row|<cell|A．样本<math|x<rsub|1>,x<rsub|2>,\<cdots\>,x<rsub|n>>的标准差>|<cell|B．样本<math|x<rsub|1>,x<rsub|2>,\<cdots\>,x<rsub|n>>的中位数>>|<row|<cell|C．样本<math|x<rsub|1>,x<rsub|2>,\<cdots\>,x<rsub|n>>的极差>|<cell|D．样本<math|x<rsub|1>,x<rsub|2>,\<cdots\>,x<rsub|n>>的平均数>>>>>\ 

  10．如图，在正方体中，<em|O>为底面的中心，<em|P>为所在棱的中点，<em|M>，<em|N>为正方体的顶点．则满足<math|M*N*\<bot\>*O*P>的是（<space|1.2em>）

  A．<image|<tuple|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
  <space|2cm>B．<image|<tuple|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  <space|2cm>D．<image|<tuple|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

  11．已知直线<math|l:a*x+b*y-r<rsup|2>=0>与圆<math|C:x<rsup|2>+y<rsup|2>=r<rsup|2>>，点<math|A*<around|(|a,b|)>>，则下列说法正确的是（<space|1.2em>）

  A．若点<em|A>在圆<em|C>上，则直线<em|l>与圆<em|C>相切\ 

  B．若点<em|A>在圆<em|C>内，则直线<em|l>与圆<em|C>相离

  C．若点<em|A>在圆<em|C>外，则直线<em|l>与圆<em|C>相离\ 

  D．若点<em|A>在直线<em|l>上，则直线<em|l>与圆<em|C>相切

  12．设正整数<math|n=a<rsub|0>\<cdot\>2<rsup|0>+a<rsub|1>\<cdot\>2+\<cdots\>+a<rsub|k-1>\<cdot\>2<rsup|k-1>+a<rsub|k>\<cdot\>2<rsup|k>>，其中<math|a<rsub|i>\<in\><around*|{|0,1|}>>，记<math|\<omega\><around|(|n|)>=a<rsub|0>+a<rsub|1>+\<cdots\>+a<rsub|k>>．则（<space|1.2em>）

  <tabular|<tformat|<twith|table-width|1par>|<twith|table-hmode|exact>|<table|<row|<cell|A．<math|\<omega\>*<around|(|2*n|)>=\<omega\><around|(|n|)>>>|<cell|B．<math|\<omega\>*<around|(|2*n+3|)>=\<omega\><around|(|n|)>+1>>>|<row|<cell|C．<math|\<omega\>*<around|(|8*n+5|)>=\<omega\>*<around|(|4*n+3|)>>>|<cell|D．<math|\<omega\>*<around*|(|2<rsup|n>-1|)>=n>>>>>>\ 

  <with|font-series|bold|三、填空题>

  13．若双曲线<math|<frac|x<rsup|2>|a<rsup|2>>-<frac|y<rsup|2>|b<rsup|2>>=1>的离心率为2，则此双曲线的渐近线方程<underline|<space|2cm>>．

  14．写出一个同时具有下列性质①②③的函数<math|f<around|(|x|)>>:
  <underline|<space|2cm>>．

  ①<math|f*<around*|(|x<rsub|1>*x<rsub|2>|)>=f<around*|(|x<rsub|1>|)>*f<around*|(|x<rsub|2>|)>>；②当<math|x\<in\>
  <around*|(|0,+\<infty\>|)>>时，<math|f<rprime|'><around|(|x|)>\<gtr\>0>；③<math|f<rprime|'><around|(|x|)>>是奇函数．

  15．已知向量<math|<wide|a|\<wide-varrightarrow\>>+<wide|b|\<wide-varrightarrow\>>+<wide|c|\<wide-varrightarrow\>>=<wide|0|\<wide-varrightarrow\>>>，<math|<around*|\||<wide|a|\<wide-varrightarrow\>>|\|>=1>，<math|<around*|\||<wide|b|\<wide-varrightarrow\>>|\|>=<around*|\||<wide|c|\<wide-varrightarrow\>>|\|>=2>，<math|<wide|a|\<wide-varrightarrow\>>\<cdot\><wide|b|\<wide-varrightarrow\>>+<wide|b|\<wide-varrightarrow\>>\<cdot\><wide|c|\<wide-varrightarrow\>>+<wide|c|\<wide-varrightarrow\>>\<cdot\><wide|a|\<wide-varrightarrow\>>=><underline|<space|2cm>>．

  16．已知函数<math|f<around|(|x|)>=<around*|\||e<rsup|x>-1|\|>,x<rsub|1>\<less\>0,x<rsub|2>\<gtr\>0>，函数<math|f<around|(|x|)>>的图象在点<math|A*<around*|(|x<rsub|1>,f<around*|(|x<rsub|1>|)>|)>>和点<math|B*<around*|(|x<rsub|2>,f<around*|(|x<rsub|2>|)>|)>>的两条切线互相垂直，且分别交<em|y>轴于<em|M>，<em|N>两点，则<math|<frac|<around|\||A*M|\|>|<around|\||B*N|\|>>>取值范围是<underline|<space|2cm>>．

  <with|font-series|bold|四、解答题>

  17．记<math|S<rsub|n>>是公差不为0的等差数列<math|<around*|{|a<rsub|n>|}>>的前<em|n>项和，若<math|a<rsub|3>=S<rsub|5>,a<rsub|2>*a<rsub|4>=S<rsub|4>>．

  （1）求数列<math|<around*|{|a<rsub|n>|}>>的通项公式<math|a<rsub|n>>；

  （2）求使<math|S<rsub|n>\<gtr\>a<rsub|n>>成立的<em|n>的最小值．<vspace|3cm>

  18．在<math|\<triangle\>A*B*C>中，角<math|A>、<math|B>、<math|C>所对的边长分别为<math|a>、<math|b>、<math|c>，<math|b=a+1>，<math|c=a+2>.

  （1）若<math|2*sin C=3*sin A>，求<math|\<triangle\>A*B*C>的面积；

  （2）是否存在正整数<math|a>，使得<math|\<triangle\>A*B*C>为钝角三角形?若存在，求出<math|a>的值；若不存在，说明理由．<vspace|3cm>

  19．在四棱锥<math|Q-A*B*C*D>中，底面<math|A*B*C*D>是正方形，若<math|A*D=2,Q*D=Q*A=<sqrt|5>,Q*C=3>．

  <center|<image|<tuple|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

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

  （2）求二面角<math|B-Q*D-A>的平面角的余弦值．<vspace|3cm>

  20．已知椭圆<em|C>的方程为<math|<frac|x<rsup|2>|a<rsup|2>>+<frac|y<rsup|2>|b<rsup|2>>=1*<around|(|a\<gtr\>b\<gtr\>0|)>>，右焦点为<math|F<around|(|<sqrt|2>,0|)>>，且离心率为<math|<frac|<sqrt|6>|3>>．

  （1）求椭圆<em|C>的方程；

  （2）设<em|M>，<em|N>是椭圆<em|C>上的两点，直线<math|M*N>与曲线<math|x<rsup|2>+y<rsup|2>=b<rsup|2>*<around|(|x\<gtr\>0|)>>相切．证明：<em|M>，<em|N>，<em|F>三点共线的充要条件是<math|<around|\||M*N|\|>=<sqrt|3>>．<vspace|3cm>

  21．一种微生物群体可以经过自身繁殖不断生存下来，设一个这种微生物为第0代，经过一次繁殖后为第1代，再经过一次繁殖后为第2代......，该微生物每代繁殖的个数是相互独立的且有相同的分布列，设<em|X>表示1个微生物个体繁殖下一代的个数，<math|P*<around|(|X=i|)>=p<rsub|i>*<around|(|i=0,1,2,3|)>>．

  （1）已知<math|p<rsub|0>=0.4,p<rsub|1>=0.3,p<rsub|2>=0.2,p<rsub|3>=0.1>，求<math|E<around|(|X|)>>；

  （2）设<em|p>表示该种微生物经过多代繁殖后临近灭绝的概率，<em|p>是关于<em|x>的方程：<math|p<rsub|0>+p<rsub|1>*x+p<rsub|2>*x<rsup|2>+p<rsub|3>*x<rsup|3>=x>的一个最小正实根，求证：当<math|E<around|(|X|)>\<leq\>1>时，<math|p=1>，当<math|E<around|(|X|)>\<gtr\>1>时，<math|p\<less\>1>；

  （3）根据你的理解说明（2）问结论的实际含义．<vspace|3cm>

  22．已知函数<math|f<around|(|x|)>=<around|(|x-1|)>*e<rsup|x>-a*x<rsup|2>+b>．

  （1）讨论<math|f<around|(|x|)>>的单调性；

  （2）从下面两个条件中选一个，证明：<math|f<around|(|x|)>>只有一个零点.

  ①<math|<frac|1|2>\<less\>a\<leq\><frac|e<rsup|2>|2>,b\<gtr\>2*a>；

  ②<math|0\<less\>a\<less\><frac|1|2>,b\<leq\>2*a>．<vspace|3cm>
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