2023届四川省绵阳市高三下学期第三次诊断性考试（三模）数学（文）PDF版含答案

  绵阳市高中2020级第三次诊断性考试

文科数学参考答案及评分意见

一、选择题：本大题共12小题，每小题5分，共60分．

    CABBA     CDDCA     CB

二、填空题：本大题共4小题，每小题5分，共20分．

13．3           14．          15．          16．

三、解答题：本大题共6小题，共70分．

17．解：（1）用平均数估计总体，

在某个销售门店春季新款的年销售额的是33万元，······················2分

用中位数估计总体，

在某个销售门店春季新款的年销售额的是31.5万元．····················4分

（2）6个销售门店分别记为A，B，C，D，E，F．

年销售额不低于40万元的有：A，D．·······························5分

从A，B，C，D，E，F中随机抽取2个，基本事件为：{A，B}，{A，C}，{A，D}，{A，E}，{A，F}，{B，C}，{B，D}，{B，E}，{B，F}，{C，D}，{C，E}，{C，F}，{D，E}，{D，F}，{E，F}，共计15个基本事件．·····8分

事件：“恰好抽到1个门店的年销售额不低于40万元”包含的基本事件为：

{A，B}，{A，C}，{A，E}，{A，F}，{B，D}，{C，D}，

{E，D}，{F，E}，············································10分

∴所求概率为．················································12分

18．解：（1）证明：如图，取的中点为O，连接BO，PO．

∵PA=PC，∴，···············································1分

∵，∴，····················································2分

∴，同理，··················································3分

又，则，

∴，·······················································4分

∵，，平面，

∴平面，····················································5分

又平面，

∴平面平面；··················································6分

（2）∵点M是线段AP上，且，

过点M作，，················································7分

∴平面，····················································8分

·························································10分

······················································11分

．··················································12分

19．解：（1）由，令n=1，

得，

∴，·······················································2分

又∵，

∴等差数列{}的公差，，···········································4分

∴．·······················································6分

（2）由（1）可知，·············································7分

当时，，····················································8分

所以当时，；················································10分

当时，也满足上式，···········································11分

所以()．····················································12分

20．解：（1）当时，，，·············································2分

因为切点为，所以切线斜率为：，··································3分

所以曲线在处切线的方程为：．···································5分

（2），·····················································6分

令得或，····················································7分

①当时，在上单调递增，

此时，，

当，即时，在区间上无零点；

当，即时，在区间上有一个零点；

当，即时，在区间上无零点；·····································9分

②当，即时，在上单调递减，

此时，在区间上无零点．·······································10分

③当时，在上单调递减，在上单调递增，

此时，，在区间上无零点．·······································11分

综上：当或时，在区间上无零点；

当时，在区间上有一个零点．·····································12分

21．解：（1）设M(x1，y1)， N(x2，y2)，直线l：y=x−2，······················1分

联立方程，整理得：，··········································2分

由韦达定理：，···············································3分

，·····················································4分

解得：，故抛物线的方程为：y2=x．································5分

（2）延长PN交x轴于点Q，设M(x1，y1)， N(x2，y2)，P(x3，y3)，

设直线MN的方程为：，···········································6分

联立直线MN与抛物线C方程可得：，整理得：，

由根与系数的关系：y1 y2=−2 ①，··································8分

同理，联立直线MP与抛物线C方程可得：，

整理得：，可得y1 y3=−3 ②，····································10分

由①②可知，，················································11分

∴．·······················································12分

22．解：（1）可得圆C的标准方程为：，

∴圆C是以C（2，0）为圆心，2为半径的圆，··························2分

∴圆C的参数方程为：（为参数）．·································5分

（2）∵，可得，··············································6分

不妨设点A所对应的参数为，则点B所对应的参数为，

∴，则，

即B，······················································7分

∴，，

∴=························································8分

==4+4，··········································9分

∵，则，

∴当=1，即=时，的最大值为．····································10分

23．解：（1）由a=1，则2b+3c=3，

由柯西不等式，得，···········································2分

∴，3分

∴，当且仅当时等号成立．·······································5分

（2）∵a+2b+3c=4，即2b+3c=4−a，

又，则，····················································6分

又由（1）可得：，············································7分

∴，即，····················································8分

令=t，所以，

解得：，即，················································9分

又2b+3c=4−a，且b>0，c>0，

∴4−a>0，即a<4，

综上可得，．················································10分