2023绵阳高三下学期第三次诊断性考试（三模）数学（文）PDF版含答案

这是一份2023绵阳高三下学期第三次诊断性考试（三模）数学（文）PDF版含答案，文件包含数学文答案docx、四川省绵阳市2023届高中毕业班三诊文科数学试题PDF版无答案pdf等2份试卷配套教学资源，其中试卷共10页， 欢迎下载使用。
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