2023湖南省五市十校教研教改共同体、三湘名校教育联盟、湖湘名校教育联合体高二上学期期中考试数学试题含答案

这是一份2023湖南省五市十校教研教改共同体、三湘名校教育联盟、湖湘名校教育联合体高二上学期期中考试数学试题含答案，共16页。试卷主要包含了已知函数，下列说法正确的是，下列说法正确的是等内容，欢迎下载使用。
绝密★启用前五市十校教研教改共同体  三湘名校教育联盟  湖湘名校教育联合体2022年下学期高二期中考试数学命题：双峰一中数学备课组  审题：南县一中  郭劲松  永州一中数学备课组本试卷共4页。全卷满分150分，考试时间120分钟。注意事项：1.答题前，考生务必将自己的姓名、准考证号填写在本试卷和答题卡上。2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应的答案标号涂黑，如有改动，用橡皮擦干净后，再选涂其他答案；回答非选择题时，将答案写在答题卡上，写在本试卷上无效。3.考试结束后，将本试卷和答题卡一并交回。一、单项选择题：本题共8小题，每小题5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.1.已知集合，，则（    ）A. B. C. D.2.已知圆C的圆心坐标为，且过坐标原点，则圆C的方程为（    ）A. B.C. D.3.党的十八大报告指出，必须坚持在发展中保障和改善民生，不断实现人民对美好生活的向往，为响应中央号召，某社区决定在现有的休闲广场内修建一个半径为4m的圆形水池来规划喷泉景观.设计如下：在水池中心竖直安装一根高出水面为2m的喷水管（水管半径忽略不计），它喷出的水柱呈抛物线型，要求水柱在与水池中心水平距离为处达到最高，且水柱刚好落在池内，则水柱的最大高度为（    ）A. B. C. D.4.已知是等比数列的前n项和，，，成等差数列，则下列结论正确的是（    ）A. B. C. D.5.已知幂函数的图象是等轴双曲线C，且它的焦点在直线上，则下列曲线中，与曲线C的实轴长相等的双曲线是（    ）A. B. C. D.6.已知函数，下列说法正确的是（    ）A.函数的最小正周期是 B.函数的最大值为C.函数的图象关于点对称 D.函数在区间上单调递增7.如图水平放置的边长为1的正方形沿x轴正向滚动，初始时顶点A在坐标原点，（沿x轴正向滚动指的是先以顶点B为中心顺时针旋转，再以顶点C为中心顺时针旋转，如此继续），设顶点的轨迹方程式，则（    ）A.0 B.1 C. D.8.已知三棱锥中，，，，若二面角的大小为120°，则三棱锥的外接球的表面积为（    ）A. B. C. D.二、多项选择题：本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分.9.下列说法正确的是（    ）A.命题“，”的否定为“，”B.在中，若“”，则“”C.若，则的充要条件是D.若直线与平行，则或210.已知各项均为正数的等差数列单调递增，且，则（    ）A.公差d的取值范围是 B.C.  D.的最小值为111.已知直线l与抛物线（）交于A，B两点，，，则下列说法正确的是（    ）A.若点D的坐标为，则B.直线过定点C.D点的轨迹方程为（原点除外）D.设与x轴交于点M，则的面积最大时，直线的斜率为112.在正方体中，，点M在正方体内部及表面上运动，下列说法正确的是（    ）A.若M为棱的中点，则直线平面B.若M在线段上运动，则的最小值为C.当M与重合时，以M为球心，为半径的球与侧面的交线长为D.若M在线段上运动，则M到直线的最短距离为三、填空题：本题共4小题，每小题5分，共20分.13.某中学高一年级有600人，高二年级有480人，高三年级有420人，因新冠疫情防控的需要，现用分层抽样从中抽取一个容量为300人的样本进行核酸检测，则高三年级被抽取的人数为___________.14.设双曲线C：（，）的左、右焦点分别为、，P是渐近线上一点，且满足，，则双曲线C的离心率为___________.15.已知动点在运动过程中总满足关系式，记，，则面积的最大值为___________.16.意大利数学家斐波那契在研究兔子繁殖问题时发现了数列1，1，2，3，5，8，13，…，数列中的每一项被称为斐波那契数，用符号表示（），已知，，（）.（1）若，则___________（2分）；（2）若，则___________（3分）.四、解答题：本题共6小题，共70分.解答应写出必要的文字说明、证明过程及演算步骤.17.（本小题满分10分）已知双曲线C：（，）的左右焦点分别为，，点M在双曲线C的右支上，且，离心率.（1）求双曲线C的标准方程；（2）若，求的面积.18.（本小题满分12分）10月9日晚，2022年世界乒乓球团体锦标赛在中国成都落幕.中国队女团与男团分别完成了五连冠与十连冠的霸业.乒乓球运动在我国一直有着光荣历史，始终领先世界水平，被国人称为“国球”，在某次团体选拔赛中，甲乙两队进行比赛，采取五局三胜制（即先胜三局的团队获得比赛的胜利），假设在一局比赛中，甲队获胜的概率为0.6，乙队获胜的概率为0.4，各局比赛结果相对独立.（1）求这场选拔赛三局结束的概率；（2）若第一局比赛乙队获胜，求比赛进入第五局的概率.19.（本小题满分12分）已知锐角三角形中，角A，B，C所对的边分别为a，b，c，向量，，且.（1）求角B的大小；（2）若，求面积的取值范围.20.（本小题满分12分）已知数列满足，且，数列是各项均为正数的等比数列，为的前n项和，满足，.（1）求数列的通项公式；（2）设，记数列的前n项和为，求的取值范围.21.（本小题满分12分）如图，在四棱锥中，，，，平面平面，E为中点.（1）求证：面；（2）求证：面；（3）点Q在棱上，设（），若二面角的余弦值为，求.22.（本小题满分12分）已知椭圆C：（）过点，A为左顶点，且直线的斜率为.（1）求椭圆C的标准方程；（2）设在椭圆内部，在椭圆外部，过Ｍ作斜率不为0的直线交椭圆C于P，Q两点，若，求证：为定值，并求出这个定值. 
五市十校教研教改共同体  三湘名校教育联盟  湖湘名校教育联合体2022年下学期高二期中考试数学参考答案、提示及评分细则一、选择题：（本题共8小题，每小题5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）1.【答案】C【解析】∵，∴.2.【答案】B【解析】圆心，半径，故圆C方程为.3.【答案】C【解析】取一截面建系如图，设抛物线方程为（），记最大高度为h，如图：，在抛物线上，故，两式相除有，解得.4.【答案】AB【解析】若公比有，，，此时，故公比，由题意，化简有，故有或，选答案AB.5.【答案】B【解析】由双曲线几何性质知，双曲线的焦点在实轴上，实轴与双曲线的交点，是双曲线的顶点，故双曲线C的实轴长，选答案B.6.【答案】D【解析】由知A，B错误.由，所以C错误.当时，，所以D正确.7.【答案】D【解析】A点运动轨迹最终构成图象如图：由图可知.故，在B→D段时，A点的轨迹方程为（），∴.8.【答案】C【解析】由题意，取中点，中点，连接，则，分别是与的外心，且，分别过，作面，面，记，则O为外接球球心，在中，，∴，故，选C.二、多选题（本题共4小题，每小题5分，在每小题给出的选项中，有多项符合要求，全部选对的得5分，有选错的得0分，部分选对的得2分）9.【答案】BC【解析】对A：否定为：，，所以A错误；对D，当时，两直线重合，所以D错误.10.【答案】AB【解析】由题意得，，∴，故A正确；由，故B正确；由，知故C错误；由有，当且仅当时取到等号，但，故不能取“=”，所以D错.11.【答案】ABC【解析】，由知方程为，联立，消去x有，记，，则，由，∴，故A正确；对选项BCD，可设：，代入有，则，由，故直线为，过定点，即，故B正确；由，得D在以为直径的圆：上运动（原点除外），故C正确；当时，面积最大，此时，有，故D错误.12.【答案】ACD【解析】易知A，D正确；对选项B：展开与到同一平面上如图.知，故B错误；对选项C：M与重合时，在侧面上的射影为，故交线是以为圆心的一段圆弧（个圆），且圆半径，故圆弧长，所以C正确.三、填空题（本题共4小题，每小题5分，共20分）13.【答案】84【解析】由分层抽样易得.14.【答案】【解析】不妨设P在第一象限，则，依题意：，∴离线率.15.【答案】18【解析】易得M在椭圆上运动，且B在椭圆上，A为左顶点，由方程：，设直线l：与椭圆相切于点M.联立，消去x得，由，依题意，时，面积最大，此时直线l与距离为，又，∴.16.【答案】（1）11（2分）  （2）（3分）【解析】（1），∴；（2）.四、解答题（本大题共6小题，共70分，解答应写出文字说明，证明过程或演算步骤）17.【答案】（1）  （2）【解析】（1）由题意，································································1分∴，···············································································2分又，···············································································3分∴，···············································································4分故双曲线C的方程为；··································································5分（2），，则由双曲线定义可得  ①，由三角形余弦定理得  ②，······························································7分有，···············································································9分∴的面积.··········································································10分18.【答案】（1）0.28  （2）0.432【解析】设“第i局甲胜”为事件，“第j局乙胜”为事件（i，，2，3，4，5），（1）记“三局结束比赛”，则，·························································2分∴；·················································································6分（2）记“决胜局进入第五局比赛”，则，··················································8分∴.·················································································12分19.【答案】（1）  （2）【解析】（1）由，···································································2分由正弦定理得，······································································4分又，∴，···········································································6分（2）解法一：在锐角中，由（1）知，，有，令，则，，由正弦定理得，的面积  ································································8分，················································································10分由得，，则，于是得，所以面积的取值范围是.································································12分解法二：由（1）可知，，故，又因为，所以，·············································································8分又因为，，所以，故，即有，则，·········································································10分又由，即，所以面积的取值范围是.································································12分20【答案】（1）  （2）【解析】（1）由，···································································1分∴（常数），········································································2分故数列是以为公差的等差数列，且首项为，··········································································3分∴，···············································································4分故；···············································································5分（2）设公比为q（），由题意：，∴，解得或（舍），∴，∴，···············································································7分∴，有，两式相减得，·················································································9分∴，··············································································10分由，知在上单调递增，································································11分∴.···············································································12分21.【答案】（1）略  （2）略  （3）【解析】（1）证明：取中点F，连接，，则，又，∴，∴四边形是平行四边形，∴，又面，面，∴面；·············································································4分（2）证明：由题意：，，∴，同理，又，∴，∴，···············································································6分又面面，∴面，∴.又且，∴面；·············································································8分（3）以D为原点，建立如图所示的空间直角坐标系，则，，，，∴，，，由，有，··········································································10分令是面的法向量，则，令，有，··········································································11分取面的法向量，由.···············································································12分22.【答案】（1）  （2）为定值4，证明略【解析】（1）由题意：，∴，故椭圆C的标准方程为；································································4分（2）设：，联立消去x，有，记，，则且，，···········································································7分若，则·············································································9分（），∴（定值），综上：为定值4.····································································...12分