2023梅州高三下学期2月总复习质检（一模）数学含答案

这是一份2023梅州高三下学期2月总复习质检（一模）数学含答案，共18页。试卷主要包含了2），已知，则，函数，设是公差为等内容，欢迎下载使用。
试卷类型：A梅州市高三总复习质检试卷（2023.2）数  学本试卷共6页，满分150分，考试用时120分钟。注意事项：1.答卷前，考生务必将自己的姓名、准考证号等填写在答题卡和试卷指定位置上。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题目的答案标号涂黑。如需要改动，用橡皮擦干净后，再选涂其它答案标号。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，只有一项符合题目要求.1.已知复数满足，是虚数单位，则在复平面内的对应点落在（    ）A.第一象限 B.第二象限 C.第二象限 D.第四象限2.已知集合，，则（    ）A. B. C. D.3.为了了解小学生的体能情况，抽取了某小学四年级100名学生进行一分钟跳绳次数测试，将所得数据整理后，绘制如下频率分布直方图。根据此图，下列结论中错误的是（    ）A.B.估计该小学四年级学生的一分钟跳绳的平均次数超过125C.估计该小学四年级学生的一分钟跳绳次数的中位数约为119D.四年级学生一分钟跳绳超过125次以上优秀，则估计该小学四年级优秀率为35%4.已知，则（    ）A. B. C. D.5.由伦敦著名建筑事务所SteynStudio设计的南非双曲线大教堂惊艳世界，该建筑是数学与建筑完美结合造就的艺术品.若将如图所示的大教堂外形弧线的一段近似看成双曲线（，）下支的部分，且此双曲线两条渐近线方向向下的夹角为60°，则该双曲线的离心率为（    ）A. B. C. D.6.若从0，1，2，3，…9这10个整数中同时取3个不同的数，则其和为偶数的概率为（    ）A. B. C. D.7.某软件研发公司对某软件进行升级，主要是软件程序中的某序列重新编辑，编辑新序列为，它的第项为，若序列的所有项都是2，且，，则（    ）A. B. C.. D.8.《九章算术》是我国古代著名的数学著作，书中记载有几何体“刍甍”。现有一个刍甍如图所示，底面为正方形，平面，四边形，为两个全等的等腰梯形，，且，则此刍甍的外接球的表面积为（    ）A. B. C. D.二、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分全科试题免费下载公众号《高中僧课堂》。9.函数（，）的部分图像如图所示，则下列结论正确的是（    ）A.B.函数的图像关于悳线对称C.函数在单调递减D.函数是偶函数10.设是公差为（）的无穷等差数列的前项和，则下列命题正确的是（    ）A若，则是数列的最大项B若数列有最小项，则C.若数列是递减数列，则对任意的：，均有D.若对任意的，均有，则数列是递增数列11.如图，在直三棱柱中，，，，为棱的中点；为棱上的动点（含端点），过点、、作三棱柱的截面，且交于，则（    ）A.线段的最小值为 B.棱上的不存在点，使得平面C.棱上的存在点，使得 D.当为棱的中点时，12.对于定义在区间上的函数，若满足：，且，都有，则称函数为区间上的“非减函数”，若为区间上的“非减函数”，且，，又当时，恒成立，下列命题中正确的有（    ）A.  B.，C. D.，三、填空题：本题共4小题，每小题5分，共20分.13.展开式中的系数为______________.14.在平面直角坐标系中，点绕着原点顺时针旋转60°得到点，点的横坐标为____________.15.甲、乙、丙三人参加数学知识应用能力比赛，他们分别来自A、B、C三个学校，并分别获得第、二、三名：已知：①甲不是A校选手；②乙不是B校选手；③A校选手不是第一名；④B校的选手获得第二名；⑤乙不是第三名.根据上述情况，可判断出丙是___________校选手，他获得的是第___________名.16.函数的最小值为___________.四、解答题；本题共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.（本小题满分10分）在中，内角，，的对边分别为，，，已知.（1.）求内角；（2）点是边上的中点，已知，求面积的最大值.18.（本小题满分12分）记是正项数列的前n项和，若存在某正数M，，都有，则称的前n项和数列有界.从以下三个数列中任选两个，①；②；③，分别判断它们的前项和数列是否有界，并给予证明.19.（本小题满分12分）如图，在边长为4的正三角形中，为边的中点，过作于.把沿翻折至的位置，连接、.（1）为边的一点，若，求赃：平面；（2）当四面体的体积取得最大值时，求平面与平面的夹角的余弦值.20.（本小题满分12分）甲、乙、丙、丁四支球队进行单循环小组赛（每两支队比赛一场），比赛分三轮，每轮两场比赛，第一轮第一场甲乙比赛，第二场丙丁比赛；第二轮第一场甲丙比赛，第二场乙丁比赛；第三轮甲对丁和乙对丙两场比赛同一时间开赛，规定：比赛无平局，获胜的球队记3分，输的球队记0分.三轮比赛结束后以积分多少进行排名，积分相同的队伍由抽签决定排名，排名前两位的队伍小组出线.假设四支球队每场比赛获胜概率以近10场球队相互之间的胜场比为参考.队伍近10场胜场比队伍甲乙甲丙甲丁乙丙乙丁丙丁（1）三轮比赛结束后甲的积分记为，求；（2）若前二轮比赛结束后，甲、乙、丙、丁四支球队积分分别为3、3、0、6，求甲队能小组出线的概率.21.（本小题满分12分）已知函数.（1）当时，求函数的单调区间；（2）若，讨论函数的零点个数.22.（本小题满分12分）已知动圆经过定点，且与圆：内切.（1）求动圆圆心的轨迹的方程；（2）设轨迹与轴从左到右的交点为点，，点为轨迹上异于，的动点，设交直线于点，连结交轨迹于点.直线、的斜率分别为、.（i）求证：为定值；（ii）证明直线经过轴上的定点，并求出该定点的坐标.  梅州市高三总复习质检（2023.2）数学参考答案与评分意见一、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.12345678CBBADDBC二、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，有选错的得0分，部分选对的得2分.9101112ABBDABDACD三、填空题：本题共4小题.每小题5分，共20分.13.40 14. 15.A；三 16.四、解答题：本题共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.（本小题满分10分）解：（1）在中，因为，由正弦定理得：，····································································1分因为，所以，于是有，·································································2分所以，即，··········································································3分因为，所以，········································································4分所以，从而.·········································································5分（2）因为点是边上的中点，所以，·······················································6分对上式两边平分得：，·································································7分因为，所以，即，··········································································8分而，有，所以，当且仅当时，等号成立.···························································9分因此.·············································································10分即面积的最大值为.18.（本小题满分12分）解：数列①②的前项和数列有界，数列③的前项和数列无界，证明如下：①若，则其前项和，···································································2分因为，所以，则，····································································4分所以存在正数1，，，即前项和数列有界.②若，当时，，······································································2分其前项和，·················································································4分因为，所以，则，····································································5分所以存在正数2，，，即前项和数列有界.····································································6分③若，其前项和为，，·················································································2分对于任意正数，取（其中表示不大于的最大整数），有，···············································································4分因此前项和数列不是有界的.····························································6分19.（本小题满分12分）（1）证明：取中点，连接，因为在正三角形中，，又因为，所以，······································································1分平面，平面，所以平面，··········································································2分又有，且，所以，而平面，平面，所以平面.·······························································3分有，所以平面平面，······································································4分又平面，因此平面.···········································································5分（2）解：因为，又因为的面积为定值，所以当到平面的距离最大时，四面体的体积有最大值，········································6分因为，，，，平面，所以平面，因为平面，所以平面平面，当时，平面平面，平面所以平面，即在翻折过程中，点到平面的最大距离是，因此四面体的体积取得最大值时，必有平面.·················································7分如图，以点为原点，为轴，为轴，为轴，建立空间直接坐标系，易知，，，，，，，·············································································8分因为平面，所以平面的法向量为，·································································9分设平面的法向量为，，由，令得：，，所以，············································································10分.·················································································11分所以平面与平面的夹角（锐角）的余弦值为.················································12分20.（本小题满分12分）解：（1）设甲的第场比赛获胜记为（，2，3），·············································1分则有···············································································2分··················································································3分.··················································································4分（2）分以下三种情况：（i）若第三轮甲胜丁，另一场比赛乙胜丙，则甲、乙、丙、丁四个球队积分变为6、6、0、6，············································5分此时甲、乙、丁三支球队积分相同，要抽签决定排名，甲抽中前两名的概率为，所以这种情况下，甲出线的概率为；······················································6分（ii）若第三轮甲胜丁，另一场比赛乙输丙，则甲、乙、丙、丁积分变为6、3、3、6，···················································7分此时甲一定出线，甲出线的概率为；······················································8分（iii）若第三轮甲输丁，另一场比赛乙输丙.则甲、乙、丙、丁积分变为3、3、3、9，···················································9分此时甲、乙、丙三支球队要抽签决定排名，甲抽到第二名的概率为，所以这种情况下，甲出线的概率为.·······················································10分综上，甲出线的概率为.································································12分21.（本小题满分12分）解：（1）首先函数的定义域为，当时，，则.················································································1分由，得，，··········································································2分所以当时，；当时，.··································································3分故的增区间为和，减区间为.·····························································4分（2），，由，得，，··········································································5分所以当时，：当时，.因此在和上单调递增，在上单调递减.······················································6分①因为当时，有.所以在上不存在零点.··································································8分②在上，由单调性知：，分以下三种悄况讨论：（i）若，在，即在上不存在零点；·······················································9分（ii）若，有，此时在有唯一零点；··································································10分（iii）若，有，而，，则在与上各有一个零点.·······························································11分综上：（i）当时，在止不存在零点；（ii）当时，在上存在一个零点；（iii）当时，在上存在两个零点.·······················································12分22.（本小题满分12分）（1）解：设动圆的半径为，由题意，得：，，···············································································1分则.················································································2分所以动点的轨迹是以，为焦点，长轴长为4的椭圆.············································3分因此轨迹方程为.······································································4分（2）（i）证法一：设，，.由题可知，，，则，，·············································································5分而，于是，··········································································6分所以，·············································································7分又，则，因此为定值.·········································································8分证法二：设，，.由题可知，，，则直线的方程为，.由，得，···········································································5分所以，即，则.················································································6分所以.··············································································7分故为定值.··········································································8分证法三：设，，.由题可知，，，财.由，得（），········································································5分所以，即，··········································································6分故，又，···········································································7分所以为定值.·········································································8分解：（ii）法一：设直线的方程为，，.由，得，···········································································9分所以.·············································································10分由（i）可知，，即，化简得：，解得或（舍去），···························································11分所以真线的方程为，因此直线经过定点.···································································12分法二：设，，①当直线的斜率存在时，设直线的方程为，由，得，所以···············································································9分由（i）已知，，即：化简得：，解得或（舍去）.····························································10分所以直线的方程为，故直线经过定点.·····································································11分②当线的斜率不存在时，则.由（i）知，，即：.又，所以，解得.所以直线的方程为，故直线经过定点.综上，直线经过定点.·································································12分法三：设，，.由题可知，，，则直线的方程为.由，得，所以，即，则，所以，同理，得.·····································································10分当，即时，直线的方程为，此时直线经过定点.·······················································11分当，即时，直线的方程为，即，此时直线经过定点.综上，直线经过定点.·································································12分