广西壮族自治区百色市贵百联考2024届高三上学期9月月考数学试题

这是一份广西壮族自治区百色市贵百联考2024届高三上学期9月月考数学试题，共10页。试卷主要包含了圆M，设，，，则，下列命题中，正确的命题是，已知抛物线C等内容，欢迎下载使用。
2024届广西名校开学考试试题数学（考试时间：150分钟    满分：150分）注意事项：1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。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题目的答案标号涂黑。如需改动，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、单选题：共8小题，每小题5分，共40分，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．若复数z满足，则在复平面内复数z所对应的点位于（    ）A．第一象限 B．第二象限 C．第三象限 D．第四象限3．函数是奇函数，则（    ）A． B． C． D．24．已知数列是公比为正数的等比数列，是其前n项和，，，则（    ）A．15 B．31 C．63 D．75．圆M：，圆N：，则两圆的一条公切线方程为（    ）A．  B．C．  D．6．某中学体育节中，羽毛球单打12强中有3个种子选手，将这12人任意分成3个组（每组4个人），则3个种子选手恰好被分在同一组的分法种数为（    ）A．210 B．105 C．315 D．6307．圆锥的底面圆半径，侧面的平面展开图的面积为，则此圆锥的体积为（    ）A． B． C． D．8．设，，，则（    ）A． B． C． D．二、多选题：共4小题，每小题5分，共20分，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，有选错得0分．9．下列命题中，正确的命题是（    ）A．数据4，5，6，7，8的第80百分位数为7B．若经验回归方程为时，则变量x与y负相关C．对于随机事件A，B，若，则A与B相互独立D．某学习小组调查5名男生和5名女生的成绩，其中男生成绩的平均数为9，方差为13；女生成绩的平均数为7，方差为10，则该10人成绩的方差为10.510．已知抛物线C：的焦点为F，过F的直线与C交于A、B两点，且A在x轴上方，过A、B分别作C的准线的垂线，垂足分别为、，则（    ）A．若A的纵坐标为3，则B．C．准线方程为D．以为直径的圆与直线相切于F11．已知四面体的四个面均为直角三角形，其中平面，，且．若该四面体的体积为，则（    ）A．平面  B．平面平面C．的最小值为3  D．四面体外接球的表面积的最小值为12．函数的两个极值点分别是，，则下列结论正确的是（    ）A．  B．C．  D．三、填空题：本题共4小题，每小题5分，共20分．13．的展开式中的系数为________．14．已知，，则________．15．函数在上恰有2个零点，则的取值范围是________．16．已知椭圆C：的左、右焦点分别为，，点P在C上，若，，则C的离心率为________．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．（本题10分）已知数列满足：，，数列是以4为公差的等差数列．（1）求数列的通项公式；（2）记数列的前n项和为，求的值．18．（本题12分）在中，角A，B，C的对边分别是a，b，c，且．（1）求的值；（2）若，，D是线段上的一点，求的最小值．19．（本题12分）四边形为菱形，平面，，，．（1）设中点为，证明：平面；（2）求平面与平面的夹角的大小．20．（本题12分）某研究小组为研究经常锻炼与成绩好差的关系，从全市若干所学校中随机抽取100名学生进行调查，其中有体育锻炼习惯的有45人．经调查，得到这100名学生近期考试的分数的频率分布直方图．记分数在600分以上的为优秀，其余为合格．（1）请完成下列列联表．根据小概率值的独立性检验，分析成绩优秀与体育锻炼有没有关系． 经常锻炼不经常锻炼合计合格25  优秀 10 合计  100（2）现采取分层抽样的方法，从这100人中抽取10人，再从这10人中随机抽取5人进行进一步调查，记抽到5人中优秀的人数为X，求X的分布列．附：，其中．0.0500.0100.001k3.8416.63510.82821．（本题12分）已知双曲线C：一个焦点F到渐近线的距离为．（1）求双曲线C的方程；（2）过点的直线与双曲线C的右支交于A，B两点，在x轴上是否存在点N，使得为定值？如果存在，求出点N的坐标及该定值；如果不存在，请说明理由．22．（本题12分）已知函数．（1）当时，讨论在区间上的单调性；（2）若当时，，求a的取值范围． 2024届广西名校开学考试试题数学参考答案一、单选题：共8小题，每小题5分，共40分，每小题给出的四个选项中，只有一项是符合题目要求的．1．，，则，故选：D．2．由，对应点坐标为，在第二象限．故选：B．3．函数为奇函数，∴，，解得，选C．4．设公比为，，，解得，，故，选A．5．设切线，，得或，则，；另两条切线与直线平行且相距为1，又由，设切线，则，得，则，选C．6．3个种子选手分在同一组的方法有种，故选：C．7．设母线为，展开图，，高，，选A．8．构造，，，在上单调递减，又，，即，又，故，选B．二、多选题：共4小题，每小题5分，共20分，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，有选错得0分．9．对于A，由于，则4，5，6，7，8的第50百分位数为，故A错；对于B，若方程为时，则变量x与y负相关，故B正确；对于C，若，则有，可得，则A与B相互独立，故C正确；对于D，10人的成绩平均，则10人的方差，故D错；选BC．10．抛物线的焦点，准线，故C正确；设直线为，，，则，，联立方程，消y得：，则，对A：∵，A错；对B：∵，，∴，∴，不互垂直，B错误；对D：∵，，∴的中点到直线的距离，又∵，故以为直径的圆与直线相切于F，D正确；故选：CD．11．如图，将四面体补全为长方体，因为平面，平面，所以，又，所以平面，故A正确；因为平面，所以，又因为平面平面，平面，，平面，则即为二面角的平面角，因为为锐角，即二面角为锐二面角，故B不对；设，，，得，，当且仅当时等号成立，，故C正确．设四面体外接球的半径为，则，当且仅当时等号成立，所以，即四面体外接球的表面积的最小值为，故D正确．故选：ACD．12．对于A，定义域为，，令，在上有两个不等实根，，，得，故A正确；对于B，由韦达得，故B错误；对于C，由，，C正确；对于D，，令，，，令，，，即函数在上单调递减，，则函数在上单调递减，于是，所以，故D正确；故选：ACD三、填空题：本题共4小题，每小题5分，共20分．13．的通项，令，则；令，，的展开式中的系数为，答案为90．14．，，答案22．15．化简得，又，得，因在上恰有2个零点，，解得．16．，，O是的中点，所以，故由得，因为，，所以，在中，，在中，，∴，即，则，离心率为．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．（1）根据题意可得；·····························································2分，················································································4分又符合上式，所以···································································5分（2）·············································································7分·················································································10分18．（1）由条件得：·································································2分所以，即，解得·····································································4分又，所以··········································································6分（2）由，则．······································································8分设的边上的高为h．的面积：，∴，．···········································································11分∵B是钝角，∴当时，垂足在边上，即的最小值是···········································12分19．（1）四边形为菱形，且，所以．因为，所以，·······································································2分因为平面，平面，所以．······························································4分又，，平面，所以平面；·········································································5分（2）设交于点，取中点，连接，所以，底面．以为原点，以，，分别为轴，轴，轴的正方向建立空间直角坐标系，因为，所以，所以，，，，，，···································································6分所以，，设平面的一个法向量为，则，令得所以；············································································8分，，平面的一个法向量为，则，令得；········································································10分所以，···········································································11分所以平面与平面的夹角的大小为．······················································12分20．（1）填列联表 经常锻炼不经常锻炼合计合格254570优秀201030合计4555100·················································································2分零假设：成绩是否优秀与是否经常体育锻炼无关．················································································5分根据小概率值的独立性检验，推断不成立，故成绩优秀与是否经常体育锻炼有关联····················································6分（2）根据直方图大于600分的频率为，小于600分的频率为，故由分层抽样知，抽取的10人中合格有人，优秀的为人········································7分则从这10人中随机抽取5人，优秀人数X服从超几何分布，由题意X的可能值为0，1，2，3··············8分，，，·················································································11分故分布列为X0123P·················································································12分21．（1）由双曲线得渐近线方程为，设，则，··············································2分∴双曲线C方程为；··································································4分（2）依题意，直线的斜率不为0，设其方程为，，···········································5分代入得，设，，，则，，············································································7分∴·················································································9分若要上式为定值，则必须有，即，······················································10分∴，·············································································11分故存在点满········································································12分22．（1），········································································1分当时，；当时，·····································································3分故在上单调递增，在上单调递减；·······················································4分（2）设，；········································································5分设，则，令，则，当，，当，，故函数在单调递增，在单调递减，所以；············································································7分令，可得，故在单调递增时，···························································8分当时，，故在上单调递增······························································9分当时，，且当时，，若，则，函数在上单调递增，因此，，符合条件···················································10分若，则存在，使得，即，当时，，则在上单调递减，此时，不符合条件．综上，实数的取值范围是······························································12分