2022南昌高三上学期摸底考试数学（理）试题含答案

这是一份2022南昌高三上学期摸底考试数学（理）试题含答案，共11页。试卷主要包含了考生必须保证答题卡整洁，直线，，则“”是“”的，已知向量，，则，已知，且，则的值为，函数的图像大致为，已知数列满足，则的前20项和等内容，欢迎下载使用。
南昌市2022届高三摸底测试卷理科数学本试卷共4页，23小题，满分150分0考试时间120分钟．注意事项：1．答卷前，考生务必将自己的姓名、准考证号填涂在答题卡上，并在相应位置贴好条形码．2．作答选择题时，选出每小题答案后，用2B铅笔把答题卡上对应题目的答案信息涂黑、如需改动，用橡皮擦干净后，再选涂其它答案．3．非选择题必须用黑色水笔作答，答案必须写在答题卡各题目指定区域内相应位置上；如需改动，先划掉原来答案，然后再写上新答案，不准使用铅笔和涂改液．不按以上要求作答无效．4．考生必须保证答题卡整洁．考试结束后，将试卷和答题卡一并交回．一．选择题：本题共12小题，每小题5分，共60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．集合的元素个数为（    ）A．3 B．4 C．5 D．62．若z为纯虚数，且，则（    ）A．i B．-i C．2i D．-2i3．为数列的前n项和，若，，则（    ）A． B． C．10 D．4．设F为抛物线焦点，直线，点A为C上任意一点，过点A作于P，则（    ）A．3 B．4 C．2． D．不能确定5．直线，，则“”是“”的（    ）A．充分不必要条件  B．必要不充分条件C．充要条件  D．既不充分也不必要条件6．已知向量，，则（    ）A． B． C． D．57．某市出台两套出租车计价方案，方案一：2公里以内收费8元（起步价），超过2公里的部分每公里收费3元，不足1公里按1公里计算；方案二：3公里以内收费12元（起步价），超过3公里不超过10公里的部分每公里收费2.5元，超过10公里的部分每公里收费3.5元，不足1公里按1公里计算．以下说法正确的是（    ）A．方案二比方案一更优惠B．乘客甲打车行驶4公里，他应该选择方案二C．乘客乙打车行驶12公里，他应该选择方案二D．乘客丙打车行驶16公里，他应该选择方案二8．已知，且，则的值为（    ）A． B． C． D．9．函数的图像大致为（    ）A．B．C．D．10．已知数列满足，则的前20项和（    ）A． B． C． D．11．已知双曲线的左、右焦点分别为、，过的直线l与C的左、右支分别相交于M、N两点，若，，则双曲线的离心率为（    ）A． B． C．2 D．12．已知函数，若，若点不可能在曲线C上，则曲线C的方程可以是（    ）A． B．C．  D．二．填空题：本题共4小题，每小题5分，共20分．13．某企业三个分厂生产同一种电子产品，三个分厂的产量分布如图所示．现在用分层抽样方法从三个分厂生产的产品中共抽取100件进行使用寿命的测试，测试结果为第一、二、三分厂取出的产品的平均使用寿命分别为1020小时、980小时、1030小时，估计这个企业生产的产品的平均使用寿命为________小时．14．若的展开式中共有7项，则常数项为________（用数字作答）．15．执行如下框图，若输出的，则输入x的取值范围为________．16．正四棱锥，底面四边形为边长为2的正方形，，其内切球为球G，平面过与棱，分别交于点M，N，且与平面所成二面角为30°，则平面截球G所得的图形的面积为________．三．解答题：共70分．解答应写出文字说明、证明过程或演算步骤．第17～21题为必考题，每个试题考生都必须作答；第22、23题为选考题，考生根据要求作答．（一）必考题：共60分．17．（12分）在中，角A，B，C所对的边分别为a，b，c，，．（Ⅰ）求的值；（Ⅱ）已知的面积为，求b边．18．（12分）如图在四棱锥中，底面为正方形，为等边三角形，E为中点，平面平面．（I）求证：平面；（II）求二面角的余弦值．19．（12分）己知椭圆，，分别为椭圆的左、右焦点，O为坐标原点，P为椭圆上任意一点．（Ⅰ）若，求的面积；（Ⅱ）斜率为1的直线与椭圆相交于A，B两点，，求直线的方程．20．（12分）已知函数．（Ⅰ）若函数在定义域上单调递增，求实数a的取值范围；（Ⅱ）若函数存在两个极值点，，求实数a的取值范围，并比较与的大小．21．（12分）甲、乙、丙、丁、戊五位同学参加一次节日活动，他们都有机会抽取奖券．墙上挂着两串奖券袋（如图），A，B，C，D，E五个袋子分别装有价值100，80，120，200，90（单位：元）的奖券，抽取方法是这样的：每个同学只能从其中一串的最下端取一个袋子，得到其中奖券，直到礼物取完为止．甲先取，然后乙、丙、丁、戊依次取，若两串都有礼物袋，则每个人等可能选择一串取．（Ⅰ）求丙取得的礼物券为80元的概率；（Ⅱ）记丁取得的礼物券为X元，求X的分布列及其数学期望．（二）选考题：共10分．请考生在第22、23题中任选一题作答，如果多做，则按所做的第一题计分．22．（10分）选修4-4：坐标系与参数方程在平面直角坐标系中，直线l的参数方程为（t为参数），以坐标原点O为极点，x轴正半轴为极轴，建立极坐标系，曲线C的极坐标方程为．（Ⅰ）求直线l的普通方程和曲线C的直角坐标方程；（Ⅱ）设，直线l与曲线C的交点为M，N，求．23．（10分）选修4-5：不等式选讲已知函数．（Ⅰ）当时，求不等式的解集；（Ⅱ）若对任意，都有恒成立，求实数a的取值范围． 2022届高三摸底测试卷理科数学参考答案及评分标准一、选择题：本大题共12个小题，每小题5分，共60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．题号123456789101112答案CCCAAACCBDBC二．填空题：本大题共4小题，每小题5分，满分20分．13．1015 14．240 15． 16．三．解答题：共70分．解答应写出文字说明、证明过程或演算步骤．第17题-21题为必考题，每个试题考生都必须作答．第22题、23题为选考题，考生根据要求作答．17．【解析】（Ⅰ）由正弦定理，（其中R为外接圆的半径），所以，，，代入已知条件可得：，····································································2分所以，即，·············································································4分，故．·················································································6分（Ⅱ）由已知可得，所以的面积为，故，解得，．···········································································9分所以，即．············································································12分18．【解析】（Ⅰ）连接交于点O，连接、，因为为等边三角形，所以，因为底面为正方形，所以，因为，所以平面，········································································3分又平面，所以，因为平面平面，平面平面，所以平面，因为E为中点，所以，则平面．······························································5分（Ⅱ）如图，以为x轴，为y轴，为z轴，建立空间直角坐标系，设，则，所以，，，，则，，，因为平面平面，所以平面的法向量为，····································································7分设平面的法向量为，则，所以，所以，···········································································9分所以,·················································································11分所以二面角的余弦值为．··································································12分19．【解析】（Ⅰ）由题意，解得，，····························································2分又，所以，即，··················································································4分所以；·················································································5分（Ⅱ）直线斜率为1，设直线方程，，，由，消元得，得··········································································7分又，知，即·············································································9分而所以，，得，满足，所以直线的方程或．·····································································12分20．【解析】（Ⅰ）由得······································································2分由题在恒成立，即在恒成立而，所以；·············································································5分（Ⅱ）由题意知，，是方程在内的两个不同实数解，令，注意到，其对称轴为直线，故只需，解得，即实数a的取值范围为；···································································8分由，是方程的两根，得，，因此，···················································································10分又，所以，即得证．··············································································12分21．【解析】（Ⅰ）由题意知，列举如下：····························································2分所以丙取得的礼物券为80元的概率；（Ⅱ）如下图，所以X的可能取值为100，80，200，90，又因为；；；；所以分布列为：··········································································8分（每个概率1分）X8090100200P所以．22．【解析】（Ⅰ）直线l的参数方程为（t为参数），转换为普通方程为········································································2分曲线C的极坐标方程为，根据转换为直角坐标方程为．·······························································5分（Ⅱ）易知直线l的参数方程标准形式为代入到中，得到；设M，N所对应的参数分别为，，则，，·················································································8分所以．················································································10分23．【解析】（Ⅰ）因为，所以，当时，，所以；当时，，所以，综上不等式的解集为．····································································5分（II）因为，···········································································8分当时，在单调递增；当时，；所以函数的最小值是a，所以．·····························································10分