江西省赣州市2023届高三数学（理）上学期1月期末考试试题（Word版附解析）

这是一份江西省赣州市2023届高三数学（理）上学期1月期末考试试题（Word版附解析），共10页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。
赣州市2022～2023学年度第一学期期末考试 高三数学（理科）试卷 2023年1月本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共150分，考试时间120分钟第Ⅰ卷一、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，，则（）A. B. C. D.2.函数则（）A. B.0 C. D.23.若数列是等比数列，且，，则（）A. B. C.62 D.644.为了研究某班学生的右手一拃长x（单位：厘米）和身高y（单位：厘米）的关系，从该班随机抽取了12名学生，根据测量数据的散点图可以看出y与x之间有线性相关关系，设其回归直线方程为，已知，，，若某学生的右手一拃长为22厘米，据此估计其身高为（）A.175 B.179 C.183 D.1875.若复数（a，，为其共轭复数），定义：.则对任意的复数，有下列命题：：；：；：；：若，则为纯虚数.其中正确的命题个数为（）A.1 B.2 C.3 D.46.程大位（1533~1606），明朝人，珠算发明家.在其杰作《直指算法统宗》里，有这样一道题：荡秋千，平地秋千未起，踏板一尺离地，送行二步与人齐，五尺人高曾记。仕女佳人争蹴，终朝笑语欢嬉，良工高士素好奇，算出索长有几？将其译成现代汉语，其大意是，一架秋千当它静止不动时，踏板离地一尺，将它向前推两步（古人将一步算作五尺）即10尺，秋千的踏板就和人一样高，此人身高5尺，如果这时秋千的绳索拉得很直，请问绳索有多长？（）A.14尺 B.14.5 尺C.15尺 D.15.5尺7.已知过抛物线C：的焦点F的直线l被C截得的弦长为8，则坐标原点O到l的距离为（）A. B. C. D.8.若展开式的各项系数和为729，展开式中的系数为（）A. B. C.30 D.909.直线与双曲线E：（，）交于M，N两点，若为直角三角形（其中O为坐标原点），则双曲线E的离心率为（）A. B. C. D.10.已知函数，的最小值为a，则实数a的值为（）A. B. C. D.111.在三棱锥中，，，且，，，，则三棱锥的外接球的表面积为（）A. B. C. D.12.已知，，，则（）A. B. C. D.第Ⅱ卷本卷包括必考题和选考题两部分，第13～21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.二、填空题：本题共4小题，每小题5分，共20分.13.已知，均为单位向量且夹角为45°，则________.14.已知tan，则__________.15.如图，、、是同一平面内的三条平行直线，与间的距离是1，与间的距离是2，等腰直角三角形的三顶点分别在、l2、上，则的斜边长可以是__________（写出一个即可）.16.斐波那契，意大利数学家，其中斐波那契数列是其代表作之一，即数列满足，且，则称数列为斐波那契数列.已知数列为斐波那契数列，数列满足，若数列的前12项和为86，则__________.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.17.（本小题满分12分）在中，角A，B，C的对边分别为a，b，c，满足.（1）求B的值；（2）若与边上的高之比为3∶5，且，求的面积.18.（本小题满分12分）如图，在四棱锥中，底面，且，，.（1）证明：平面平面；（2）求平面与平面夹角的余弦值.19.（本小题满分12分）设有标号为1，2，3，…，n的n个小球（除标号不同外，其余均一样）和标号为1，2，3，…，n的n个盒子，将这n个小球任意地放入这n个盒子，每个盒子放一个小球，若i（，2，3，…，n）号球放入了i号盒子，则称该球放对了，否则称放错了.用表示放对了的球的个数.（1）当时，求的概率；（2）当时，求的分布列与数学期望.20.（本小题满分12分）已知椭圆C：（）过点且离心率为，过点作两条斜率之和为0的直线，，交C于A，B两点，交C于M，N两点.（1）求椭圆C的方程；（2）是否存在实数使得？若存在，请求出的值，若不存在，说明理由.21.（本小题满分12分）已知函数（其中e为自然对数的底数），且曲线在处的切线方程为.（1）求实数m，n的值；（2）证明：对任意的，有.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.[选修4-4：坐标系与参数方程]在平面直角坐标系中，已知直线l的参数方程为（t为参数），以坐标原点O为极点，x轴的正半轴为极轴，取相同的单位长度建立极坐标系，曲线C的极坐标方程为.（1）求直线l的普通方程和曲线C的直角坐标方程；（2）设点，直线l与曲线C的交点为A，B，求的值.23.[选修4-5：不等式选讲]已知函数的最小值为m.（1）求m的值；（2）设a，b，c为正数，且，求证：. 
赣州市2022～2023学年度第一学期期末考试高三理科数学参考答案一、选择题题号123456789101112答案ADCCBBCDBDCA12.解：由，得，由得，即，当时，令，，则，则在上单调递增，则，即，即，则.或由对数平均不等式得，即，即.二、填空题13.1；14.；15.或或（三个任选一个）；16.8.16.解析：斐波那契数列：1，1，2，3，5，8，13，21，34，55，89，144，…….（特征：每三项中前两项为奇数后一项为偶数）由得：，，，则，同理：，，，，，得：，，，，则，，则，则.三、解答题17.解：（1）由，由内角和定理得：····································································1分进而得：···········································································2分解得：，（舍去）····································································4分从而得.·············································································6分（2）由题设知，不妨设，······························································7分由余弦定理得：······································································8分联立得：···········································································9分即，，············································································10分故，从而的面积·····································································12分18.解：（1）连接，由题设，，得.························································1分又，故，由余弦定理可得·······························································2分从而有·············································································3分又面且面，得········································································4分，故面·············································································5分又面，所以面面······································································6分（2）法一：由（1）同理可得面，························································6分以为坐标原点，,分别为x，y轴的正方向建立如图所示的空间直角坐标系，则，，，，，········································································7分由（1）知面，故面的一个法向量为·······················································8分设面的法向量为，由，，结合，令，得········································································9分故的一个法向量为···································································10分记平面与平面夹角为，则有··············································································12分（注：取的中点，连接，易证面，则为平面的一个法向量）法二：以为坐标原点，，分别为，轴的正方向建系，可得面的一个法向量为，面的一个法向量为19.证：（1）由题设知：································································4分（2）的所有可能取值为0，1，2，3，5·····················································5分··················································································6分··················································································7分··················································································8分··················································································9分·················································································10分故的分布列为01235P故的数学期望为·····································································12分20.解：（1）解得·····································································3分∴椭圆C的方程为·····································································4分（2）∵直线与的斜率之和为0，令直线的斜率为，则直线的斜率为，令，，则的方程为，··················································································6分则，···············································································7分则直线的斜率为，令，，同理：，···········································································8分，，·················································································10分同理：·················································································11分，所以存在实数·····································································12分方法二（2）∵直线与的斜率之和为0，令直线的倾斜角为，则直线的斜率为倾斜角为··················5分令直线的参数方程为（t为参数）·························································6分代入得：即·················································································7分令点，对应的参数分别为，，则，则·················································································9分同理：令直线的参数方程为，（为参数）代入得：即令点，对应的参数分别为，，则·························································10分则················································································11分则，所以存在实数···································································12分21.解：（1）由，得···································································1分由题意知：··········································································3分··················································································4分（2）记············································································5分则记记·················································································6分显然在上单调递增，且当时，；当时，；所以在上递减，在上递增，故·················································································7分（注：这里的符号也可以通过来说明.）又，故存在唯一，，使得···································································8分故当时，；当时，；即在与上单调递增，在上单调递减························································9分且，，故存在唯一，使································································10分故当时，；当时，.所以在与单调递减，在与上递增·························································11分故是与中的较小值，又且，故恒成立.证毕·······································································12分22.解：（1）直线l的普通方程为.·························································2分由得，则曲线C的直角坐标方程为································································5分（2）直线l的普通方程为（t为参数）······················································6分设点，对应的参数分别为，，将代入得···········································································7分则，，则，··········································································8分则················································································10分23.解：（1）由题意得·································································3分从而函数在递减，在上递增·····························································4分故，即.·············································································5分（2）由（1）知：，又a，b，c为正数，由，，·············································································6分法一：·············································································7分而·················································································8分所以··············································································10分法二：由，，，，，···············································································8分得················································································10