2022邢台高三上学期入学考试数学试题含答案

这是一份2022邢台高三上学期入学考试数学试题含答案，共16页。试卷主要包含了本试卷主要考试内容，在的展开式中，项的系数是，已知函数，若，则，已知，，且，则的最小值是，已知向量，，则下列结论正确的是等内容，欢迎下载使用。
河北省邢台市2022届高三上学期入学考试数学注意事项：1．答题前，考生务必将自己的姓名、考生号、考场号、座位号填写在答题卡上。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，再选涂其他答案标号。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。4．本试卷主要考试内容：高考全部内容。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．复数，则复数z的实部与虚部之和是（    ）A． B． C．10 D．182．已知集合，，若，则a的取值范围是（    ）A． B． C． D．3．如果双曲线的离心率为，我们称该双曲线为黄金分割双曲线，简称为黄金双曲线．现有一黄金双曲线，则该黄金双曲线C的虚轴长为（    ）A．2 B．4 C． D．4．六氟化硫，化学式为，在常压下是十种无色、无臭、无毒、不燃的稳定气体，有良好的绝缘性，在电器工业方面具有广泛用途．六氟化硫分子结构为正八面体结构（正八面体是每个面都是正三角形的八面体），如图所示，硫原子位于正八面体的中心，6个氟原子分别位于正八面体的6个顶点．若相邻两个氟原子间的距离为2a，则六氟化硫分子中6个氟原子构成的正八面体的体积是（不计氟原子的大小）A． B． C． D．5．在的展开式中，项的系数是（    ）A．280 B．-280 C．560 D．-5606．已知函数，若，则（    ）A．-7 B．-3 C．3 D．77．已知，，且，则的最小值是（    ）A．1 B．2 C． D．8．已知函数恰有两个零点，则a的取值范围是（    ）A． B． C． D．二、选择题：本题共44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．已知向量，，则下列结论正确的是（    ）A．若，则 B．若，则C．若，则 D．若，则10．旅游是人们为寻求精神上的愉快感受而进行的非定居性旅行和游览过程中所发生的一切关系和现象的总和．随着经济生活水平的不断提高，旅游已经成为人们生活的一部分．某地旅游部门从2020年到该地旅游的游客中随机抽取部分游客进行调查，得到各年龄段游客的人数和旅游方式如图所示，则下列结论不正确的有（    ）A．估计2020年到该地旅游的游客选择自助游的中年人的人数少于选择自助游的青年人人数的一半B．估计2020年到该地旅游的游客选择自助游的青年人的人数占总游客人数的13.5%C．估计2020年到该地旅游的游客选择自助游的老年人和中年人的人数之和比选择自助游的青年人多D．估计2020年到该地旅游的游客选择自助游的比率为25% 11．已知函数，若函数的三个相邻零点分别为，，，且，则的值可能是（    ）A． B． C．4 D．612．已知三棱柱的6个顶点全部在球O的表面上，，，三棱柱的侧面积为，则球O的表面积可能是（    ）A．4π B．8π C．16π D．32π三、填空题：本题共4小题，每小题5分，共20分．把答案填在答题卡中的横线上．13．已知函数则________．14．小华、小明、小李小章去A，B，C三个工厂参加社会实践，要求每个工厂都有人去，且这四人都在这三个工厂实践，则小华和小李都没去B工厂的概率是________．15．已知数列的前n项和为，且，则________．16．已知抛物线的焦点为F，直线与抛物线C交于A，B两点（其中点A在x轴上方），则________．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、证明过程或演算步聚．17.（10分）在①②的前n项和，③这三个条件中任选一个，补充在下面的问题中并解答．问题：在等差数列中，，且________．（1）求数列的通项公式；（2）若，求数列的前n项和．注：如果选择多个条件分别解答，按第一个解答计分．18.（12分）在中，角A，B，C所对的边分别为a，b，c，且．（1）求角A的大小；（2）若的面积为，求外接圆面积的最小值．19．（12分）如图，在多面体中四边形是正方形，平面，平面，．（1）证明：平面平面．（2）求平面与平面所成锐二面角的余弦值．20．（12分）北京冬季奥运会将于2022年2月4日至2022年2月20日在中华人民共和国北京市和河北省张家口市联合举行．这是中国历史上第一次举办冬季奥运会，北京、张家口同为主办城市，也是中国继北京奥运会、南京青奥会之后第三次举办奥运赛事．北京冬奥组委对报名参加北京冬奥会志愿者的人员开展冬奥会志愿者的培训活动，并在培训结束后进行了一次考核．为了解本次培训活动的效果，从中随机抽取80名志愿者的考核成绩，根据这80名志愿者的考核成绩得到的统计图表如下所示．    女志愿者考核成绩频率分布表分组频数频率[75,80]20.050[80,85]130.325[85,90]180.450[90,95]am[95,100]b0.075男志愿者考核成绩频率分布直方图若参加这次考核的志愿者考核成绩在[90,100]内，则考核等级为优秀．（1）分别求这次培训考核等级为优秀的男、女志愿者人数；（2）若从样本中考核等级为优秀的志愿者中随机抽取3人进行学习心得分享，记抽到女志愿者的人数为X，求X的分布列及期望．21．（12分）已知椭圆的左、右焦点分别为，，且，点在椭圆C上．（1）求椭圆C的标准方程；（2）已知点为椭圆C上一点，过点的直线l与椭圆C交于异于点P的A，B两点，若的面积是，求直线l的方程．22．（12分）已知函数．（1）求函数图象在处的切线方程．（2）证明：．高三入学数学考试参考答案1．B【解析】本题考查复数，考查运算求解能力．由题意可得，则复数z的实部是7，虚部是-11，故复数z的实部与虚部之和是．2．A【解析】本题考查集合的运算，考查逻辑推理能力．由题意可得，．因为，所以，即．3．D【解析】本题考查双曲线的性质，考查运算求解能力．由题意可得，解得，则，故该黄金双曲线C的虚轴长为．4．B【解析】本题考查化学分子结构与正八面体的体积，考查空间想象能力与阅读理解能力．如图，连接，，，连接．因为，，所以，，所以平面．因为，所以．因为四边形是正方形，所以，则，故该正八面体的体积为．5．C【解析】本题考查二项式定理，考查逻辑推理能力与运算求解能力．展开式中，通项．令，得，则，故项的系数是560．6．B【解析】本题考查函数的性质，考查运算求解能力．设，则，即，故．7．C【解析】本题考查基本不等式，考查运算求解能力．由题意可得，当且仅当，时，等号成立．8．D【解析】本题考查导数与函数的零点问题，考查逻辑推理能力．令，得．设，则．由，得；由，得．所以在上单调递减，在上单调递增，故，即．9．AC【解析】本题考查平面向量，考查运算求解能力．由，得，则A正确，B错误；由，得，即，则C正确，D错误．10．ACD【解析】本题考查统计图表，考查数据处理能力．设2020年到该地旅游的游客总人数为a，由题意可知游客中老年人、中年人、青年人的人数分别为0.2a，0.35a，0.45a，其中选择自助游的老年人、中年人、青年人的人数分别为0.04a，0.0875a，0.135a．因为，所以A错误；2020年到该地旅游的游客选择自助游的青年人的人数与总游客人数的比值为，则B正确；因为，所以C错误；2020年到该地旅游的游客选择自助游的比率为，则D错误．11．AD【解析】本题考查三角函数的图象与性质，考查数形结合的数学思想．由，得，则．当时，，，则，，从而，，故；当时，，则，，从而，，故．综上，或．12．CD【解析】本题考查简单几何体及其外接球，考查空间想象能力．设三棱柱的高为h，．因为，所以，则该三棱柱的侧面积为，故．设的外接圆半径为r，则．设球O的半径为R，则，故球O的表面积为．13．【解析】本题考查分段函数求值，考查运算求解能力．由题意可得，则．14．【解析】本题考查古典概型，考查运算求解能力．由题意可知总的分配情况有种，其中满足条件的情况有种，故所求概率．15．1022【解析】本题考查等比数列，考查运算求解能力．因为，所以，所以，即．因为，所以，则是首项为2，公比为2的等比数列，故．16．【解析】本题考查抛物线的性质，考查数形结合的数学思想与运算求解能力．由题意可知直线l经过焦点F，设其倾斜角为，则．如图，直线是抛物线C的准线，作，，，则，，故，．因为，所以，则．17．解：选①（1）设的公差为d．由题意可得．···········································································2分因为，所以，···········································································4分则．··················································································5分选②（1）设的公差为d．因为，所以，则．··················································································2分因为，所以，所以．······································································4分当时，满足上式，故．····································································5分选③（1）设的公差为d．因为，所以，即．········································································2分因为，所以，···········································································4分则．··················································································5分（2）由（1）可得，则．···································································7分故···················································································10分评分细则：（1）在第一问中，若选择条件②解答，没有考虑，扣1分；（2）在第二问中，也可以由，得到；（3）若用其他解法，参照评分标准按步骤给分．18．解：（1）因为，所以，·················································································1分所以，即．·············································································3分因为，所以，所以．······································································4分因为，所以．···········································································5分（2）由（1）可知，则．···································································6分因为的面积为，所以，所以．·······························································7分由余弦定理可得，则．····································································8分设外接圆的半径为r，则，即，······························································10分故外接圆的面积，当且仅当时，等号成立．····················································11分即当时，外接圆面积的最小值为．···························································12分评分细则：（1）在第一问中，也可以通过把角化为边，得到，再由余弦定理得到，从而求出角A；（2）在第二问中，没有写出取等条件，只要计算正确，不予扣分；（3）若用其他解法，参照评分标准按步骤给分．19．（1）证明：因为平面，平面，所以．·················································································1分因为平面，平面，所以平面．·············································································2分因为四边形是正方形，所以．·······························································3分因为平面，平面，所以平面．·············································································4分因为平面，平面，且，所以平面平面．5分（2）解：由题意可知，，两两垂直，则以D为原点，分别以，，的方向为x，y，z轴的正方向，建立如图所示的空间直角坐标系．设，则，，，，从而，．···············································································7分设平面的法向量为，则，令，得．···········································································9分平面的一个法向量为．···································································10分故，即平面与平面所成锐二面角的余弦值为．······················································12分评分细则：（1）在第一问中，也可以建立空间直角坐标系，分别求出平面和平面的法向量，通过证明平面和平面的法向量平行，从而得到平面平面；（2）在第二问中，也可以先找出平面和平面所成的锐二面角，再通过余弦定理求出；（3）若用其他解法，参照评分标准按步骤给分．20．解：（1）由女志愿者考核成绩的频率分布表可知被抽取的女志愿者的人数为．···································1分因为，所以·············································································2分所以这次培训考核等级为优秀的女志愿者人数为．···············································3分因为被抽取的志愿者人数是80，所以被抽取的男志愿者人数是．·····································4分由男志愿者考核成绩频率分布直方图可知男志愿者这次培训考核等级为优秀的频率为，····················5分则这次培训考核等级为优秀的男志愿者人数为．·················································6分（2）由题意可知X的可能取值为0，1，2，3．···················································7分，，··················································································8分，····················································································9分X的分布列为X0123P故．·················································································12分评分细则：（1）在第一问中，也可以先求出被抽取的女志愿者的人数，再求出的值，即考核等级为优秀的女志愿者的人数；（2）在第二问中，没有分别求出X对应取值的概率，直接写出X的分布列，扣1分；（3）若用其他解法，参照评分标准按步骤给分．21．解：（1）设椭圆的半焦距为c，由题意可得，···········································································1分解得，．···············································································3分故椭圆C的标准方程为．···································································4分（2）因为在椭圆C上，所以，解得．①当直线l的斜率为0时，，则的面积为．因为的面积是，所以直线l的斜率为0不符合题意．················································6分②当直线l的斜率不为0或斜率不存在时，设直线l的方程为，，．联立整理得．则，．·················································································7分故．··················································································8分因为点P到直线l的距离，所以．·················································································9分因为的面积是，所以，整理得，解得，即．·····································································11分故直线l的方程为，即．···································································12分评分细则：（1）在第一问中，可以先根据，求出c的值，从而求出的值，再把点M的坐标代入椭圆C的方程，从而求出a，b的值，最后得到椭圆C的标准方程；（2）在第二问中，没有考虑直线l的斜率为0的情况，扣1分；（3）在第二问中，也可以按直线l的斜率存在和不存在分类讨论计算；（4）若用其他解法，参照评分标准按步骤给分．22．（1）解：因为，所以，·················································································1分则．··················································································2分因为，·················································································3分所以所求切线方程为，即．··················································································4分（2）证明：设，则．·····································································5分由，得；由，得．所以在上单调递减，在上单调递增，··························································6分故，即，当且仅当时取等号．·······························································7分因为，所以，所以，所以．···········································································9分当时，，··············································································10分所以，················································································11分则，即．··············································································12分评分细则：（1）在第一问中，所求切线方程写成，不予扣分；（2）在第二问中，也可以将转化为，然后构造函数和，得到，从而得到；（3）若用其他解法，参照评分标准按步骤给分．