湖北省恩施市清江外国语学校2021届第一学期高三数学周考试卷（10月21日）

清江外国语学校高三年级2018级高三数学周考试卷（10月21日）1．本试卷共4页，22小题，满分150分，考试用时120分钟．2．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上．3．考试结束后，考生上交答题卡．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．复数的共轭复数是A． B． C． D．2．已知集合，，则A． B． C． D．3．已知抛物线的准线为，圆与相切．则A． B． C． D．4．某学校组织学生参加数学测试，某班成绩的频率分布直方图如图，数据的分组依次为，，，．若不低于分的人数是人，则该班的学生人数是A． B． C． D．5．中国古代数学名著《周髀算经》记载的“日月历法”曰：“阴阳之数，日月之法，十九岁为一章，四章为一部，部七十六岁，二十部为一遂，遂千百五二十岁，…．生数皆终，万物复苏，天以更元作纪历”．某老年公寓住有位老人，他们的年龄（都为正整数）之和恰好为一遂，其中最年长者的年龄在之间，其余人的年龄依次相差一岁，则最年长者的年龄为A． B． C． D．6．已知，，则A． B． C． D．7．已知直三棱柱的个顶点都在球的球面上．若，，，，则球的表面积为A． B． C． D．8．对于定义在上的函数，且为偶函数．当时，，设，，，则，，的大小关系为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，有选错的得0分，部分选对的得3分)9．设，，为正实数，且，则A．  B．C．  D．10．将函数的图象向左平移个单位得到函数的图象，则a可以取的值为A.  B.  C.  D. 11．对，，若函数同时满足：（1）当时，有；（2）当时，有，则称为函数．下列是函数的有A．  B． C．  D．12．在长方体中，，是平面内不同的两点，，是平面内不同的两点，且，，，，，分别是线段，的中点，则下列结论正确的是A．若，则           B．若，重合，则C．若与相交，且，则可以与相交D．若与是异面直线，则不可能与平行三、填空题：(本题共4小题，每小题5分，共20分)13．函数的图象在点处的切线方程为________．14．已知，且，则________．15．已知向量，（，），若，则的最小值为________．16．已知，是双曲线的左，右焦点，以为直径的圆与的左支交于点，与的右支交于点，，则的离心率为________． 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．（10分）在①，②，③这三个条件中任选一个，补充在下面问题中，若问题中的三角形存在，求的值；若问题中的三角形不存在，说明理由．问题：是否存在，它的内角，，的对边分别为，，，且，，________？注：如果选择多个条件分别解答，按第一个解答计分． 18．（12分）设是公比大于的等比数列，，且是，的等差中项．（1）求数列的通项公式；（2）若，数列的前项和． 19．（12分）如图，在圆柱中，为圆的直径，，是弧上的两个三等分点，是圆柱的母线．（1）求证：平面；（2）设，，求二面角的余弦值． 20．（12分）习近平总书记一直十分重视生态环境保护，十八大以来多次对生态文明建设作出重要指示，在不同场合反复强调，“绿水青山就是金山银山”，随着中国经济的快速发展，环保问题已经成为一个不容忽视的问题，而与每个居民的日常生活密切相关的就是水资源问题．某污水处理厂在国家环保部门的支持下，引进新设备，污水处理能力大大提高．已知该厂每月的污水处理量最少为150万吨，最多为300万吨，月处理成本万元与月处理量万吨之间的函数关系可近似地表示为，且每处理一万吨污水产生的收益价值为万元．
该厂每月污水处理量为多少万吨时，才能使每万吨的处理成本最低；
该厂每月能否获利？如果获利，求出最大利润． 21．（12分）已知椭圆的两个焦点分别是，，并且经过点．（1）求椭圆的标准方程；（2）已知点，若上总存在两个点、关于直线对称，且，求实数的取值范围． 22．（12分）已知函数，．（1）讨论的单调性；（2）设，若函数有两个不同的零点，，求的取值范围． 一、选择题：本题共12小题，每小题5分，共60分．1～8小题为单项选择题，在每小题给出的四个选项中，只有一项是符合题目要求的；9～12小题为多项选择题，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3分．题号123456789101112答案ACBBBDCCADBDBCBD8．解，因为函数为偶函数，所以，即函数的图象关于直线对称，即．又因为当时，，所以函数在上单调递减，因而在上单调递增，因为，所以，即，即．故选C．12．解：若，则、、、四点共面，当时，平面、、两两相交有三条交线，分别为、、，则三条交线交于一点，则与平面交于点，则与不平行．故A错误：若，两点重合，则，、、﹑四点共面，平面、、两两相交有三条交线，分别为、、，由，得，故B正确；若与相交，确定平面，平面、、两两相交有三条交线，分别为、、，因为，所以，所以与不可能相交．故C错误；当与是异面直线时，如图，连接，取中点，连接，．则，因为平面，平面，则平面．假设，因为平面，平面，所以平面．又，∴平面平面，同理可得，平面平面，则平面平面，与平面平面矛盾．所以假设错误，不可能与平行，故D正确，故选BD．二、填空题：本题共4小题，每小题5分，共20分．13． 14．      15． 16．16．解：由题意知，，所以，即，易得．设，，．由双曲线的定义得：，解得：，所以，因为，所以离心率 ．三、解答题；本大题共6小题，满分70分．解答应写出文字说明，证明过程或演算步骤．17．（10分）解：已知，由正弦定理，得，···········································································2分因为为三角形内角，，·····························································3分所以，即········································································4分所以，·········································································· 5分因为，所以······································································6分选择条件①的解析：解法一：由及正弦定理，可得，······················································7分由余弦定理，则，············································································ 9分解得．·········································································10分解法二：由，又因为，所以，························································7分则，展开得，，···································································8分所以，所以，··········································································9分所以．·········································································10分选择条件②的解析：解法一：由，可得，·······························································7分由余弦定理得，··································································· 8分，············································································· 9分解得．·········································································· 0分解法二：由得．···································································7分因为，所以，是以为顶角的等边三角形．所以，所以．····································································· 8分由正弦定理得，，·································································9分解得．·········································································10分选择条件③的解析：解法一：由，又因为，则，··························································8分与矛盾，则问题中的三角形不存在．··················································10分解法二：由，则，则，···········································································8分与三角形内角和等于矛盾，因而三角形不存在．·········································10分18．（12分）解：（1）设等比数列的公比为．依题意，有·············································································1分将代入得，得．···········································································2分联立得两式两边相除消去得，解得或（舍去），·································································3分所以．·········································································· 4分所以．··········································································5分（2）解法一：因为································································6分所以，……①····································································7分……②··········································································8分①②，得········································································9分．············································································11分所以，数列的前项和．····························································12分解法二：因为所以···········································································8分进而得··············································································11分所以数列的前项和为······························································12分19．（12分）解：（1）连接，，································································ 1分因为，是半圆上的两个三等分点，所以，又，所以，，均为等边三角形．所以，··········································································2分所以四边形是平行四边形．·························································· 3分所以，·········································································· 4分因为平面，平面，所以平面．······································································5分（2）因为是圆柱的母线，所以平面，平面，所以·····························································6分因为为圆的直径，所以，在中，，，所以，所以在中，······································································7分（方法一）因为，，，所以平面，又平面，所以．在内，作于点，连接．因为，，平面，所以平面，······································································8分又平面，所以，所以就是二面角的平面角．··························································9分在中，，．············································································10分在中，，所以，·········································································11分所以．所以，二面角的余弦值为．·························································12分（方法二）以为坐标原点，分别以，，所在直线为，，轴，建立如图所示的空间直角坐标系， 则，，，所以，··········································································8分设平面的法向量为，则即令，则,所以平面的一个法向量为．··························································9分因为平面的一个法向量，···························································· 10分所以．·········································································11分所以结合图形得，二面角的余弦值为．················································12分【答案】解：由题意可知，每万吨污水的处理成本为：
，
当且仅当时等号成立；
所以该厂每月污水处理量为200万吨时，
才能使每万吨的处理成本最低，最低成本为万元．
设该厂每月获利为Z万元，则
，
因为，所以，
当时，Z有最大值，
所以该污水处理厂每月能获利；
且当月处理量为250万吨时，利润最大，为万元．【解析】由题意知每万吨污水的处理成本为，用基本不等式求出它的最小值以及对应的x值；
设该厂每月获利为Z万元，列出函数解析式，利用函数关系求出的取值范围和最大值即可．
本题考查了二次函数模型的实际应用问题，也考查了利用基本不等式求最值问题，是中档题．
21．（12分）解：（1）因为椭圆的焦点在轴上，所以设它的标准方程为．····························································1分由椭圆的定义得，所以．·········································································· 2分因为，所以．····································································· 3分因此，椭圆的标准方程为．·························································· 4分（2）根据题意可设直线的方程为，联立，整理得，········································································ 5分由，得．········································································6分设，，则，．··········································································7分又设的点为，则，．由于点在直线上，所以，得，······································································8分代入，得，所以……①．····························································9分因为，，所以．············································································· 10分由，得，得，得，所以……②······························································11分由①②得，故实数的取值范围为．····················································12分22．（12分）解：（1）函数的定义域为．·························································1分则．···········································································2分（i）当，即时，令得，，得，又因为，所以，所以函数在上单调递增，在上单调递减．················································ 3分（ⅱ）当，即时，，又由得对任意的恒成立．所以函数在上单调递增．····························································4分综上，当时，函数在单调递增，在上单调递减；当时，函数在上单调递增．·····················5分（2）解法一：，函数的定义域为，．（i）当时，，函数在上是增函数，不可能有两个零点；····································6分（ii）当时，在上，．在上，．······················································7分所以函数在单调速增，在上单调递减．此时为函数的最大值．·····························································8分若，则最多有一个零点，不合题意．所以，解得．··········································································9分此时，且，··············································································10分令，则所以在上单调递增．所以，即．·····································································11分故函数有两个不同的零点，，．综上，的取值范围是．····························································12分解法二：因为···········································································6分所以“函数有两个零点”等价于“直线与函数的图象有两个交点”····························7分则；···········································································8分得函数在上单调递增，在上单调递减，所以函数的最大值为，································9分又因为函数在其定义域上连续不断，（这个理由可以不写）且易知当时，，当时，﹐当时，·········································································10分所以当函数有两个零点时，只需满足，················································· 11分即的取值范围为．································································· 12分注：求最大值后也可以画图象说明：