2020青岛胶州高二下学期期末考试数学试题含答案

这是一份2020青岛胶州高二下学期期末考试数学试题含答案，共11页。试卷主要包含了作答选择题时，若，使得成立，则实数的最大值为，已知，则的值为，设全集，集合，集合，则等内容，欢迎下载使用。
2019-2020学年度第二学期期末学业水平检测高二数学本试卷4页，22小题，满分150分．考试用时120分钟． 注意事项：1．答题前，考生务必将自己的姓名、考生号、考场号和座号填写在试题卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置；2．作答选择题时：选出每小题答案后，用2B铅笔在答题卡上对应题目选项的答案信息点涂黑；如需要改动，用橡皮擦干净后，再选涂其它答案，答案不能答在试卷上；非选择题必须用黑色字迹的专用签字笔作答，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，先划掉原来的答案，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；3．考生必须保证答题卡的整洁，考试结束后，请将答题卡上交． 一、单项选择题：本大题共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．“”是“”成立的（    ） A．充分非必要条件   B．必要非充分条件   C．充要条件  D．既非充分也非必要条件2．函数的零点所在区间为（    ） A．  B．            C．  D． 3．已知数列的前项和为，，则（    ） A．  B．  C．  D． 4．若，使得成立，则实数的最大值为（    ） A． B． C． D．5．已知，则的值为（    ） A． B．  C． D．6．已知函数的部分图象如图所示，则的解析式可能为（    ）A．   B．  C．  D．  7．为了普及环保知识，增强环保意识，某中学随机抽取30名学生参加环保知识竞赛，得分（分制）的频数分布表如下：得分频数设得分的中位数，众数，平均数，下列关系正确的是（    ） A． B． C． D．8．已知函数的定义域为，且是偶函数，是奇函数，在上单调递增，则（    ）A． B．C． D． 二、多项选择题：本大题共4小题，每小题5分，共20分。在每小题给出的四个选项中，有多项符合题目要求。全部选对的得5分，部分选对的得3分，有选错的得0分。 9．设全集，集合，集合，则（    ） A．  B．  C． D． 10．已知复数满足为虚数单位，复数的共轭复数为，则（    ）A．  B．C．复数的实部为 D．复数对应复平面上的点在第二象限11．若函数，则下述正确的是（    ） A．在单调递增 B．的值域为 C．的图象关于直线对称  D．的图象关于点对称12．若，则（    ） A．  B． C．      D． 三、填空题：本大题共4个小题，每小题5分，共20分。 13．已知直线与函数的图象相切，则            ．14．已知数列的前项和为，，，，则            ．15．若是函数的极值点，则的极小值为            ．16．一袋中装有个大小相同的黑球和白球．已知从袋中任意摸出个球，至少得到个白球的概率是，则袋中白球的个数为            ；从袋中任意摸出个球，则摸到白球的个数的数学期望为            ．（本题第一个空分，第二个空分） 四、解答题：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①，②，③三个条件中任选两个，补充到下面问题中，并解答．已知等差数列的前项和为，满足：                    ，．（1）求的最小值；（2）设数列的前项和，证明：． 18．（12分）某足球运动员进行射门训练，若打进球门算成功，否则算失败．已知某天该球员射门成功次数与射门距离的统计数据如下： 射门距离不超过米射门距离超过米总计射门成功射门失败总计       （1）请问是否有的把握认为该球员射门成功与射门距离是否超过米有关？参考公式及数据：． （2）当该球员距离球门米射门时，设射门角（射门点与球场底线中点的连线和底线所成的锐角或直角）为，其射门成功率为，求该球员射门成功率最高时射门角的值．  19．（12分）已知数列的前项和为，，．（1）证明：数列为等比数列；（2）若数列满足：，，证明：． 20．（12分）已知函数，，为自然对数的底数．（1）若，求的零点；（2）讨论的单调性；（3）当时，，求实数的取值范围． 21．（12分）探索浩瀚宇宙是全人类的共同梦想，我国广大科技工作者、航天工作者为推动世界航天事业发展付出了艰辛的努力，为人类和平利用太空、推动构建人类命运共同体贡献了中国智慧、中国方案、中国力量．（1）某公司试生产一种航空零件，在生产过程中，当每小时次品数超过件时，产品的次品率会大幅度增加，为检测公司的试生产能力，同时尽可能控制不合格品总量，抽取几组一小时生产的产品数据进行次品情况检查分析，已知在（单位：百件）件产品中，得到次品数量（单位：件）的情况汇总如下表所示，且（单位：件）与（单位：百件）线性相关：（百件）（件）根据公司规定，在一小时内不允许次品数超过件，请通过计算分析，按照公司的现有生产技术设备情况，判断可否安排一小时试生产件的任务？（2）“战神”太空空间站工作人员需走出太空站外完成某项试验任务，每次只派一个人出去，且每个人只派出一次，工作时间不超过分钟，如果有人分钟内不能完成任务则撤回，再派下一个人．现在一共有个人可派，工作人员各自在分钟内能完成任务的概率分别依次为，且，，各人能否完成任务相互独立，派出工作人员顺序随机，记派出工作人员的人数为，的数学期望为，证明：．（参考公式：用最小二乘法求线性回归方程的系数公式;．）（参考数据：，．） 22．（12分）已知函数，，为自然对数的底数．（1）若，证明：；（2）讨论的极值点个数．
2019-2020学年度第二学期期末学业水平检测高二数学参考答案一、单项选择题：本大题共8小题，每小题5分，共40分。  1-8:  A B A C  D B D B  二、多项选择题：本大题共4小题，每小题5分，共20分。9：AB；    10：BD ；    11：AD；    12：ACD三、填空题：本大题共4个小题，每小题5分，共20分。13. ；           14. ；           15.  ；         16.  （1）；（2）；四、解答题：共70分。解答应写出文字说明，证明过程或演算步骤。17.（10分）解：（1）若选择②③；由题知：····························································1分又因为所以··························································2分所以·······························································3分所以·······························································.4分所以，·····························································5分所以·······························································6分若选择①②；由题知：····························································1分又因为所以··························································2分所以·······························································3分所以·······························································.4分所以，·····························································5分所以·······························································6分若选择①③；由题知：，所以······················································1分由题知：，所以······················································2分所以，·····························································4分所以，·····························································5分所以.·······························································6分（2）因为，所以······················································8分所以.······························································10分 18.（12分）解：（1）由题知：····················································3分所以有的把握认为该球员射门成功与射门距离是否超过米有关····················4分（2）由题知：·······················································7分因为，得····························································8分所以当时，；当时，···················································9分所以在上单调递增；在上单调递减········································10分所以，即球员射门成功率最高时射门角····································12分 19.（12分）解: （1）由题知：·····················································1分两式相减得··························································2分所以，·····························································4分又因为，所以························································5分因为，所以数列是首项为，公比为的等比数列·····································6分（2）由（1）知：，得·················································7分所以·······························································8分所以，····························································10分所以······························································12分 20.（12分）解：（1）若，则，····················································1分因为，当时，；当时，；所以在上单调递增；···················································2分又因为，所以的零点为·················································3分（2）由题意知，因为··················································4分①若，由得：当时，，在上单调递减；当时，，在上单调递增；··············································5分②若，由得：或，且当时，，在上单调递增；当时，，在上单调递减；当时，，在上单调递增；··············································6分③若，由（1）知：在上单调递增·········································7分④若，由得：或，且当时，，在上单调递增；当时，，在上单调递减；当时，，在上单调递增；··············································8分综上，当时，在上单调递减，在上单调递增      当时，在，上单调递增；在上单调递减；      当时，在上单调递增；       当时，在，上单调递增；在上单调递减（3）由（2）知，当时，，不满足题意当时，，，不满足题意当时，，不满足题意所以·······························································9分当时，，在上单调递增；在上单调递减；在上单调递增；所以对恒成立所以·······························································10分当时，，在上单调递增；在上单调递减；所以，所以··························································11分综上知：···························································12分 21.（12分）解：（1）由已知可得：；；·································································2分又因为；；由回归直线的系数公式知：···································································3分···································································4分所以当（百件）时，,符合有关要求所以按照公司的现有生产技术设备情况，可以安排一小时试生产件的任务. ··········5分（2）由题意知：，，；·······························································7分···································································8分所以·······························································9分 两式相减得： ························································10分 ··················································11分故································································12分22.（12分）解:（1）法一：若，则，····························································1分令，则当时，，在上单调递减；当时，，在上单调递增；···············································2分因此，即；也有······················································3分所以当时，··························································4分所以在上单调递增；···················································5分又因为，所以，当时，；当时，；所以·······························································6分法二：若，则，····························································1分令，则令，则所以在上单调递增·····················································3分又因为所以当时，，在上单调递减；当时，，在上单调递增； 因此，即对恒成立所以在上单调递增·····················································5分又因为，所以，当时，；当时，；所以·······························································6分（2）由题意知令，则当时，所以在上单调递增，无极值点；··········································7分当时，，且在上单调递增故存在满足因此·······························································8分当时，，所以在上单调递减；当时，，所以在上单调递增；所以·······························································9分再令，所以在上单调递减且，即···············································10分因为，又知，所以所以存在，满足······················································11分所以当时，，在上单调递增；当时，，在上单调递减；当时，，在上单调递增；所以，当时， 存在两个极值点综上可知：当时，不存在极值点；当时，存在两个极值点········································12分