2022青岛地区高二下学期期中考试数学试题含答案

这是一份2022青岛地区高二下学期期中考试数学试题含答案，文件包含山东省青岛地区2021-2022学年高二下学期期中考试数学答案docx、山东省青岛地区2021-2022学年高二下学期期中考试数学试题pdf等2份试卷配套教学资源，其中试卷共12页， 欢迎下载使用。
2021—2022学年度第二学期期中学业水平检测高二数学答案及评分标准一、单项选择题：本大题共8小题．每小题5分，共40分．A D C A     B A C D  二、多项选择题：本大题共4小题．每小题5分，共20分．9．AC；     10．BC；     11．BD；     12．ACD.三、填空题：本大题共4小题，每小题5分，共20分．13．；     14．；     15．；     16．；.四、解答题：本大题共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．（本小题满分10分）解：选择条件①：（1）设切点为·······················································1分因为，所以··························································3分所以切线的斜率为·····················································4分解得·······························································5分所以切点为··························································6分（2）由（1）知切点为，因为点在直线上······················································8分所以，即···························································10分选择条件②：（1）设切点设切点为··················································1分因为，所以··························································3分所以切线的斜率为·····················································4分解得·······························································5分所以切点为··························································6分（2）由（1）知切点为因为点在直线上······················································8分所以，即···························································10分18. （本小题满分12分）解：（1）根据图可得，女生中得分不超过分的人数，女生得分超过分的人数，男生中得分不超过分的人数，男生得分超过分的人数，则列联表为：性别了解航空航天知识程度合计得分不超过分的人数得分超过分的人数女生男生合计········································3分（2）因为··························································6分所以，依据的独立性检验，认为该学校高中生了解航空航天知识程度与性别有关联，此推断犯错误的概率不大于．······7分（3）由（1）可得，得分超过分的学生中男生：女生，从得分超过分的同学中采用分层抽样的方法抽取人，则男生占人，女生占人·················································8分则取值可能为，，，，，所以随机变量的分布列为················································11分所以．····························································12分19．（本小题满分12分）解：（1）设“取出的球中有红球”， “第次取出的是红球” ···················1分（ⅰ）由题意知，随机试验为从号箱中不放回的进行次取球则·································································2分又·································································3分所以由古典概型的概率公式得， 故故所求的概率为·····················································4分（ⅱ）因为··························································5分所以由条件概率公式得，故所求的概率为······················································6分（2）设“最后从号箱中取出的球是红球”，“从号箱中取出的球是红球” ············································7分则，，·································································9分由全概率公式得所以，先随机从号箱中取出一球放入号箱中，再从号箱中随机取出一球，最后从号箱中取出的球是红球的概率为······12分 20．（本小题满分12分）解：（1）根据散点图可知，更适合作为关于的经验回归方程；···················3分（2）令，则·························································4分所以·······························································6分所以·······························································7分所以，故关于的经验回归方程为··········································8分（3）一天的利润为····················································9分··································································11分当且仅当即时等号成立，所以预计每吨定价为万元时，该产品一天的销售利润最大，最大利润是万元．······················································12分 21．（本小题满分12分）解：（1）设考生的成绩为，则由题意可得应服从正态分布，即，令，则··························································2分由分及以上高分考生名可得··············································3分即·································································4分即有，则，可得，可得·······························································6分设最低录取分数线为，则···············································7分即有，即有，可得，即最低录取分数线为··············································8分（2）考生甲的成绩，所以能被录取········································9分··································································11分表明不低于考生甲的成绩的人数大约为总人数的，即考生甲大约排在第名，排在前名之前，所以能被录取为高薪职位···············12分 22. （本小题满分12分）解：（1）由题意得每轮游戏获得分的概率为，获得分的概率为···················1分可能取值为，，，，，所以的分布列： ·····························································3分所以·······························································4分（2）（ⅰ）证明：，即累计得分为分，是第次掷骰子，向上点数不超过点的概率，则·5分累计得分为分的情况有两种：①，即前一轮累计得分，又掷骰子点数超过点得分，其概率为②，即前一轮累计得分，又掷骰子点数没超过点得分，其概率为所以·······························································7分所以所以数列是首项为，公比为的等比数列．···································8分（ⅱ）因为数列是首项为，公比为的等比数列所以·······························································9分所以，，……，，各式相加，得：，所以  ·····························································11分所以活动参与者得到纪念品的概率为：··································································12分