2022青岛地区高二下学期期中考试数学试题含答案

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：

········································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