福建省宁德市2021-2022学年高二下期期末数学质量检测数学试题（含答案）

宁德市2021-2022学年度第二学期期末高二质量检测

数学试题

本试卷共6页，22题．考试时间120分钟，满分150分．

注意事项：

　　1.答题前，考生务必在试题卷、答题卡规定的地方填写自己的准考证号、姓名，考生要认真核对答题卡上粘贴的条形码的“准考证号，姓名”与考生本人准考证号、姓名是否一致．

　　2.选择题每小题选出答案后，用2B铅笔把答题卡上对应的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；填空题和解答题用0.5毫米黑色签字笔在答题卡上书写作答，在试题卷上作答，答案无效．

3.考试结束，考生必须将试题卷和答题卡一并交回．

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，有且只有一个是符合题目要求的．

1．对于x，y两变量，有四组样本数据，分别算出它们的线性相关系数r（如下），则线性

相关最强的是

A．0.82          B．0.78            C．0.69           D．0.87

2．函数的单调递减区间是

A．          B．          C．          D．

3．若由一个列联表中的数据计算得，则有（    ）把握认为两个变量有关系.

A．              B．          C．          D．

4．某校一次高二年级数学检测，经抽样分析，成绩近似服从正态分布，且

．若该校有800人参加此次检测，估计该校此次检测数学成绩不低于分的人数为

A．100             B．125           C．150          D．160

5．某高校有智能餐厅A、人工餐厅B，甲第一天随机地选择一餐厅用餐，如果第一天去A餐厅，那么第二天去A餐厅的概率为0.6；如果第一天去B餐厅，那么第二天去A餐厅的概率为0.8．则甲第二天去A餐厅用餐的概率为

A．0.75               B．0.7               C．0.56          D．0.38

6．如图，在空间四边形中，两两垂直，

则点到直线的距离为

A．      B．      C．      D． 

7．已知，，若存在，R，使得成立，则实数的取值范围为

A．           B．       C．          D．

8．已知a，bR，且，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， 有多项是符合题目要求，全部选对得5分，部分选对得2分，有选错得0分.

9．设离散型随机变量的分布列如右表，若离散型随机变量

满足，则下列结果正确的有

A．             B．

C．                 D．

10．  已知，，，则下列结论正确的是

A．                 B．

C. 为钝角             D．在方向上的投影向量为

11．设是定义域为R的偶函数，其导函数为，若时 ，图象如图所示，则可以使成立的的取值范围是  

A．                 B． 　

C．                    D．

12．如图所示，在正方体中，，分别是，的中点，是线段上的动点，是直线与平面的交点，则下列判断正确是 

A．   

B．三棱锥的体积是定值

C．唯一存在点使得

D．与平面所成的角为定值

三、填空题: 本大题共4小题，每小题5分，共20分. 把答案填在答题卡的相应位置，其中第15题，第一空2分，第二空3分.

13．已知，，，则的坐标为___________．

14．如图，在一段线路中并联两个自动控制的常用开关，只要其

中有一个开关能够闭合，线路就能正常工作．假定在某段时间

内每个开关能够闭合的概率都是0.8，则这段时间内线路正常

工作的概率为____________．

15．在正四棱台中，,,

，设，

则向量=_________（用表示）,

=　           　．(第1空2分，第2空3分）．       

16．已知函数有两个零点，则正实数a的取值范围为___________．

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）

已知函数．

(1) 当时，求曲线在处的切线方程；

(2) 讨论函数的单调性．

18．（本小题满分12分）

  一袋中装有4个黄球2个红球，现从中随机不放回地抽取3个球．

(1) 求至少抽到一个红球的概率；

(2) 求取出的黄球个数X的分布列和数学期望．

19．（本小题满分12分）

2022年上半年受新冠疫情的影响，国内车市在上半年累计销量相比去年同期有较大下降，国内多地在3月开始陆续发现促进汽车消费的政策，开展汽车下乡活动，这也是继2009年首次汽车下乡之后开启的又一次大规模汽车下乡活动．某销售商在活动的前2天大力宣传后，从第3天开始连续统计了6天的汽车销售量（单位：辆）如下：

(1) 证明：；

(2) 根据上表中前4组数据，求关于的线性回归方程；

(3) 用(2)中的结果分别计算第7、8天所对应，再求与当天实际销售量的差，若差的绝对值都不超过，则认为所求得的回归方程“可靠”，若“可靠”则可利用此回归方程预测以后的销售量.请根据题意进行判断，该回归方程是否可靠？若可靠，请预测第12天的销售量；若不可靠，请说明理由．

参考公式及数据：

， ．

20．（本小题满分12分）

如图，四棱锥的底面是平行四边形，平面，，，．

(1) 证明：平面平面；

(2)若直线与平面所成角的正弦值为，

求二面角的余弦值．

21．（本小题满分12分）

2022北京冬奥会和冬残奥会吉祥物冰墩墩、雪容融亮相上海展览中心. 为了庆祝吉祥物在上海的亮相，某商场举办了一场赢取吉祥物挂件的“定点投篮”活动，方案如下：

方案一：共投9次，每次投中得1分，否则得0分，累计所得分数记为；

方案二：共进行三轮投篮，每轮最多投三次，直到投中两球为止得3分，否则得0分，

三轮累计所得分数记为．

累计所得分数越多，所获得奖品越多．现在甲准备参加这个“定点投篮”活动，已知甲每次投篮的命中率为，每次投篮互不影响．

(1) 若，甲选择方案二，求第一轮投篮结束时，甲得3分的概率；

(2) 以最终累计得分的期望值为决策依据，甲在方案一，方案二之中选其一，应选择那个方案？

22． (本小题满分12分)

已知函数．

(1) 求函数的最小值；

(2) 若不等式对于恒成立，求a的取值范围．

宁德市2021-2022学年度第二学期期末高二质量检测

参考答案及评分标准

说明:

一、本解答指出了每题要考查的主要知识和能力，并给出了一种或几种解法供参考，如果考生的解法与本解法不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则.

二、对计算题，当考生的解答在某一部分解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分.

三、解答右端所注分数，表示考生正确做到这一步应得的累加分数.

四、只给整数分数，选择题和填空题不给中间分.

一、单项选择题: 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，只有一个选项是符合题目要求的.

1. D    2.  D     3．C    4．D     5. B     6.  A     7. B    8. D

二、多项选择题：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，部分选对的得2分，有选错的得0分）

9.  ABC        10. BD        11. ABD       12. BC  

三、填空题:（本大题共4小题，每小题5分，共20分. 把答案填在答题卡的相应位置） (本小题第一个空2分，第二个空3分)

13.     14. 0.96   15. （1） ,（2） .  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：(1) 当，，·······················································1分

则，即切点为（1,）.··················································2分

，································································3分

即切线斜率k=1.······················································4分

所以切线方程为：····················································5分

(2) ·······························································6分

令，解得：·························································7分

①若，显然单调递增．················································8分

②若，当时，，在区间单调当时，，在区间单调递减；························9分

③若，当时，，在区间单调递增；当时，，在区间单调递减；··················10分

综上，若时，在上单调递增；若时，在区间单调递增，在区间单调递减；若时，在区间单调递增；区间单调递减.

（备注：没有结论不扣分。）

18.  (本小题满分12分)

解：(1)解法一

设取出的3个球至少一个红球为事件A，····································1分

则取到三个都是黄球的的事件为·········································3分

 ............................................................................................................................5分

解法二:设取出的3个球至少一个红球为事件A，······························1分

 ...........................................................................................................3分

 .............................................................................................................................5分

(2)随机变量X可能取值为1,2,3············································6分

，································································7分

，································································8分

， ································································9分

故X的分布列为：

·································································10分

期望·····························································11分

·····························································12分

 备注：（i）写成：,酌情给分,

   （ii）利用公式计算数学期望，酌情给分 .

19.  （本小题满分12分）

（1）证明：

··································································1分

··································································2分

（说明：未用统计学符号证明，过程正确不扣分）

（2）·····························································3分

·································································5分

，································································6分

得································································7分

（3）当时，；······················································8分

当时，；···························································9分

，·······························································11分

可知回归方程“可靠”；预测时，·······································12分

20. （本小题满分12分）

解法一：

解：(1)在三角形ABC中，

由余弦定理得：,·····················································1分

所以  

所以,······························································2分

在中,,

所以,······························································3分

平面, 平面,

,·································································4分

又

平面,  ·····························································5分

平面

平面平面···························································6分

（2）由(1)得,如图建立空间直角坐标系，···································7分

设，则,,,

，,，······························································8分

设面的法向量为，

由,  得是面的一个法向量；·············································9分

设直线与平面所成角为

则·······························································10分

设面的法向量为，

由得是面的一个法向量；··············································11分

，

二面角的余弦值为. ··················································12分

解法二：（1）在三角形ABC中，

由余弦定理得：,·····················································1分

所以  

所以,······························································2分

又平面, 如图建立空间直角坐标系，

设，

则,,,·······························································3分

，,，

设面的法向量为，

得

则是面的一个法向量；················································4分

由平面,则是面的一个法向量；···········································5分

平面平面···························································6分

（2）证明：过点作,垂足为,过点作,垂足为;

因为

所以，又,，

所以,又

所以又,,所以；

又,所以

所以为二面角平面角··················································9分

在中，,,

所以·····························································12分

21.  (本小题满分12分)

解：(1)若,设甲选择方案二，第一轮投篮结束时甲得分为X1，···················1分

则 ································································4分

(2)方案一满足，所以方案一的数学期望为··································6分

方案二进行,每一轮能得3分的概率为， ····································7分

每一轮的得分的期望为，  进行三轮总得分,

所以选择方案二的得分期望为···········································8分

甲如何选择，由两种方案的期望大小决定，所以：

··································································9分

当，，两种方案期望相同，所以方案一，二都可以；·························10分

当，，方案二期望大，所以甲应该选方案二；······························11分

当，，方案一期望大，所以甲应该选方案一.·······························12分

22.   (本小题满分12分)

解法一：

(1) 求导：··························································1分

即································································2分

当解得当解得

的单调递减区间为；单调递增区间为······································4分

函数的最小值为·····················································5分

(2)由(1)得，所以要使得恒成立，必须满足：

··································································6分

下面证明：当时恒成立

··································································8分

只需证明

设 ，

则································································9分

由(1)得且只在取等号

当时，，单调递减

当时，，单调递增

。

综上.·····························································12分

解法二：

(1)同解法一·························································5分

(2)（变量分离）整理得：··············································6分

只需······························································7分

先证明：，

构造，，

当时，，单调递增

从而证明得·························································8分

当仅且当即处取得等号。

。·······························································12分

解法三：

（1）同解法一······················································5分

（2）不分离

得

下面证明当时，

··································································8分

只需证明

设 ，

则································································9分

由(1)得且只在取等号

当时，，单调递减

当时，，单调递增

。

综上.·····························································12分