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）

这是一份福建省宁德市2021-2022学年高二下期期末数学质量检测数学试题（含答案），共14页。试卷主要包含了82 B．0， ABC 10，024，635等内容，欢迎下载使用。
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．0.872．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.0.250.150.100.050.0250.0100.0050.0011.3232.0722.7063.8415.0246.6357.87910.828 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．1605．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．0.386．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.010.60.49．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：第天345678销售量辆250300400450522598(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：X123P ·································································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