【数学】山东省青岛市胶州市2019-2020学年高二下学期期中学业水平检测试题

山东省青岛市胶州市2019-2020学年高二下学期期中学业水平检测试题本试卷6页，22小题，满分150分．考试用时120分钟．   注意事项：1．答卷前，考生务必将自己的姓名、考生号等填写在答题卡和试卷指定位置上，并将条形码粘贴在答题卡指定位置上。2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应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。回答非选择题时，将答案写在答题卡上。写在本试卷上无效。3．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．某物体的运动方程为，其中位移的单位是米，时间的单位是秒，那么该物体在秒末的瞬时速度大小是（    ）                   A．米/秒 B．米/秒 C．米/秒 D．米/秒 2．命题“函数是偶函数”的否定可表示为（    ） A．  B．  C． D． 3．在下列区间上函数单调递减的是（    ）    A． B． C． D．4．若复数，为虚数单位，则“”是“为纯虚数”的（    ）A．充分不必要条件  B．必要不充分条件 C．充要条件         D．既不充分也不必要条件5．已知函数的图象如图所示，则可以为（    ） A．  B．  C．  D．6．分别独立的扔一枚骰子和硬币，并记下骰子向上的点数和硬币朝上的面，则结果中含有“点或正面向上”的概率为（   ） A． B． C． D．7．由于受到网络电商的冲击，某品牌的洗衣机在线下的销售受到影响，承受了一定的经济损失，现将地区家实体店该品牌洗衣机的月经济损失统计如图所示．估算月经济损失的平均数为，中位数为，则（    ）      A． B． C． D．8．在直角坐标系中，曲线与圆的公共点个数为（    ） A．个 B．个 C．个 D．个二、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．随机变量服从正态分布，则下述正确的是（    ） A．  B．  C． D．10．若复数满足（其中是虚数单位），复数的共轭复数为，则（    ） A．  B．的实部是  C．的虚部是  D．复数在复平面内对应的点在第一象限11．某市坚持农业与旅游融合发展，着力做好旅游各要素，完善旅游业态，提升旅游接待能力。为了给游客提供更好的服务，旅游部门需要了解游客人数的变化规律，收集并整理了年月至年月期间月接待游客量(单位：万人)的数据，绘制了如图所示的折线图．      根据该折线图，下列结论正确的是（    ） A．月接待游客量逐月增加 B．年接待游客量逐年增加C．各年的月接待游客量高峰期大致在7,8月 D．各年1月至6月的月接待游客量相对于7月至12月，波动性更小，变化比较平稳12．关于的方程在上有个解．则实数可以等于（    ）    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（其中是虚数单位，），则         ． 14．若“，”是假命题，则实数的取值范围是         ． 15．已知是的导函数（），，则         ．16．若函数在区间上不是单调函数，则实数的取值范围是          ．   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）如图，已知长方形的周长为，其中点分别为的中点，将平面沿直线向上折起使得平面平面，连接，得到三棱柱．设，记三棱柱体积为．                                            （1）求函数的解析式；（2）求函数的最大值．  18．（12分）已知函数恒过定点．（1）当时，求在点处的切线方程；（2）当时，求在上的最小值．   19．（12分）已知函数，．（1）当时，证明：在上单调递增；（2）当时，讨论的极值点．        20．（12分）数学是研究数量、结构、变化、空间以及信息等概念的一门科学．在人类历史发展和社会生活中，数学发挥着不可替代的作用，也是学习和研究现代科学技术必不可少的基本工具．（1）为调查大学生喜欢数学命题是否与性别有关，随机选取名大学生进行问卷调查，当被调查者问卷评分不低于分则认为其喜欢数学命题，当评分低于分则认为其不喜欢数学命题，问卷评分的茎叶图如下：    依据上述数据制成如下列联表： 喜欢数学命题人数不喜欢数学命题人数总计女生男生总计     请问是否有的把握认为大学生是否喜欢数学命题与性别有关？参考公式及数据：．（2）在某次命题大赛中，同学要进行轮命题，其在每轮命题成功的概率均为，各轮命题相互独立，若该同学在轮命题中恰有次成功的概率为，记该同学在轮命题中的成功次数为，求．   21．（12分）近期，某超市针对一款饮料推出刷脸支付活动，活动设置了一段时间的推广期，由于推广期内优惠力度较大，吸引越来越多的人开始使用刷脸支付．该超市统计了活动刚推出一周内每一天使用刷脸支付的人次，用表示活动推出的天数，表示每天使用刷脸支付的人次，统计数据如下表所示：（1）在推广期内， 与（均为大于零的常数）哪一个适宜作为刷脸支付的人次关于活动推出天数的回归方程类型？（给出判断即可，不必说明理由）；（2）根据（1）的判断结果及表中的数据，求关于的回归方程，并预测活动推出第天使用刷脸支付的人次；（3）已知一瓶该饮料的售价为元，顾客的支付方式有三种：现金支付、扫码支付和刷脸支付，其中有使用现金支付，使用现金支付的顾客无优惠；有使用扫码支付，使用扫码支付享受折优惠；有使用刷脸支付，根据统计结果得知，使用刷脸支付的顾客，享受折优惠的概率为，享受折优惠的概率为，享受折优惠的概率为．根据所给数据估计购买一瓶该饮料的平均花费．参考数据: 其中，   参考公式: 对于一组数据，其回归直线的斜率和截距的最小二乘估计公式分别为: ．  22．（12分）已知函数，．（1）若，证明：；（2）若，有且只有个零点，求实数的取值范围；（3）若，，，求正整数的最小值．
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： BBDA  ACCA  二、多项选择题：本大题共4小题，每小题5分，共20分。9．AC；    10．ABD；    11．BCD；    12．CD．三、填空题：本大题共4个小题，每小题5分，共20分。13．；         14．；        15．；       16．．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）由题知：····················································2分因为，长方形的周长为，所以，所以，·····························································5分（2）由（1）知：，···················································6分所以令,解得··························································7分当在单调递增；······················································8分当在单调递减；······················································9分所以······························································10分18．（12分）解：（1）由题知：所以定点·············································2分若，，所以··························································4分所以在点处的切线方程：，即············································5分　　　　　（2）若，，所以······················································6分令,解得······························································7分当在单调递增························································8分当在单调递减························································9分又因为，···························································11分所以的最小值为······················································12分19．（12分）解：（1）由题知：，所以···············································2分因为，所以··························································3分又因为·····························································4分所以 所以在上单调递增·····················································5分（2）由题知：·······················································6分所以·······························································7分若，则在上单调递增，无极值点··········································8分若，由，解得························································9分当时，，在上单调递减················································10分当时，，在上单调递增················································11分所以的极小值点为····················································12分 综上，当时，无极值点；当时，有一个极小值点，无极大值点20．（12分）解：（1）由题知：····················································4分则·································································5分所以没有的把握认为大学生是否喜欢数学命题与性别有关·······················6分（2）由题知：·······················································8分依据二项分布知：,所以··················································9分令··································································10分当在单调递减；当在单调递增；因此，所以·························································11分所以······························································12分21．（12分）解：（1）根据散点图判断，适宜作为扫码支付的人数关于活动推出天数的回归方程类型·····················2分（2）因为，两边同时取常用对数得：，设所以·····························································3分因为所以·······························································4分把样本中心点代入,得: ，所以，所以关于的回归方程式：···············································5分把代入上式，，所以活动推出第天使用刷脸支付的人次为···································6分（3）记购买一瓶该饮料的花费为（元），则的取值可能为： ···································································7分 ···································································8分···································································9分··································································10分分布列为：因为所以估计购买一瓶该饮料的平均花费为（元）·······························12分22．（12分）解：（1）由题知，，··················································1分所以，当时，，在上单调递增；      当时，，在上单调递减；·············································2分所以·······························································3分（2）因为，当时，，在上单调递增，不可能有个零点···································4分当时，令，解得，所以，当时，，在上单调递增；      当时，，在上单调递减；所以·······························································6分若只有个零点，则，解得：··············································7分由（1）知：，所以，令，解得：或所以，存在，满足；      存在，满足；所以在和上个恰有个零点，符合题意综上，所求实数的取值范围为············································8分（3）令，所以，当时，因为，所以所以在上是递增函数，又因为，所以关于的不等式不能恒成立············································9分当时，．令，得，所以当时，，当时，．因此函数在上是增函数，在是减函数，所以······························································10分令，因为，又因为在上是减函数，所以当时，········································11分所以整数的最小值为．················································12分