【数学】贵州省遵义市南白中学2019-2020学年高二上学期第三次月考（文） 试卷
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贵州省遵义市南白中学2019-2020学年高二上学期第三次月考（文）注意事项：1.本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分.2.答题前，考生务必将自己的姓名，准考证号填写在本试题相应的位置.3.全部答案在答题卡上完成，答在本试题上无效.4.考试结束后，将本试题和答题卡一并交回.第Ⅰ卷（选择题）一、选择题：（本大题共12个小题,每小题5分,共60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.已知集合,则（ ）A. B. C. D.2.己知等差数列中，，则（ ）A.8 B.7 C.14 D.163.椭圆的离心率为（ ）A. B. C. D.4.命题“”的否定为（ ）A. B.C. D.5.若一个正方体截去一个三棱锥后所得的几何体如图所示.则该几何体的正视图是（ ） A. B. C. D.6.已知，，，则（ ） A. B. C. D.7.已知函数是定义在上的偶函数，且函数在上是减函数，如果，则不等式的解集为（ ）A. B. C. D.8.已知命题：直线与直线垂直，：原点到直线的距离为，则（ ）A.为假 B.为真 C.为真 D.为真9.已知，，且是的必要不充分条件，则实数的取值范围为（ ）A. B. C.或 D.或10.明代数学家程大位(1533-1606 年)所著《算法统宗》中有这样一个问题:“旷野之地有个桩，桩上系着一腔羊，团团踏破三亩二.试问羊绳几丈长”.意思是“一条绳索系着一只羊，羊踏坏一块面积为亩的圆形庄稼，试求绳的长度(即圆形半径)”(明代度量制:步=尺，亩=平方步，丈=尺，圆周率.)（ ）A.丈 B.丈 C.丈 D.丈11.函数的图像大致为 ( )A. B. C. D.12.已知函数是定义域为的奇函数，且满足，若函数有两个零点.其中，分别记为，则的取值范围是 ( )A. B. C. D.第Ⅱ卷（非选择题）二、填空题（本大题共4个小题，每小题5分，共20分，将答案填在答题纸上）13.向量，若，则实数 .14.平行直线与之间的距离为 .15.在△ABC中，如果，那么等于 .16.已知，分别为椭圆的左、右焦点，若直线上存在点，使为等腰三角形，则椭圆离心率的范围是 .三、解答题 （本大题共6个小题，第17题10分，其余每个题12分，共70分.解答应写出文字说明、证明过程或演算步骤.）17.已知等差数列的前项和为，有，.（Ⅰ）求数列的通项公式；（Ⅱ）令，记数列的前项和为，证明：. 18.《中华人民共和国道路交通安全法》第条的相关规定：机动车行经人行道时，应当减速慢行；遇行人正在通过人行道，应当停车让行，俗称“礼让斑马线”， 《中华人民共和国道路交通安全法》第条规定：对不礼让行人的驾驶员处以扣分，罚款元的处罚.下表是某市一主干路口监控设备所抓拍的5个月内驾驶员“礼让斑马线”行为统计数据：月份违章驾驶员人数（Ⅰ）请利用所给数据求违章人数与月份之间的回归直线方程；（Ⅱ）预测该路口月份的不“礼让斑马线”违章驾驶员人数.参考公式： ,参考数据：. 19.已知四面体中面，，垂足为，，为中点，,.（Ⅰ）求证： 面；（Ⅱ）求三棱锥的体积. 20.三个内角A,B,C对应的三条边长分别是，且满足.（Ⅰ）求角的大小；（Ⅱ）若，，求. 21.如图，四棱锥中，平面，底面是正方形，且,为中点.（Ⅰ）求证：平面；（Ⅱ）求点到平面的距离. 22.在平面直角坐标系中，已知椭圆的离心率为，点在椭圆上.（Ⅰ）求椭圆的方程；（Ⅱ）设直线与圆相切，与椭圆相交于两点，求证：是定值.
参考答案一、选择题：（共12个小题,每小题5分,共60分）123456789101112CBADACDBACBD二、填空题：（共4个小题，每小题5分，共20分）13． 14． 15． 16．三、解答题：（共6个小题，共70分）17.（本大题10分）解：（Ⅰ）设数列的公差为，有解得，有············································3分数列的通项公式为··················································5分（Ⅱ）由（1）知········································7分有由，，由上知····································10分18.（本大题12分）解：（Ⅰ）由表中数据知， ∴ ··············································4分∴所求回归直线方程为·······························6分（Ⅱ）令，则人该路口月份的不“礼让斑马线”违章驾驶员人数预计为49人····················12分19.（本大题12分）解：（Ⅰ）因为,所以为中点,又因为是中点所以·····················································3分而面,面所以面················································6分（Ⅱ）由已知得,,,三角形的面积···········································9分··········································12分20.（本大题12分）解：（Ⅰ）由正弦定理得····························································3分由已知得，，因为，所以····················································6分（Ⅱ）由余弦定理，得···················································9分即，解得或，负值舍去，所以·····································································12分21.（本大题12分）解：（Ⅰ）证明：平面，又正方形中，平面·························································3分又平面，，，是的中点，平面·························································6分（Ⅱ）过点作于点，由（1）知平面平面，又平面平面，平面，线段的长度就是点到平面的距离·····································8分，，···············································12分22.（本大题12分）解：（Ⅰ）由题意得：即 ············································2分 椭圆方程为将代入椭圆方程得： 椭圆的方程为：············································5分（Ⅱ）①当直线斜率不存在时，方程为：或当时，，，此时 当时，同理可得·········································7分②当直线斜率存在时，设方程为：，即直线与圆相切 ，即联立得：设， ，············································9分代入整理可得： 综上所述：为定值·····················································12分
