山东省青岛市2019-2020学年高一下学期期末考试数学试题

这是一份山东省青岛市2019-2020学年高一下学期期末考试数学试题，共10页。试卷主要包含了作答选择题时，在中，，，，则，如图，在四棱锥中，，，点等内容，欢迎下载使用。
2019-2020学年度第二学期期末学业水平检测高一数学本试卷4页，22小题，满分150分．考试用时120分钟．   注意事项：1．答题前，考生务必将自己的姓名、考生号、考场号和座号填写在试题卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置；2．作答选择题时：选出每小题答案后，用2B铅笔在答题卡上对应题目选项的答案信息点涂黑；如需要改动，用橡皮擦干净后，再选涂其它答案，答案不能答在试卷上；非选择题必须用黑色字迹的专用签字笔作答，答案必须写在答题卡各题目指定区域内相应位置上；如需改动，先划掉原来的答案，然后再写上新答案；不准使用铅笔和涂改液．不按以上要求作答无效；3．考生必须保证答题卡的整洁，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．已知复数的共轭复数为，且（其中是虚数单位），则（    ） A． B．  C． D．2．某制药厂正在测试一种减肥药的疗效，有名志愿者服用此药，体重变化结果统计如下：体重变化体重减轻体重不变体重增加人数如果另有一人服用此药，估计这个人体重减轻的概率约为（    ）   A． B．  C． D．3．若圆锥的底面半径与高均为，则圆锥的表面积等于（    ）    A． B．  C． D．4．随机掷两枚骰子，记“向上的点数之和是偶数”为事件，记“向上的点数之差为奇数”为事件，则（    ）    A．  B．  C．互斥但不对立 D．对立5．在中，，，，则（    ）    A． B．  C． D．6．在三棱柱中，上下底面均为等腰直角三角形，且平面，若该三棱柱存在内切球，则（    ） A． B．  C． D．7．甲、乙两人独立地破译一份密码，破译的概率分别为，则密码被破译的概率为（    ） A． B．  C． D．8．设是两条不同的直线，是两个不同的平面，下述错误的是（    ）    A．若，，，则；  B．若，，，则    C．若，，，则； D．若，，则 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．如图，在四棱锥中，，，点分别为的中点，若，，则下述正确的是（    ）    A．             B．直线与异面        C． D．三点共线    10．某地区公共部门为了调查本地区中学生的吸烟情况，对随机抽出的编号为的名学生进行了调查．调查中使用了两个问题，问题：您的编号是否为奇数？问题：您是否吸烟？被调查者随机从设计好的随机装置（内有除颜色外完全相同的白球个，红球个）中摸出一个小球：若摸出白球则回答问题，若摸出红球则回答问题，共有人回答“是”，则下述正确的是（    ）    A．估计被调查者中约有人吸烟 B．估计约有人对问题的回答为“是”             C．估计该地区约有的中学生吸烟 D．估计该地区约有的中学生吸烟11．如图，在平行四边形中，分别为线段的中点，，则（    ）    A．        B．              C．      D．12．如图，线段为圆的直径，点，在圆上，，矩形所在平面和圆所在平面垂直，且，，则下述正确的是（    ）    A．平面    B．平面    C．点到平面的距离为       D．三棱锥外接球的体积为  三、填空题：本大题共4个小题，每小题5分，共20分。13．已知向量的夹角为，，，则             ． 14．在三棱锥中，若平面平面，且．则直线与平面所成角的大小为             ．15．设角是的三个内角，已知向量，，且．则角的大小为             ．16．某人有把钥匙，其中把能打开门，如果随机地取一把钥匙试着开门，把不能打开门的钥匙扔掉，那么第二次才能打开门的概率为             ；如果试过的钥匙又混进去，第二次才能打开门的概率为             ．（本题第一个空分，第二个空分） 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）已知是虚数单位，复数．（1）求；          （2）随机从复数中有放回的先后任取两个复数，求所取两个复数的模之积等于的概率．  18．（12分）如图，在几何体中，四边形为平行四边形，为的中点，平面平面，为线段上的一点，，是等边三角形．（1）证明：平面；（2）证明：；（3）证明：平面平面．       19．（12分）在①；②这两个条件中任选一个，补充到下面问题中，并进行作答．在中，内角的对边分别为，，，              ．（1）求角的大小；（2）求的周长和面积．   20．（12分）如图，在半圆柱中，为上底面直径，为下底面直径，为母线，，点在上，点在上，， 为的中点． （1）求三棱锥的体积；（2）求直线与直线所成角的余弦值；（3）求二面角的正切值．       21．（12分）有一种鱼的身体吸收汞，当这种鱼身体中的汞含量超过其体重的（即百万分之一）时，人食用它，就会对人体产生危害．现从一批该鱼中随机选出条鱼，检验鱼体中的汞含量与其体重的比值（单位：），数据统计如下：   （1）求上述数据的中位数、众数、极差，并估计这批鱼该项数据的分位数；（2）有，两个水池，两水池之间有个完全相同的小孔联通，所有的小孔均在水下，且可以同时通过条鱼．    （ⅰ）将其中汞的含量最低的条鱼分别放入水池和水池中，若这条鱼的游动相互独立，均有的概率进入另一水池且不再游回，求这两条鱼最终在同一水池的概率；    （ⅱ）将其中汞的含量最低的条鱼都先放入水池中，若这条鱼均会独立地且等可能地从其中任意一个小孔由水池进入水池且不再游回水池，求这两条鱼由不同小孔进入水池的概率．    22．（12分）某学校高一名学生参加数学竞赛，成绩均在分到分之间．学生成绩的频率分布直方图如右图： （1）估计这名学生分数的中位数与平均数；（精确到）（2）某老师抽取了名学生的分数：，已知这个分数的平均数，标准差，若剔除其中的和两个分数，求剩余个分数的平均数与标准差．（参考公式：）（3）该学校有座构造相同教学楼，各教学楼高均为米，东西长均为米，南北宽均为 米．其中号教学楼在号教学楼的正南且楼距为米，号教学楼在号教学楼的正东且楼距为米．现有种型号的考试屏蔽仪，它们的信号覆盖半径依次为米，每个售价相应依次为元．若屏蔽仪可在地下及地上任意位置安装且每个安装费用均为元，求让各教学楼均被屏蔽仪信号完全覆盖的最小花费．（参考数据：）
2019-2020学年度第二学期期末学业水平检测高一数学参考答案一、单项选择题：本大题共8小题，每小题5分，共40分。  1-8: B D A D    A B B C  二、多项选择题：本大题共4小题，每小题5分，共20分。9：BCD；  10：BC；  11：AB；     12：ABC；三、填空题：本大题共4个小题，每小题5分，共20分。13. ；     14. ；      15. ；     16. （1） ；（2）；      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）由题意知：··················································1分···································································2分···································································3分···································································4分（2）设随机从复数中有放回的任取两个复数的样本点为，则该随机试验的样本空间为········································6分所以·······························································7分设事件“所取两个复数的模之积等于”，则事件，所以························································9分             所以······························································10分 18.（12分）解：（1），在平行四边形中，连接交于点，则为的中点，连接·······················1分因为点为中点，所以为的中位线，所以·································3分所以平面， 平面，所以平面······························4分（2）因为，平面，平面，所以平面····························································7分因为平面，平面平面所以·······························································8分（3）因为为正三角形，点为中点，所以 又因为，所以平面···························································11分又因为平面所以，平面平面······················································12分 19.（12分）解：（1）若选择①：因为，，所以························································2分所以因为，所以··························································4分又因为，所以，·····························································6分若选择②：（法一）由题意知，，所以·······························································2分因为当且仅当时，上式的等号成立，且·····································3分所以·······························································5分所以·······························································6分（法二）设为方程，的两根··············································2分解得，且····························································4分所以·······························································5分所以·······························································6分（2）由正弦定理知：··················································7分因为，，所以·······························································9分所以的周长为·······················································10分所以的面积·························································12分 20．（12分）解：（1）由题意知，为正三角形 ，所以·································2分因为为圆柱的母线，所以平面 ······························3分所以·································4分（2）过点作圆柱的母线交于因为与均为圆柱的母线，所以且，所以四边形为平行四边形，所以且，所以为正三角形又因为为正三角形，所以，所以，所以为直线与所成的角············································6分在中， ·····························································7分所以由余弦定理知：所以直线与直线所成角的余弦值为········································9分（3）因为平面，平面，所以又因为，所以平面···························································10分所以，因此为二面角的平面角···········································11分在中，，，所以二面角的正切值为················································12分 21．（12分）解：（1）由题意知，数据的中位数为······································1分数据的众数为························································2分数据的极差为························································3分估计这批鱼该项数据的百分位数约为·······································4分（2）（ⅰ）记“两鱼最终均在水池”为事件，则·······························5分记“两鱼最终均在水池”为事件，则·········································6分因为事件与事件互斥，所以两条鱼最终在同一水池的概率为·······································8分（ⅱ）记“两鱼同时从第一个小孔通过”为事件，“两鱼同时从第二个小孔通过”为事件，依次类推．因为两鱼的游动独立，所以··············································9分因为事件，事件，互斥，所以······························································10分记“两条鱼由不同小孔进入水池”为事件，则与对立，所以······························································12分 22.（12分）解：（1）因为所以中位数为满足·····················································1分由，解得····························································3分设平均分为，则·································································4分（2）由题意，剩余个分数的平均值为······································5分因为个分数的标准差所以·······························································6分所以剩余个分数的标准差为··········································8分（3）将座教学楼完全包裹的球的最小直径为：因此若用一个覆盖半径为米的屏蔽仪则总费用为元；···························9分将一座教学楼完全包裹的球的最小直径为因此若用个覆盖半径为米的屏蔽仪则总费用为元；···························10分将号教学楼与号教学楼完全包裹的球的最小直径为：又因为因此若用个覆盖半径为米和个覆盖半径为米的屏蔽仪则总费用为元；所以，让各教学楼均被屏蔽仪信号完全覆盖的最小花费为元····················12分