新教材高一数学第二学期期中质量监测试卷（原卷版+教师版）

这是一份新教材高一数学第二学期期中质量监测试卷（原卷版+教师版），共10页。试卷主要包含了每小题选出答案后，填入答题卡中，ACD，CD等内容，欢迎下载使用。
第二学期期中质量监测高一数学试题（满分150分，120分钟完卷）注意事项：1.答卷前，考生务必将班级、姓名、座号填写清楚。          2.每小题选出答案后，填入答题卡中。          3.考试结束，考生只将答题卡交回，试卷自己保留。一 、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，只有一项是符合题目要求的． 1．已知复数满足，则A． B．  C． D．2．已知向量，若，则A． B． C． D．3．，是两个平面，，是两条直线，下列四个命题中正确的是A．若，，则 B．若，，则C．若，，，则 D．若，，则4．中，点M为边AC上的点，且，若，则的值是A． B． C． D．5．已知正方形的边长为，按照斜二测画法作出它的直观图，直观图面积为，则值为A．          B．2           C．1           D．6．在中，点D是边的中点，，，，则的值为A．            B．           C．           D．7．体育锻炼是青少年生活学习中非常重要的组成部分.某学生做引体向上运 动，处于如图所示的平衡状态，若两只胳膊的夹角为，每只胳膊的拉 力大小均为380N，则该学生的体重（单位kg）约为(参考数据：取重力加速度大小为10m/s2，)A． B． C．            D．8．将边长为2的正方形纸片折成一个三棱锥，使三棱锥的四个面刚好可以组成该正方形纸片，若三棱锥的各顶点都在同一球面上，则该球的表面积为A． B． C． D． 二、 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，有多项符合要求．全部选对的得5分,有选错的得0分,部分选对的得2分．9．设向量，则    A．                 B．     C．             D．在上的投影向量为10．已知,下列结论正确的为    A．        B．    C．       D．11．在中，，角所对的边，下列结论正确的为A．若，有一个解          B．若，无解       C．若，有两个解        D．若，有一个解12．在正方体中,点满足,其中,,则A.当时,直线与直线异面  B.当时,的周长为定值C.当时,直线 D.当时,三棱锥的体积为定值三、 填空题:本题共4小题，每小题5分，共20分．13．已知点，，，若，则值为         .14．已知复数的实部为, 且,则复数的虚部为         .15．若一个圆锥的侧面展开图是面积为的半圆面，则该圆锥的体积为         .16．如图放置的边长为1的正方形的顶点，分别在轴的  正半轴、轴的非负半轴上滑动，则的取值范围为         .四、 解答题:本题共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．(10分)    已知复数，复数在复平面内对应的向量为,(1)若为纯虚数，求的值；(2)若在复平面内对应的点在第四象限，求的取值范围.  18．(12分) 如图，某中学在实施“五项管理”中，将学校的“五项管理”做成宣传牌（CD），放置在教学楼的顶部（如图所示），该中学研究性学习小组在山坡的坡脚A处测得宣传牌底部D的仰角为60°，沿该中学围墙边坡AB向上走到B处测得宣传牌顶部C的仰角为45°．已知．(1)分别求AE、BH的长；(2)求宣传牌CD的高度（结果保留根号）．   19．（12分）如图，在直三棱柱中，，，，为的中点．(1) 证明：；(2) 过三点的一个平面，截三棱柱得到一个截面，画出截面图，说明理由并求截面面积.   20．（12分）的内角的对边分别为，且．(1)求A；(2)若，三角形面积，求边上的中线的长．   21.（12分）如图，在三棱台中，，，， 为线段中点，为线段上的点，．(1)求证：点为线段的中点；(2)求三棱台的表面积．    22．（12分）如图，A，B是单位圆上的相异两定点（为圆心），（），点C为单位圆上的动点，线段AC交线段于点M（点M异于点、B），记的面积为．(1)记，求的表达式；(2)若①求的取值范围； ②设，记，求的最小值. 
高一数学参考答案及评分标准（1）本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可参照本答案的评分标准的精神进行评分．（2）解答右端所注分数表示考生正确作完该步应得的累加分数.（3）评分只给整数分，选择题和填空题均不给中间分．一、单选题1.B2.C3.D4.D5.B6.A7.B8.A二、多选题9.ACD10.CD11.BCD12.CD三、填空题13. -214． 15． 16．[1，2]四、 解答题:本题共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．解(1) ,·························································2分由为纯虚数得:，·····················································4分解得.·····························································5分(2)，·····························································6分在复平面内的对应点在第四象限··································································8分解得.·····························································10分18．解（1）由于所以,·················································1分设，则，···························································3分所以······························································4分在中， 所以······························································6分（2）过点作，垂足为.·················································7分则,又··································································9分所以······························································11分故宣传牌CD的高度为··················································12分19．解（1）连接，设，连接∵M是的中点，N是的中点··············································2分∴································································3分∵∴································································5分（2）作图过程：取中点P，连接AP 、MP、MC，则四边形APMC即为截面图形··································································6分证明如下：∵M是的中点，P是的中点∴∵∴∴A、P、M、C四点共面，四边形APMC即为所得截面··································································9分此时，四边形APMC为等腰梯形，，面积为····························································12分20．解(1) 由正弦定理得，···················································1分又，则，化简得．···························································3分又，所以，则．······················································5分因为，所以．·······················································6分(2) 由得，   ………………………………………8分法一：由得   ……………………………………………10分边上的中线的长为.  …………………………………………………………12分法二：由余弦定理得：，  ………………9分由得，   ………………11分解得，，即边上的中线的长为. …………………………………12分21．解（1）连接CD，设，连接HO、DG·····································1分∵平面FGH，平面CBD，平面平面，∴································································3分∵四边形DFCG是正方形，O是CD的中点∴点H是BC的中点·····················································5分（2）三棱台中 ∵为等边三角形∴为等边三角形，．··················································6分上底面为等边三角形，其边长为1，面积为下底面为等边三角形，其边长为2，面积为 ·································8分侧面ADFC和侧面EFCB为直角梯形，面积为侧面ADEB为等腰梯形，，面积为·········································10分所以，三棱台的表面积为．·············································12分22．解（1）：因为，； ··································································2分所以（）. ··························································4分（2）①设，，则，，，··································································6分又，则.····························································8分②设，则，因为，所以，所以，····························································9分因为，所以，即，化简得，，·························································10分所以，当且仅当，即时，等号成立，故的最小值为．······················································12分