福建省福宁古五校联合体2022-2023学年高一下学期期中质量监测数学试题及答案

这是一份福建省福宁古五校联合体2022-2023学年高一下学期期中质量监测数学试题及答案，文件包含福宁古五校教学联合体2022-2023学年第二学期期中质量监测高一数学试卷docx、福宁古五校教学联合体2022-2023学年第二学期期中质量监测高一数学答案docx等2份试卷配套教学资源，其中试卷共11页， 欢迎下载使用。
福宁古五校教学联合体2022-2023学年第二学期期中质量监测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