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

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

二、多选题

三、填空题

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