2023届山东省青岛地区西海岸、平度、胶州、城阳四区高三上学期期中考试数学试题 PDF版

2022-2023学年度第一学期期中学业水平检测高三数学评分标准

一、单项选择题：本大题共8小题．每小题5分，共40分．

1-8：D C B A  A C D A  

二、多项选择题：本大题共4小题．每小题5分，共20分．

9．ACD；     10．AD；     11．AB；     12．ACD．

三、填空题：本大题共4小题，每小题5分，共20分．

13．；           14．；           15．；    

16．(1)；(2)．

四、解答题：本大题共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）

解：（1）由题知，····················································1分

所以·······························································2分

所以·······························································3分

结合正弦定理，所以···················································4分

（2）由（1）知：·····················································5分

所以，即，所以······················································7分

解得或（舍）························································8分

所以········································10分

18．（12分）

解：（1）由题知：····················································3分

设点到平面的距离为，则，

因为 ，所以·························································5分

（2）由题知：，······················································6分

以为坐标原点，直线，，分别为，，轴，建立空间直角坐标系，

则，·······························································7分

设，则，

则直线的单位方向向量为···············································8分

则点到直线的距离为 ··················································10分

··········································11分

所以的面积

所以面积的取值范围为················································12分

19．（12分）

解：（1）在中，由正弦定理知···········································1分

所以，即····························································2分

又因为，所以························································3分

所以（舍）··························································4分

（2）在中，，所以····················································5分

又因为·····························································6分

所以，·····························································7分

又因为，所以························································8分

在中，由余弦定理知：·················································9分

所以，即···························································10分

解得或（舍）·······················································11分

所以，即···························································12分

20．（12分）

解：（1）由题知：平面，所以···········································1分

因为平面平面，平面平面，平面，

所以平面····························································4分

因为平面，所以······················································5分

（2）若选择①

因为平面，平面，平面平面

所以，因此四边形为平行四边形，即为中点·································6分

若选择②

因为平面，平面，所以，

所以四边形为平行四边形，即为中点·······································6分

所以，

因为直线平面，

所以直线与平面所成角为，所以··········································7分

所以·······························································8分

以为坐标原点，分别以所在直线为轴建立空间直角坐标系

设，则·····························································9分

，为平面的一个法向量················································10分

设平面的一个法向量为，且，

由，令，则，

解得······························································11分

设平面与平面所成锐二面角为，

则 12分

21．（12分）

解:（1）由题知：，且·················································2分

①当时，有，

所以，在上单调递增，

在上单调递减，

在上单调递增···················································4分

②当时，有，

所以在上单调递增·····················································5分

③当时，有，

所以，在上单调递增，

在上单调递减，

在上单调递增···················································7分

（2）由（1）知：若，当时，，

所以·······························································9分

所以

······························································10分

······························································11分

综上，命题得证······················································12分

22．（12分）

解:（1）若，则，·····················································1分

所以·······························································2分

令，所以

当时，，；

当时，，，；

所以，对恒成立

所以，在上单调递增···················································3分

又因为

所以，当时，，在上单调递减；

当时，，在上单调递增；

又因为，

所以·······························································4分

（2）若，则·························································5分

由，

得，·······························································6分

令

再令，则··························································7分

若，令，则

所以，当时，，在上单调递减；

当时，，在上单调递增；

所以，，得和

则，满足题意···················································8分

若，则，不合题意·················································9分

若，因为在上单调递增，

且···························································10分

所以存在，使得，

即，即·······················································11分

所以，当时，，在上单调递减；

当时，，在上单调递增；

所以

综上，数的取值范围是················································12分