山东省青岛市胶州市2022-2023学年高三上学期期中考试数学试题（含答案）

这是一份山东省青岛市胶州市2022-2023学年高三上学期期中考试数学试题（含答案），共8页。试卷主要包含了单项选择题，多项选择题，填空题，解答题等内容，欢迎下载使用。
                                             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