2023龙岩一中高三上学期第二次月考数学试题PDF版含答案

这是一份2023龙岩一中高三上学期第二次月考数学试题PDF版含答案，文件包含福建省龙岩市第一中学2023届高三上学期第二次月考数学答案doc、福建省龙岩市第一中学2023届高三上学期第二次月考数学试题PDF版pdf等2份试卷配套教学资源，其中试卷共11页， 欢迎下载使用。
龙岩一中2023届高三上学期第二次月考数学答案1-8：DDCC BABD   9.ABD   10.AB   11.AC   12.BC13.14. 和 15. 16. 1317．（1）因为，由余弦定理可得.故···································································4（2）因为，所以.  由余弦定理得， 则.···································································6因为的周长为，所以，解得.  ·······································8所以的面积为···························································1018.（1）因为函数·····································································3令，解得，即对称中心····················································5当时，则，再结合三角函数图像可得所以，函数对称中心：，，值域：.···········································7（2）因为函数的图像与函数的图像关于y轴对称，则，··································································9令，，解得····························································11当时，即为所以当时，的单调递增区间：.·····································12 19.（1）由题表知，随着时间x的增大，y的值随的增大，先减小后增大，而所给的函数，和在上显然都是单调函数，不满足题意，故选择．·····3（2）把，，分别代入，得解得，，∴，．································································5∴当时，y有最小值，且．故当该纪念章上市10天时，市场价最低，最低市场价为每枚70元．···················7（3）令，·····························································8因为存在，使得不等式成立，则．··································································9又∴ 当时，取得最小值，且最小值为，∴．··································································1220.解：（1）由，可得．因为，，所以切点坐标为，切线方程为：，因为切线经过，所以，解得．···············································4（2）解：由题可知的定义域为，，令，则，解得或，·······················································6因为所以，所以，令，即，解得：，令，即，解得：或，······················································8又的定义域为，所以，增区间为，减区间为．因为，所以函数在区间的最大值为，·········································9函数在上单调递增，故在区间上，···········································10所以，即，故，所以的取值范围是．·········································1221.（1）取BC中点O，连接AO，,,因为,所以,············································2因为，，所以，所以，所以，·········································4因为，平面，所以平面，因为平面，所以；··············································· 6（2）连接，则平面即为平面，由（1）知平面，因为平面ABC，且平面，故平面平面ABC，平面平面，过O作于M，则平面ABC，过作于H，则平面，因为知，在中：，所以，所以，所以，································································8法一：设，则，在中，所以，又，所以点M为线段的中点，以O为原点，分别以分别为x，y，z轴正方向建立空间直角坐标系，，，设面的法向量为，则有，两式相减得：，所以，令，可得：，所以，设面的法向量为，则有，解得：，令，解得：所以，设锐二面角为，则有.·····················································12法二：过H做，连接，面，，则面，，则即为所求二面角.在中，，则，在中，，由可得：，，则，.·····································································1222.解：(1)，····························································1①当时，因为，所以，，，在上单调递增，没有极值点，不合题意，舍去；②当时，令，则，因为，所以，所以在上递增，又因为，，所以在上有唯一零点，且，所以，；，，所以在上有唯一极值点，符合题意.综上，.································································ 4(2)由（1）知，所以时，，所以，，单调递减；，，单调递增，所以时，，则，又因为，所以在上有唯一零点，即在上有唯一零点.····································· 6因为，由（1）知，所以，································································7则，构造，····························································8所以，记，则，显然在上单调递增，所以，········································· 9所以在上单调递增，所以，所以，所以在上单调递增，所以，······················10所以，由前面讨论可知：，，且在单调递增，所以.······························12