2023龙岩一中高三上学期第二次月考数学试题PDF版含答案

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.BC

13.14. 和 15. 16. 13

17．（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···························································10

18.（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，

∴．··································································12

20.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．·········································12

21.（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，

.·····································································12

22.解：(1)，····························································1

①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