2021~2022学年度苏锡常镇四市高三教学情况调研（二）数学参考答案练习题

这是一份2021~2022学年度苏锡常镇四市高三教学情况调研（二）数学参考答案练习题，共6页。试卷主要包含了 14，12分等内容，欢迎下载使用。
2021~2022学年度苏锡常镇四市高三教学情况调研（二）           数学（参考答案）        2022．5   一、选择题：本题共8小题，每小题5分，共40分．题号12345678答案BACBACDC二、选择题：本题共4小题，每小题5分，共20分．题号9101112答案BCD BDADAC三、填空题：本题共4小题，每小题5分，共20分．13．     14．（答案不唯一）    15．    16．2；四、解答题：本题共6小题，共70分．17．（10分）解：（1）由正弦定理，得，···························································1分所以，化简得，所以．·························································3分又因为，所以．··················································5分（2）由余弦定理，得，将代入上式，得，················································8分所以的面积．····················································10分18．（12分）解：（1）由，得，···················································1分又因为，所以公差，···············································2分所以．·························································3分设数列的公比为，则．若选①， 因为，，所以，因为，所以，····················································5分又，所以，所以．·························································6分若选②，，所以，，即，所以或，因为，所以，····················································5分所以．·························································6分若选③，由，得，又，解得，因为，所以，····················································5分所以．·························································6分（2）由（1）得，·················································8分所以，·························································9分因为，·························································11分所以当或2时，；当时，；当时，．所以取得最大值时的值为3或4．·····································12分19．（12分）解：（1）法一：如图，取的中点，连结．因为为正三角形，所以．又因为平面平面，，，所以．·························································2分因为，所以．因为，，所以为正三角形，所以．····································4分如图，以为原点，所在直线为轴，建立空间直角坐标系．设，则，，，所以．因为，所以是平面的一个法向量．设平面的一个法向量为，由得不妨取，得．····················································6分设二面角的大小为，则，显然是锐二面角，所以二面角的大小为．·······························8分法二：取的中点，连结．因为为正三角形，所以．又因为平面平面，，，所以，·························································2分所以． 因为，，所以为正三角形，所以．因为，所以．····················································4分又，且，所以，所以．从而为二面角的平面角．···········································6分因为，，所以，在中，因为，所以，即二面角的大小为．···············································8分（2）不存在．证明如下：法一：在空间直角坐标系中中，因为，，，，，所以，．若线段（端点除外）上存在一点，使得．则存在，使得，即又因为，所以，从而，············································10分将代入上式可得，这与矛盾．故线段（端点除外）上不存在点，使得．······························12分法二：若线段（端点除外）上存在一点，使得．因为菱形中，且，，所以，从而．···················································10分又由（1）可得，且，所以，所以，这与矛盾．故线段（端点除外）上不存在点，使得．······························12分20．（12分）解：（1）设“超市甲的月需求量为400件”，“超市甲的月需求量为600件”，设“超市乙的月需求量为400件”，“超市乙的月需求量为600件”．由题意知，且与，与均为对立事件，所以．·························································2分设“两超市的月需求总量为1000件”，则，因为与互斥，且与，与相互独立， 所以．·························································4分答：两超市的月需求总量为1000件的概率为．···························5分（2）设“两超市的月需求总量为800件”，“两超市的月需求总量为1200件”，因为与相互独立，所以．因为与相互独立，所以．①若月生产量，则（元）；·······················································7分②若月生产量，则（元）；······························································9分③若月生产量，则（元）．······················································11分综上所述，当时，利润Y的数学期望最大．·····························12分21．（12分）解：（1）因为，所以． 因为，所以，所以，所以函数在上为增函数．···········································3分（2）令，所以，则，所以在R上单调递增，且， 所以当时，，；当时，，．·········································5分由，得，设，则．令，由上述推理可得或．···········································6分①当时，，因为，当且仅当，所以在R上单调递增，又因为，所以的零点有且仅有一个为0．································8分②当时，列表如下：0+++0++00+↗极大值↘极小值↗·······················································9分首先，下证：．事实上，当时，，因为，所以，又，所以，所以，所以．从而在上有且仅有一个零点．综上所述，曲线与曲线有且仅有一个公共点．···························12分22．（12分）解：（1）线段PM与NQ的长度相等．·····································1分证明：设，则．因为直线AB过点，所以，化简得，因为，所以．·············································2分联立与，得，所以．·························································4分同理，又因为，所以．故线段PM与NQ的长度相等．········································6分（2）由题意知由，得，因为，所以化简得，因为，所以化简可得 ①．···········································9分由，得，因为，所以化简可得 ②．由①②知，，即直线AB斜率的取值范围是．····························12分