2026全国卷一卷6月高考适应性考试数学试题二（含答案解析）

II卷非选择题（92分）三、填空题：本题共3小题，每小题5分，共15分．12．已知9个数据的平均数为6，方差为4，现又加入一个新数据6，此时这10个数据的方差为______．13．已知的展开式中含项的系数为16，则_____.14．已知椭圆的左、右焦点分别为、，椭圆上存在一点，使得为等腰三角形，且为钝角，则椭圆的离心率的取值范围为__________.四、解答题：本题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤．15．（13分）中国的非遗项目丰富多样，涵盖广泛，体现了中华民族的智慧和独特的文化魅力．春节期间某地为充分宣扬该地非物质文化，加大非遗传承人的技艺展示．该地市场开发与发展机构统计了非遗传承人的技艺展示量与市场消费收入的6组数据如下表：技艺展示量x（单位：个）212324272932市场消费收入y（单位：万元）61120275777(1)若用线性回归理论进行统计分析，求市场消费收入y关于技艺展示量x的回归方程（精确到0.1）；(2)若用非线性回归模型求得市场消费收入y关于技艺展示量x的回归方程为，且决定系数，与（1）中的线性回归模型相比，应用决定系数说明哪种模型的拟合效果更好．附：一组数据，，…，，其回归直线的斜率和截距的最小二乘估计为，；决定系数参考数据：，，，线性回归模型的残差平方和为（其中，分别为非遗传承人的技艺展示量和市场消费收入，）．16．（15分）已知数列的前项和为，．(1)求；(2)求．17．（15分）如图甲，多边形是由一个等腰三角形和一个菱形组成，其中.现将沿翻折，点翻折到点的位置，得到四棱锥如图乙所示.(1)求证:；(2)如图乙，若二面角的大小为点为的重心，点在线段上，且.(i)求证:平面;(ii)求平面与平面夹角的正弦值.18．（17分）已知函数(1)证明不等式：；(2)记，证明：；(3)已知，证明：.19．（17分）箱子中有个小球，标号为1，2，…，，现从中摸2次球，每次摸一个小球，记录标号后放回．已知每次摸出小球的标号服从概率分布，其中，．当不全为0时，该箱子中的小球是不均匀的，否则是均匀的．(1)记第一次摸出的小球标号为，第二次摸出的小球标号为．（i）当时，设事件“”的概率为，试比较与的大小，并说明理由；（ii）设事件“”的概率为，证明：．(2)证明：对于任意的，其中，都有．参考答案一．选择题题号12345678答案ADDCCBAA二．选择题题号91011答案CDACDBD三．填空题12．13．214．四．解答题15．解：（1）由题意，则························1分，···················································2分···························································4分，·································································6分y关于x的线性回归方程为．··················································7分（2）对于线性回归模型，，，····························9分决定系数为，·····································11分因为，所以用非线性回归模型拟合效果更好．····································13分16．解：（1）因为，所以当时，······················································1分，因为，所以解得；···························································3分当时，，，解得.················································5分（2）因为①，所以②，所以②—①可得，时，上式也成立，·······················7分所以用代换可得，··································9分所以，································12分这说明数列是以为首项，18为公差的等差数列.·········································13分所以.················································15分17．（1）证明：取的中点，连接.·············································1分因为为等腰三角形，点为的中点，所以，因为四边形为菱形，所以，所以.······························································2分因为四边形为菱形，所以为等边三角形，所以，进而.又，所以平面，······················································3分又平面，所以.···························································4分（2）（i）以为原点，以及垂直于平面的直线分别为轴，建立空间直角坐标系.因为，二面角的大小为120°，所以.····················5分则，，，，，.所以.·······································7分设平面的法向量为，则，所以，令，则.··································9分所以.所以与平面的法向量垂直，所以平面.····································10分（ii），，设平面的法向量为，则，所以，令，则，所以.·····················································12分所以.·········································14分所以平面与平面夹角的正弦值为.·································15分18．解：（1）令··········································1分则，·························································2分在，在故在上递减，在上递增················································3分所以即.·········································4分（2）由（1）知············································5分所以································6分令得；··········································7分（3）要证只要证··········9分令，则，···············································11分在，在故在上递增，在上递减，··········13分所以，故·················································15分所以····························16分令则；即.·······················17分19．解：（1）（i）事件“”的概率为·················2分································4分因为，可得，当且仅当小球均匀时取等号．···················5分（ii）证明：事件“”的概率：，····················7分因为，则，当且仅当小球均匀时取等号，················8分根据概率的对称性，事件“”的概率等于事件“”的概率，所以．···················9分所以，当且仅当小球均匀时取等号．···········································10分（2）设，则，··············································11分令，则，··························································12分要证，···············