2026届福建省高三数学高考一模模拟试卷（含答案详解与评分标准）

2026届福建省高三数学高考一模模拟试卷（含答案详解与评分标准）学校：________________班级：____________姓名：____________考号：________________考试时间：120分钟满分：150分题型选择题填空题解答题总分分值30分18分102分150分注意事项：1．本卷用于2026届福建省高三数学高考一模考前综合检测，重点覆盖函数与导数、三角与向量、数列、概率统计、立体几何、解析几何等主干内容。2．答题前，请将学校、班级、姓名、考号填写清楚；选择题答案填入客观题作答栏，填空题只写最终结果。3．解答题须写出必要的文字说明、证明过程或演算步骤；只写结论而无过程的题目，按评分标准酌情给分。4．全卷满分150分，考试时间120分钟。请合理分配时间，保持卷面整洁。一、选择题：本大题共10小题，每小题3分，共30分。在每小题给出的四个选项中，只有一项符合题目要求。1．已知集合，，则等于（）A．B．C．D．2．设复数，其中为虚数单位，则的虚部为（）A．B．C．D．3．已知平面向量满足，，且与的夹角为，则（）A．B．C．D．4．函数的最大值为（）A．B．C．D．5．等差数列中，若，则前9项和等于（）A．B．C．D．6．一个袋中有3个红球、2个白球，这5个球除颜色外完全相同。从中不放回地随机取出2个球，取到一红一白的概率为（）A．B．C．D．7．若函数在区间上单调递增，则实数的取值范围是（）A．B．C．D．8．在棱长为2的正方体中，点到平面的距离为（）A．B．C．D．

9．椭圆的左、右焦点分别为。若点在椭圆上且，则（）A．B．C．D．10．若方程恰有两个不同的实数根，则实数的取值范围是（）A．B．C．D．二、填空题：本大题共6小题，每小题3分，共18分。11．二项式的展开式中的系数为__________。12．在区间内，方程的所有解之和为__________。13．曲线在点处的切线与轴的交点纵坐标为__________。14．随机变量表示连续抛掷3枚质地均匀硬币所得正面向上的次数，则__________。15．一个直圆锥的底面半径为3，高为4，其内切球半径为__________。16．若不等式对一切恒成立，则实数的取值范围为__________。客观题作答栏题号12345678910答案题号111213141516答案三、解答题：本大题共6小题，共102分。解答应写出文字说明、证明过程或演算步骤。17．（14分）已知函数。（1）求的最小正周期和单调递增区间；（2）在中，角所对的边分别为，若，，，求边的长和的面积。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18．（16分）数列满足，（）。（1）证明数列是等差数列，并求；（2）设，求；（3）结合一模复习中对递推数列的处理方法，说明本题中“同除以的量级”为什么能化复杂递推为基础数列。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19．（17分）某校为分析2026届高三数学高考一模阶段复习效果，随机抽取100名学生的模拟检测成绩，得到如下频数分布表。成绩区间频数481523251573（1）用组中值估计这100名学生的平均分，并估计成绩不低于110分的比例；（2）从成绩不低于120分的学生中随机抽取2人进行试卷复盘，求至少有1人成绩在的概率；（3）若全校共有760名高三学生参加同类检测，估计成绩不低于110分的人数，并说明估计的依据。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20．（17分）四棱锥的底面是边长为2的正方形，平面，且。点是棱的中点。（1）证明平面；（2）求直线与平面所成角的正切值；（3）求二面角的余弦值。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21．（18分）已知椭圆，左、右焦点分别为，右顶点为。过点作斜率为（）的直线与椭圆交于另一点。（1）求椭圆的离心率；（2）用表示点的坐标；（3）求面积的最大值，并求此时的值。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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参考答案与解析一、选择题答案与关键理由评分标准：本大题每小题3分，共30分。选对得3分，选错、多选或不选均得0分。1．B由得，与整数集合取交集，得。2．C，，虚部为。3．C，故。4．B，故最大值为。5．C等差数列中，所以，。6．C总取法数为，一红一白的取法数为，概率为。7．B。若在上单调递增，则对任意成立，最小值在处给出。8．B建立坐标系，，，，平面的方程为，点到该平面的距离为。9．C椭圆长轴长为，故，又。由余弦定理，，故。10．D令。若，方程无两个交点；若，在处取最小值。方程有两个不同实根，当且仅当，即。二、填空题答案与解析评分标准：本大题每小题3分，共18分。答案正确得3分；只写等价正确形式得3分；结果不完整或化简错误不得满分。11．6通项为，令，得，系数为。12．方程化为。在内，解为，和为。13．曲线导数，在处斜率为，切点为，切线方程，与轴交点纵坐标为。14．，，，，故。15．圆锥轴截面为底边、高、腰的等腰三角形。内切球半径等于轴截面内切圆半径，。16．由得。函数在处取最大值，故。三、解答题答案详解与评分标准17．（14分）解：（1）。所以最小正周期。由，得，即单调递增区间为，。（2）由，得。又，可得。由余弦定理，，所以。三角形面积。评分标准：化简三角函数表达式3分；周期2分；单调区间3分；由求出2分；余弦定理求2分；面积计算2分。18．（16分）解：（1）令。由，两边同除以，得。又，所以是首项为1、公差为1的等差数列，，从而。（2）。设，则。两式相减，。（3）本递推中齐次部分的放大倍数为2，非齐次项为，与同量级。将除以后，递推中的乘2效应被抵消，只剩，于是问题转化为等差数列。评分标准：构造2分；推出3分；求得3分；错位相减求和5分；方法说明3分。19．（17分）解：（1）用各组组中值估计平均分：。成绩不低于110分的频数为，比例估计为。（2）成绩不低于120分的学生共有人，其中在的有3人。随机抽取2人，至少有1人在的概率为。（3）用样本比例估计总体比例，全校成绩不低于110分的人数约为人。依据是随机抽取的100名学生可视为总体的代表样本，用样本频率估计总体频率。评分标准：平均分列式与计算6分；比例估计3分；组合概率6分；总体人数估计与依据2分。20．（17分）解：（1）因为平面，所以。又正方形中对角线，且，平面，故平面。（2）建立空间直角坐标系：，，，，。点为中点，故。点在平面上的射影为，于是直线与底面所成角的正切值为。（3）取平面的法向量，平面的法向量。沿棱取同向半平面内的垂线，可得二面角的余弦值为。评分标准：证明和各2分，推出线面垂直1分；建系和求中点3分；线面角正切2分；法向量或垂线向量计算5分；二面角余弦2分。21．（18分）解：（1）椭圆中，，，所以。（2）过且斜率为的直线为。代入椭圆方程，得。除去交点对应的根，另一交点的横坐标为，纵坐标。（3），故。令，则。由，得，当且仅当，即时取等号。因此最大面积为，此时。评分标准：求离心率4分；直线方程与代入3分；求坐标4分；面积表达式3分；最值及等号条件4分。22．（20分）解：（1）当时，。由基本不等式型结论（），得，等号当且仅当。（2）。若，则，函数在上单调递增，且时，时，故恰有1个零点。若，