2026届杭州市高三数学高考一模模拟试卷（含答案详解与评分标准）

2026届杭州市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分注意事项：1．本试卷为2026届高三数学高考一模阶段检测卷，重点考查阶段复习后的基础知识、关键能力与综合运用水平。2．作答前请将学校、班级、姓名、考号填写清楚；选择题答案填入答题栏，填空题答案写在相应横线上。3．解答题须写出必要的文字说明、证明过程或演算步骤；只写最后结果不得满分。4．全卷共22题，选择题10题30分，填空题6题18分，解答题6题102分，合计150分。选择题答题栏12345678910填空题答题栏111213141516一、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，只有一项是符合题目要求的。1．（3分）已知全集，集合，，则等于（）A．B．C．D．2．（3分）复数在复平面内对应的点位于（）A．第一象限B．第二象限C．第三象限D．第四象限3．（3分）已知，且，则的值为（）A．B．C．D．4．（3分）函数的图象性质正确的是（）A．在区间（-1，0）上单调递增B．在区间（0，+∞）上单调递减C．对任意x>-1，均有f(x)≥0D．对任意x>-1，均有f(x)≤05．（3分）等差数列的前项和为。若，，则等于（）A．25B．29C．33D．376．（3分）从5名男生和4名女生中任意选出3人参加一模试卷讲评小组，则选出的3人中至少有2名女生的概率为（）A．B．C．D．7．（3分）在直三棱柱中，，，则点到平面的距离为（）A．B．C．D．8．（3分）已知函数在区间内有两个不同的极值点，则实数的取值范围是（）A．B．C．D．9．（3分）椭圆的左、右焦点分别为。若点在椭圆上，且，则三角形的面积为（）A．B．C．D．10．（3分）已知。若对任意，都有，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。请将正确答案写在相应横线上。11．（3分）方程的解为答：________________12．（3分）展开式中的常数项为答：________________13．（3分）已知向量满足，，，则的值为答：________________14．（3分）一个袋中有2个红球和3个蓝球，随机不放回地逐个取球，第一次取到红球恰好发生在第二次取球的概率为答：________________15．（3分）曲线在点处的切线方程为答：________________16．（3分）实数满足，则的最大值为答：________________三、解答题：本大题共6小题，每小题17分，共102分。解答应写出文字说明、证明过程或演算步骤。得分：__________评卷人：__________17．（17分）已知函数。

（1）化简，并求其最小正周期和最大值；

（2）求在上的零点；

（3）若，且，求。________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________得分：__________评卷人：__________18．（17分）为检测一模阶段复习效果，某校随机抽取20名高三学生的数学模拟成绩（满分150分），按分数段整理如下表。分数段[90,100)[100,110)[110,120)[120,130)[130,140]人数23564（1）用组中值估计这20名学生的平均分；

（2）用样本频率估计该校高三学生数学成绩不低于120分的概率；

（3）从“低于100分”和“130分及以上”的6名学生中任取2名作访谈，求至少有1名来自“130分及以上”这一组的概率。________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________得分：__________评卷人：__________19．（17分）如图形情境所述：四棱锥的底面为边长为2的正方形，平面，且。点为的中点，点为的中点。

（1）证明：平面；

（2）求平面与底面所成锐二面角的正弦值。________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________得分：__________评卷人：__________20．（17分）已知椭圆的焦距为2，且过点。

（1）求椭圆的标准方程；

（2）过点作椭圆的切线，求切线斜率；

（3）当切线斜率为正时，求切点坐标。________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________得分：__________评卷人：__________21．（17分）已知函数。

（1）求的单调区间与最大值；

（2）证明：对任意，都有，并指出等号成立条件；

（3）设均为正数，且，请利用第（2）问结论证明。________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________得分：__________评卷人：__________22．（17分）已知正项数列满足，。

（1）证明数列是等差数列；

（2）求数列的通项公式；

（3）设，求，并求使成立的最小正整数。________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

参考答案与解析评分标准总则项目判定要求客观题选择题每小题3分，选项唯一；多选、错选或未选均不得分。填空题每小题3分，答案形式等价且数值正确即可得分。计算步骤解答题以关键步骤、核心公式、运算结果共同计分；只写最终答案但无必要过程，不能获得对应过程分。证明表达证明题应写明使用的判定定理、向量关系或函数性质，逻辑跳步但结论正确时按评分要点酌情给分。参数讨论含参数题必须说明范围来源；若范围筛选缺失，虽有计算结果，也应扣除相应分类讨论分。作图与设元几何与解析几何题可用坐标法、向量法或综合法，设元、建系、方程建立只要自洽均可按同等标准评分。运算误差单处非关键运算错误导致后续结果偏差时，保留已完成的正确方法分；关键公式错误则影响后续关联步骤。等号条件涉及最值、不等式和概率结果时，需写出必要条件或事件解释；等号条件属于完整性评分内容。一、选择题答案与解析12345678910CCACCBBBCA1．答案C。由得，结合全集取整数，；又，所以。2．答案C。，故，对应点为，在第三象限。3．答案A。令，则。4．答案C。，所以函数在上递减，在上递增，最小值，故。5．答案C。等差数列中，所以。又，得，公差，从而。6．答案B。所求概率为。7．答案B。建立空间直角坐标系，令，平面的方程为，距离为。8．答案B。。要在内有两个不同极值点，需，故。9．答案C。椭圆中，由定义，得。又，得；代入椭圆得。面积为。10．答案A。当时，由得。函数在上的下确界为1，且处条件恒成立，故。二、填空题答案与解析111213141516-1+2√3-160√373/10y=3x-21011．由对数性质得，即。结合，解为。12．通项为，令得，常数项为。13．由，得，故答案为。14．第一次取蓝球、第二次取红球，概率为。15．函数导数，在处斜率为3，切线为，即。16．可行域顶点为，代入得最大值为，在处取得。三、解答题答案、解析与评分标准17．答案与解析（1），所以最小正周期，最大值为。（2）令，即。当时，，故，零点为。（3）设，则。由得，即。又，故，，所以。评分要点分值正确化简为sin2x+cos2x，并写出周期、最大值5分求出[0,π]上两个零点5分建立t=2α并确定唯一角4分求出sin2α的准确值3分评分细则补充：第（1）问若只写出sin2x+cos2x而未化为√2sin(2x+π/4)，可给化简部分分；周期与最大值须同时写明。第（2）问需说明2x的取值范围，再列出两个角；若只写一个零点，则本问不得满分。第（3）问关键在于利用α的范围确定唯一角，若出现两个候选角但未筛选，应扣除范围判断分。三角恒等变形中允许使用等价表达，如√2cos(2x-π/4)，只要周期、最大值、零点与后续结论一致即可。过程判分边界：写出f(x)=√2sin(2x+π/4)后，若周期误写为2π，应只扣周期分，不影响后续零点的独立评分。若第（2）问采用单位圆或诱导公式求零点，只要两个零点均在[0,π]内且无多余根，可按正确结论给分。第（3）问若先求出2α=7π/12，再用和角公式求sin(7π/12)，计算过程完整即可给满分。答案中角度统一使用弧度制，若混用角度制导致结果不一致，应按对应步骤扣分。本题评分强调范围筛选，所有由三角方程产生的候选角都应回到原定义域核验。阅卷时应区分“表达式化简错误”和“由错误表达式继续计算”的连带影响，后续若逻辑自洽，可按步骤酌情给分。零点答案若写成集合形式{3π/8,7π/8}或分行列出，均视为等价表达。若三角函数值化简到sin75°，但未继续写成根式，可保留角值分，结果准确性分不满。18．答案与解析（1）用各组组中值估计平均分：。（2）不低于120分的人数为，样本频率为，故估计概率为。（3）低于100分有2人，130分及以上有4人。任选2人的总方法数为，没有来自130分及以上的情况为两人均来自低于100分，方法数，所以所求概率为。评分要点分值准确使用组中值并列出平均分计算式6分求出不低于120分的频率并解释为估计概率4分列出总方法数与对立事件方法数4分求得概率14/153分评分细则补充：第（1）问平均分属于分组数据估计，必须使用各组组中值乘以对应频数后再除以样本容量。第（2）问应明确“以样本频率估计总体概率”的含义，只写人数10人不等同于完整结论。第（3）问可直接列举、可用对立事件，也可用分类求和；总样本空间必须为从6人中任取2人的所有情况。若将“至少1名来自130分及以上”误算为“恰好1名”，只能给出部分分类计数分。过程判分边界：平均分计算式中的分子可以不完全展开，但组中值与频数的对应关系必须正确。样本容量为20，若误以组数5作为分母，平均分估计与概率估计均不得相应结果分。第（3）问若使用正向分类，可列为一人来自130分及以上和两人都来自130分及以上两类。访谈抽取不区分先后顺序，因此使用组合数更直接；使用排列数时若分子分母同时一致，也可给分。概率结果可以写成14/15或等值小数，但小数近似需能看出准确来源。统计题书写应保留必要的样本容量说明，本题所有概率分母均来自清楚的总体或样本空间。第（2）问中的概率为估计值，若表述为“样本中比例为1/2”，可视为已经体现估计意义。第（3）问若把两类学生人数看错为2和6，应先扣人数读取分，再按后续计数情况给分。19．答案与解析建立空间直角坐标系：取，则。由中点得。（1），平面的方程为。向量的x分量为0，且点不在平面内，故平面。（2）取，，则平面的一个法向量为。底面的法向量可取。设锐二面角为，则，故。评分要点分值建立合适坐标系并写出关键点坐标4分求出MN方向向量并说明与平面PAD平行5分求出平面MND的法向量4分用法向量求出锐二面角正弦值4分评分细则补充：第（1）问采用坐标法时，应先说明坐标系建立方式并给出M、N坐标，再由方向向量与平面方向关系作出结论。若不用坐标法，也可通过构造平面内平行向量完成线面平行证明，但必须符合线面平行判定定理。第（2）问若求的是二面角余弦值，应再转化为题目要求的正弦值；只写cosθ