2026届安徽省高三数学二模模拟原创模拟试卷第004套（含答案详解与评分标准）

2026届安徽省高三数学二模模拟原创模拟试卷第004套（含答案详解与评分标准）考试时间：120分钟满分：150分题号范围：1—23题姓名：________________班级：________________准考证号：________________注意事项：1．本卷为2025—2026学年高三数学二模模拟训练卷，试题范围覆盖函数与导数、三角与向量、数列、立体几何、解析几何、概率统计等核心考点。2．选择题作答时，请将唯一正确选项或全部正确选项填入答题栏；多项选择题全部选对得满分，部分选对得相应分，有错选不得分。3．填空题只写最终结果；解答题应写出必要的文字说明、证明过程或演算步骤。4．本卷试题后另起新页附“参考答案与解析”，教师讲评或学生自评时可独立使用。试卷结构与分值题型题号每题分值小计单项选择题1—85分40分多项选择题9—125分20分填空题13—165分20分解答题17—2310/12/14分70分合计1—23150分选择题与填空题答题栏12345678910111213141516一、单项选择题：本大题共8小题，每小题5分，共40分。每小题只有一个选项符合题意。1.（5分）已知集合A={x∈R｜x²−3x≤0}，B={x∈R｜log₂(x+1)>1}，则A∩B=（）A.[0，1]B.(1，+∞)C.(1，3]D.[0，3]2.（5分）复数z=(1+i)²/(1−i)，则z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.（5分）数列{aₙ}满足a₁=2，aₙ₊₁=3aₙ−2，则a₅=（）A.41B.80C.82D.1224.（5分）函数f(x)=lnx−x+2在区间(0，+∞)内的零点个数为（）A.1B.2C.3D.05.（5分）已知向量a=(1，2)，b=(m，−1)，若(a+b)⊥(a−2b)，则m的值为（）A.(−1+√41)/4或(−1−√41)/4B.(1+√41)/4或(1−√41)/4C.2或−5/2D.−2或5/26.（5分）在区间[0，π]内，方程sin(2x+π/6)=1/2的解的个数为（）A.0B.1C.2D.37.（5分）一个袋中有5个红球、3个白球，除颜色外完全相同，从中不放回任取2个球，则取出的2个球颜色相同的概率为（）A.5/14B.3/8C.13/28D.15/288.（5分）一个底面半径为3、高为4的直圆锥内切球半径为（）A.1B.3/2C.2D.12/5二、多项选择题：本大题共4小题，每小题5分，共20分。每小题有多个选项符合题意，全部选对得5分，部分选对得2分，有错选得0分。9.（5分）已知二次函数f(x)=x²−4x+3，下列结论正确的是（）A.图象的顶点坐标为(2，−1)B.f(x)在(−∞，2)上单调递增C.方程f(x)=0的两个根为1和3D.不等式f(x)≥0的解集为(−∞，1]∪[3，+∞)10.（5分）某小组5次数学限时训练成绩为8，10，10，12，15（单位：分），下列说法正确的是（）A.中位数为10B.平均数为11C.方差为28/5D.极差为611.（5分）函数f(x)=eˣ+e⁻ˣ，下列结论正确的是（）A.f(x)为偶函数B.f(x)的最小值为2C.f(x)在(0，+∞)上单调递增D.对任意x，f(x+1)=f(x)+f(1)12.（5分）圆C：x²+y²=4，点P(3，0)，从P向圆C作两条切线，切点分别为T₁，T₂，下列结论正确的是（）A.PT₁=√5B.切点的横坐标均为4/3C.切点的纵坐标为±2√5/3D.两条切线所成角的余弦值为3/5三、填空题：本大题共4小题，每小题5分，共20分。13.（5分）(x²−2/x)⁶的展开式中的常数项为__________．14.（5分）椭圆x²/9+y²/4=1的离心率为__________．15.（5分）已知等比数列{aₙ}的公比q>0，a₂=6，a₅=48，则其前6项和S₆=__________．16.（5分）抛物线y²=4x在点A(1，2)处的切线与焦点F的距离为__________．四、解答题：本大题共7小题，共70分。解答应写出必要的文字说明、证明过程或演算步骤。17.（10分）在△ABC中，角A，B，C所对的边分别为a，b，c，已知cosA=3/5，b=6，c=4．

（1）求边a的长；

（2）求△ABC的面积．答：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18.（10分）已知数列{aₙ}满足a₁=2，aₙ₊₁=aₙ+2n+2（n∈N*）．

（1）求数列{aₙ}的通项公式；

（2）设Tₘ=1/a₁+1/a₂+…+1/aₘ，求Tₘ．答：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19.（10分）某校高三年级围绕“二模后数学错题整理方式”进行问卷，随机调查100名学生，得到如下列联表：

类别赞成规范整理不赞成规范整理合计男生282250女生321850合计6040100（1）从这100名学生中随机选1名，求该学生为女生且赞成规范整理的概率；

（2）从赞成规范整理的学生中随机选1名，求该学生为男生的概率；

（3）用样本频率估计总体概率，从该校高三学生中随机抽取5名，求其中至少4名赞成规范整理的概率．答：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20.（12分）正方体ABCD-A₁B₁C₁D₁的棱长为2，F为BC的中点．

（1）证明：A₁C⊥BD；

（2）求直线A₁F与平面BDD₁B₁所成角的大小．答：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21.（14分）已知椭圆E：x²/a²+y²/b²=1（a>b>0）过点P(2，1)，且离心率为1/2．

（1）求椭圆E的方程；

（2）设椭圆E的右焦点为F，直线l：y=t（−2<t<2）与椭圆E交于M，N两点，求△FMN面积的最大值．答：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22.（14分）已知函数f(x)=lnx−ax+1（x>0，a∈R）．

（1）当a=1时，求f(x)的单调区间和最大值；

（2）讨论方程f(x)=0在(0，+∞)内实根的个数．答：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________23.（14分）已知数列{aₙ}满足a₁=1/3，aₙ₊₁=aₙ/(1+2aₙ)（n∈N*）．

（1）求数列{aₙ}的通项公式；

（2）设Pₙ=(1+2a₁)(1+2a₂)…(1+2aₙ)，求Pₙ；

（3）求证：对任意n∈N*，a₁a₂+a₂a₃+…+aₙaₙ₊₁<1/6．答：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

参考答案与解析评分总说明：本答案用于教师阅卷、课堂讲评和学生自评。选择题按题型规则给分，填空题以最终结果为主，解答题按关键步骤累计给分。学生采用不同方法，只要推理严密、结论正确，应按等价步骤给分；若出现符号、计算或化简小错，但核心思路正确，可保留相应方法分。书写中缺少必要论证、未说明定义域或遗漏分类讨论的，应在对应步骤内扣分。解答题评分补充：证明题应先给出所用定义、定理或坐标设定，再完成推导；计算题应保留关键代入式、化简式和最终结论；概率统计题应说明样本空间、事件含义或随机变量模型；解析几何题应写清参数范围，避免只给最终数值。若答案中出现同一处错误引起后续连锁失分，阅卷时以首次错误为主要扣分点，后续步骤在不改变题意的情况下可按思路给分。客观题讲评补充：单项选择题应重点回看定义域、端点、单调性、数量积、组合计数等易错点；多项选择题应逐项验证，不宜只凭特殊值判断；填空题要检查结果是否需要化简、是否遗漏正负号或参数范围。教师讲评时可先让学生说明错因，再对照解析补全关键步骤，从而形成二模后针对性复习记录。复核提示：全卷分值合计150分，解答题合计70分。答案页中每道题号与试题正文题号一一对应，阅卷时可按“先看结论、再看过程、最后核算分值”的顺序处理。对含参数、含绝对值或含区间限制的问题，应特别检查分类是否完整；对概率、统计与几何问题，应检查模型假设与题干条件是否一致。对压轴题可先给通性方法分，再依据运算质量和表达完整度细分。所有结论均应回扣题设，避免因漏写范围造成失分。若学生给出更简洁的等价解法，阅卷时以数学正确性和表达清楚为准，保持标准统一、规范、严谨、公平。一、客观题答案速查12345678910111213141516CBCBADCBACDABCABCABC240√5/3189√2二、逐题解析与评分标准1.（5分）由x²−3x≤0得0≤x≤3，即A=[0，3]；由log₂(x+1)>1得x+1>2，即x>1，所以A∩B=(1，3]．故选C．

评分提示：本题考查一元二次不等式解集与对数不等式的交集运算，答对选项得5分；若把端点1误取入，说明未注意严格不等号，应扣除本题分数。2.（5分）(1+i)²=2i，z=2i/(1−i)=2i(1+i)/2=−1+i．对应点为(−1，1)，在第二象限．故选B．

评分提示：本题考查复数代数运算与几何意义，答对选项得5分；关键在于分母实数化后判断实部、虚部符号。3.（5分）设bₙ=aₙ−1，则bₙ₊₁=3bₙ，b₁=1，所以bₙ=3ⁿ⁻¹，aₙ=3ⁿ⁻¹+1，a₅=3⁴+1=82．故选C．

评分提示：本题考查线性递推数列的转化，答对选项得5分；讲评时应强调构造bₙ=aₙ−1后成为等比数列。4.（5分）f'(x)=1/x−1，当0<x<1时f'(x)>0，当x>1时f'(x)<0，f(x)在x=1处取最大值f(1)=1．又x→0⁺时f(x)→−∞，x→+∞时f(x)→−∞，故图象与x轴有两个交点．故选B．

评分提示：本题考查导数判断函数零点个数，答对选项得5分；只判断端点极限而不找极值，容易漏判两个零点。5.（5分）a+b=(1+m，1)，a−2b=(1−2m，4)．由垂直得(1+m)(1−2m)+4=0，即2m²+m−5=0，解得m=(−1±√41)/4．故选A．

评分提示：本题考查平面向量数量积与一元二次方程，答对选项得5分；若只写一个根，视为未完整求解。6.（5分）由sin(2x+π/6)=1/2，得2x+π/6=π/6+2kπ或5π/6+2kπ，于是x=kπ或x=π/3+kπ．在[0，π]内有x=0，π/3，π，共3个解．故选D．

评分提示：本题考查三角方程在给定区间内的解，答对选项得5分；应将一般解代回区间筛选，端点x=0与x=π均要保留。7.（5分）所求概率为[C₅²+C₃²]/C₈²=(10+3)/28=13/28．故选C．

评分提示：本题考查古典概型与组合计数，答对选项得5分；事件“颜色相同”应分红红、白白两类相加。8.（5分）直圆锥母线长l=5，轴截面为底边6、高4的等腰三角形，内切球半径等于轴截面内切圆半径r=面积/半周长=(1/2×6×4)/(5+5+6)/2=12/8=3/2．故选B．

评分提示：本题考查圆锥轴截面与内切圆半径，答对选项得5分；也可使用r=Rh/(l+R)直接求解。9.（5分）f(x)=x²−4x+3=(x−2)²−1，顶点为(2，−1)，在(−∞，2)上单调递减，零点为1和3，f(x)≥0的解集为(−∞，1]∪[3，+∞)．故选ACD．

评分提示：本题为多项选择，全部选对得5分，部分选对得2分，有错选得0分；B项错在二次函数开口向上时对称轴左侧为递减。10.（5分）数据排序后仍为8，10，10，12，15，中位数为10；平均数为(8+10+10+12+15)/5=11；方差为[(−3)²+(−1)²+(−1)²+1²+4²]/5=28/5；极差为15−8=7．故选ABC．

评分提示：本题为多项选择，全部选对得5分，部分选对得2分，有错选得0分；方差按总体方差公式计算，极差为最大值与最小值之差。11.（5分）f(−x)=e⁻ˣ+eˣ=f(x)，所以为偶函数；由eˣ+e⁻ˣ≥2知最小值为2；f'(x)=eˣ−e⁻ˣ，当x>0时f'(x)>0；D式不满足指数函数运算规律．故选ABC．

评分提示：本题为多项选择，全部选对得5分，部分选对得2分，有错选得0分；D项可用x=0代入快速排除。12.（5分）OP=3，圆半径为2，切线长PT=√(OP²−r²)=√5；切点T(x，y)满足OT⊥PT，即(x，y)·(x−3，y)=0，结合x²+y²=4，得4−3x=0，x=4/3，y=±2√5/3．两条切线夹角余弦值为1/9，不是3/5．故选ABC．

评分提示：本题为多项选择，全部选对得5分，部分选对得2分，有错选得0分；切点坐标可由半径垂直切线建立数量积方程。13.（5分）通项为C₆ᵏ(x²)⁶⁻ᵏ(−2/x)ᵏ=C₆ᵏ(−2)ᵏx¹²⁻³ᵏ．令12−3k=0，得k=4，常数项为C₆⁴(−2)⁴=15×16=240．

评分提示：结果240正确得5分；若列出通项但指数方程求错，可得2分；若只写组合数而未算出常数项，可得3分。14.（5分）椭圆中a=3，b=2，c=√(a²−b²)=√5，离心率e=c/a=√5/3．

评分提示：结果√5/3正确得5分；若求出c=√5但离心率写错，可得2分；需注意椭圆长半轴a=3。15.（5分）由a₅/a₂=q³=48/6=8，又q>0，得q=2，a₁=a₂/q=3，S₆=a₁(q⁶−1)/(q−1)=3(64−1)=189．

评分提示：结果189正确得5分；求出q=2得2分，求出a₁=3得1分，前6项和计算正确再得2分。16.（5分）抛物线y²=4x的焦点F(1，0)，点A(1，2)处切线为yy₁=2p(x+x₁)，即2y=2(x+1)，得y=x+1．点F到直线x−y+1=0的距离为|1−0+1|/√2=√2．

评分提示：结果√2正确得5分；写出焦点F