2026届南京市高三数学高考一模模拟试卷（含答案详解与评分标准）

2026届南京市高三数学高考一模模拟试卷（含答案详解与评分标准）数学学校：________________班级：____________姓名：____________考号：____________考试时间：120分钟满分：150分考试节点：高考一模适用对象：2026届高三注意事项1．本试卷按高考一模检测要求命制，重点考查一轮复习后的基础运算、核心概念、综合建模与规范表达能力。2．选择题每小题只有一个选项符合题意；填空题只写最后结果；解答题须写出必要的文字说明、演算过程或证明步骤。3．全卷共22题，考试时间120分钟，满分150分。答题时请保持书写清楚，步骤完整，结果化简。4．本卷所用常数、符号与高中数学通常约定一致；未特殊说明时，所有角均以弧度计。一、选择题（本大题共10小题，每小题3分，共30分）1．已知集合，，则为（）A．B．C．D．2．复数的虚部为（）A．B．C．D．3．函数在点处的切线方程为（）A．B．C．D．4．一组数据的平均数为12，方差为4。若，则数据的方差为（）A．4B．8C．16D．215．方程在区间内的解的个数为（）A．1B．2C．3D．46．函数的最小值为（）A．4B．5C．D．87．已知向量满足，，且与的夹角为，则等于（）A．5B．C．D．78．已知等比数列各项均为正，公比，且，，则等于（）A．B．1C．9D．279．一个不透明袋中有3个红球和2个蓝球，从中不放回地任取2个球。设随机变量表示取到红球的个数，则为（）A．B．1C．D．10．抛物线的焦点为，直线与抛物线交于两点，则等于（）A．4B．C．6D．8选择题答题栏题号12345678910答案二、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常数项为________。12．圆上过点所作切线的切线长为________。13．5名同学排成一排，其中甲、乙不相邻的排法共有________种。14．已知函数在区间上单调递增，则实数的最大值为________。15．椭圆的离心率为________。16．圆锥的底面半径为3，高为4，则该圆锥的侧面积为________。三、解答题（本大题共6小题，共102分）17．（本小题满分15分）在中，角的对边分别为。已知，，。（1）求的值；（2）求的面积。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18．（本小题满分17分）已知数列满足，。（1）设，证明是等比数列；（2）求与前项和；（3）求使成立的最小正整数。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19．（本小题满分17分）某高三年级在一模前进行数学专项训练。袋中有4张“函数”卡、3张“几何”卡、2张“概率”卡，卡片除类别外完全相同。从中不放回地随机抽取3张，设随机变量表示抽到“函数”卡的张数。（1）求的分布列；（2）求与；（3）若抽到不少于2张“函数”卡，则安排学生参加函数专题加练，求该安排发生的概率，并说明其计算依据。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（每小题3分，共30分）题号12345678910答案DBCCBACDCC1．答案D。由得；由得，即。两集合取交集为。2．答案B。，所以，虚部为。3．答案C。，，故，切线方程为。4．答案C。线性变换不改变离散程度的平移部分，方差乘以系数平方，即新方差为。5．答案B。若，原式不为0；故可除以，得，即。在内分别对应一个解，共2个解。6．答案A。由均值不等式，，当时取等号。7．答案C。，所以。8．答案D。由，，得，即。因，得，故。9．答案C。不放回抽取2个球，红球个数的期望可按超几何分布公式计算：。10．答案C。抛物线焦点为，直线与抛物线交于、。两点到焦点的距离均为3，故和为6。评分标准：选择题每题3分，选对得3分，未选、错选或多选得0分。二、填空题答案与解析（每小题3分，共18分）11．答案：。通项为。令，得，常数项为。12．答案：4。圆化为，圆心，半径。，切线长为。13．答案：72。5名同学全排列有种；甲、乙相邻时可将二人看作一个整体，有种；不相邻排法为种。14．答案：。。函数在上单调递增需恒成立，即对所有成立。右端最小值为，故的最大值为。15．答案：。椭圆中，，故，，离心率。16．答案：。母线长，圆锥侧面积。评分标准：填空题每题3分，只写结果且结果正确得3分；结果形式等价、化简正确可得满分；有多余错误结果不得分。三、解答题参考答案、解析与评分标准（共102分）解答题评分以关键数学关系、运算过程和结论完整性为核心。若考生采用不同方法，只要逻辑严密、计算正确、结论等价，可按相应步骤给分。涉及证明的问题，必须写明依据与推理链条；只给出结论而缺少必要过程时，不得获得该步骤的全部分值。涉及参数、最值、概率分布和空间距离的问题，应关注定义域、取值范围、样本空间或几何位置关系。若前一步计算错误但后续方法正确，可根据后续过程的独立价值酌情给分。评分表中的分值为建议切分，阅卷时应坚持同题同标、等价给分、分步评价的原则。17．解析：由正弦定理，，设，。由余弦定理，。代入，，得，所以。故，。面积。评分要点分值利用正弦定理得到并合理设元3分正确使用余弦定理建立5分求出且化简正确3分使用面积公式并求得4分补充说明：本题的关键是把正弦比转化为边长比，再用余弦定理把未知比例系数确定下来。若学生直接令，但未说明，不影响主要得分；若面积公式中角选错，应扣除相应面积步骤分。等价解法：也可先由结合余弦定理求出边长，再用海伦公式核验面积；只要结果为且过程完整，应按满分处理。18．解析：由，得，且。因此是首项为4、公比为2的等比数列。于是，所以。前项和。检验，。由于随增大而增大，故使成立的最小正整数为。评分要点分值写出并说明等比数列性质5分求得4分正确求得5分比较并确定最小正整数3分补充说明：构造的目的在于消去递推式中的常数项。若学生写出，即已抓住核心变形，可给构造分。在求最小正整数时，不能只估算