2026届河北省高三数学高考一模模拟试卷（含答案详解与评分标准）

—第页—2026届河北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分注意事项：1．本试卷依据高三数学高考一模考前检测要求命制，重点考查基础知识、核心方法、综合运用与规范表达。2．答题前填写学校、班级、姓名和考号；选择题用规定方式作答，填空题只写最终结果，解答题须写出必要的文字说明、演算步骤或证明过程。3．全卷共22题，满分150分；选择题10题共30分，填空题6题共18分，解答题6题共102分。题型选择题填空题解答题合计题量106622分值3018102150客观题答题栏题号12345678910答案填空题答题栏题号111213141516答案一、选择题：本题共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．设集合，，则A．B．C．D．2．若复数满足，则A．B．C．D．3．已知等差数列的首项，且前5项和，则A．B．C．D．4．某学习小组有4名男生、3名女生，从中不放回地随机抽取2人参加板演，则抽到的2人性别相同的概率为A．B．C．D．5．已知向量，，若与垂直，则的取值集合为A．B．C．D．6．若，且，则A．B．C．D．7．抛物线的焦点为，点在该抛物线上且横坐标为4，则A．B．C．D．8．二项式展开式中的系数为A．B．C．D．9．某次阶段复习测评中，变量与的四组数据为，，，。以最小二乘法求线性回归直线，则斜率A．B．C．D．10．函数有三个不同的零点，则实数的取值范围是A．B．C．D．二、填空题：本题共6小题，每小题3分，共18分。请把答案填写在题中横线上。11．若，则__________。12．直线与圆相切，则实数的较大值为__________。13．在中，，，，则边__________。14．函数（）的最小值为__________。15．等比数列满足，公比，若，则__________。16．双曲线的离心率为，且经过点，则__________。三、解答题：本题共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）求的最小正周期和最大值；（2）求方程在区间内的解。学生作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18．（16分）某校高三数学一模复习中，从10名完成“函数与导数专项突破”的学生中随机抽取3名进行讲题展示。已知这10名学生中女生4名、男生6名，设被抽到的女生人数为随机变量。（1）求事件“恰有1名女生被抽到”的概率；（2）写出的分布列，并求。学生作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19．（16分）如图形条件所述，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弦值。学生作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20．（18分）已知椭圆（）的离心率为，且过点。（1）求椭圆的方程；（2）直线与椭圆交于、两点，其中。若线段的中点横坐标为，求及。学生作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21．（18分）已知函数，定义域为。（1）当时，求的单调区间与最小值；（2）求实数的取值范围，使得对任意恒成立；（3）当时，写出由本题得到的一个对数不等式，并说明等号成立条件。学生作答区：_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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参考答案与解析一、选择题答案与关键理由题号12345678910答案BBCBBBCCBB1．由得，又，故交集为，选B。2．，所以，选B。3．，代入得，故，，选C。4．所求概率为，选B。5．，。由垂直得，故或，选B。6．，选B。7．抛物线中，焦点为。抛物线上点到焦点距离等于到准线距离，故，选C。8．的系数为，选C。9．，，，，故，选B。10．，极大值点，极小值点。三不同零点需且，即且，故，选B。二、填空题答案与解析题号111213141516答案11．，故，。12．圆心为，半径为2。相切条件为，即，较大值为。13．由余弦定理，，故。14．当时，，当且仅当时等号成立。15．，故，。16．，故，即，从而。点代入得，故。三、解答题答案详解与评分标准17．（14分）解：。（1）因为角变量为，所以最小正周期为；最大值为。（2）由得，即。当时，，故或，于是或。评分标准：化简为得3分；写成辅助角形式得3分；周期与最大值各2分；建立方程并限定角范围得2分；求出两个解得2分。细化评分：若只写出周期但未说明由决定周期，可给1分；若辅助角写成等等价形式，结果正确同样给分。规范要求：第（2）问必须结合限定角的范围，不能只写通解后直接取值；最后两个解均需写全。18．（16分）解：从10名学生中任取3名，共有种等可能结果。（1）恰有1名女生被抽到的结果数为，故概率。（2）的可能取值为。。评分标准：列出总数得2分；第（1）问计数正确3分，概率正确2分；写出的取值2分；四个概率每个1.5分，共6分；数学期望正确3分。细化评分：分布列中的四个概率应满足和为1，若某一概率计算错误但取值范围正确，可按对应概率分扣分，其余部分按步骤给分。规范要求：本题属于一模概率统计基础题，答题时应先说明等可能基本事件总数，再使用组合数列式，不能只给最终小数。19．（16分）解：（1）因为是正方形，所以。又平面，而平面，故。直线与相交于点，且均在平面内，所以平面。（2）以为原点，分别以所在直线为轴建立空间直角坐标系，则，，，。点为的中点，故。于是。其在平面上的投影向量为，竖直分量长度为1，。直线与平面所成角满足。评分标准：指出得3分；利用底面推出得3分；根据线面垂直判定完成证明得2分；建立坐标系并写出关键点坐标得3分；求出与得2分；正确求出正弦值得3分。细化评分：第（1）问的关键是找到平面内两条相交直线与，并分别证明它们都与垂直；只证明其中一条垂直不能推出线面垂直。规范要求：第（2）问若不用坐标法，也可作出在底面上的射影并利用直角三角形求线面角；只要线面角定义、投影长度和结果正确，均按相应步骤给分。20．（18分）解：（1）由离心率，得。又。点在椭圆上，故。由得，所以，。椭圆方程为。（2）将代入椭圆方程，得，整理为。一个交点为，另一个交点的横坐标。由中点横坐标为，得，所以。于是，即，所以，故。此时，所以。评分标准：由离心率得到得4分；代入点求出得4分；椭圆方程正确得2分；联立直线与椭圆并得到关于的方程得3分；利用中点横坐标求出得3分；求出弦长得2分。细化评分：第（1）问若先由得，与本解析等价；方程中的必须同时求出。规范要求：第（2）问联立后得到的二次方程含有已知根，另一根可由根与系数关系得到。使用中点条件时要注意另一点横坐标为，不能误写为。弦长计