1780701468924-2026届山东省高三数学高考一模模拟试卷（含答案详解与评分标准）第055套

2026届山东省通用高三数学高考一模模拟试卷（含答案详解与评分标准）第055套学校：________________班级：________________姓名：________________考号：________________考试时间：120分钟满分：150分试卷类型：高考一模适用地区：山东省通用注意事项：1．答题前，考生务必将学校、班级、姓名、考号填写清楚。2．本试卷共22题，满分150分，考试时间120分钟。选择题、填空题答案按要求填涂或填写，解答题须写出必要的文字说明、证明过程或演算步骤。3．允许使用黑色签字笔、2B铅笔、直尺、圆规等常规考试工具；作图题应保留必要辅助线。4．本卷体现高考一模阶段综合检测要求，重点考查函数与导数、三角与数列、立体几何、解析几何、概率统计及数学建模等主干内容。考生答题记录栏：本栏便于核对，正式填涂或书写以答题卡为准。题号123456789101112选项题号13141516答案一、选择题（本大题共12小题，每小题5分，共60分。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A∩B＝（）A．[1,3]B．(2,3]C．[2,3]D．[3,+∞)2．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．已知则tanα＝（）A．1B．√3/3C．√3D．34．等差数列{a_n}的公差d>0，且则a10＝（）A．17B．18C．19D．215．函数在x=0处切线斜率为0，则f(ln2)＝（）A．2-ln2B．2+ln2C．1-ln2D．ln2-16．盒A中有3个红球、2个白球，盒B中有2个红球、3个白球。先从盒A中任取1球放入盒B，再从盒B中任取1球，则最后取到红球的概率为（）A．7/15B．13/30C．1/2D．8/157．平面向量a，b满足则|2a-b|＝（）A．√13B．5C．√37D．78．函数的最大值为（）A．ln2-1B．ln2C．1-ln2D．09．正四棱锥的底面边长为2，高为√2，则侧棱与底面所成角为（）A．30°B．45°C．60°D．75°10．一组数据为6，8，9，10，12，则这组数据的方差为（）A．2B．3C．4D．511．抛物线的焦点为F，过F且斜率为1的直线交抛物线于M，N两点，则弦MN的长为（）A．4B．4√2C．8D．8√212．若不等式对一切x>0恒成立，则实数k的值为（）A．0B．1/2C．1D．2二、填空题（本大题共4小题，每小题5分，共20分。请把答案填写在题中横线上。）13．二项式展开式中x²项的系数为__________。14．若直线与圆(x-1)²+(y+2)²=5相切，则m的值为__________。15．已知tanα=2，tanβ=3，且α，β均为锐角，则tan(α+β)=__________。16．设函数若f(x)在区间[-2,a]上的最大值为2，则实数a的取值范围为__________。三、解答题（本大题共6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．（本小题10分）在△ABC中，角A，B，C所对的边分别为a，b，c。已知A=60°，b=2，c=√3+1。（1）求边a的长；（2）求△ABC的面积；（3）若AD为∠A的角平分线，D在BC上，求BD∶DC。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18．（本小题12分）数列{a_n}满足a1=1，且对任意n∈N*，有（1）设b_n=a_n/2^{n-1}，证明{b_n}为等差数列；（2）求数列{a_n}的通项公式；（3）求前n项和T_n。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19．（本小题12分）某校高三年级为了解一模前数学限时训练情况，从全年级随机抽取50名学生的最近一次训练成绩，按分数段整理如下表（满分100分）：分数段[60,70)[70,80)[80,90)[90,100]人数6142010（1）用各组中点估计这50名学生的平均成绩；（2）估计该校学生成绩不低于90分的概率；（3）已知抽取的10名优秀学生中女生6人、男生4人，现从这10人中任选3人参加经验分享，求恰有2名女生的概率，并写出被选女生人数X的分布列与数学期望。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20．（本小题12分）在四棱锥P-ABCD中，底面ABCD为边长2的正方形，O为底面中心，PO⊥平面ABCD，PO=2，E为PC的中点。（1）证明：BD⊥平面PAC；（2）求直线BE与平面ABCD所成角的正弦值；（3）求点B到平面PAC的距离。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21．（本小题12分）已知椭圆C的中心在原点，焦点在x轴上，离心率为√3/2，且点M(1，√3/2)在椭圆C上。（1）求椭圆C的标准方程；（2）设直线l：y=kx+1（k≠0）与椭圆C交于点P(0，1)和另一点Q，O为坐标原点，求△OPQ面积的最大值。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________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参考答案、详解与评分标准评分总说明：选择题每小题5分，选对得5分，错选、多选或不选均不得分；填空题每小题5分，结果等价可得满分，含多个答案的题目须全部写全；解答题按步骤给分，关键结论正确但缺少必要说明时酌情扣分。解答题评分坚持“过程与结果并重”。方法正确、运算偶有小误但未影响后续主线的，可在相应评分点内扣1分；若因前一步计算错误导致后续结果连带错误，但思路合理，应保留后续可迁移的步骤分。涉及证明的题目，须写出判定依据、相交直线或垂直关系等必要条件；涉及概率统计的题目，须写明样本空间、组合数或分布列来源；涉及导数的题目，须明确单调性、极值与端点或极限的关系。一、选择题答案题号123456789101112答案CCBDABCABCCC1．由x²-5x+6≤0得2≤x≤3；由ln(x-1)≥0得x-1≥1，即x≥2，故交集为[2,3]。2．化简z=(1-2i)/(1+i)=(-1-3i)/2，实部、虚部均小于0，对应点在第三象限。3．α+π/6∈(π/6,2π/3)，满足正弦为√3/2的只有π/3，故α=π/6，tanα=√3/3。4．a1+a5=2a3=14，得a3=7；a2a4=(7-d)(7+d)=45，得d²=4，又d>0，故d=2，a10=a3+7d=21。5．f′(x)=e^x-a，x=0处斜率为0，得a=1，故f(ln2)=2-ln2。6．若从A取红球，概率3/5，此时B中红球概率1/2；若从A取白球，概率2/5，此时B中红球概率1/3，总概率为3/5×1/2+2/5×1/3=13/30。7．a·b=|a||b|cos120°=-3，|2a-b|²=4|a|²+|b|²-4a·b=16+9+12=37。8．f′(x)=1/x-1/2，x=2时取极大值，也是最大值，最大值为ln2-1。9．底面中心到顶点的距离为√2，侧棱长为√((√2)²+(√2)²)=2，所成角θ满足sinθ=√2/2，故θ=45°。10．平均数为9，方差为[(6-9)²+(8-9)²+0²+(10-9)²+(12-9)²]/5=4。11．抛物线焦点F(1,0)，直线为y=x-1。代入y²=4x得x²-6x+1=0，两个根差为4√2，弦长为√2×4√2=8。12．由lnx≤x-1对一切x>0成立可知k=1可行。若x>1，需k≥lnx/(x-1)，极限趋近1；若0<x<1，除以负数后需k≤lnx/(x-1)，极限也趋近1，故只能k=1。客观题复核提示：第1题注意对数定义域与不等式方向；第6题是典型的分步条件概率问题，不可直接用盒B原始红球比例；第12题需同时考察x>1与0<x<1两侧极限，只有k=1能兼顾两侧要求。二、填空题答案题号13141516答案601或-9-1[-1,2]13．通项为C(6,k)x^{6-k}(-2/x)^k=C(6,k)(-2)^kx^{6-2k}。令6-2k=2，得k=2，系数为C(6,2)×4=60。14．圆心为(1,-2)，半径为√5。相切条件为|2×1-(-2)+m|/√5=√5，即|m+4|=5，所以m=1或m=-9。15．tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)=5/(1-6)=-1。16．f′(x)=3x²-3，f(x)在x=-1处取局部最大值2，在x=1处取局部最小值-2，且f(-2)=-2，f(2)=2。区间[-2,a]上的最大值为2，必须且只需包含x=-1且