1780727117043-2026届广东省高三数学高考二模模拟试卷（含答案详解与评分标准）第056套

2026届广东省通用高三数学高考二模模拟试卷（含答案详解与评分标准）第056套学校：________________班级：________________姓名：________________考号：________________考试时间：120分钟满分：150分试卷类型：闭卷适用地区：广东省通用注意事项：1．答题前，考生务必将学校、班级、姓名、考号填写清楚；请使用黑色字迹签字笔规范作答。2．选择题每小题只有一个正确选项；填空题只需填写最终结果；解答题应写出必要的推理、计算过程和结论。3．本卷突出高考二模阶段的综合检测，覆盖函数与导数、数列、立体几何、解析几何、概率统计、三角与向量等主干内容。4．保持卷面整洁，答案写在相应作答区域内；无过程或过程不完整的解答题按评分标准酌情给分。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．（5分）已知复数z=(1−2i)/(1+i)，则z的虚部为A．−3/2B．−1/2C．1/2D．3/22．（5分）设集合A={x|log₂(x+1)<2}，B={x|x²−5x+6≤0}，则A∩B=A．(−1,3]B．[2,3]C．[2,3)D．(−1,2]3．（5分）在(x−2/x)⁶的展开式中，常数项为A．−160B．160C．−120D．1204．（5分）已知向量a=(1,2)，b=(m,1)。若(a+b)·a=0，则m=A．−7B．−3C．1D．75．（5分）一个袋中有4个红球、2个蓝球，这些球除颜色外完全相同。从中不放回地任取2个球，恰有1个红球的概率为A．4/15B．7/15C．8/15D．2/36．（5分）函数f(x)=ln((2+x)/(2−x))的定义域为(−2,2)，则不等式f(x)>0的解集为A．(−2,0)B．(0,2)C．(−2,2)D．(1,2)7．（5分）方程sin(2x+π/6)=1/2在区间[0,π]内的解的个数为A．1B．2C．3D．48．（5分）一个长方体的三条棱长分别为1，2，2。若以该长方体所有顶点为球面上的点作外接球，则该球的表面积为A．6πB．9πC．12πD．18π9．（5分）曲线y=x³−3x²在点(1,−2)处的切线方程为A．y=3x−5B．y=−3x+1C．y=−3x−1D．y=3x+110．（5分）数列{aₙ}满足a₁=2，aₙ₊₁=3aₙ+1，则aₙ=A．(5·3ⁿ⁻¹−1)/2B．(3ⁿ−1)/2C．2·3ⁿ⁻¹+1D．5·3ⁿ⁻¹11．（5分）椭圆x²/a²+y²/b²=1（a>b>0）的离心率为1/2，且经过点(2,√3)，则a²=A．4B．6C．8D．1212．（5分）若函数f(x)=eˣ−ax在R上恰有两个零点，则实数a的取值范围是A．a≥eB．a>eC．0<a<eD．a≤1二、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．（5分）等差数列{aₙ}中，a₂=7，a₆=19，则前10项和S₁₀=__________。14．（5分）直线x+y=t与圆x²+y²=5相交所得弦长为4，则t²=__________。15．（5分）若α∈(π/2,π)，且sinα+cosα=√2/2，则sin2α=__________。16．（5分）函数f(x)=x³−3mx（m>0）的极大值为4，则m=__________。三、解答题：本大题共6小题，共70分。解答应写出文字说明、证明过程或演算步骤。17．（10分）在△ABC中，角A，B，C的对边分别为a，b，c，且sinC=2sinAcosB。

（1）证明：A=B；

（2）若c=4，△ABC的面积为4√3，求a，b的值。________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18．（12分）某校高三年级在二模前进行一次数学限时训练，随机抽取80名学生的成绩情况，按“是否达到阶段目标”和“性别”统计如下表。已知男生40人，女生40人，达到阶段目标的男生为22人，未达到阶段目标的女生为10人。类别达到阶段目标未达到阶段目标合计男生221840女生301040合计522880（1）完成上表，并估计该校高三学生数学限时训练达到阶段目标的比例；

（2）从达到阶段目标的学生中随机抽取2人，求抽到的2人均为女生的概率；

（3）依据列联表，判断在0.05的显著性水平下，能否认为“是否达到阶段目标”与“性别”有关？参考公式：K²=n(ad−bc)²/[(a+b)(c+d)(a+c)(b+d)]，临界值为3.841。________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19．（12分）已知正项数列{aₙ}满足a₁=2，aₙ₊₁=2aₙ/(aₙ+2)。

（1）证明数列{1/aₙ}为等差数列，并求aₙ；

（2）设Sₙ=a₁a₂+a₂a₃+⋯+aₙaₙ₊₁，求Sₙ；

（3）求满足Sₙ>3.9的最小正整数n。________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20．（12分）如图意描述：四棱锥P−ABCD的底面ABCD为矩形，AB=4，AD=3，PA⊥平面ABCD，PA=6。取空间直角坐标系，使A(0,0,0)，B(4,0,0)，D(0,3,0)，P(0,0,6)。

（1）证明：CD∥平面PAB；

（2）求点A到平面PBD的距离；

（3）求平面PBD与底面ABCD所成锐二面角的余弦值。________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21．（12分）已知抛物线C：y²=4x，直线l：x=my+4与C交于A，B两点，O为坐标原点。

（1）证明：OA⊥OB；

（2）若△OAB的面积为24，求m的值；

（3）设AB的中点为M，求点M的轨迹方程。________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22．（12分）设函数fₐ(x)=lnx−ax+1，x>0。

（1）讨论fₐ(x)的单调性与极值；

（2）求实数a的取值范围，使fₐ(x)≤0对一切x>0恒成立；

（3）当a=1时，证明方程f₁(x)=−1恰有两个正根，并分别指出两个根所在的区间。________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

参考答案与解析客观题答案表题号123456789101112答案ACAACBCBBACB分值555555555555题号13141516答案1752−1/22^(2/3)分值5555评分矩阵题号题型分值给分说明1选择题5选对得5分，错选、多选或不选得0分。2选择题5选对得5分，错选、多选或不选得0分。3选择题5选对得5分，错选、多选或不选得0分。4选择题5选对得5分，错选、多选或不选得0分。5选择题5选对得5分，错选、多选或不选得0分。6选择题5选对得5分，错选、多选或不选得0分。7选择题5选对得5分，错选、多选或不选得0分。8选择题5选对得5分，错选、多选或不选得0分。9选择题5选对得5分，错选、多选或不选得0分。10选择题5选对得5分，错选、多选或不选得0分。11选择题5选对得5分，错选、多选或不选得0分。12选择题5选对得5分，错选、多选或不选得0分。13填空题5结果正确得5分；形式等价可给满分。14填空题5结果正确得5分；形式等价可给满分。15填空题5结果正确得5分；形式等价可给满分。16填空题5结果正确得5分；形式等价可给满分。17解答题10按证明、计算与结论分步给分。18解答题12按建模、推理、计算与结论分步给分。19解答题12按建模、推理、计算与结论分步给分。20解答题12按建模、推理、计算与结论分步给分。21解答题12按建模、推理、计算与结论分步给分。22解答题12按建模、推理、计算与结论分步给分。一、选择题解析1．答案：A解析：z=(1−2i)/(1+i)=(1−2i)(1−i)/[(1+i)(1−i)]=(−1−3i)/2，因此虚部为−3/2。2．答案：C解析：由log₂(x+1)<2得0<x+1<4，即−1<x<3；由x²−5x+6≤0得2≤x≤3。两集合交集为[2,3)。3．答案：A解析：通项为C₆ᵏx⁶⁻ᵏ(−2/x)ᵏ=C₆ᵏ(−2)ᵏx⁶⁻²ᵏ。常数项需6−2k=0，得k=3，常数项C₆³(−2)³=20×(−8)=−160。4．答案：A解析：a+b=(m+1,3)，(a+b)·a=(m+1)×1+3×2=m+7。由题意m+7=0，故m=−7。5．答案：C解析：不放回任取2个球的总数为C₆²=15，恰有1红1蓝的取法为C₄¹C₂¹=8，概率为8/15。6．答案：B解析：在定义域(−2,2)内，2−x>0。f(x)>0等价于(2+x)/(2−x)>1，即2+x>2−x，解得x>0。7．答案：C解析：令2x+π/6=π/6+2kπ或2x+π/6=5π/6+2kπ。结合x∈[0,π]，可得x=0，π/3，π，共3个解。8．答案：B解析：长方体外接球直径为体对角线√(1²+2²+2²)=3，所以半径r=3/2，球表面积为4πr²=9π。9．答案：B解析：y′=3x²−6x，x=1时斜率为−3。过点(1,−2)的切线为y+2=−3(x−1)，即y=−3x+1。10．答案：A解析：由aₙ₊₁=3aₙ+1，两边加1/2得aₙ₊₁+1/2=3(aₙ+1/2)。故aₙ+1/2=(2+1/2)3ⁿ⁻¹=(5/2)3ⁿ⁻¹，整理得aₙ=(5·3ⁿ⁻¹−1)/2。11．答案：C解析：离心率e=c/a=1/2，故c²=a²/4，b²=a²−c²=3a²/4。点(2,√3)在椭圆上，代入得4/a²+3/(3a²/4)=4/a²+4/a²=1，所以a²=8。12．答案：B解析：f(x)=eˣ−ax。若a≤0，函数无两个零点；若a>0，f′(x)=eˣ−a，在x=lna处取得极小值f(lna)=a−alna=a(1−lna)。要有两个零点，需极小值小于0，即lna>1，故a>e。二、填空题解析13．答案：175解析：由a₂=7，a₆=19得公差d=(19−7)/4=3，首项a₁=4。S₁₀=10(a₁+a₁+9d)/2=5(8+27)=175。14．答案：2解析：圆半径r=√5，弦长为4，则半弦长为2。圆心到直线x+y=t的距离d满足d²+2²=5，得d=1。又d=|t|/√2，故t²=2。15．答案：−1/2解析：因为(sinα+cosα)²=1+sin2α，且sinα+cosα=√2/2，所以1+sin2α=1/2，得sin2α=−1/2。区间条件保证符号与结论一致。16．答案：2^(2/3)解析：f′(x)=3x²−3m。m>0时，x=−√m处为极大值点，极大值f(−√m)=2m√m。由2m√m=4得m^(3/2)=2，所以m=2^(2/3)。三、解答题答案详解与评分标准17．答案：（1）A=B；（2）a=b=4。解析：（1）由C=π−A−B，得sinC=sin(A+B)=sinAcosB+cosAsinB。又sinC=2sinAcosB，所以cosAsinB=sinAcosB，即sin(B−A)=0。因为A，B均为三角形内角，故A=B。

（2）由A=B得a=b，三角形为以AB对应的两腰相等的等腰三角形，底边c=4。设底边上的高为h，则面积S=ch/2=4√3，所以h=2√3。底边一半为2，故腰长a=b=√(2²+(2√3)²)=4。评分标准：写出sinC=sin(A+B)并化简得sin(B−A)=0，得4分；由角的范围推出A=B，得2分；利用面积求高，得2分；由勾股定理求出a=b=4，得2分。18．答案：（1）表中比例为52/80=0.65；（2）145/442；（3）不能认为二者有关。解析：（1）由题意，男生未达到为40−22=18，女生达到为40−10=30，达到阶段目标人数为52，总人数80，估计比例为52/80=0.65。

（2）从52名达到阶段目标的学生中抽2人，基本事件数为C₅₂²；抽到的2人均为女生的取法为C₃₀²。因此所求概率P=C₃₀²/C₅₂²=435/1326=145/442。

（3）列联表中取a=22，b=18，c=30，d=10，n=80。K²=80(22×10−18×30)²/(40×40×52×28)≈3.52。因为3.52<3.841，所以在0.05显著性水平下，没有足够证据认为“是否达到阶段目标”与“性别”有关。评分标准：正确补全列联表并求比例，得3分；列出组合概率并化简，得3分；正确代入K²公式，得4分；比较临界值并作出结论，得2分。19．答案：（1）aₙ=2/n；（2）Sₙ=4n/(n+1)；（3）最小n=40。解析：（1）因为aₙ>0，令bₙ=1/aₙ。由aₙ₊₁=2aₙ/(aₙ+2)，得bₙ₊₁=(aₙ+2)/(2aₙ)=bₙ+1/2。又b₁=1/2，所以bₙ=n/2，故aₙ=2/n。

（2）aₖaₖ₊₁=(2/k)·[2/(k+1)]=4/[k(k+1)]=4(1/k−1/(k+1))。所以Sₙ=4(1−1/(n+1))=4n/(n+1)。

（3）由4n/(n+1)>3.9，得4n>3.9n+3.9，故0.1n>3.9，即n>39，因此最小正整数n为40。评分标准：构造bₙ=1/aₙ并证明等差，得4分；求出通项aₙ=2/n，得2分；把乘积化为裂项