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第1页共4页第2页共4页中学生标准学术能力诊断性测试2024年3月测试数学试卷本试卷共150分，考试时间120分钟。一、单项选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合

题目要求的．1．已知集合22,4,10,10,24,25AaaBaa=−−=−+−−，且10AB=−，则A．8,2,10A=−−C．2a=或20B．10,78,25B=−−D．800,16,10,82

,25AB=−−2．已知函数()233,3log,>3xxxfxxx−=，若0xR，使得()20104fxmm+成立，则实数m的取值范围为A．91,44−−C．91,,44−−−+B．5,02−D．)5,0,2−−+

3．已知21sin75+=，那么31tan14−=A．15−B．26C．265D．264．已知数列na的前n项和为nS，且23nSnn=+，若首项为12的数列nb满足111nnnabb+−=，则数列nb的前2024项和为A．1012202

3B．20252024C．20232024D．202420255．已知点()()()72,6,2,3,0,1,,62ABCD−−，则与向量2ABCD+同方向的单位向量为A．31010,1010B．10310,1010C．25

5,55−D．43,55−6．已知圆()22:20>0Mxyaxa+−=的圆心到直线22xy+=距离是5，则圆M与圆()()22:211Nxy−++=的位置关系是A．外离B．相交C．内含D．内切7．已知213nxx+的展开式的各项系数和为4096，

则展开式中6x的系数为A．15B．1215C．2430D．818．设aR，若复数23i2ia+−（i为虚数单位）在复平面内对应的点在直线yx=−上，则a=A．2−B．10−C．25D．2二、多项选择题：本题共4小题，每

小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求．全部选对得5分，部分选对但不全得2分，有错选的得0分.9．下列说法正确的是A．不等式2451>0xx−+的解集是1><14xxx或B．不等式2260xx−−的解集是322xxx−或C．若不等式2821

<0axax++恒成立，则a的取值范围是D．若关于x的不等式223<0xpx+−的解集是(),1q，则pq+的值为12−10．已知mn、为两条不重合的直线，、为两个不重合的平面，则下列说法正确的是A．若,mn⊥⊥且，则mnC．若,,mn

n，则mB．若,,mnmn⊥⊥⊥，则⊥D．若,,,mnnm⊥⊥，则m11．设椭圆22:12516xyC+=的左、右焦点分别为12,FFP、是C上的动点，则下列结论正确的是A．125PFPF+=B．离心率3e5=C．12PFF面积

的最大值为12D．以线段12FF为直径的圆与圆()()22434xy−+−=相切第3页共4页第4页共4页12．已知函数()3,1log,>1kxkxfxxx−=，下列关于函数()()2yffx=−的零点个数的

判断，其中正确的是A．当>0k时，有2个零点C．当>0k时，有1个零点B．当<0k时，至少有2个零点D．当<0k时，可能有4个零点三、填空题：本题共4小题，每小题5分，共20分．13．若变量,xy满足约束条件21024020xyxyy+−−++，则xy的

最大值是．14．在平面直角坐标系中，已知点()()1,22,4,ABEF−−、、是直线3yx=+上的两个动点，且32EF=，则AEBF的最小值为．15．设nS是等差数列na的前n项和，若761311

SS=，则1511SS=．16．若,ab是两个夹角为120的单位向量，则向量53ab−在向量ab+方向上的投影向量为．四、解答题：本题共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）在锐角ABC中，角,,ABC所对的边分别为,,abc

，已知3tan24C=−．（1）求cosC；（2）若4c=，求ABC面积的最大值．18．（12分）设数列na满足：112,244nnaaan+==+−．（1）求数列na的通项公式；（2）求数列3nnna+的前n项和nS．19．（

12分）已知过点()1,0的动直线l与圆221:40Cxyx+−=相交于不同的两点,AB．（1）求圆1C的圆心坐标；（2）求线段AB的中点M的轨迹C的方程．20．（12分）某中外合作办学学院为了统计学院往届毕业生薪酬情况，面向学院部分毕业生发放问卷统计了其薪资情况，共有2

00名毕业生进行了问卷填写．毕业生年薪（单位：万元），以)))))))10,20,20,30,30,40,40,50,50,60,60,70,70,80分组的频率分布直方图如图所示，年薪在)50,60的毕业生人数比年薪在

)10,20的毕业生人数多22人．（第20题图）（1）求直方图中x，y的值；（2）①用样本估计总体，比较学院毕业生与同类型合作办学高校毕业生薪资水平，如果至少77%的毕业生年薪高于同类型合作办学高校毕业生平均薪资水平，则说明同类型

合作办学高校毕业生平均年薪最高为多少；②若将频率视为概率，现从该学院毕业生中随机抽取4人，其中年薪高于50万的人数为，求的分布列及数学期望()E．21．（12分）已知函数()22exxaxafx−+=，其中aR．（1）

当1a=时，求曲线()yfx=在()()0,0f处的切线方程；（2）求证：()fx的极大值恒为正数．22．（12分）在平面直角坐标系xOy中，已知椭圆22:14xEy+=的左、右焦点分别为12FF、，点A在椭

圆E上且在第一象限内，12AFAF⊥，点A关于y轴的对称点为点B．（1）求A点坐标；（2）在x轴上任取一点P，直线AP与直线3y=相交于点Q，求OPOQ的最大值；（3）设点M在椭圆E上，记OAB与MAB的面积分别为12,SS，若122SS=，求点

M的坐标．1020304050607080毕业生年薪情况（单位：万元）第1页共5页中学生标准学术能力诊断性测试2024年3月测试数学参考答案一、单项选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的．12345678DCBDACBC二、多项选择

题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，部分选对但不全的得2分，有错选的得0分．9101112CDABDBCDABD三、填空题：本题共4小题，每小题5分，共20分．13．1614．

1858−15．64545116．ab+四、解答题：本题共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）（1）22tan3tan21tan4CCC==−−········································

·································2分解得tan3C=，故10cos10C=··································································4分（2）222102

cos1625abcabCabab+=+=+··············································6分解得808109ab+·····································

·············································8分由（1）知3sin10C=故14410sin23ABCSabC+=···································

·····························10分故ABC面积的最大值为44103+．第2页共5页18．（12分）（1）对于2n时，()()1412824nnnananan+++=+=+······································2分

112,416aa=+=，432,324nnnnanan+==−···································································3分经验算，1a符合上述结果，故324nnan=−

·················································4分（2）设33643nnnnnbnann=+=+−，则()()1816143216nnnnnSn−+=+−−···································

······················6分设43nnTn=，123438312343nnTn=++++，23413438312343nnTn+=++++··········································

···········7分作差得到123124343434343nnnTn+=−−−−−+······························8分故()()1161323213313nnnnTnn++−−=+=−

+−··········································10分()()18161216nnnnnST−+=+−−故()21183362132255nnnnnSn+=++−−−·······················

·························12分19．（12分）（1）圆1C的方程可变形为()2224xy−+=··············································

··············2分故1C的圆心坐标为()2,0，半径为2·······························································4分（2）设(),

MMMxy，因为点M是AB的中点，1CMAB⊥，11CMABkk=−············································································

············6分故121MMMMyyxx=−−−················································································8分第3页共5页由此可得22320MMMxxy−++=···

································································10分故轨迹方程为223124MMxy−+=，轨迹是以圆心为3,02，半径为12的圆·······12分20．（12

分）（1）解：10100.005100.0110100.019100.02100.0271xy++++++=，故0.019xy+=·····················································

····································1分200102001022yx−=，故0.011yx−=····································································

·····················2分解得0.004,0.015xy==····························································

················3分（2）①学院毕业生年薪在)30,80区间的人数比例为：()0.020.0270.0150.010.005++++1077%=，故同类型合作办学高校毕业生平均年薪最高为30万元························

················5分②对于单个毕业生，其年薪高于50万的概率()0.0050.010.015100.3P=++=，故随机变量3~4,10B，故()()4010.30.2401P==−=··

································································6分()()314110.30.30.4116PC==−=··················

······································7分()()2224210.30.30.2646PC==−=······················································

8分()()334310.30.30.0756PC==−=·······················································9分()440.30.0081P=

==·········································································10分的分布列为：01234P0.24010.41160.26460.07560.0081第4页共5页的数学期望()0.341.2E=

=·······························································12分21．（12分）（1）()()()()()2224ee2242eexxxxxaxaxaxaxafx−−−+−++−==··

·····················2分当1a=时，()2252exxxfx−+−=，()02f=−············································4分又()01f=，故曲线()yfx=在()()0,0f处的切线方程为21

yx=−+············5分（2）()()()()2242220eexxxaxaxaxfx−++−−+−===，解得知122,2axx==············································

·······································7分若()>4,afx在(),2,,2a−+递减，2,2a递增························

···············8分极大值2>02eaaaf=···············································································9分若=4a，函数单调递减，无极大值··

···························································10分若()<4,afx在(),,2,2a−+递减，,22a递增·····

································11分极大值()282>0eaf−=······························································

··············12分综上，()fx的极大值恒为正数．22．（12分）（1）椭圆22:14xEy+=的左，右焦点分别为()()123,0,3,0FF−，设()12,,AmnAFAF⊥，故()()123,3,

0AFAFmnmn=−−−−−=···············1分即223mn+=······················································································

·····2分又2214mn+=，解得263,33mn==·························································3分A点坐标为263,33·······

·····································································4分第5页共5页（2）设P点坐标为(),0p，则可得Q点坐标为()262,3p−·································5

分()()226,0262,3226232OPOQppppp=−=−+=−−+·················7分当62p=时，OPOQ取最大值，最大值为3················································8分（

3）A点坐标为263,33，B点坐标为263,33−，点O到线段AB的距离133h=·····································································9分若122SS=

，则点M到线段AB的距离应为236h=，故M点的纵坐标为36或32，代入椭圆方程，解得M点的横坐标为333或1····················································

···········11分故M点的坐标为：333,36或31,2··············································12分