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长治市第二中学校2021届高三9月质量调研考试文科数学试卷注意事项：1．本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分。2．答题前，考生务必将自己的姓名、准考证号填写在答题卡上相应的位置。3．全部答案写在答题卡上，写在本试卷上无效。4．本试题满分

150分，考试时间120分钟。5．考试范围：高考全部内容。第Ⅰ卷一、选择题（本大题共12小题，每小题5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{|}11Mxx=−Z，()2{|}0NxZxx

=−，则MN=（）A．{}1,2−B．0,1C．{}1,0,1−D．1,0,{}1,2−2．已知aR，若()2i1aa−+−（i为虚数单位）是实数，则a=（）A．2B．1C．1−D．2−3．已知圆22:24200Mxyxy+−+−

=，直线:21lyx=+交圆M于P，Q两点，则PQ=（）A．25B．8C．45D．104．若点P为抛物线24yx=上的动点，F为该抛物线的焦点，则PF的最小值为（）A．2B．1C．18D．1165．“勾股定理”在西方被称为“毕达哥拉斯定理”，三国时期吴国的数学家

赵爽创制了一幅“勾股圆方图”，用数形结合的方法给出了勾股定理的详细证明．如图所示的“勾股圆方图”中，四个相同的直角三角形与中间的小正方形拼成一个大正方形，若直角三角形中较小的锐角6=，现在向该大正方形区域内随机地投掷一枚飞镖，则飞镖落

在阴影部分的概率是（）A．232−B．12C．134−D．346．某几何体的三视图如图，ABC△是边长为2的等边三角形，P为线段AB的中点，三视图中的点C，P分别对应几何体中的点M，N，则在几何体侧面展开图中M，N之间的距离为（）A

．3B．3C．5D．57．数列na，nb均为等比数列，前n项和分别为nS，nT，若21nnnSTn−=，则53ab=（）A．18B．16C．14D．128．已知函数()sin24fxx=−图像向右平移()0

个单位，所得图像关于y轴对称，则的最小值是（）A．38−B．8C．4D．389．在ABC△中，内角A、B、C的对边分别为a，b，c，满足6A=，1b=，ABC△的面积为32，则ABC△的外接圆半径为（）A．32B．3C．72D．71

0．已知函数()23logfxx=+，4[]1,x，则()()()22gxfxfx−=的值域为（）A．18,2−−B．11,6−−C．18,6−−D．11,2−−11．已知1F，2

F分别是双曲线2222:1(0,0)xyCabab−=的左、右焦点，AB是右支上过2F的一条弦，234AFAB=且1212AFAFAB=+，则C的离心率为（）A．52B．5C．102D．1012．在菱形ABCD中，3A=，||43AB=，将ABD△沿BD折起到PBD△的位置，二

面角PBDC−−的大小为23，则三棱锥PBCD−的外接球的表面积为（）A．23B．27C．72D．112第Ⅱ卷二、填空题：（本大题共4小题，每小题5分，共20分，把答案填在答卷纸的相应位置上）13．已知x，y满足约束条件5503xyxyx+−+，使()0zxaya=

+取得最小值的最优解有无数个，则a的值为________．14．已知(1,3)a=，(1,)bm=，若向量a，b的夹角为3，则实数m=________．15．直线ykb=+与曲线31yxax=++相切于点(2,3)

，则b的值为________．16．已知函数()()2020cos,[0,]2log,,xxxxfx=−+，若有三个不同的实数a，b，c，使得()()()fafbfc==，则abc++的取值范围是________．三、解答题：共70分，解

答应写出文字说明、证明过程或演算步骤。第17~21题为必考题，每个试题考生都必须作答。第22、23题为选考题，考生根据要求作答。必考题：共60分。17．（本题满分12分）已知等差数列na的前n项和为nS，47a=，525S=．（1）求nS的表达式；（2）设11n

nnbaa+=，求数列nb的前n项和nT．18．（本题满分12分）在全面抗击新冠肺炎疫情这一特殊时期，学生线上学习。某校数学教师为了调查高三学生数学成绩与线上学习时间之间的相关关系，对高三年级随机选取45名学生进行跟踪问卷，其

中每周线上学习数学时间不少于5小时的有19人，余下的人中，在检测考试中数学平均成绩不足120分的占813，统计成绩后得到如下列联表：分数不少于120分分数不足120分合计线上学习时间不少于5小时419线上学习时间不足5小时合计45（1）请完成上面列联表；并判断是否有99%的把握认

为“高三学生的数学成绩与学生线上学习时间有关”；（2）（i）按照分层抽样的方法，在上述样本中从分数不少于120分和分数不足120分的两组学生中抽取9名学生，再从这9名学生中随机抽取2名学生，求抽取的两名学生分数都不足120的概率．（下面的临界值表供参考）()2PK

k0.100.050.0250.0100.0050.001k2.7063.8415.0246.6357.87910.828（参考公式22()()()()()nadbcKabcdacbd−=++++，其中nabcd=++

+）19．（本题满分12分）如图，四边形ABCD是边长为2的正方形，AE⊥平面BCE，且1AE=．（1）求证：平面ABCD⊥平面ABE．（2）线段AD上是存在一点F，使三棱锥CBEF−的高65h=？若存在，请求||||DFAF的值：若不存在，请说明理由．2

0．（本题满分12分）设椭圆2222:1(0)xyEabab+=的左右焦点分别为1F，2F椭圆上点M到两焦点的距离之和为42，椭圆的离心率为32．（1）求椭圆E的方程；（2）直线:2lx=与椭圆E在第一

象限交于点N，点A是第四象限的点且在椭圆E上，线段AB被直线l垂直平分，直线NB与椭圆E交于点D（异于点N），求证直线AD的斜率为定值．21．（本题满分12分）已知函数2()(2)ln()fxxaxaxa=+−−R．（1）当2a=时，求()fx的图象

在1x=处的切线方程；（2）当3a时，求证：()fx在[1,)+上有唯一零点．（二）选考题，共10分。请考生在第22、23题中任选一题作答。如果多做，则按所做的第一题计分。22．[选修4—4：坐标系与参数方程]（本题满分10分）已知平面直角坐标系xOy，以O为极点，x轴的非负半轴为极

轴建立极坐标系，P点的极坐标为3,3，曲线C的极坐标方程2cos3=−．（1）写出点P的直角坐标及曲线C的直角坐标方程；（2）若Q为曲线C上的动点，求PQ的中点M到直线l

：cos2sin23+=的距离的最小值。23．[选修4—5：不等式选讲]（本题满分10分）设函数()12fxxx=++−．（1）求不等式()4fx的解集；（2）设*,,abcR，函数()fx的最小值为m，且111234mabc++=，求证：2343abc++．文科数学

答案一、选择题1．B2．A3．C4．D5．A6．C7．B8．B9．D10．B11．C12．D二、填空题：13．114．015．15−16．()2,2021三、解答题：17．解：(1)设等差数列}{na的首

项为1a，公差为d，由题意得：=+=+2524557311dada····················································································

······1分解得：==211da···························································································

················2分12)1(1−=−+=ndnaan······················································································

·········3分212)121(2)(nnnaanSnn=−+=+=···············································································．5分（2）由（1）可知1

2−=nan111111(21)(21)22121nnnbaannnn+===−−+−+··························································7分nnbbbT+++=21111111111

2323522121nn=−+−++−−+111111123352121nn=−+−++−−+111221n=−+12+=nn···································

··················································································11分nT的表达式为12+=nnTn···················

···········································································12分18．解：（1）分数不少于120分分数不足120分合计每周线上学习数学时间不少于5小时1541

9每周线上学习数学时间不足5小时101626合计252045由列联表可知：2245(1516104)7.2876.63525201926K−=，所以有99%的把握认为“高三学生的数学成绩与学

生线上学习时间有关”；····································5分（2）由分层抽样知，需要从分数不足120分的学生中抽取920445=，则分数不少于120分的抽取5人．·······································

····························································································7分设分数不少于120分的5人为

：12345,,,,.aaaaa分数不足120分的4人为：1234,,,.bbbb则基本事件为：121314151112131423(,)(,)(,)(,)(,)(,)(,)(,)(,)aaaaaaaaababababaa、、、、、、、、、242521222324343

5313233(,)(,)(,)(,)(,)(,)(,)(,)(,)(,)(,)aaaaababababaaaaababab、、、、、、、、、、、3445414243445152535412(,)(,)(,)(,)(,)(,)

(,)(,)(,)(,)(,)abaaababababababababbb、、、、、、、、、、1314232434(,)(,)(,)(,)(,)bbbbbbbbbb、、、、·········································

·······································10分两名学生分数都不足120分的概率61366p==．·······························································12分19．解：（

1）∵AE⊥平面BCE，BE平面BCE，BC平面BCE．∴AEBE⊥,AEBC⊥··············································································

······················2分又∵BCAB⊥，AEABA=，BC⊥平面ABE································································4分又BE面ABCD，ABCD⊥平

面ABE············································································5分（2）∵1AE=，2AB=，AEBE⊥，∴3BE=，假设线段AD上存在一点F满足题意，························

·············································································································6分由（1）知，平面ABCD⊥平面ABE，平面A

BCD平面ABEAB=．又∵DAAB⊥，∴DA⊥平面ABE，则DABE⊥·······························································8分∵BEAE⊥，BEAD⊥，AEADA=，∴BE⊥平面ADE，又EF平面ADE，BE

EF⊥，116333255CBEFVEFEF−==．∵//ADBC，AD平面BCE，BC平面BCE，∴//AD平面BCE，∴点F到平面BCE的距离与点A到平面BCE的距离相等．又BCBE⊥．∴113(23)1323FBCEV−==．

···························································10分又FBCECBEFVV−−=，∴53EF=．∵222EFAFAE=+，∴43AF=，∴1

2DFAF=···································································12分20．解：（1）设)0,(cF，由条件知，32ca=，

242a=·······················································1分所以6,22==ca，所以2222=−=cab···············································

··························2分故椭圆E的方程为22182xy+=．··········································································

··············4分（2）由题得N的坐标为)1,2(N，················································································

·····5分直线ND不与x轴垂直，设直线)2(1:−=−xkyND，则直线)2(1:−−=−xkyNA，················································6分设),,(),,(A

ADDyxAyxD将直线ND方程代入椭圆12822=+yx整理得：041616)21(8)14(222=−−+−++kkxkkxk，·······8分,14288,144161622222+−−=+−−=kkkxkkkxDD同理可得，,1428822

+−+=kkkxA又12DDykxk=+−，12AAykxk=−++，······································································10分.2114164144164)(222=+−−+−=−−+=−−=k

kkkkkxxkxxkxxyykADADADADAD所以直线AD的斜率为定值12．·····················································································.2112分21．解：（

1）当2a=时，函数2()2lnfxxx=−，定义域为(0,)+··········································1分则2()2fxxx=−，则()01f=，(1)1f=．·············

····························································2分故()fx的图象在1x=处的切线方程为10y−=，即1y=····························

·····························3分（2）证明：22(2)(1)(2)()22axaxaxxafxxaxxx+−−+−=+−−==．······························4分因为3a，令()0fx，得02ax；令()0f

x，得2ax．又322a，()fx在1,2a上单调递减，在,2a+上单调递增．··············································5分22min()(2)ln

ln1ln24224242aaaaaaaafxfaaaaa==+−−=−−=−−···························6分令()1lnln1ln2(3)424aaagaaa=−−=−−++．显然()ga在(3,)+上单调递减．·······

················································································7分又114413333(3)lnlnelnln4lnln2ln042222g=−=−−=−．所以()0ga，即min()0

fx·····························································································8分（其它方法证得也给分，但直接判断小于0不给分）()()222

ee(2)elnee(2)eaaaaaafaaaa=+−−=+−−令22()e(2)e(3)aahaaaa=+−−，······································

·········································9分（找其它合理的点或用放缩法构造函数找点只要合理也给分）则()()22()2ee(1)2ee1e2aaaaahaaaaa=+−−=−++−．令()e(3)aaaa

=−，则()e10aa=−，所以()a在(3,)+上单调递增，则()(3)0a，所以e0aa−，2e20aa−，故()0ha，········································10分所以()ha在(3,)+上单调递增，()()633333

()(3)ee9ee1922190hah=−−=−−−−，所以()e0af．·······································································

··········································11分（若用极限的思想得出结论只得1分（9分—11分之间））又(1)30fa=−，结合单调性可知()fx在[1,)+

上有唯一零点，命题得证．·························12分（二）选考题：22．解：（1）点P的直角坐标为333,22；·······························

·····································1分由s32co=−得2co3ssin=+①···························································2分将222xy=+，c

osx=，siny=代入①，可得曲线C的直角坐标方程为2211223xy−+−=．····················································4分（2）直线:l3c

os2sin2+=的直角坐标方程为0322xy+−=，·······························6分设点Q的直角坐标为cos,si123n2++，则cossi

n1,232M++，··································7分那么M到直线l的距离：()2233cossin12225sin2cos2sin22252521d

+++−++++===+，所以M到直线:cos2sin23l+=的距离的最小值为0················································10分23．解：（1）12,1()123,1221

,2xxfxxxxxx−−=++−=−−·····························································2分()4fx，1241xx−−或2142xx−，32x−或

52x，············································4分不等式的解集为35,,22−−+；····························

···············································5分（2）证明：由（1）知()3minfx=，3m=，1113234mabc++==，········

························································································7分1113(234)(234)234abcabcabc

++=++++2324433324234abaccbbacabc=++++++23244332229324234abaccbbacabc+++=，2343abc++，当且仅当2341abc===，即12a

=，13b=，14c=时取等号，2343abc++··········································································································10分