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长治市第二中学校2021届高三9月质量调研考试数学试卷注意事项：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aR，若()2i1aa−+−（i为虚数单位）是实数，则a=（）A．2B．1C．1−D．2−2．已知集合(

)lg2|3Axyx==−，2|4Bxx=，则=AB（）A．322xx−B．2xxC．322xx−D2xx3．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．54．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12=，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．12B．58C．34D．786．已知点(

)1,0A，(2,3)B，,()2Cm−−，向量AC，AB的夹角为56，则实数m=（）A．23B．3C．0D．3−7．已知各项为正数的等比数列na满足2589aaa=﹐则3334353637logloglogloglogaaaaa++++的值为（）A．73B．83C．3D．1038．已知

函数22()(sincos)2cosfxxxx=++的图像向右平移4个单位长度后得到函数()gx的图像，则函数()gx在[0,]上的单调递减区间是（）A．37,88B．3,44C．30,,,44D．3

70,,,889．在()()51231xx−+的展开式中，含3x项的系数为（）A．80−B．40−C．40D．12010．在ABC△中，内角A、B、C的对边分别为a，b，c，满足33c=，

sinsin2CcAa=，则ABC△面积的最大值为（）A．348B．336C．324D．31211．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12．定义函数()348,1221,222xxfxxfx−−=，则函数()()6gxxfx=−在区间*[](1,2nnN）内的所有零点的和为（）A．()3

212n−B．2nC．3(21)4n−D．n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．函数2()lnfxaxbx=−在点()2,2Pf处的切线方程为32ln22yx=−++，则ab+=________．15．已知1F，2F分别是双曲线()2222:10,0xyCa

bab−=的左、右焦点，AB是右支上过2F的一条弦，234AFAB=且1212AAFABF+=，则C的离心率为________．16．在一个棱长为12的正方体形状的铁盒内放置一个正四面体（四个面都是正三角形的三棱锥），且能使该正四

面体在铁盒内任意转动，该正四面体的体积的最大值是________．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=，数列nb满足113b=，113nnnbbn++=．（1）求数列na，nb的通项公式；（2）求数列nb的前n项和nT．18．（本题满分12分）如图，四棱锥SABCD−

中，22SDCDSCABBC====，平面ABCD⊥底面SDC，ABCD，90ABC=，E是SD中点．（1）证明：直线AE平面SBC；（2）点F为线段AS的中点，求二面角FCDS−−的大小．19．（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，设抽到不足120分且每周线上学习时间不足5小时的人数是X，求X的分布列（概率用组合数算式表示）；（ii）若将频率视为概率，从全校高三该次检测

数学成绩不少于120分的学生中随机抽取20人，求这些人中每周线上学习时间不少于5小时的人数的期望和方差。（下面的临界值表供参考）()2PKk0.100.050.0250.0100.0050.001k2.7063.8415.0246.635

7.87910.828（参考公式22()()()()()nadbcKabcdacbd−=++++，其中nabcd=+++）20．（本题满分12分）已知点()0,1N，椭圆2222:1(0)xyEabab+=的离心率为32，F是椭圆E的右焦点，直线NF的斜率为33−．（1）求椭圆E

的方程；（2）过点1(2,)P−作直线l，交椭圆E于异于点N的A，B两点，直线NA，NB的斜率分别为1k，2k，证明12kk+为定值．21．（本题满分12分）已知函数()xfxemx=−．（1）讨论(

)fx的单调区间与极值：（2）已知函数()fx的图象与直线ym=−相交于11(),Mxy，()22,Nxy两点()12xx，证明：124xx+．（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++．理科数学答案一、选择

题1．A2．D3．C4．D5．A6．B7．D8．A9．C10．B．11．D12．A二．填空题13．114．_3___5．10216．643三．解答题17．解：(1)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分又nnbbbnnbnnnn313111+=+=++······························

····························································4分1211211213(1)3(2)3133nnnnnnbbbnnnbbbbbnn−−−−===−−····························

·····················6分（2）由（1）可得：13nnbn=nnbbbT++=211211112333nn=+++······························

···················································7分2311111123333nnTn+=+++··············

·····························································8分1212111133333nnnTn+=+++−

1111133111111323313nnnnnn++−=−=−−−································

·················10分131313231323114323443443nnnnnnnTn+++=−−=−=−····

······················11分nT的表达式为3231443nnnT+=−··············································································12

分18．证明：（1）如图，取SC中点G，连接BG，EG，因为EG为SDC△中位线，所以//EGCD且12EGCD=，···········································

···············································1分又因为//ABCD且12ABCD=，所以//EGCD且EGAB=，所以四边形AEGB为平行四边形，所以//AE

BG，·································································3分又因为BG平面SBC，AE平面SBC，所以//AE平面SBC···········

·····························································································5分（2）设1AB=，则1BC=，2CD=，取CD中点O，所

以12COCDAB==又因为//ABCD，90ABC=，所以四边形ABCO为矩形，所以AOCO⊥，AOCD⊥，平面ABCD平面SDCCD=，所以AO⊥平面SDC，AOSO⊥，又因为三角形SDC为正三

角形，所以SOCD⊥，··············································································································7分故如图建立空间直角坐标

系Oxyz−，可得，(0,0,1)A，(3,0,0)S，(0,1,0)C，(0,1,0)D−，31,0,22F，所以31,1,22FC=−−，31,1,22FD=−−−，设平面FCD的一个法向量为(,,)mabc=，则00FCmFDm=

=，可取(1,0,3)m=−，易知平面SDC的一个法向量(0,0,1)n=，所以3cos,||||2mnmnmn==，又知二面角FCDS−−为锐角，则二面角FCDS−−的大小为6·····

··················································································12分19．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15419每周线上学习数学时间不足5小时101626合计252045由22列联表可知：2245(1516104)7.2876.63525201926K−=·····························

···········································4分所以，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”；（2）(i)由分层抽样知，需要从分数不足120分的学生中抽取920445=，又分数不足120分且每周线上学习时间不足

5小时的人数为16人．设抽取的20人中每周线上学习时间不足5小时的人数为X，所以X的可能取值为1642040,1,2,3,4.()(0,1,2,3,4)4CkCkPXkkC−===X的分布列为：X01234P04164420C

CC13164420CCC22164420CCC31164420CCC40164420CCC··································································································

···································9分（ii）从全校数学成绩不少于120分的学生中随机抽取1人此人每周上线时间不少于5小时的概率为150.625=，设从全校数学成绩不少于120分的学生中随

机抽取20人，这些人中每周线上学习时间不少于5小时的人数为Y，则()~20,0.6YB，所以()200.612,()200.6(10.6)4.8EYDY===−=···················································12分20

．解：（1）设)0,(cF，由条件知，,331−=−c得3c=．又23=ac，所以2a=，2221bac=−=．故E的方程为2214xy+=．·······························································

································4分（2）当直线l的斜率不存在时，直线l与椭圆只有一个交点，不满足题意。当直线l的斜率存在时，设其方程为11221(2),(,),(,)ykxA

xyBxy+=−，将直线l方程代入椭圆1422=+yx，整理得01616)12(8)14(:222=+++−+kkxkkxk··································································6分则1228(

21)41kkxxk++=+，2122161641kkxxk+=+，······································································8分由题知21,

xx不为零，从而212121221121))(22(211xxxxkxkxxyxykk++−=−+−=+.1)12(21616)12)(22(822−=+−=+++−=kkkkkkkk综上，恒有121kk+=−．·························

·······································································12分21．解：（1）'()xfxem=−，·······································

····················································1分①当0m时，'()0fx，此时()fx在R上单调递增，无极值；②当0m时，由'()0fx=，得lnxm=．所以(,ln)xm−时，'()0fx，()fx单调递减；(ln,)x

m+时，'()0fx，()fx···················································································3分此时函数有极小值为(ln)lnfmm

mm=−，无极大值·····························································4分（2）方法一：由题设可得12()()fxfxm==−，所以1212(1)(1)xxemxemx=−=−5分且由

（1）可知1lnxm，2lnxm，2em．由11(1)xemx=−，可知110x−，所以120111xx−−．设211(1)xxt−=−+（0t），由212111xxxeex−=−，得11(1)e1txtx−+=−，所以11e1ttx−=−，即11e1ttx=+−，所以2

e1e1tttx=+−··································································································

··········6分124xx+e2e1e1ttttt+−−e2e2tttt+−2ee20tttt−−−．设()2ee2tthttt=−−−（0t），·················································

····································8分则'()ee1tthtt=−−，设()'()ee1ttgthtt==−−，则'()etgtt=−，所以'()0gt．所以'()ht在(0,)+单调递减，'()'(0)1010hth=−−=····

·················································10分所以()ht在(0,)+单调递减，()(0)20020hth=−−−=．··················

·····························11分所以124xx+······································································································

········12分方法二：由题设可得12()()fxfxm==−，所以1212(1)(1)xxemxemx=−=−，························································5分且由（1）可知1lnxm，2lnxm，2em

．由11(1)xemx=−，可知110x−，所以120111xx−−．由1212(1)(1)xxemxemx=−=−，得1122lnln(1)lnln(1)xmxxmx=+−=+−，··········································

···························6分作差得22111ln1xxxx−=−−设211(1)xtx−=−（1t），由22111ln1xxxx−=−−，得1ln(1)(1)ttx=−−，所以1ln11txt−=−，即1

ln11txt=+−，所以2ln11ttxt=+−，·····························································································

·············8分124xx+lnln211ttttt+−−(1)ln21ttt+−4ln201tt+−+．设4()ln21httt=+−+（1t），····························································

····························9分则22(1)'()0(1)thttt−=+．所以()ht在(1,)+单调递增，()(1)0220hth=+−=·········································

··············11分所以124xx+············································································

··································12分22．解：（1）点P的直角坐标为333,22；····································································1分由s32co

=−得2co3ssin=+①···························································2分将222xy=+，cosx=，siny=代入①，可得曲线C的直角坐标

方程为2211223xy−+−=．····················································4分（2）直线:l3cos2sin2+=的直角坐标方程为0322xy+−=，··········

····················6分设点Q的直角坐标为cos,si123n2++，则cossin1,232M++，·····························

·····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min()3f

x=，3m=，1113234mabc++==，·································7分1113(234)(234)234abcabcabc++=++++2324433324234abaccbbacabc=++++++2

3244332229324234abaccbbacabc+++=，2343abc++，当且仅当2341abc===，即12a=，13b=，14c=时取等号，2343abc++···············

···························································································10分