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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3334353637loglogloglogloga

aaaa++++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．370,,,889．在()()51231xx−+的展开式中，含3x项的系数为（）A．80−B．40−C．40D．12010．在A

BC△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()()6gxx

fx=−在区间*[](1,2nnN）内的所有零点的和为（）A．()3212n−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()2222:10,0xyCabab−=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．1

9．（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0.100.050.0250.0100.0050.001k2.7063.8415.0246.6357.87910.828（参考公式22()()()()()nadbcKabc

dacbd−=++++，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()1

2xx，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//AEB

G，·································································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Ox

yz−，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F

CD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2

0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：X01234P04164420CCC13164420CCC22164420CCC31164420CCC40164

420CCC·······························································································································

······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),(,),(,)ykxAxyBxy+=−，

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,)xm+时，'()0fx，()fx···················································································3分此时函数有极小值为(ln)lnfmmmm=−，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，2e

m．由11(1)xemx=−，可知110x−，所以120111xx−−．设211(1)xxt−=−+（0t），由212111xxxeex−=−，得11(1)e1txtx−+=−，所以11e1t

tx−=−，即11e1ttx=+−，所以2e1e1tttx=+−············································································

································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1lnx

m，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1ln11txt=+−，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cos

sin1,232M++，··································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()3fx=，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