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·1·福清市2020届高三年“线上教学”质量检测文科数学试卷（满分：150分考试时间：120分钟）本试题卷分第Ⅰ卷（选择题）和第Ⅱ卷（必考题和选考题两部分）．第Ⅰ卷（选择题共60分）一、选择题：本题共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260Axxx，ln1Bxyx，则AB（A）12xx（B）12xx（C）13xx（

D）13xx(2)已知复数z满足113zii，其中i为虚数单位，则在复平面内z对应的点位于（A）第一象限（B）第二象限（C）第三象限（D）第四象限(3)已知圆222:10Cxyr

r，直线:3420lxy．若圆C上恰有三个点到直线的距离为1，则r的值为（A）2（B）3（C）4（D）6(4)执行如图所示的程序框图，则输出的S是（A）3（B）1（C）1（D）3(5)甲、

乙、丙、丁、戊五人乘坐高铁出差，他们正好坐在同一排的A、B、C、D、F五个座位．已知：(1)若甲或者乙中的一人坐在C座，则丙坐在B座；(2)若戊坐在C座，则丁坐在F座．如果丁坐在B座，那么可以确定的是：（A）甲坐在A座（B）乙坐在D座（C）丙坐在C座（D）戊坐在F座(6)如图，如

图，网格纸上小正方形的边长为1，粗实线画出的是某几何体的三视图，已知其俯视图是正三角形，则该几何体的表面积是（A）225（B）425（C）235（D）435（第4题图）俯视图侧视图正视图（第6题图）·2·(7)下列图象中，函数sinxxfxeex，,x图象的

是yxπ-πOyxπ-πOyxπ-πOyxπ-πO（A）（B）（C）（D）(8)已知0,2x，0,2y，cossin1cos2cossinsin2xxyxxy，则（

A）4yx（B）24yx（C）2yx（D）22yx(9)将函数sin3fxx的图象横坐标变成原来的12（纵坐标不变），并向左平移3个单位，所得函数记为gx．若12,0,2xx，12xx

，且12gxgx，则12gxx（A）12（B）32（C）0（D）32(10)已知正方体1111ABCDABCD的棱长为2，1AC平面．平面截此正方体所得的截面有以下四个结论：①截面形状可能是正三角形②截面的形状可能是正方形③截面形状可能是正五边

形④截面面积最大值为33则正确结论的编号是（A）①④（B）①③（C）②③（D）②④(11)若函数1xfxkx有两个零点，则k的取值范围是（A）0,（B）0,11,（C）0,1（D）1,(12)已知抛物

线220ypxp的焦点为F，与双曲线2222:10,0xyCabab的一条渐近线交于P（异于原点）．抛物线的准线与另一条渐近线交于Q．若PQPF，则双曲线的渐近线方程为（A）yx（B）2yx（C）3yx（D）2y

x第Ⅱ卷（非选择题共70分）本卷包括必考题和选考题两部分．第13题第21题为必考题，每个试题考生都必须做答．第22题~第24题为选考题，考生根据要求做答．二、填空题：本大题共4小题，每小题5分．·3·(13)已知aab，abab，则a与b的夹角为(14)

已知实数,xy满足约束条件30,240,20,xyxyxy则2xy的最小值为(15)《九章算术》是我国古代数学名著，也是古代东方数学的代表作．书中有如下问题：“今有勾八步，股十五步．文勾中容圆径几何？”其意思是：“已知直角三角形两直角边长分别为8步和15步，问

其内切圆的直径是多少？”现若向此三角形内投豆子，则落在其内切圆内的概率是(16)设△ABC的内角A，B，C所对的边分别为a，b，c，3b，2cos3cosacBC，则△ABC面积的最大值是三、解答题

：解答应写出文字说明、证明过程或演算步骤.(17)（本小题满分12分）据历年大学生就业统计资料显示：某大学理工学院学生的就业去向涉及公务员、教师、金融、公司和自主创业等五大行业.2020届该学院有数学

与应用数学、计算机科学与技术和金融工程等三个本科专业，毕业生人数分别是70人，140人和210人.现采用分层抽样的方法，从该学院毕业生中抽取18人调查学生的就业意向．（Ⅰ）应从该学院三个专业的毕业生中分别抽取多少人？（Ⅱ）国家鼓励大学生自主创业，在抽取的18人中，就业意向恰有三个行业的学

生有5人.为方便统计，将恰有三个行业就业意向的这5名学生分别记为A，B，C，D，E，统计如下表:ABCDE公务员○○×○×教师○×○×○金融○○○×○公司××○○○自主创业×○×○×其中“○”表示有该行业就业意向，“×”表示无该行业就业意向．现从A，

B，C，D，E这5人中随机抽取2人接受采访.设M为事件“抽取的2人中至少有一人有自主创业意向”，求事件M发生的概率．(18)（本小题满分12分）已知数列na的前n项和为nS，满足22nnaS（Ⅰ）求na；（Ⅱ）若数列nb满足

*14nnnnabnNSS，求nb的前n项和nT．·4·(19)（本小题满分12分）在三棱柱111ABCABC中，已知AB侧面11BBCC，2BC，12ABBB，14BCC，E为1BB中点，（Ⅰ）求证：1ACB

C（Ⅱ）求C到平面1ACE的距离．(20)（本小题满分12分）已知椭圆2222:10xyCabab的右焦点为F，离心率22e，过原点的直线（不与坐标轴重合）与C交于P，Q两点，且4PFQF（Ⅰ）

求椭圆C的方程；（Ⅱ）过P作PEx轴于E，连接QE并延长交椭圆于M，求证：以QM为直径的圆过点P．B1A1C1CBAE·5·(21)（本小题满分12分）已知函数2lnfxxmxmR的最大值是0，（Ⅰ）求m的值；（Ⅱ）若212fxxaxbe，求b

a的最小值．请考生在第（22）、（23）两题中任选一题作答。注意：只能做所选定的题目。如果多做，则按所做第一个题目计分，作答时请用2B铅笔在答题卡上将所选题号后的方框涂黑。(22)（本小题满分10分）选修4-4：坐标系与参数方程在直角

坐标系xOy中，直线l的参数方程为232212xtyt(t为参数)，在以坐标原点为极点，x轴正半轴为极轴的极坐标系中，曲线C的方程为4cos6sin.（Ⅰ）求C的直角坐标方程

；（Ⅱ）设C与l交于点MN，，点A的坐标为3,1，求AMAN.(23)（本小题满分10分）选修4-5：不等式选讲已知函数211,fxxax（Ⅰ）当2a时,求不等式1fx的解集；（Ⅱ）当1,2x时,不等式1fxx恒成立,求实数a的取值范围.·6·福清市202

0届高三年“线上教学”质量检测文科数学参考答案及评分细则评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则。2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响

的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分。3．解答右端所注分数，表示考生正确做到这一步应得的累加分数。4．只给整数分数。选择题和填空题不给中间分。一、选择题：本大题考

查基础知识和基本运算．每小题5分，满分60分．（1）C（2）A（3）A（4）B（5）C（6）B（7）D（8）A（9）D（10）A（11）C（12）D二、填空题：本大题考查基础知识和基本运算．每小题5分

，满分20分．（13）3（14）4（15）320（16）334三、解答题：解答应写出文字说明、证明过程或演算步骤．(17)··············································································

·······················本小题主要考查分层抽样、古典概型等基础知识，考查数据处理能力、运算求解能力、应用意识，考查统计与概率思想．满分12分．(Ⅰ)由已知，数学与应用数学、计算机科学与技术和金融工程

三个专业的毕业生人数之比为1:2:3,由于采取分层抽样的方法抽取18人,因此应从数学与应用数学、计算机科学与技术和金融工程三个专业分别抽取3人6人9人，··················································4分（Ⅱ）从

这5个人中随机抽取2人的所有结果有：,,,,,,,,,,,,ABACADAEBCBD,,,,,,,BECDCEDE，共10种···················································

·8分由统计表可知，事件M包含的结果有：,,,,,,,,,,,,ABBCBDBEADCDDE，，共7种··············································································

·····················10分所以事件M发生的概率为710PM····················································12分(18)·······

······························································································本小题考查等比·7·数列的通项公式、前n项和公式、数列求和等基础知识

，考查运算求解能力，考查函数与方程思想、分类与整合思想等．满分12分．(Ⅰ)当1n时，1122aS，故12a·················································

···1分由22nnaS……①得11222nnaSn②····································································3分①-②得，11220nnnnaa

SS，即1220nnnaaa整理得122nnaan故na是以2为首项，2为公比的等比数列，··········································5分所以1222n

nna，········································································6分（Ⅱ）由（Ⅰ）得，12122212nnnS········

····································7分112142211212122222121nnnnnnnnnb·····················

·····10分故1212231111111212121212121nnnnTbbb11121n·································

···················································12分(19)····················································

·················································本小题主要考查几何体的体积及直线与直线、直线与平面、平面与平面的位置关系等基础知识，考查空间想象能力、推理论证能力、运算求解能力，考查数形结合思想、化归与转化思想等．满分12分．(Ⅰ)∵

AB⊥侧面11BBCC，∴1ABBC……①··········································1分∵2BC，12CC，14BCC，∴2211112cos2BC

BCCCBCCCBCC∵22211BCBCCC，∴1BCBC……②··············································3分由①②及ABBCB

，故1BC平面ABC∵AC平面ABC，∴1BCAC··························································5分（Ⅱ）设C到平面1ACE的距离为d由11CACEACCEVV得，111133ACECCE

SdSAB……（*）···························7分·8·∵E为1BB中点，∴1111122122CCEBCCBSS································8分又1112BCBC，所以11CEBB，1111

2CEBB∵AB⊥侧面11BBCC，∴1ABCE又1ABBBB，故1CE平面11ABBA又AE平面11ABBA，所以1CEAE∵2,1,ABBEABBE，∴5AE故111522CEASAECE············

························································11分由（*）得522d，故455d，即C到平面1ACE的距离为455············12分(20)······························

·······································································本小题主要考查坐标法、椭圆的定义及标准方程、直线与椭圆的位置关系、圆的性质等基础知识，考查推理论证能力、运算求解

能力，考查数形结合思想、函数与方程思想、化归与转化思想等．满分12分．(Ⅰ)设F为椭圆的左焦点，由对称性可知，,OPOQOFOF故顶点为,,,PFQF的四边形是平行四边形，·························

·················2分故24aPFPFQFPF，2a又22cea，故2,2cb··················································

·········4分故所求椭圆方程为22142xy·······························································5分（Ⅱ）设过原点（不与坐标轴重合）的直线方程为0ykxk，1111

1,,,,,0PxkxQxkxEx则111012QEkxkkxx故1:2QEklyxx·······························································

·············7分与椭圆方程22142xy联立得，22222112280kxkxxkx又直线QE与椭圆C交于Q，M两点，所以21222PMkxxxk，即211222Mkxxxk故311222MMkxk

yxxk···································································10分所以112,2PQxkx，2112222,22kxkxPMkk·9·所以2221122

44022kxkxPQPMkk故PQPM，即90MPQ故以QM为直径的圆过点P．·······························································1

2分(21)·····································································································本小

题主要考查导数及其应用、不等式等基础知识，考查推理论证能力、运算求解能力、抽象概括能力等，考查函数与方程思想、化归与转化思想、分类与整合思想、数形结合思想等．满分12分．(Ⅰ)由已知得212120mxfxmxxxx·························

················1分当0m时，0fx，fx在0,上单调递增，不存在最大值，不符合题意舍去；·················································

·····················································2分当0m时，0fx解得12xm当102xm时，0fx，当12xm时，0fx故fx在10,2m上单调递增，1,2m

上单调递减·······················4分故2max111ln0222fxfmmmm解得12me·········

···········································································5分（Ⅱ）由已知条件得lnxaxb……（*）设ln

gxxaxb，（*）等价于证明0gx则1gxax①当0a时，则0gx，gx在0,上单调递增，当max1,bxa时，ln0gxxaxbaxb

故0a<不符合题意；·········································································7分②当0a时，当10xa时，0gx，当1xa时，

0gx故gx在10,a上单调递增，1,a上单调递减故gx有最大值111lnln1gababaaa······························9分·10·所以212fxxaxb

e等价于ln1ba，因此ln1baaa设ln1ahaa，则ha2lnaa当01a时，0ha，当1a时，0ha故ha在0,1上单调递减，在1,上单调递增故ha在1a处取得最小值，即11hah

，1ba·························11分故当1a，1b时，212fxxaxbe成立，综上ba的最小值为1．······································

································12分(22)····················································································

·················本小题主要考查参数方程、极坐标方程等基础知识，考查运算求解能力，考查数形结合思想、化归与转化思想、函数与方程思想等．满分10分．(Ⅰ)曲线C的方程4cos6sin，∴24cos6sin

，∴2246xyxy，即C的直角坐标方程为222313xy·········································4分（Ⅱ）设点MN，对应的参数分别为12tt，.把直线2322

12xtyt(t为参数)代入C得，2222121322tt，整理得，23280tt.23232500，1232tt，128tt，

∴12tt，为异号，··········8分又∵点3,1A在直线l上，∴21212121245052AMANtttttttt.···················10分(23)·····

································································································本小题主要考查绝对值不等式等基础知识，考查运算求

解能力、推理论证能力,考查化归与转化思想、分类与整合思想、数形结合思想等．满分10分．(Ⅰ)当2a时,21211fxxx，当12x„时,不等式122121fxxx成立；当1122x时，122141fxxxx

，1124x；·11·当12x…时,212121fxxx不成立，综上，不等式1fx的解集为14xx.············································5分（Ⅱ）当1,2x时，1fxx

化为2111xaxx,321xax，23132,xaxx1333axx，13yx在1,2单调递减，故522y；33yx在1,2单调递增，故302y

，所以20a剟，所以a的取值范围是2,0.··································································10分