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·1·绝密★启用前福清市2020届高三年“线上教学”质量检测数学（理科）试卷（完卷时间120分钟；满分150分）本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分．第Ⅰ卷1至2页，第Ⅱ卷3至5页．满分150分.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5Uxx，327xAx„，则AðUA．3xx„B．5xxC．35xxD．035xxx或„2.已知复数z满足0zz，且9zz，则zA．3B．3iC．3D．3i3.已知两个力14,

2F，22,3F作用于平面内某静止物体的同一点上，为使该物体仍保持静止，还需给该物体同一点上再加上一个力3F，则3FA．2,5B．2,5C．5,2D．5,24.已知等比数列na

的前n项和为nS，若24132aaaa，且135512aaa，则10SA.1022B.2046C.2048D.40945.如图1为某省2019年1~4月快递业务量统计图，图2是该省2019年1~4月快递业务收入统计图，下列对统计图理解错误的是A．2019年1~4月的业务量，

3月最高，2月最低，差值接近2000万件B．2019年1~4月的业务量同比增长率超过50%，在3月最高C．从两图来看2019年1~4月中的同一个月快递业务量与收入的同比增长率并不完全一致D．从1~4月来看，该省在2019年快递业务收入同比增长率逐月增长

6.已知222fxxxf，则曲线yfx在点1f1，处的切线方程为·2·A．490xyB．610xyC．1010xyD．610xy7.若412(1)xax展开式中2x的系数为7

8，则整数a的值为A．3B．2C．2D．38.已知函数()eexxfx，若0.50.6a，0.5log0.6b，0.6log5c，则A．()()()fafbfcB．()()()fcfbfaC．()()()fbfa

fcD．()()()fcfafb9.如图，网格纸上的小正方形的边长为1，粗线画出的是某几何体的三视图，则该几何体的体积为A.4B.163C.323D.1610.将曲线12sin124yx向左平移4个单位长

度，得到曲线的对称中心为A.2,0,kkZB.2,0,4kkZC.2,1,4kkZD.2,1,4kkZ11.已知双曲线2222:10,0xyEabab的右焦点F的坐标为,0c，过F作与E的

两条渐近线平行的直线12,ll，若12,ll与E的渐近线分别交于,AB两点，且四边形OAFB（O为坐标原点）的面积为bc，则E的离心率为A．3B．2C．3D．212.半正多面体亦称“阿基米德多面体”，是由边数不全相同的正多边形为面的多面体，体现了数学的对称美．如图，将正方体

沿交于一顶点的三条棱的中点截去一个三棱锥，如此共可截去八个三棱锥，得到一个有十四个面的半正多面体，它们的棱长都相等，其中八个为正三角形，六个为正方形，称这样的半正多面体为二十四等边体.若棱长为2的二十四等边体的各个顶点都在同一个球面上，则该球的表面积为A．16πB．32π3C．8πD．4π第Ⅱ

卷注意事项：用0.5毫米黑色签字笔在答题卡上书写作答．在试题卷上作答，答案无效．二、填空题：本大题共4小题，每小题5分，共20分．把答案填在题中的横线上．·3·13.已知3sintan80,,2，则tan______.14.某电

视台的夏日水上闯关节目中的前三关的过关率分别为542,,655，只有通过前一关才能进入下一关，且通过每关相互独立．某选手参加该节目，则该选手能进入第四关的概率为_________.15.已知等差数列na的前n项和为nS，且1310aa，972S.数列n

b的首项为3，且13nnbb，则210020ab________.16.过点1,0M的直线l与抛物线C：24yx交于,AB两点（A在,MB之间），F是C的焦点，点N满足6NFAF，则ABF△与AMN△的面积之和的最小值是____

__.三、解答题：本大题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤．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．（一）必考题：共60分．17.（本小题满分12分）已知ABC△的内角,,ABC的对边分别为,,abc，设7coscosa

BbAac，且sin2sinAA.（1）求A及a；（2）若2bc，求BC边上的高.18.(本小题满分12分)如图，四边形ABCD是梯形，四边形CDEF是矩形，且平面ABCD平面CDEF，90BADCDA，122ABADDECD

，M是AE的中点.（1）证明：AC∥平面MDF；（2）求平面DMF与平面ABCD所成锐二面角的余弦值.19.(本小题满分12分)2019年9月24日国家统计局在庆祝中华人民共和国成立70周年活动新闻中心举办新闻发布会

指出，1952年～2018年，我国GDP查679.1亿元跃升至90.03万亿元，实际增长174倍；人均GDP从119元提高到6.46万元，实际增长70倍.全国各族人民，砥砺奋进，顽强拼搏，实现了经济社

会的跨越式发展.如图是全国2010年至2018年GDP总量y（万亿元）的折线图.·4·注：年份代码1～9分别对应年份2010～2018.（1）由折线图看出，可用线性回归模型拟合y与年份代码t的关系，请用相关系数加以说明；（2）建立y关于t的回归方程（

系数精确到0.01），并预测2021年全国GDP的总量.附注：参考数据：9922111199582.01,64.668,3254.80,345.900iiiiiiiiittyyyyty

.参考公式：相关系数12211niiinniiiittyyrttyy；回归方程yabt$$$中斜率和截距的最小二乘法估计公式分别为121niiiniittyybtt$，ayb

t$$.20.(本小题满分12分)已知椭圆2222:10xyCabab过点2,1P，且离心率为32.（1）求C的方程；（2）已知直线l不经过点P，且斜率为12，若l与C交于两个不同点,AB，且直线,PAPB的倾斜角分别为,，试判断

是否为定值，若是，求出该定值；否则，请说明理由．21.(本小题满分12分)已知函数ln2sinfxxxx，证明：（1）fx在区间0,存在唯一极大值点；（2）fx有且仅有2个零点.（二）选考题：共10分．请考生在第22，23两题中任选一题作答．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44：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.·5·（1）求C的直角坐标方程；（2）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45：不等式选讲已知函数211,fxxax

（1）当2a时，求不等式1fx的解集；（2）当1,2x时，不等式1fxx成立，求实数a的取值范围.福清市2020届高三年“线上教学”质量检测理科数学参考答案及评分细则评分说明：1．本解答给出了解法供参考，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.C2.C3.A4.B5.D6.D7.A

8.A9.B10.C11.B12.A二、填空题：本大题共4小题，每小题5分，共20分．13.2214.41515.1316.8三、解答题：本大题共6小题，共70分．17.【命题意图】本题主要考查解三角形、正

弦定理和余弦定理等基础知识，意在考查逻辑推理、数学运算、直观想象的数学核心素养．满分12分．【解析】（1）因为7coscosaBbAac，根据正弦定理得，·6·7sincossincossin,7

ABBAaC································································2分7sinsin,7CaC又因为sin0,C，7.a

·································································································3分sin2sin,

2sincossin,AAAAA······························································4分因为sin0,A所以1cos2A，······································

································5分(0,),.3AA··············································································

······6分（2）由（1）知，7,.3aA由余弦定理得2222cos,abcbcA2227,7(),bcbcbcbc·······························

·································8分因为2bc，所以74,bc所以3.bc·······················································9分设BC边上的高为h.11333sin3.2224ABCSbcA

△··························································10分11337,224ABCSahh△，321.14h即BC边上的

高为32114.············································································12分18.【命题意图】本题主要考查直

线与直线、直线与平面的位置关系，二面角等基础知识，意在考查直观想象、逻辑推理与数学运算的数学核心素养．满分12分．【解析】（1）连结CE，交DF于N，连结MN，如图所示，因为四边形CDEF是矩形，所以N是CE的中点，···························

·················1分由于M是AE的中点，所以//MNAC，·························································································2分由于MN平面MDF，

AC平面MDF，所以//AC平面MDF.………………………………………4分（2）因为平面ABCD平面CDEF，平面ABCD平面CDEFCD，DECD，所以DE平面ABCD，可知,,ADCDDE两两

垂直，········································································5分以点D为原点，分别以,,DADCDEuuuruuuruuur的方向为x轴、y轴、z轴的正方向，建立空间直角

坐标系Oxyz.因为2AB，则(1,0,1)M，(0,4,2)F，(1,0,1)DM，(0,4,2)DF，设平面MDF的法向量为1(,,)nxyz，·7·则110,0,nDMnDF所以0,420,xzyz··············

····················································7分取1y，则1(2,1,2)n，······································

····································8分依题意，得平面ABCD的一个法向量为2(0,0,1)n，········································9分12121222cos,34141nnnn

nn，··················································11分故平面MDF与平面ABCD所成锐二面角的余弦值为23.·················

···················12分19.【命题意图】本题主要考查线性回归、相关系数等基础知识，意在考查数学建模、数学抽象、数学运算、数据分析的数学核心素养．满分12分．【解析】（1）由折线图中的数据

和附注中参考数据得12345678959t，·····························································1分92160iitt，············

··········································································2分999111iiiiiiiittyytyty3254.8055

82.01344.75，·······················4分所以344.750.997345.90r，···································································

·········5分因为y与t的相关系数近似为0.997，说明y与t的线性相关程度相当高，从而可以用线性回归模型拟合y与t的关系.···················································6分（2）由已知及（1）得91921

ˆiiiiittyybtt344.755.74660，··························8分64.6685.74ˆ653ˆ5.94aybt，···················································

········9分所以y关于t的回归方程为35.945.75yt.····················································10分将2021年对应的年份代码12t代入回归方

程，得35.945.7512104.94y，所以预测2021年全国GDP总量约为104.94万亿元.·······································12分20.【命题意图】本题主要考查椭圆方程、

直线与椭圆的位置关系等基础知识，意在考查的数学核心素养是直观想象、逻辑推理与数学运算．满分12分.【解析】(1)由题意得2222411312abcbeaa，··········································

·······2分解得228,2ab，····················································································3分所以C的方程为

22182xy.········································································4分·8·(2)设直线1:02lyxmm，11

22,,,AxyBxy，由2212182yxmxy，得222240xmxm，由22481600mmm，解得20m或02m，···································5分则212122,24xxm

xxm，····································································6分依题意，易知PA与PB的斜率存在，所以,22，设直线PA与PB的斜率分

别为12,kk，则1tank，2tank，··························7分欲证，只需证tantantan，即证120kk.······

··············8分12121211,,22yykkxx故1221121122121212112222yxyxyyxxxxkk.··

································9分又111,2yxm2212yxm，所以12211212yxyx122111121222xmxxmx

··················································10分121212241xxmxxmxx2242241mmmm0,···························

············································································11分120,kk所以.·······························

·········································12分21.【命题意图】本题主要考查函数和导数及其应用、三角函数等基础知识，意在考查直观想象、逻辑推理与数学运算的数学核心素养．满

分12分.【解析】(1)设112cosgxfxxx，···················································1分当0,x时，212sin0gxxx

，所以gx在0,上单调递减，····································································2分又因为331103g，

2102g，········································3分所以gx在,32上有唯一的零点，即函数fx在0,上

存在唯一零点．··························································4分当0,x时，0fx，fx在0,上单调递增;

·9·当,x时，0fx，fx在,上单调递减;所以fx在0,上存在唯一的极大值点.32····································5分(2)①由(1)知：fx在

0,上存在唯一的极大值点.32所以ln2202222ff，又因为2222111122sin220eeeef，所以fx在0,上恰有一个零点

．·····························································6分又因为ln20f，所以fx在,上也恰有一个零

点．··························································7分②当,2x时，sin0x≤，lnfxxx≤，设lnhxxx，110hxx，所以hx在,2上单调递减，所以0hxh„

，·····································8分所以当,2x时，0fxhxh≤≤恒成立，所以fx在,2上没有零点．················

··················································9分③当2,x时，ln2fxxx≤，设ln2xxx，110xx，所以x在2,上单调递减，所以2

ln222222420x≤···································10分所以当2,x时，20fxx≤≤恒成立所以

fx在2,上没有零点．·······························································11分综上，fx有且仅有两个零点．····················

·············································12分22.【命题意图】本题主要考查直线的参数方程和参数的几何意义，直角坐标方程和极坐标方程的互化，直线和圆的位置关系等基础知识，意在考

查直观想象、逻辑推理与数学运算的数学核心素养．满分10分.【解析】(1)曲线C的方程4cos6sin，∴24cos6sin，····················································

······················2分∴2246xyxy，·················································································4分即C的直角坐标

方程为222313xy.··················································5分(2)设点MN，对应的参数分别为12tt，.·10·把直线232212xtyt(t为参数)代入C得，2

222121322tt，整理得，23280tt.23232500，1232tt，128tt，·········································7分∴12tt，为异

号，························································································8分又∵点3,1A在直线l上，∴21212121245052AMANtttttttt.

···························10分23.【命题意图】本题主要考查解绝对值不等式和不等式恒成立等基础知识，意在考查直观想象、逻辑推理与数学运算的数学核心素养．满分10分.【解析】1当2a时，21211

fxxx，当12x„时，不等式122121fxxx成立；··································1分当1122x时，122141f

xxxx，1124x；·················································································

······2分当12x…时，212121fxxx不成立，·······································3分综上，不等式1fx的解集为14xx

.····················································5分(2)当1,2x时，1fxx化为2111xaxx，321xax，·········································

············································6分23132,xaxx1333axx，····················································7分13yx在1

,2单调递减，故522y；················································8分33yx在1,2单调递增，故302y，·········

·············································9分所以20a剟，所以a的取值范围是2,0.·····················································

····················10分