福建省 2022届高三数学三模试题（PDF版带解析）

2021—2022学年福州第一中学高三校质检数学试题（完卷时间120分钟；满分150分）第Ⅰ卷一、单项选择题：本题共8小题，每小题5分，共40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M，N是R的子集，且MN，则M(ðRN)A．MB．NC．D．R6242.(2xy)的展开式中，xy项的系数是A．30B．30C．60D．601tan3323.若sin，且(,)，则521tan211A．B．C．2D．2224.以下四组向量在同一平面的是A．(1,1,0)，(0,1,1)，(1,0,1)B．(3,0,0)，(1,1,2)，(2,2,4)C．(1,2,3)，(1,3,2)，(2,3,1)D．(1,0,0)，(0,0,2)，(0,3,0)25.已知函数fxln(xx1)cos(x).则当[0,]时，f(x)的图象不可能是A.B.C.D.π16.已知函数fxsinx(0,0)的图象过点P(0,)，现将yf(x)的22π图象向左平移个单位长度得到的函数图象也过点P，则3A.的最小值为2B.的最小值为6C.的最大值为2D.的最大值为67.已知AB，CD分别是圆柱上、下底面圆的直径，且ABCD．O，O分别为上、下底面1的圆心，若圆柱的底面圆半径与母线长相等，且三棱锥ABCD的体积为18，则该圆柱的侧面积为A．9B．12C．16D．181\n8.数学赋予建筑美感与活力，许多建筑融入数学元素，更具神韵.图（1）是单叶双曲面（由双曲线绕虚轴旋转形成的立体图形）型建筑.图（2）是其最细处的部分截面图象，上、下底面与地面平行.现测得下底直径AB2010米，上底直径CD202米，AB与CD间的距离为80米，与上下底面等距离的E点处的直径等于CD，则该建筑最细处的直径为A．10B．20C．103D．105图（1）图（2）二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，部分选对的得2分，有选错的得0分．19．设复数z(aR)，当a变化时，下列结论正确的是a2iA．|z|=|z|恒成立B．z可能是纯虚数11C．z可能是实数D．|z|的最大值为z210.某人投掷骰子5次，由于记录遗失，只有数据平均数为3和方差不超过1，则这5次点数中可能出现的结果是A．众数为3B．中位数为2C．极差为2D．最大点数为511.已知曲线C是平面内到定点F(0,1)和定直线l:y1的距离之和等于4的点的轨迹，若P(x,y)在曲线C上，则下列结论正确的是00A．曲线C关于x轴对称B．曲线C关于y轴对称C．2≤x≤2D．1≤PF≤40112.已知函数f(x)lnxx，则下列结论正确的是xA．f(x)为偶函数B．f(x)有且仅有两个零点C．f(x)既无最大值，也无最小值D．若xx0且f(x)f(x)0，则xx11212122\n第Ⅱ卷三、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．13．已知等比数列a的前n项和为S，a1，a8a，若S31，则n.nn152n2214.过点M(2,3)的直线与eC:(x3)y16交于A，B两点，当M为线段AB中点时，uuruurCACB.15.产品质量检验过程主要包括进货检验（IQC），生产过程检验（IPQC），出货检验（OQC）3三个环节.已知某产品IQC单独通过率为，IPQC单独通过率为p(0p1)，规定上一4类检验不通过则不进入下一类检验，未通过可修复后再检验一次（修复后无需从头检验，通过率不变且每类检验最多两次），且各类检验间相互独立.若该产品能进入OQC的概5率为，则p.616.在三棱锥PABC中，AP平面PBC，PBPC，PAPC2PB4，则三棱锥PABC外接球的表面积为；若动点M在该三棱锥外接球上，且MPBMPC，则点M的轨迹长为.（第一空2分，第二空3分）四、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．17.（本题满分10分）AB已知△ABC的内角A，B，C所对的边分别为a，b，c，bsincsinB.2（1）求角C；（2）若AB边上的高线长为23，求△ABC面积的最小值.18.（本题满分12分）新能源汽车是指除汽油、柴油发动机之外的所有其他能源汽车，被认为能减少空气污染和缓解能源短缺的压力.在当今提倡全球环保的前提下，新能源汽车越来越受到消费者的青睐.某车企调查了近期购车的200位车主的性别与购车种类的情况，得到如下数据：购置新能源汽车购置传统燃油汽车总计男性8020100女性6535100总计14555200（1）根据表中数据，判断能否有95%的把握认为是否购置新能源汽车与性别有关；3\n（2）已知该车企某汽车销售服务4S店现有5种款式不同的汽车在销售，每种款式的汽车均有新能源和传统燃油两种类型各1辆.假设某单位从这10辆汽车中随机购买4辆汽车，设其中款式相同的汽车的对数为，求的分布列与数学期望.22nadbc附：K，nabcd.abcdacbd2PKk00.100.050.0250.010k2.7063.8415.0246.635019.（本题满分12分）已知空间几何体ABCDE中，△ABE与△BCD是全等的正三角形，平面ABE平面ABC，平面BCD平面ABC.（1）求证：AC∥DE；（2）若ABBC，求直线BD与平面ADE所成角的正弦值.20.（本题满分12分）设数列an的前n项和为Sn，a10，a21，nSn1(2n1)Sn(n1)Sn110(n≥2).（1）证明：an为等差数列；ban，在b和bd的等差数列，（2）设n2nn1之间插入n个数，使这n2个数构成公差为n1求的前n项和.dn21.（本题满分12分）22xy3已知椭圆C:1(ab0)的右顶点为A(2,0)，离心率为.22ab2（1）求C的方程；（2）设斜率为1的直线l与C交于P，Q两点，点P关于x轴的对称点为M，若△PQM的外接圆恰过坐标原点，求直线l的方程.22.（本题满分12分）x已知函数f(x)easinx1在区间(0,)内有唯一极值点x1.2（1）求实数a的取值范围；（2）证明：f(x)在区间(0,)内有唯一零点x，且x2x.2214\n2021—2022学年福州第一中学高三校质检数学答案与评分标准一、单项选择题：本题共8小题，每小题5分，共40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．12345678CCDBDADB二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，部分选对的得2分，有选错的得0分．9101112ABDACBDBCD三、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．213.514.815.16.36；343四、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．17.解：（1）由已知ABC，ABCC所以bsinbsinbcos，··························································1分222CC所以bcoscsinB，由正弦定理得sinBcossinCsinB，····························2分22C显然sinB0，则cossinC，··································································3分2CCC则cos2sincos，·············································································4分222CC1C因为cos0，所以sin，因为0，22222C所以，则C.···············································································5分26311（2）由Sc23absinC，得ab4c，···········································7分△ABC22222222由余弦定理cab2abcosC，得cabab≥2ababab，···············8分2即c≥4c，得c≥4，当且仅当abc4取等，············································9分此时△ABC面积的最小值为43.································································10分22200(80356520)180018.（1）依题意可得K5.6433.841，·············3分14555100100319所以有95%的把握认为是否购置新能源汽车与性别有关.····································5分（2）的可能取值有0，1，2，·································································6分\n44C285则P(0)，··········································································7分4C2110122CC2454P(1)，············································································8分4C7102C15P(2)，·················································································9分4C2110所以的分布列为012P84121721············································································································10分8412所以E()012.····························································12分21721319.（1）证明：设M，N分别为边AB，边BC的中点，连接EM，DN，因为△ABE为等边三角形，所以EMAB，··················································1分因为平面ABE平面ABC，且平面ABEI平面ABCAB，所以EM平面ABC，···············································································2分同理可证DN平面ABC，所以EM∥DN，······················································································3分因为△ABE与△BCD是全等的正三角形，所以EMDN，所以四边形EMND为平行四边形，·······························································4分所以DE∥MN，因为MN为△ABC的中位线，所以MN∥AC，所以AC∥DE.·································································5分uuuruur（2）因为ABBC，以B为坐标原点，BC方向为x轴，BA方uuur向为y轴，ME方向为z轴建立空间直角坐标系，如图所示...............................................................................................6分设AB2，则B(0,0,0)，A(0,2,0)，D(1,0,3)，E(0,1,3)，................................................................................................7分uuuruuur所以AD(1,2,3)，DE(1,1,0)，ur设平面ADE的法向量为m(x,y,z)，uuururADm0x2y3z0ur则uuurur，所以，取m(3,3,3)，·································9分DEm0xy0\nuuurBD(1,0,3)，设直线BD与平面ADE所成角为，uuururuuurBDm621则sincosBD,muuurur，·····································11分BDm421721所以直线BD与平面ADE所成角的正弦值为.··········································12分720.（1）证明：因为n≥2时，nS(2n1)S(n1)S10，n1nn1则n(SS)(n1)(SS)10，n1nnn1即na(n1)a10,n≥2，···································································2分n1n因为a2a10，··················································································3分21*则na(n1)a10,nN①，n1n所以(n1)ana10,n≥2②，nn1则①－②得na2nana0,n≥2，n1nn1即aa2a,n≥2，············································································5分n1n1n所以a为等差数列.·················································································6分n（2）a的首项为a0，公差为aa1，所以an1，n121nn1所以b2，··························································································7分nnn1n1bn1bn2221n1所以d，则，·········································8分nn1n1n1n1d2n1记的前n项和为Tn，dn1011121n1则T2()3()4()(n1)()①，n222211112131n11n所以T2()3()4()n()(n1)()②，··············9分n22222211121n11n则①－②得T2()()()(n1)()，·······························10分n222221n1()121n1n所以T1(n1)()3(n3)()，········································11分n2122121n1所以T6(n3)().············································································12分n2\na2,c321.解：（1）依题意,···································································2分a2222abc.a2解得，····························································································3分b12x2所以椭圆C的标准方程为y1.·····························································4分4（2）设l的方程为yxm，设P(x,y)，Q(x,y)，则M(x,y).112211yxm,22由x2消去y得，5x8mx4m40，2y1.422依题意64m20(4m4)0，即5m5，······································5分8mxx125所以，················································································6分24m4xx1258m2m4mm所以yyxx2m2m，所以线段PQ的中点坐标为(,)，12125555m4m3m所以线段PQ的中垂线方程为y(x)，即yx，······················7分5553m依题意，线段PQ的中垂线与x轴的交点E(,0)即为△PQM外接圆的圆心，52m点E到直线l的距离为d，5228m216(m1)422PQ2(xx)4xx2()5m，·············8分1212555PQ2222406m设△PQM外接圆的半径为r，则rd()，··························9分22523m22406m所以△PQM外接圆的方程为(x)y，·································10分525因为△PQM外接圆恰过原点O(0,0)，23m2406m26所以()，解得m，·····················································11分525326所以直线l的方程为yx.·································································12分3\nx22.解：（1）f(x)eacosx，···································································1分xx(0,)时，cosx(0,1)，1ee2，2①当a≤1时，f(x)0，f(x)在(0,)上单调递增，没有极值点，不合题意，舍去.·································2分2②当a1时，显然f(x)在(0,)上递增，又因为f(0)1a0，f()e20，22所以f(x)在(0,)上有唯一零点x，所以x(0,x)，f(x)0；x(x,)，f(x)0，11122所以f(x)在(0,)上有唯一极值点，符合题意.···············································3分2综上，a(1,).······················································································4分x（2）由（1）知a1，所以x,时，f(x)eacosx0，·····················5分2所以x(0,x)，f(x)0，f(x)单调递减；x(x,)，f(x)0，f(x)单调递增，11所以x(0,x)时，f(x)f(0)0，······························································6分1则f(x)0，又因为f()e10，··························································7分1所以f(x)在(x,)上有唯一零点x，12即f(x)在(0,)上有唯一零点x.··································································8分2因为f(2x)e2x1asin2x1e2x12asinxcosx1，1111由（1）知f(x)0，所以ex1acosx，则f(2x)e2x12ex1sinx1，11112tt构造p(t)e2esint1,t(0,)，···························································9分22tttt所以p(t)2e2e(sintcost)2e(esintcost)，tt记(t)esintcost,t(0,)，则(t)ecostsint，显然(t)在(0,)上单调递增，22所以(t)(0)0，所以(t)在(0,)上单调递增，所以(t)(0)0，2所以p(t)0，所以p(t)在(0,)上单调递增，所以p(t)p(0)0，2所以f(2x)0f(x)，···········································································11分12由前面讨论可知：x2x，xx，1112且f(x)在x(x,)单调递增，所以2xx.················································12分112