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绝密★启用前试卷类型：A2020年茂名市高三级第二次综合测试数学试卷(理科）2020.5本试卷分选择题和非选择题两部分，共5页，23小题，满分150分，考试时间120分钟.注意事项：1．答题前，先将自己的姓名、准考证号填写在试题

卷和答题卡上.2．选择题的作答：每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑.写在试题卷、草稿纸和答题卡上的非答题区域均无效.3．填空题和解答题的作答：用签字笔直接答在答题卡上对应的答题区域内.写在试题卷草稿纸和答题卡上的非答题区域均无效.

4．选考题的作答：先把所选题目的题号在答题卡上指定的位置用2B铅笔涂黑.答案写在答题卡上对应的答题区域内，写在试题卷、草稿纸和答题卡上的非答题区域均无效.5．考试结束后，请将答题卡上交.第一部分选择题(共60分）1.若()2,,xiiyixyR，则复数yi

x的虚部为（）A.2B.1C.iD.-12．已知集合UR,2lg(4)Axyx，21xxB，则AB（）A．(2,2)B．(2,1)C．[2,2]D．[2,2)3.已知π1sin62，且π0,

2则πcos3（）A.0B.12C.1D.324.下列命题错误的是（）A．“x＝2”是“x2−4x+4＝0”的充要条件B．命题“若14m，则方程x2+x−m＝0有实根”的逆命题为真命题C．

在△ABC中，若“A＞B”，则“sinA＞sinB”D．命题p：“x0∈R，x02−2x0+4＞0”，则﹁p：“x∈R，x2−2x+4＜0”5.《易·系辞上》有“河出图，洛出书”之说，河图、洛书是中国古代流传下来的两幅神秘图案，蕴

含了深奥的宇宙星象之理，被誉为“宇宙魔方”，是中华文化阴阳术数之源。河图的排列结构如图所示，一与六共宗居下，二与七为朋居上，三与八为友居左，四与九同道居右，五与十相守居中，其中白圈为阳数，黑点为阴数，若从阳数和阴数中各取一数，则其差的绝对值为5的概率为（）A.51

B.256C.257D.2586.“辗转相除法”是欧几里得《原本》中记录的一个算法，是由欧几里得在公元前300年左右首先提出的，因而又叫欧几里得算法.如图所示一个当型循环结构的“辗转相除法”程序框图.当输入m=1995，n=228，输出的m是()A.3B.19C.57D.1147.如图，某沙

漏由上下两个圆锥组成，圆锥的底面直径和高均为8cm,细沙全部在上部时，其高度为圆锥高度的21（细管长度忽略不计）.当细沙全部漏入下部后，恰好堆成一个盖住沙漏底部的圆锥形沙堆，则此沙堆的侧面积为（）A.45B.85C

.317D.4178.设偶函数()fx满足1()()2(0)2xfxx，则使不等式914fx成立的x取值范围是（）A.(,1)(3,)B.(1,3)C.(0,2)D.(,0)(2,)9.圆M:224xmy与双曲线C:22221xy

ab（0a，0b）的两条渐近线相切于A、B两点，若32AB，则C的离心率为（）A.332B.3C.2D.310.某贫困县为了实施精准扶贫计划，使困难群众脱贫致富，对贫困户实行购买饲料优惠政策如下：否结束输出m是r＞0?r=1开始输入m,n求m除以n的余数rm=nn=r(1)

若购买饲料不超过2000元，则不给予优惠；(2)若购买饲料超过2000元但不超过5000元，则按标价给予9折优惠；(3)若购买饲料超过5000元，其5000元内的给予9折优惠，超过5000元的部分给予7折优惠．某贫穷户购买一批饲料，有如下两种方案：方案一：分两次付款购买，分别为2880元和

4850元；方案二：一次性付款购买.若取用方案二购买此批饲料，则比方案一节省（）元A.540B.620C.640D.80011.已知六棱锥PABCDEF的底面是正六边形，PA平面ABC，2PAAB.则下列命题中正确的有()．①平面PAB⊥平面PAE；②PB⊥AD；③直线CD与PF

所成角的余弦值为55；④直线PD与平面ABC所成的角为45°;⑤CD∥平面PAE.A.①④B.①③④C.②③⑤D.①②④⑤12.若关于x的方程1123042xxmmm在

,1上有唯一实数解，则实数m的取值范围（）43]320.A或，（]320.B，（41]920.C或，（]920.D，（第Ⅱ部分非选择题（共90分）二．填空题：本大题共4小题，每小题5分，共20分.13．已知向量42(,)a，11(,)b，若b(

akb)，则k.14．62()xx的展开式中，常数项是.15.已知曲线21ln(1)2fxxx在点1,1f处的切线的倾斜角为,则22sinsincos.16．在ABC中，角A，B，C所

对应的边分别为a，b，c，且6cos(2cos),6AaCc，则ABC的a，b的等量关系式为，其面积的最大值为.(本题第一空2分，第二空3分）三.解答（本大题共5小题，每题12分共60分.解答应写出文字说明，证明过程或演算步骤.）17.设*nN，数列na的前n项和为nS，

已知nnnass221，且211a，正项的等差数列nb的首项为2，且321,1,bbb成等比数列.(1)求na和nb的通项公式；(2)求证：1227nbbbaaa.18.如图，已知ABC内接于

圆O，AB是圆O的直径，四边形DBCE为平行四边形，F是CD的中点，（1）证明：//OF平面ADE；（2）若四边形DBCE为矩形，且四边形DBCE所在的平面与圆O所在的平面互相垂直，22ACAB，AE与圆O所在的平面的线面角为600．求二面角D-AE-B的平面角的余弦

值.19.已知椭圆)0(1:2222babyaxC右焦点与抛物线xy342的焦点重合，以原点为圆心、椭圆短半轴长为半径的圆与直线1l:2xy相切.（1）求椭圆的方程（2）若直线2l：02ybx与y

轴交点为P，A、B是椭圆上两个动点，它们在y轴两侧,PBPA,APB的平分线与y轴重合，则直线AB是否过定点，若过定点，求这个定点坐标，若不过定点说明理由.20.2020年初全球爆发了新冠肺炎疫情，为了防控疫情，某医疗科研团队攻坚克难研发出一种

新型防疫产品，该产品的成本由原料成本及非原料成本组成，每件产品的非原料成本y（元）与生产该产品的数量x（千件）有关，根据已经生产的统计数据，绘制了如下的散点图.观察散点图，两个变量不具有线性相关关系，现考虑用函数byax对两个变量的关系进行拟合。参考数据

（其中1iiux）：u2u621uii61iiy621iiy61iiiuy0.48345252.440.410.16811.49230620858.44173.850.39（1）求y关于x的回归方程，并求y关于u的相关系数（精确到0.01）.（

2）该产品采取订单生产模式（根据订单数量进行生产，即产品全部售出）.根据市场调研数据，若该产品单价定为80元，则签订9千件订单的概率为0.7，签订10千件订单的概率为0.3；若单价定为70元，则签订10千件订单的概率为0.3，签订11千件订单的概率为0.7.已知每件产品的原料成本为30元

，根据（1）的结果，要想获得更高利润，产品单价应选择80元还是70元，请说明理由.参考公式：对于一组数据11,u，22,u，…，,nnu，其回归直线u的斜率和截距的最小二乘估计分别为：1221ˆniiiniiunuunu，ˆˆ

u，相关系数1222211niiinniiiiunurunun.21.已知函数ln1afxxx，aR.（1）若ea1,求证：)(xf有且只有两个零点（2

）axaxxxafxg222)()(有两个极值点1x,212()xxx,且不等式21)(mxxg恒成立，试求实数m的取值范围.(二)选考部分：共10分请考生在第22、23两题中任选一题作答，如果多做，则按所做的第一题计分，做答时，请用2B铅笔在答题

卡上把所选题目对应的题号涂黑．22.(本小题满分10分)选修4-4：坐标系与参数方程在平面直角坐标系xOy中，已知曲线C:2cos,3sin,xy(为参数)，以原点O为极点，x轴正半轴为极轴建立极坐标系，直线l的极坐标方程cos()4

a，点M(2,4).在直线l上，直线l与曲线C交于A，B两点.（1）求曲线C的普通方程及直线l的参数方程；（2）求△OAB的面积.23．（本小题满分10分）选修4-5：不等式选讲已知函数f(x)

=|x+1|−|x−2|.（1）若f(x)≤1，求x的取值范围；（2）若f(x)最大值为M，且a+b+c=M，求证：a2+b2+c2≥3.2020年茂名市高三级第二次综合测试数学试题参考答案和评分标准(理科数学

）一、选择题：本大题共12小题，每小题5分，共60分.123456789101112BDCDACDAACBA二．填空题：本大题共4小题，每小题5分，共20分.13．【答案】314．【答案】6015.【答案】2413

16．【答案】b=2a;12三.解答（本大题共5小题，每题12分共60分.解答应写出文字说明，证明过程或演算步骤.）17.解：（1）由nnnass221得，nnnnaass1122-2„„„„„„„„„„„„

„„„„1分即211nnaa„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„2分na是以首项21，公比为21的等比数列„„„„„„„„„„„„„„„„„„„„„„„„3分nna）（21„„„„„„„„„„„„„„„„„„„„„„

„„„„„„„4分设等差数列nb的公差为d，由21b，且321,1,bbb成等比数列.)22(2)1(2dd即03-2-2dd„„„„„„„„„„„„„„„„„„„„„„„„5分0d3d„„„„„„„„„„

„„„„„„„„„„„„„„„„„„„„„„6分13nbn„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„7分（2）由（1）得1313)21(nnbaan„„„„„„„„„„„„„„„„„„„„„„„„„8分nbbbaaa

21=1352naaa„„„„„„„„„„„„„„„„„„„„„„„„9分=1352)21()21()21(n„„„„„„„„„„„„„„„„„„„10分=33132)21(-1)21()21(-)21(n„„„„„„„„„„„„„„„„„„„„„„11分=31211

2-()7727n„„„„„„„„„„„„„„„„„„„„„„12分18.证明：（1）连结BE，DBCE平行四边形且F为CD中点F为BE中点„„„„„„„„„„„„„„„„„„„„„„„„1分又O为AB的中点//OFAE„„„„„„„„„„„„„„„2分AE平面ADE，OF平

面ADE„„„„„„„„„„„„„„3分//OF平面ADE„„„„„„„„„„„„„„„„„„„„„4分（2）矩形DBCE平面ABC，平面DBCE平面ABC=BC，ECBC，EC平面DBCE

EC平面ABC„„„„„„„„„„„5分又AB为圆O的直径ACBC以C点为原点建立如图所示的空间直角坐标系22ACABBC=3，AC=1由EC平面ABC得，∠EAC就是AE与平面ABC所成的角由tan600=ACCE得，

CE=3„„„„„„„„„„„„„6分A（1，0，0），E（0，0，3），D（0，3，3），B（0，3，0）„„„„„„„„„7分AE=（-1，0，3），AD=(-1,3,3),AB=(-1,3,0)„„„„„„„„„„„

8分设平面AED的一个法向量),,(m111zyx，由，，ADmAEm得，，0ADm0AEm即1111130330xzxyz，所以)1,0,3(m„„„„„„9

分同理可得，平面AEB的一个法向量)1,1,3(n„„„„„„„„„„„„„„„„„„„10分5522513nmnmn,mcos„„„„„„„„„„„„„„„„„„„11分二面角D-AE-B二面角D-AE-B的平面角的余弦值为55

2„„„„„„„„„„„„„„12分19.解：（1）抛物线xy342的焦点为)0,3(，所以c=3„„„„„„„„„„„„„„„1分∵直线:02yx与圆222byx相切，∴bd1112„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„2分∴42

22cba„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„3分∵椭圆C的方程是2214xy„„„„„„„„„„„„„„„„„„„„„„„„4分（2）b=1，直线:2l02yx与y轴交点P（0，2），„„

„„„„„„„„„„„„„„„„5分设椭圆上A、B两个动点的坐标为：1122(,)(,)AxyBxy、.AB方程为：,mkxy由0448)41(4)(4142222222mkmxxkmkxxyxmk

xy,4144,4182221221kmxxkmkxx得：„„„„„„„„„„„„„„„„„„„„„„„„6分2111112,222xmkKxmkxmkxxyKPBPA同理得：„

„„„„„„„„„„8分又APB的平分线在y轴上0441684428222222121mmkkmmmkkxxxxmkKKPBPA)())((.........10分,0,kPBPAm=21,„„„„„„„„„„„„„„

„„„„„„„„„„„„„„„11分直线21:kxyAB恒过定点)21,0(„„„„„„„„„„„„„„„„„„„„„„„12分20.【解析】（1）令1ux，则byax可转化为yabu，因为306516y，„„„„„„„„„„„

„„„„„„„„„„„„„„„„„„„„1分所以6162216173.860.41511001.492648.34ˆ0.4830.168416iiiiiuyuybuu，„„

„„„„„„„„„„„„2分则51ˆˆ1000.4110aybu，„„„„„„„„„„„„„„„„„„„„„„„„3分所以10100ˆyu，因此y关于x的回归方程为10010ˆyx；„„„„„„„„„„„„„4分y与

u的相关系数为：6126622221148.3448.3460.960.48345252.4450396.6iiiiiiiuyuyruuyy，„„„„„„6分

（2）法一：（i）若产品单价为80元，记企业利润为X（元），订单为9千件时，每件产品的成本为10010010304099元，企业的利润为100804090002600009[()]（元），„„„„„„„„„„„„„„„„„„„7分订单为10千件时，每件产

品的成本为10100103050元，企业的利润为805010000300000()（元），„„„„„„„„„„„„„„„„„„„„„„8分企业利润X（元）的分布列为X260000300000P0.70.3所以2600000.73000000.3272000EX

（元）；„„„„„„„„„„„„„„„9分（ii）若产品单价为70元，记企业利润为Y（元），订单为10千件时，每件产品的成本为10100103050元，企业的利润为705010000200000()（元），订单为11千件时，每件产品的成本为10010010

30401111元，企业的利润为10070401100023000011[()]（元），„„„„„„„„„„„„„„„„„10分企业利润Y（元）的分布列为Y200000230000P0.30.7所以2000000.32300000.7221000EY（元

），„„„„„„„„„„„„„„„„11分>EXEY又故企业要想获得更高利润，产品单价应选择80元.„„„„„„„„„„„„„12分法二：（i）若产品单价为80元，记企业的产量为X（千件），其分布列为X910P0.70.3所以90.7100.39.3EX

„„„„„„„„„„„„„„„„„„„„„„„„„„„8分企业的利润为：1008040930027200093[()].„„„„„„„„„„„„„„„„„„„9分（ii）若产品单价为70元，记企业的产量为Y（千件），其分布列

为X1011P0.30.7所以100.3110.710.7EY„„„„„„„„„„„„„„„„„„„„„„„„„10分企业的利润为：100704010700221000107[()].„„„„„

„„„„„„„„„„„„„„11分272000>221000又故企业要想获得更高利润，产品单价应选择80元.„„„„„„„„12分21.解:(1)1ln)(xaxxf定义域为),0(，2/12)(xa

xxaxxf，又ea1„„„„„„„„„„„„„„„„„„„„„„1分所以()fx在)1,0(e是减函数，在),1(e是增函数„„„„„„„„„„„„„„„„2分又01111ln)1(eef，0411l

n)1(2233eeeef所以()fx在)1,1(3ee有唯一零点，且在)1,0(e也有且只有唯一零点，„„„„„„„„„„3分同理01111ln)1(eef，01112eeeeefln)(所以()fx在),(ee1有唯一零

点，且在),1(e也有且只有唯一零点„„„„„„„„„„4分所以)(xf有且只有两个零点„„„„„„„„„„„„„„„„„„„„„„„„„„„„5分（2）xxxaxg2ln)(2定义域为),0(

，)(xg有两个极值点1x，212xxx，即02222)(2/xxxaxxaxg，0222axx有两不等实根)0(,2121xxxx„„6分1002a，且212111,22xxaxx，„„„„„„„

„„„„„„„„„„„„„„„„„„7分从而121012xx,„„„„„„„„„„„„„„„„„„„„„„„„„„„„8分由不等式21)(mxxg恒成立，得11211121211212112222xxxxxxxxaxxxxgmln)(

ln)(1111112ln1xxxx恒成立„„„„„„„„„„„„„„„„„„„10分令11()12ln012htttttt,当102t时21()12ln0(1)httt恒成立，所以函

数ht在10,2上单调递减，13()ln222hth，„„„„„„„„„„„„„„„„„„„„„„„„„11分故实数m的取值范围是)2ln23,(„„„„„„„„„„„„„„„„„„„„„„12

分(二)选考部分：22.解：（Ⅰ）将曲线C:2cos,3sin,xy消去参数得，曲线C的普通方程为：22143yx.„„„„„1分∵点M(2,4)在直线cos()4a上，∴a=2co

s()44=2.„„„„„„„„„„2分∴cos()24，展开得22(cos+sin)=2,又x＝cos，y＝sin，∴直线l的直角坐标方程为x+y−2=0,„„„„„„„„„„„„„„„„„„„„„„„„„4分显然l过

点(1,1),倾斜角为34.∴直线l的参数方程为21,221,2xtyt(t为参数).„„„„„„„„„„„„„„„„„„„„„5分（Ⅱ）法一：由（Ⅰ），将直线l的参数方程代入曲线C的普通方程得：222211(1)(1)14232tt，„„„„„„

„„„„„„„„„„„„„„„„„„„6分整理得2722100tt，显然△＞0设A,B对应的参数为t1,t2,则由韦达定理得12227tt，12107tt.„„„„„„„„„7分由参数t的几何意义得|AB|=|t1−

t2|=21212()4tttt=22212210()4777，„„„„8分又原点O到直线l的距离为d=|002|22.„„„„„„„„„„„„„„„„„„„9分因此，△OAB的面积为S=1221112||22277ABd.„„

„„„„„„„„„„„„„„10分法二：由（Ⅰ）可知，直线l的直角坐标方程为x+y−2=0,联立2220143xyxy,整理得271640xx，显然△＞0„„„„„„„„„„„„„„„„6分设A,B对应的坐标为11

22(,),(,)xyxyt1则由韦达定理得12167xx，1247xx.„„„„„„„7分所以2221212122164||(1)[()4]2[()4777ABkxxxx„„„„„„„„„„„„8分又原点O到直线

l的距离为d=|002|22.„„„„„„„„„v„„„„„„„„„„9分因此，△OAB的面积为S=1221112||22277ABd.„„„„„„„„„„„„„„„„10分法三：由（Ⅰ）可知，直线l的直

角坐标方程为x+y−2=0,联立2220143xyxy,整理得271640xx，显然△＞0„„„„„„„„„„„„„„„„6分设A,B对应的坐标为1122(,),(,)xyxy则由

韦达定理得12167xx，1247xx.„„„„„„„7分因为直线l过椭圆右顶点（2,0），所以71622x，722x„„„„„„„8分把722x代入直线l的方程得7122y„„„„„„„9分因此

，△OAB的面积为S=712712221.„„„„„„„„„„„„„„„„„„„„„10分23．解：(Ⅰ)由已知3,2,21,12,(,1.)3xxfxxx＜＜„„„„„„„„„„„„„„„„„„„„„„1分当x≥2时，f(

x)=3，不符合；„„„„„„„„„„„„„„„„„„„„„„„„„„„„2分当−1≤x＜2时，f(x)=2x−1，由f(x)≤1，即2x−1≤1，解得x≤1；„„„„„„„„„„„„„3分当x＜−1时，f(x)=−3，f(x)≤1恒成立.„„„„„„„„„„„„„„„„„„„„

„„„„4分综上，x的取值范围是(,1].„„„„„„„„„„„„„„„„„„„„„„„„„„„„5分(Ⅱ)()|1||2||12|3fxxxxx,由(Ⅰ)知当且仅当x≥2时，f

(x)=3，„„„„„„„„„„„„„„„„„„„„„„„„„6分所以M=f(x)Max=3.即a+b+c=3，„„„„„„„„„„„„„„„„„„„„„„„„„„„„7分因为a2+b2≥2ab，a2+c2≥2ac，c2+b2≥2cb，„„„„„„„„„„„„

„„„„„„„„„„8分所以2(a2+b2+c2)≥2(ab+ac+cb)所以3(a2+b2+c2)≥a2+b2+c2+2ab+2ac+2cb=(a+b+c)2=9„„„„„„„„„„„„„„„„9分因此(a2+b2+c2)≥3„„„„„„„„„„„„„„„„„„„„„„„„„„„

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