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绝密★启用前卷类型：A2020年茂名市高三级第二次综合测试文科数学2020.5本试卷分选择题和非选择题两部分，共6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.一、选择题：(本大题共12小题，每小题5分，共60分．在每小题给出的四个选项中，有且只有一项是符合题目要求的)1．已知集合U={1,2,3,4,5}，A={2,3,5}，B={2,5}

，则()A.ABB.∁UB={1,3,4}C.A∪B={2,5}D.A∩B={3}2．若()i2i,,Rxiyxy，则复数ixy的虚部为()A.2B.1C.iD.−13．已知函数f(x)在点(1,f(1))处的切线方程为x+2y−2=0，则f(

1)+f′(1)=()A．32B．1C．12D．04．函数()=sin()(0,0,||)2fxAxA的图象如图所示，则()3f的值为()A．12B．1C．2D．35．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．若等比数列{an}公比为q，则―q＞1‖是―{an}为递增数列‖的

充要条件6．《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，三与八同道居左，四与九为友居右

，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.15B.625C.725D.825xOy2623–2第4题图第6题图7．“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.如图所示是一个当型循环结构的“辗转相除法”程序框图.当输入m=2020，n=303时，则输出的m是()A.2B.6C.101D.2028．已知双曲线22221yxab(a＞0,b＞0)的离心率为2，其一条渐近线被圆

(x−m)2+y2=4(m＞0)截得的线段长为2，则实数m的值为()A．3B．2C．2D．19．已知函数()fx是定义在R上的偶函数，当0x时，1()22xfx．则使不等式9(1)4fx成立的x取值范围是()A.(,1)(3,)∪B.(1,3)C.(0,2)D.(

,0)(2,)∪10．函数1+e()cos1exxfxx在[−5,5]的图形大致是()11．已知三棱锥PABC中，2,3,5,4,3APBPAPBACBC且平面PAB⊥平面ABC，则该三棱锥的外接球的表面积

为()A．16B．28C．24D．3212．已知函数+1()=e1xxfxx，对于函数()fx有下述四个结论：①函数()fx在其定义域上为增函数；②对于任意的0a，都有()1fa成立；③()fx有且仅有两个零点；④若y=ex在点0

0(,e)xx0(1)x处的切线也是y=lnx的切线，则x0必是()fx零点．其中所有正确的结论序号是A．①②③B．①②C．②③④D．②③二、填空题：(本大题共4小题，每小题5分，共20分.把答案填在答题卡的相应位置)13．已知向量(4,2)a，(1,1)b，

若()bakb，则k.14.为了贯彻落实十九大提出的“精准扶贫”政策，某地政府投入16万元帮助当地贫困户通过购买机器办厂的形式脱贫，假设该厂第一年需投入运营成本3万元，从第二年起每年投入运营成本比上一年增加

2万元，该厂每年可以收入20万元，若该厂n（n∈N*）年后，年平均盈利额达到最大值，则n等于.（盈利额=总收入−总成本）15．在棱长为2的正方体ABCD–A1B1C1D1中，E是棱DD1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第7题图AOy5−5COy5−5xxBOy5−5xDOy5−5xBCAB1C1A1DD1EF第15题图的中点，则平面A1EC截该正方体所得截面面积为：.16．过点11,2P作圆221xy的切线l，已知A，B分别为切点，直线AB恰好经过椭

圆的右焦点和下顶点，则直线AB方程为；椭圆的标准方程是.（第一空2分，第二空3分）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)

(一)必考题：共60分17．(12分)在ABC△中，角A，B，C的对边分别为a，b，c，已知2BC，34bc．(1)求cosC；(2)若3c，求ABC△的面积．18．(12分)某种治疗新型冠状病毒感染肺炎的复方中药产品的质量以其质量指标值

衡量，质量指标越大表明质量越好，为了提高产品质量，我国医疗科研专家攻坚克难，新研发出A、B两种新配方，在两种新配方生产的产品中随机抽取数量相同的样本，测量这些产品的质量指标值，规定指标值小于85时为废品，指标值在

[85,115)为一等品，大于115为特等品.现把测量数据整理如下,其中B配方废品有6件.A配方的频数分布表质量指标值分组[75,85)[85,95)[95,105)[105,115)[115,125)频数8

a36248(1)求a,b的值；(2)试确定A配方和B配方哪一种好？(说明：在统计方法中，同一组数据常用该组区间的中点值作为代表)组距频率B配方的频频率分布直方图758595105115125质量指标值O0.0080.006b0.0220.038第18题图19．(

12分)如图1，在□ABCD中，AD=4，AB=22，∠DAB=45°，E为边AD的中点，以BE为折痕将△ABE折起，使点A到达P的位置,得到图2几何体P−EBCD.(1)证明：PDBE；(2)当BC⊥平面PEB时，求三棱锥C−PBD的体积.20．(12分)已知抛物线C:y2=2px

(p＞0)与直线l:x+y+1=0相切于点A，点B与A关于x轴对称.(1)求抛物线C的方程，及点B的坐标；(2)设M、N是x轴上两个不同的动点，且满足∠BMN=∠BNM，直线BM、BN与抛物线C的另一个

交点分别为P、Q，试判断直线PQ与直线l的位置关系，并说明理由.如果相交，求出的交点的坐标.EBCAD第19题图1PEBCD第19题图221．(12分)设函数2()(+)exfxxm.(1)讨论()fx的单调性；(2)若()2e1()xgxnxfx，当m

=1，且0x时，()0gx，求n的取值范围.(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，则按所做的第一题计分，作答时，请用2B铅笔在答题卡上把所选题目对应的题号涂黑．22．[选修4−4：坐标系与参数方程](10分)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()4a，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．[选修4-5：不等式选讲](10分)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.绝密★启用前卷类型：A20

20年茂名市高三级第二次综合测试文科数学参考答案及评分标准一、选择题（本大题共12小题，每小题5分，共60分）题号123456789101112答案BBDBDACCAABC二、填空题：本大题共4小题，每小题5分，共20分.把答

案填在答题卡的相应位置.13．314.415.2616.2x−y−2=0(2分);22154yx(3分).提示：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

.(一)必考题：共60分17.解：(1)依题意，由正弦定理得：3sin4sinBC．·····································1分∵2BC，∴3sin24sinCC，······

·················································2分∴3sincos2sinCCC，················································

··············3分∴(0,)C，sin0C，··················4分∴2cos3C．···············5分(2)解法一：由题意得：3,4cb.··················································

6分∵(0,)C，∴25sin1cos3CC，··········································7分∴45sinsin22sincos9BCCC，···············································

8分221coscos2cossin9BCCC，···············································9分∴sinsin()sin()sincoscossinABCBCBCBC.···········10分4557

521.939327················································11分∴7514511sin4322279ABCSbcA.·····

································12分解法二：由题意及(1)得：3,4cb.2cos3C··································6分∵

(0,)C，∴25sin1cos3CC，···········································7分由余弦定理222=+2coscababC得：229=+1683aa，························8分即231621=

0aa，解得7=3=3aa或.···············································9分若=3a，又3,c则A=C，又B=2C，得△ABC为直角三角形，而三边为=3,4,3abc的三角形不构成直角三角形

，矛盾.∴7=3a.·················11分∴5145711sin422339ABCSabC.·······································12分18.解：(1)依题意，A、B配方样本

容量相同，设为n，又B配方废品有6件.由B配方的频频率分布直方图，得废品的频率为60.00610n，·················1分解得n=100.···················2分∴a=100−(8+36+24+8)=24.···············3分由(0.00

6+b+0.038+0.022+0.008)10=1······························4分解得b=0.026.因此a,b的值分别为24,0.026；································5分(2)由(1)及A配方的频数分布表得，A配方质量指标值

的样本平均数为808902410036110241208=100Ax20082002410036==100.100····7分质量指标值的样本方差为21=100As[(−20)28+(−10)224+036+102

24+2028]=112.···8分由B配方的频频率分布直方图得，B配方质量指标值的样本平均数为Bx=800.06+900.26+1000.38+1100.22+1200.08=100.··············9分质量指标值的样本方差为5

221()Biiisxxp=(−20)20.06+(−10)20.26+00.38+1020.22+2020.08=104.········10分综上ABxx，2As＞2Bs，···································11分即两种配方质量指标值的样本

平均数相等，但A配方质量指标值不够稳定，所以选择B配方比较好.···········································································12分(2)当BC⊥平面PEB时，求三棱锥C−PBD的体积.EBCAD

PECD19.证明：（1）依题意，在△ABE中（图1），AE=2，AB=22，∠EAB=45°，由余弦定理得EB2=AB2+AE2−2AB·AEcos45°=8+4−222222=4，···············································

················2分∴AB2=AE2+EB2，···········································································3分即在□ABCD中，EB⊥AD.·························

···········································4分以BE为折痕将△ABE折起，由翻折不变性得，在几何体P−EBCD中，EB⊥PE，EB⊥ED.又ED∩PE=E，∴BE⊥平面PED，···························5分又BE平面PEB，

∴PDBE；·······················································6分(2)∵BC⊥平面PEB，PE平面PEB，∴BC⊥PE.····································7分由

(1)得EB⊥PE，同理可得PE⊥平面BCE，·············································8分即PE⊥平面BCD，PE就是三棱锥P−CBD的高.·························

···············9分又∠DCB=∠DAB=45°，BC=AD=4，CD=AB=22，PE=AE=2，∴S△CBD=12BCCDsin45°=1242222=4.·································

10分VC−PBD=VP−CBD=13S△BCDPE=1342=83.因此，三棱锥C−PBD的体积为83.··························································12分（写出VC−PB

D=VP−CBD得1分，结果正确并作答得1分）20.解：(1)联立22,10,ypxxy·········································1分消去x得y2+2py+2p=0，················

···························2分∵直线与抛物线相切，∴△=4p2−8p=0，又p＞0，解得p=2，∴抛物线C的方程为y2=4x．·········3分由y2+4y+4=0，得y=−2，∴切点为A(1,−2)，∵点B与A关于x轴对称，

点B的坐标B(1,2).···········4分(2)直线PQ∥l.····························5分理由如下：依题意直线BM的斜率不为0,设M(t,0)(t≠1),直线BM的方程为x=my+t,·····6分由(1)B(1,2)，

1=2m+t，∴直线BM的方程为x=12ty+t，·························7分代入y2=4x．解得y=2(舍)或y=−2t，∴P(t2,−2t).·······························8分∵∠BMN=∠BNM，∴M、N关于AB对称，

得N(2−t,0).·····················9分同理得BN的方程为x=12ty+2−t，代入y2=4x．得Q((t−2)2,2t−4).···········10分224444144(2)PQttkttt

，·······················································11分直线l的斜率为−1，因此PQ∥l.····················································

···12分21.解：(1)依题得，()fx定义域为R，2()(+2+)exfxxxm，e0x，··········1分令2()2hxxxm，=44m△.xOyNBMPQA①若0△，即1m，则()

0hx恒成立，从而()0fx恒成立，当且仅当1m，1x时，()0fx.所以()fx在R上单调递增.································································2分②若0△＞，即1m＜，令()0hx，得11xm

或11xm.当(11,11)xmm时，()0fx；····································3分当(,11)(11,)xmm时，()0fx

.·····················4分综合上述：当1m时，()fx在R上单调递增；当1m＜时，()fx在区间(11,11)mm上单调递减，()fx在区间(,11),(11,)mm上单调递增.·

··················5分(2)依题意可知：2()21()1xxxgxenxfxexenx···················6分令0x，可得(0)0g,··············································

·············7分2()(12)(R)xgxxxenx.设2()(12)xhxxxen，则2()(41)xhxxxe.·····························8分当0x时,()0hx，()gx单调递减,·······

·······························9分故()(0)1gxgn.······················································10分要使()0gx在0x时恒成立,需要()gx在[

0,)上单调递减，所以需要()10gxn.······················································11分即1n,此时()(0)0gxg,故1n.综上所述,n的取值范围是[1,).················

······················12分(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，则按所做的第一题计分，作答时，请用2B铅笔在答题卡上把所选题目对应的题号涂黑．22.解:(1)将曲线C:2c

os,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，∴=2cos()=244a.············2分∴cos()=24，展开

得2(cossin)=22,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(2)解法一：由(1)，将直线l的参数方程代入曲线C的普通方程得：22221

1(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(0,0)到直线l的距离为|002|22d

.····································9分因此，△OAB的面积为1221112||27722SABd.···················10分(2)解法二:由(1),联立221,43

+20,yxxy消去y得:271640xx,显然△＞0.····6分设1122(,),(,)AxyBxy，则由韦达定理得12167xx，1247xx.······················7分由弦长公式得|AB|=2212121+()4kxxxx=

21221642()4777，····················8分又原点O(0,0)到直线l的距离为|002|22d.··································9分因此，△OAB的面积为1221112||2772

2SABd.··················10分(2)解法三：由(1),联立221,43+20,yxxy消去y得:271640xx,显然△＞0.····6分设1122(,),(,)

AxyBxy，则由韦达定理得12167xx，1247xx.·····················7分∵直线l过椭圆右顶点(2,0)，∴21627x，∴227x······················8分把227x代入直线l的方程得，2127y······

················9分因此，△OAB的面积为2111212||27722SOAy.··························10分23．解：(1)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)=2

x−1,由f(x)≤1,即2x−1≤1,解得x≤1,∴−1≤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x≤1

.·····························································5分(2)由(1)知f(x)≤3，当且仅当x≥2时，f(x)=3，····································

····6分∴M=f(x)Max=3.即a+b+c=3，·······················································7分·∵a2+b2≥2ab，a2+c2≥2ac，c2+b2≥2cb，·············

································8分∴2(a2+b2+c2)≥2(ab+ac+cb)∴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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