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共5页第1页巴中市高2018级“零诊”考试文科数学参考答案及评分细则一、选择题（每小题5分，共60分）题号123456789101112答案DBDAABCCBCBA二、填空题（每小题5分，共20分）13

.0000,sinxxx≥14.315.3216.②③④三、解答题（共70分）17、解：（1）由1122nnnaa变形得：12211nnnnaa·························

···································3分又21a，故112a··························································································4

分∴数列{}2nna是以1为首项1为公差的等差数列。·················································5分（2）由（1）知：nabnnn2·············

·····································································6分∴11111(1)1nnbbnnnn············

································································8分∴1223111111111(1)()()2231nnbbbbbbnn

·······························10分=1111n···························································11分∴122311111nnbbbbbb

·········································································12分18、解：（1）这一天小王朋友圈中好友走路步数的平均数6014010060201822610141822309.044004004

00400400400400千步所以这一天小王400名好友走路的平均步数约为9.04千步。····································3分（2）因频率约等概率可得()19.048(60140

100)0.5654004AP所以事件A的概率约为0.565·············································································7分（3）健步达人非健步达人合计40岁以上150

50200不超过40岁50150200合计20020040022224400(15050)10010.828200K所以有99.9%以上的把握认为健步达人与年龄有关。····················

··························12分19、解：（1）证明：共5页第2页方法一取AB的中点M，连结1,DMMB（如图）∵,ADDCAMMB∴1//,2DMBCDMBC·····················································

···························1分由棱柱的性质知：1111//,BCBCBCBC································································2分又11BEEC∴11//,DMBEDMBE·

················································································3分∴四边形1DMBE为平行四边形∴1//DEMB·

································································································4分∵1MB平面11ABBA，

DE平面11ABBA∴//DE平面11ABBA··············································································

········6分方法二取11AB的中点N，连结,ENAN（如图）∵1111,BEECANNB∴11111//,2ENACENAC··········································

·····································1分由棱柱的性质知：1111//,ACACACAC································································2分又ADDC∴//,NEAD

NEAD····················································································3分∴四边形ADEN为平行四边形∴//DEAN·····························

····································································4分∵AN平面11ABBA，DE平面11ABBA∴//DE平面11ABBA···········

···········································································6分方法三取BC的中点F，连结,EFDF（如图）∵,AD

DCBFFC∴//DFAB··································································································1分∵AB平面11ABBA，DF平面11ABB

A∴//DF平面11ABBA······················································································2分由棱柱的性质知

：1111//,BCBCBCBC又11,BEECBFFC∴11//,BEBFBEBF∴四边形1BFEB为平行四边形∴1//EFBB·············································

·····················································3分∵1BB平面11ABBA，EF平面11ABBA∴//EF平面11ABBA···················

···································································4分∵,EFDF平面DEF，且DFEFF∴平面//DEF平面11ABBA·····································

·········································5分∵DE平面DEF∴//DE平面11ABBA································································

······················6分方法四取11AC的中点G，连结,EGDG仿方法三同理证明（2）方法一∵D是AC的中点∴D到平面ABE的距离为C到平面ABE的距离的一半········································7分DBACE1B1A1

CMDBACE1B1A1CNFBACE1B1A1CDDBACE1B1A1CH共5页第3页过点C作CHBE交BE于H在直三棱柱111ABCABC中，1BBAB又ABBC且1BCBBB∴AB⊥平

面B1BCC1，又CH平面B1BCC1······················································8分∴AB⊥CH又CH⊥BE，BEABB∴CHABE平面···········

·············································································9分∴D到平面ABE的距离为12CH············································

·························10分在正方形B1BCC1中，又BB1=BC=2∴1122BCESBCCC又152BCESCH∴455CH·········································

····················11分∴所求距离为255···········································································

············12分方法二设点D到平面ABE的距离为d∵D是AC的中点，且,ABBC2ABBC∴111221222ABDABCSS△△····················

··············································7分由E平面111ABC及直棱柱的性质知，E到平面ABD的距离＝12AA····················

····8分∴12233EABDABDVS△···········································································

···9分由直棱柱的性质知：111,BBBEBBAB又,ABBC且1BCBBB∴AB⊥平面B1BCC1····································································

·················10分又BE平面B1BCC1故ABBE∴22221111221522ABESABBEBBBE△····································11分∵13EABDDABEABEVVdS△∴325255EA

BDABEVdS△·············································································12分20、解：（1）设椭圆1C的焦点坐标为0)0,(cc∴2c，又22ce

a············································································1分∴2a，又2222cab，∴2b···

··············································3分∴1C的方程为22142yx···············································································4分（2）

法一：设(1,)Nm由已知得，0PQk，设1122(,),(,)PxyQxy∴12122,2xxyym················································································6分又

12412422222121yxyx两式相减得：2112121212xxyyxxyy········································8分共5页第4页∴12PQkm

·······························································································9分∴直线l的方程为2(1)y

mxm，即(21)ymx·············································10分取21x，则0y，故点)0,21(在直线l上·························································11分

∴直线l过1(,0)2·························································································12分法二：由已知得，0PQk，设112

2(,),(,)PxyQxy设直线PQ的方程为xmyt············································································6分由已知有22142

xyxmyt∴222(2)240mymtyt··············································································8分由0得

22240mt∴12222mtyym，1212()22xxmyyt∴222mt∴12,(1,)2myymN··········································

···········································10分∴直线PQ的中垂线l的方程为(1)2mymx即1()2ymx∴直线l过1(,0)2··································

·······················································12分法三：当直线PQ斜率存在且不为0时，设PQ直线方程为ykxm由已知有22142xyykxm∴222(21)4240kykmxm···········

·······························································8分由0得22420km212242112,2221kmkx

xmkkkk12121()22()yykxxmkmk·································································10分1(1,)2Nk∴直线l：11(1)2yxkk即11()2yxk∴直线l过

1(,0)2·························································································11分当直线PQ斜率不存在时，1(,0)2也在l上综上：

直线l过1(,0)2····················································································12分21、

解：（1）当0a时，()1xfxex，xR；()1xfxe··················································1分由()0fx得0x·························

··································································2分当(,0)x时()0fx，()fx单调递减······················

········································3分共5页第5页当(0,)x时()0fx,()fx单调递增····································

··························4分∴min()(0)0fxf······················································································

·5分（2）由已知得：()12xfxeax，(0)0f，(0)0f∴()2xfxea，0x≥·········································································

·······6分①当12a≤时，[0,)x，()0fx≥，()fx在单增··········································7分∴()0fx≥，故()fx单增····································

·····································8分∴()(0)0fxf≥恒成立···········································································9分②当12a时，若[0,ln2)x

a，则()0fx，此时()fx单调递减又(0)0f∴当[0,ln2)xa时()0fx·················································10分故()fx在[0,ln2)a上单调递减，此时()(0)0

fxf≤∴()0fx≥在[0,)不能恒成立····························································11分综上可知，实数a的取值范围为1(,]2··································

····························12分22、解：（1）∵2sincos∴22sincos····································

···················································1分又cos,sinxy······································

················································2分∴C的直角坐标方程为2yx········································

··································4分（2）将32:12xatlyt（t为参数）代入2yx得：22340tta由0a知：12160a△······················

··························································5分设12,tt是方程22340tta的两根，则12122340tttta

··········································6分∴1212121212||||||1111||||||||||||ttttPMPNtttttt··········

··········································7分=2121212()412161||4ttttatta····················································8分∴12a或32a

···························································································9分0a又∴32a···········································

···························································10分23、解：（1）∵()|1||3||13|4fxxxxx≥······················

·········································1分由题意可知，即246mm≤，化简得：220mm≤···········································2分解得：12m≤≤··

··························································································3分∴m的取值范围为[1,2]·····························

··················································4分（2）由（1）知：02m，故332ab当0b时，由32a得：32a此时，32ab符合题意··········

········································································5分当0b时，∵3322223()()()[()]24bababaabbabab∴当0b时，223()024ba

b·······································································6分共5页第6页故由3320ab知0ab··········

····································································7分∴332222()()()[()3]ababaabbababab22331()[()()]

()44abababab≥··························································8分变形得：3()8ab≤∴2ab≤·················································

·················································9分综上可知：02ab≤···································································

················10分