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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6014010060201822610141822309.044

00400400400400400400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(60140100)

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15050200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,BEEC

ANNB∴11111//,2ENACENAC···············································································1分由棱

柱的性质知：1111//,ACACACAC································································2分又ADDC∴//,NEADNEAD····················

································································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）∵,ADDCBFFC∴//DFAB··

································································································1分∵AB平面11ABBA，DF平面11ABBA∴//

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DBACE1B1A1CMDBACE1B1A1CNFBACE1B1A1CDDBACE1B1A1CH共5页第3页过点C作CHBE交BE于H在直三棱柱111ABCABC中，1BBAB又ABB

C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B

1BCC1故ABBE∴22221111221522ABESABBEBBBE△····································11分∵13EABDDABEABEVVdS△

∴325255EABDABEVdS△·············································································12分20、解：（1）设椭圆1C的焦点坐标为0)0,(cc∴2c

，又22cea············································································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)ymxm，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1122(,),(,)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kmkxxmkkkk12121()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)xa

，则()0fx，此时()fx单调递减又(0)0f∴当[0,ln2)xa时()0fx·················································1

0分故()fx在[0,ln2)a上单调递减，此时()(0)0fxf≤∴()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，∵33

22223()()()[()]24bababaabbabab∴当0b时，223()024bab···························································

············6分共5页第6页故由3320ab知0ab··············································································7分∴332

222()()()[()3]ababaabbababab22331()[()()]()44abababab≥··························································8分变形得：3()8ab≤∴

2ab≤··································································································9分综上可知：02ab≤···

················································································10分