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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60140100602018226101418223

09.04400400400400400400400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,BEECANNB∴11111//,2ENACENAC··············

·································································1分由棱柱的性质知：1111//,ACACACAC··········································

······················2分又ADDC∴//,NEADNEAD····················································································3分∴四边形A

DEN为平行四边形∴//DEAN·································································································4分∵AN平面11ABBA，DE平面11ABBA∴//DE平面11A

BBA······················································································6分方法三取BC的中点F，连结,EF

DF（如图）∵,ADDCBFFC∴//DFAB··································································································1分∵AB平面11ABBA，DF平面1

1ABBA∴//DF平面11ABBA······················································································2分由棱柱的性质知：1111//,BCBCBCBC又11,BEECBF

FC∴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ABBC且1BCBBB∴AB⊥平面B1BCC1，又CH平面B1BCC1······················································8分∴AB⊥C

H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∴22221111221522ABESABBEBBB

E△····································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∴12

PQkm·······························································································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km212

242112,2221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·················································10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：3

2a此时，32ab符合题意··················································································5分当0

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

······························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