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2021年哈三中高三学年第一次模拟考试文科数学答案一、选择题BDCABCCBADAD二、填空题13.214.1515.616.162三、解答题17.（1）()()8,1266412111=+++=++

dadadaa······························3分整理得()()09311=−−aa又5,3,121==daaa25−=nan·······················································6分（2）()()

+−−=+−=3512515135251nnnnbn()11111535315553nTnn=−=−++··································12分18.（1）平面EDAF⊥平面ABC

D，DE平面EDAF平面EDAF平面ABCDAD=，DEAD⊥DE⊥平面ABCD···················································2分AC平面ABCDDEA

C⊥四边形ABCD是正方形ACBD⊥···································3分DEBD、平面BDE，DEBDD=，AC⊥平面BDE·············4分

AC平面ACE平面AEC⊥平面BDE·····························5分（2）6519.（1）8.05.0ˆ2+=xy··································

·····················6分（2）模型二的残差平方和为0.423.7，模型二的拟合效果更好··················6分20.（1）22:14xMy+=···········································

············3分（2）()23,0F，设直线:3lxmy=+，()11,Axy，()22,Bxy，224403xyxmy+−==+，代入得：()2242310mymy++−=由于()216

10m=+恒成立则有122234myym−+=+，12214yym=+·································5分()()22221212122411144mABmyymyyyym+=+−=++−=+·········6分点O到直线l的距离231dm=+······

································7分则2222123123132411AOBmSABdmmm+===++++·············10分当且仅当：22311mm+=+，即2m=时取等号，又由于32OCOA=−，知5522ABC

AOBSS=，此时:23lxy=+.···12分21.(1)21yx=+…………4分(2)当0x=时，()00e01sin00g=−−+=，0x=是()gx的一个零点，…………6分由()ecosxgxax=

−+，可得()esinxgxx=−.因为12a，①当()0,x+时，e1x，()1sin0gxx−，()gx在()0,+单调递增，()()020gxga=−，()

gx在()0,+单调递增，()()00gxg=，此时()gx在()0,+无零点……8分②当(,x−−时，ax−，有()esin1esin10xxgxaxxx=−+−++−，此时()gx在(,−−无零点.…………10分③

当(),0x−时，sin0x，()esin0xgxx=−，()gx在(),0−单调递增，又()020ga=−，()e10ga−=−−，由零点存在性定理知，存在唯一()0,0x−，使得()00gx=.当()0,xx−时，

()0gx，()gx在()0,x−单调递减；当()0,0xx时，()0gx，()gx在()0,0x单调递增；又()e10ga−−=+−，()()000gxg=，所以()gx在(),0−上有1个零点.综上，当12a时，()gx有2

个零点.…………12分22.（1）曲线C的普通方程为1422=+yx，直线l直角坐标方程为0832=−+yx······························5分（2）由参数方程设点)sin,cos2(P，则点P到直线l的距离为13|)sin(58|13|8sin3cos4

|+−=−+=d其中34tan=所以13133d，此时54cos,53sin==所以点P的坐标为)53,58(P·······································10分23.（1）①当时，，所以无解②当时，，所以③当时，，所以无解综上，不等式的解

集为···································5分（2）因为，所以，所以，···························10分1−x31322−+−xxx31−x1x31x3x5322+−xxx51|xx4|31|

|3||1|)(=+−+−++=xxxxxf4=m4==++mcba89)111)](()()[(81111++++++++++=+++++accbbaaccbbaaccbba