2023.4西城区初三一模数学答案

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这是一份2023.4西城区初三一模数学答案，共6页。试卷主要包含了5，4等内容，欢迎下载使用。
北京市西城区九年级统一测试试卷数学答案及评分参考 2023.4一、选择题（共16分，每题2分）题号12345678答案BCCBDAC D二、填空题（共16分，每题2分）9．x≥1． 10．． 11．9． 12．x=-1．13．． 14．1.3． 15．． 16．；1，5，1．三、解答题（共68分，第17-20题，每题5分，第21题6分，第22-23题，每题5分，第24-26题，每题6分，第27-28题，每题7分）17．解：. =····························································4分 =．······························································5分18．解：解不等式①，得．···················································2分解不等式②，得．·····················································4分所以原不等式组的解集为．··············································5分19．解： =····························································2分 =····························································3分∵a是方程的一个根， ∴，即．··························································4分∴原式．······················································5分 20．方法一证明：如图，过点E作MN∥AB. ∴ ∠A=∠AEM．···············································2分 ∵ AB∥CD， ∴ MN∥CD. ∴ ∠C=∠CEM．······················4分 ∵ ∠AEC＝∠AEM＋∠CEM， ∴ ∠AEC＝∠A＋∠C． ·················5分方法二证明：如图，延长AE，交CD于点F. ∵ AB∥CD， ∴ ∠A=∠AFC．·······················2分 ∵ ∠AEC＝∠AFC＋∠C，···············4分 ∴ ∠AEC＝∠A＋∠C． ·················5分21．（1）证明：∵ CE∥FB， ∴ ∠BFE＝∠CEF． ∵ AD是BC边上的中线， ∴ BD=DC． ∵ ∠BDF＝∠CDE， ∴ △BDF≌△CDE． ∴ FB＝CE． ∴四边形BFCE是平行四边形．···································3分（2）①依题意补全图2，如图； ②证明：∵ ∠ABC＝∠ACB， ∴ AB=AC． ∵ AD是BC边上的中线， ∴ AD⊥BC． ∵ 四边形BFCE是平行四边形， ∴ 四边形BFCE为菱形．·······································6分
22．解：（1）4.5，4.5；················································2分（2）＜＜；························································3分（3）推荐乙，理由略，答案不唯一，合理即可．·························5分23．解：（1）∵ 一次函数的图象由函数的图象平移得到， ∴ ，得到一次函数的解析式为． ∵一次函数的图象过点A（-2，1），∴ ，得到． ∴一次函数的解析式为．············································3分（2）m≥1．······················································5分24．（1）证明：连接OD，如图1．∵ DE是⊙O 的切线，切点是D， ∴ OD⊥DE． ∴ ∠ODE＝90°． ∵ AB是⊙O的直径， ∴ ∠ACB＝90°． ∵ ∠ACB的平分线交⊙O于点D， ∴ ∠ACD＝∠BCD＝45°． ∴ ∠AOD＝90°． ∴ ∠AOD＝∠ODE． ∴ DE∥AB．····················································3分（2）解：作BH⊥DE于H，如图2． ∴ ∠BHD＝∠BHE＝90°． ∵ OD⊥DE，∠AOD＝90°， ∴ ∠BOD＝∠ODH＝90°． ∴ 四边形OBHD是矩形．∵ OA＝OB＝OD＝5，∴ 四边形OBHD是正方形． ∴ BH＝OD＝DH＝5． ∵ 在Rt△BHE中，sinA=， ∴ tanA=． ∵ ∠ACB=90°， ∴ ∠A＋∠ABC＝90°． ∵ ∠EBH＋∠ABC＝90°， ∴ ∠A＝∠EBH． ∴ tan∠EBH＝tanA＝． ∴ HE＝BH∙tan∠EBH＝＝． ∴ DE＝HE＋DH＝．··············································6分25．解：（1）①由题意可设所求的的函数关系式为．因为点（0，0）在该函数的图象上，所以．解得．所求的的函数关系为．即．····················································2分 ②喷水头喷出的水柱能够越过这棵树．理由如下：因为当x＝8时的函数值与当x＝16时的函数值相等，所以当x＝8时，．所以喷水头喷出的水柱能够越过这棵树．·························4分（2）（A）（C）．···············································6分26．解：（1）∵ 点（2，4）在抛物线上， ∴ 4a＋2b＋4＝4． ∴ b＝－2a． ∴ ． 2分（2）①当t＝1时，b＝-2a，所以． ∵ 点（，3），（，6）在抛物线上，∴ 当a＞0时，有a -2a＋4≤3．得4-a≤3，得a≥1．当a＜0时，有a -2a＋4≥6．得4-a≤6，得a≤-2．综上，的取值范围是a≤-2或a≥1．································4分 ②的取值范围是0＜a≤3． ·········································6分27．（1）证明：作EH⊥CD，EK⊥AB，垂足分别是H，K，如图1． ∵ OE是∠BOC的平分线， ∴ EH＝EK． ∵ ME＝NE，∴ Rt△EHN≌Rt△EKM．∴ ∠ENH＝∠EMK．记ME与OC的交点为P， ∴ ∠EPN＝∠OPM． ∴ ∠MEN＝∠AOC．·············································3分（2）OM＝ NF＋OG．证明：在线段OM上截取OG1＝OG，连接EG1，如图2． ∵ OE是∠BOC的平分线，∴ ∠EON＝∠EOB．∵ ∠MOF＝∠DOB，∴ ∠EOM＝∠EOD．∵ OE=OE，∴ △EOG1≌△EOG．∴ EG1=EG，∠EG1O=∠EGF． ∵ EF＝EG，∴ EF=EG1，∠EFG=∠EGF．∴ ∠EFG =∠EG1O．∴ ∠EFN =∠EG1M．∵ ∠ENF =∠EM G1． ∴ △ENF≌△EM G1． ∴ NF＝M G1． ∵ OM＝M G1＋O G1，∴ OM＝NF＋OG．·······································7分28．解：（1）① ，；···················································2分②由题意可得线段OA的所有相关点都在以OA为直径的圆上及其内部，如图．设这个圆的圆心是H． ∵ A（2，0）， ∴ H（1，0）．当直线与⊙H相切，且b＞0时，将直线与x轴的交点分别记为B，则点B的坐标是（-b，0）． ∴ BH＝1＋b． ∵ BH＝， ∴ ，解得．当直线与⊙H相切，且b＜0时，同理可求得．所以b的取值范围是≤b≤．··········································5分（2）d≥．·······················································7分