2023.4西城区初三一模数学答案

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       数学答案及评分参考       2023.4

一、选择题（共16分，每题2分）

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