2017-2018学年靖江市九年级上数学期末试卷及答案(2)

28. (14分) 如图，抛物线y＝mx―2mx―3m(m＞0)与x轴交于A、B两点， 与y轴交于C点. （1）请求抛物线顶点M的坐标（用含m的代数式表示），A，B

y 两点的坐标；

（2）经探究可知，△BCM与△ABC的面积比不变，试求出这个比值；

（3）是否存在使△BCM为直角三角形的抛物线？若存在，请求出；如果不存在，请说明理由.

C M A O B x 2

2017-2018学年度第一学期期末试卷九年级数学答案

一.C D B A B B D C

二.9. 2 10.外切 11.3200(1-x)2=2500 12. x?3或x?4 13.k<1 14.15? 15.-2 16. 2 17.9 18. 3或6或9

三.19.(1)原式=32-42+2 ····················································································· 3分

=0 ································································································· 4分

(2) 原式=9?2?21·································································· 3分 ?1??22?1·

22 =9 ········································································································ 4分 20.(1)x=2或x=5 ······························································································· 4分 12,(x?1)2?2,x?1??2 ·(2) 解：x2－2x＋＝························································· 2分

∴x1?1?2；x2?1?2 ·················································································· 4分 21. 解：(1) x甲=

____1(82+81+79+78+95+88+93+84)=85， ············································· 1分 81···························································· 2分 x乙=(92+95+80+75+83+80+90+85)=85． ·

8 这两组数据的平均数都是85．

这两组数据的中位数分别为83，84． ·········································································· 4分 (2) 派甲参赛比较合适．理由如下：由(1)知x甲=x乙，

____12s甲?[(78?85)2?(79?85)2?(81?85)2?(82?85)2?(84?85)2 ······························ 5分 8?(88?85)2?(93?85)2?(95?85)2]?35.512s乙?[(75?85)2?(80?85)2?(80?85)2?(83?85)2?(85?85)2 ······························ 6分 8?(90?85)2?(92?85)2?(95?85)2]?41∵x甲=x乙，s甲2?s乙2，

∴甲的成绩较稳定，派甲参赛比较合适． ···································································· 8分 22. 解:(1)因为点A（1，1）在二次函数y?x?2ax?b图像上,所以1=1-2a+b 可得b=2a ┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄3分 (2)由题意,方程x2-ax+b=0有两个相等的实数根, 所以4a2-4b=4a2-8a=0

解得a=0或a=2 ┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄5分

2

当a=0时,y=x, 这个二次函数的图像的顶点坐标为(0,0); ┄┄┄┄┄┄┄┄┄┄6分

22

当a=2时,y=x-4x+4=(x-2), 这个二次函数的图像的顶点坐标为(2,0); 所以, 这个二次函数的图像的顶点坐标为(0,0) ,(2,0). ┄┄┄┄┄┄8分 23.

G

H

2____F

解：（1）如图，扇形BFG和扇形CGH为羊活动的区域．………………………2分 （2）S扇形BFG120?62??12?m2……………………………………………4分

360S扇形CGH60?222???m2………………………………………………6分

3603238???m2…………………………………8分 33∴羊活动区域的面积为：12??24. 已知：①③，①④，②④，③④均可,其余均不可以. ………………………………4分 已知：在四边形ABCD中，①AD∥BC，③?A??C. 求证：四边形ABCD是平行四边形． 证明：∵ AD∥BC

∴?A??B?180?，?C??D?180? ∵?A??C，∴?B??D

∴四边形ABCD是平行四边形. ……………………………………………10分

25. 解：设AB、CD的延长线相交于点E ∵∠CBE=45o CE⊥AE ∴CE=BE…………（2分） ∵CE=26.65-1.65=25 ∴BE=25

∴AE=AB+BE=30 ………………………………（4分） 在Rt△ADE中，∵∠DAE=30o 3∴DE=AE×tan30 o =30× =103 ……………（7分）

3

∴CD=CE-DE=25-103 ≈25-10×1.732=7.68≈7.7(m) ……………（9分）

答：广告屏幕上端与下端之间的距离约为7.7m ……………………（10分） （注：不作答不扣分）

(1) 26. 证明:连接OD

∵DE为⊙O的切线, ∴OD⊥DE┄┄┄┄┄┄┄┄2分 ∵O为AB中点, D为BC的中点

∴OD‖AC ┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄3分 ∴DE⊥AC ┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄4分 (2)过O作OF⊥BD,则BF=FD ┄┄┄┄┄┄┄┄┄5分 在Rt△BFO中, ∠ABC=30°

∴OF=OB, BF=

123OB ┄┄┄┄┄┄┄┄┄7分 2∵BD=DC, BF=FD,

∴FC=3BF=33OB ┄┄┄┄┄┄┄┄┄8分 2在Rt△OFC中,

1OBOF32??tan∠BCO=. ┄┄┄┄┄┄┄┄┄10分 FC339OB227. 解：（1）∵AE=MC，∴BE=BM， ∴∠BEM=∠EMB=45°， ∴∠AEM=135°，

∵CN平分∠DCP，∴∠PCN=45°，∴∠AEM=∠MCN=135°·················································· 2分

??AEM??MCN,??AE?MC,??EAM=?CMN,在△AEM和△MCN中：∵?

∴△AEM≌△MCN，··································································································· 4分 ∴AM=MN ················································································································ 5分 （2）仍然成立． ···································································································· 6分 在边AB上截取AE=MC，连接ME ·················································································· 7分 ∵△ABC是等边三角形， ∴AB=BC，∠B=∠ACB=60°， ∴∠ACP=120°． ∵AE=MC，∴BE=BM ∴∠BEM=∠EMB=60°

∴∠AEM=120°． ···································································································· 8分 ∵CN平分∠ACP，∴∠PCN=60°， ∴∠AEM=∠MCN=120°

∵∠CMN=180°—∠AMN—∠AMB=180°—∠B—∠AMB=∠BAM

∴△AEM≌△MCN，··································································································· 9分 ∴AM=MN ··············································································································· 10分

(n?2)180?（3）

n ··································································································· 12分

28. 解：（1）∵y＝mx2―2mx―3m＝m(x2―2x―3)＝m(x－1)2―4m,

∴抛物线顶点M的坐标为（1，―4m） ······································································· 2分 ∵抛物线y＝mx2―2mx―3m(m＞0)与x轴交于A、B两点， ∴当y＝0时，mx2―2mx―3m＝0，

∵m＞0，

∴x2―2x―3＝0， 解得x1＝－1，x,2＝3，

∴A，B两点的坐标为（－1,0）、（3,0）. ································································· 4分 （2）当x＝0时，y＝―3m， ∴点C的坐标为（0，－3m）,

1∴S△ABC＝×|3－(－1)|×|－3m|＝6|m|＝6m， ····························································· 5分

2过点M作MD⊥x轴于D，

则OD＝1，BD＝OB－OD＝2，MD＝|－4m |＝4m.

∴S△BCM＝S△BDM ＋S梯形OCMD－S△OBC 111

＝BD·DM＋(OC＋DM)·OD－OB·OC 222

111

＝×2×4m＋(3m＋4m)×1－×3×3m＝3m， ······················································ 7分 222∴ S△BCM：S△ABC＝1∶2. ···················································································· 8分 （3）存在使△BCM为直角三角形的抛物线.

过点C作CN⊥DM于点N，则△CMN为Rt△，CN＝OD＝1，DN＝OC＝3m, ∴MN＝DM－DN＝m, ∴CM2＝CN2＋MN2＝1＋m2，

在Rt△OBC中，BC2＝OB2＋OC2＝9＋9m2， 在Rt△BDM中，BM2＝BD2＋DM2＝4＋16m2.

①如果△BCM是Rt△，且∠BMC＝90°时，CM2＋BM2＝BC2, 即1＋m2＋4＋16m2＝9＋9m2， 解得 m＝±

2

, 22. 2

2232x－2x－使得△BCM是Rt△; ··········································· 10分 22

C N M A O D B x y ∵m＞0，∴m＝

∴存在抛物线y＝

②①如果△BCM是Rt△，且∠BCM＝90°时，BC2＋CM2＝BM2. 即9＋9m2＋1＋m2＝4＋16m2，

解得 m＝±1, ∵m＞0，∴m＝1.

∴存在抛物线y＝x2－2x－3使得△BCM是Rt△; ··················································· 12分 ③如果△BCM是Rt△，且∠CBM＝90°时，BC2＋BM2＝CM2. 即9＋9m2＋4＋16m2＝1＋m2， 1

整理得 m2＝－，此方程无解,

2

∴以∠CBM为直角的直角三角形不存在.

（或∵9＋9m2＞1＋m2，4＋16m2＞1＋m2，∴以∠CBM为直角的直角三角形不存在.） 综上的所述，存在抛物线y＝

2232x－2x－和y＝x2－2x－3使得△BCM是Rt△. ··········· 14分 22

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