2009年广西梧州市初中毕业升学考试试题卷数学试题（含参考答案和评分标准）

数学参考答案及评分标准

一、填空题（本大题共10小题，每小题3分，共30分．）

题号 1 2 3 4 5 二、选择题（本

答案 题号?6 11 ＞ 12 13 3.5 14 2（x＋315 ）（x－3） 16 100 17 18 题号 答案6 B 7 C A 8 D 9 B C 10 A D 答案 ?15 4 10 3π 2n(n?1) 分．）

三、解答题（本大题共8小题，满分66分．） 19．解:原式＝23?2?2?32 ························································································ 3分 ＝23?2?3 ···························································································· 4分 ＝3?2 ·········································································································· 6分 20．解:(x?3)(x?3?2x)?0···························································································· 2分 (x?3)(3x?3)?0 ································································································· 3分 x?3?0或3x?3?0 ···························································································· 4分 即x1?3或x2?1 ··································································································· 6分 21．解：（1） 600 ········································································································· 2分 （2）在右图上补全条形图如图． ······················································································· 4分 台 1000 900 1000 700

600

600 700 2005 2006 2007 2008 年

图(6)-2

（3）500÷100×1000×10%＝500 ·························································································· 6分

22．解：（1）y?600x?1000(150?x) ············································································· 2分

y??400x?150000 ·················································································· 3分

（2）依题意得，150?x≥2x ················································································ 5分 x≤50 ································································································· 6分 因为－400＜0，由一次函数的性质知，当x=50时，y有最小值 ························ 7分

所以150－50=100

答: 甲工种招聘50人，乙工种招聘100人时可使得每月所付的工资最少． （8分） 23．（1）证明：∵MN是AC的垂直平分线 ······························· 1分

A∴OA＝OC ∠AOD＝∠EOC=90° ······················ 3分

D第 5 页 共 7 页

MOENB图(7)

C大题共

8小题，

每小题

3分，共24

∵CE∥AB ∴∠DAO＝∠ECO ··········································· 4分 ∴△ADO≌△CEO ············································· 5分 ∴AD＝CE ···················································· 6分

（2）四边形ADCE是菱形． ··········································· 8分 （填写平行四边形给1分）

24．解:（1）设甲队单独完成此项工程需x天，由题意得 ················································· 1分

66??1 ············································································································ 3分 x2x3 解之得x?15 ·············································································································· 4分

经检验，x?15是原方程的解． ············································································· 5分

所以甲队单独完成此项工程需15天，

乙队单独完成此项工程需15×=10（天） ······························································· 6分

（2）甲队所得报酬:20000?乙队所得报酬:20000?231?6?8000（元） ······················································· 8分 151?6?12000（元） ·································································· 10分 1025．（1）证明：连接OC········································································································· 1分 ∵OA＝OC

∴∠OAC＝∠OCA ∵CE是⊙O的切线

∴∠OCE＝90° ····················································· 2分 ∵AE⊥CE

∴∠AEC＝∠OCE＝90°

∴OC∥AE ·························································· 3分 ∴∠OCA＝∠CAD

∴∠CAD＝∠BAC ················································ 4分

图（8）

ECDB· OA??BC? ∴DC∴DC＝BC ··················································································································· 5分 （2）∵AB是⊙O的直径 ∴∠ACB＝90° ∴BC?······························································· 6分 AB2?AC2?52?42?3 ·

∵∠CAE＝∠BAC ∠AEC＝∠ACB＝90°

∴△ACE∽△ABC ········································································································ 7分 ∴ ∴ECAC? BCABEC412? EC? ··························································································· 8分 355 ∵DC＝BC＝3

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∴ED?129······························································ 9分 DC2?CE2?32?()2? ·

559ED53 ∴tan?DCE?············································································· 10分 ?? ·

12EC4526．（1）解：把A（?1，0），C（3，?2）代入抛物线 y?ax2?3ax?b 得

?(?1)2a?3?(?1)a?b?0 ? ························································································ 1分

9a?9a?b??2? 整理得

1??4a?b?0?a? ? ?????? 2分 解得?2??????3分

b??2???b??2123 ∴抛物线的解析式为 y?x?x?2 ·········································································· 4分

22123 （2）令x?x?2?0 解得 x1??1，x2?4

22 ∴ B点坐标为（4，0）

又∵D点坐标为（0，?2） ∴AB∥CD ∴四边形ABCD是梯形．

1 ∴S梯形ABCD ＝(5?3)?2?8 ······························· 5分

2设直线y?kx?1(k?0)与x轴的交点为H，

与CD的交点为T，

y 31，0）， T（?，?2） ···················· 6分 kk∵直线y?kx?1(k?0)将四边形ABCD面积二等分

1∴S梯形AHTD ＝S梯形ABCD＝４

2113∴(??1?)?2?4 ········································· 7分 2kk4∴k?? ·································································· 8分

3（3）∵MG⊥x轴于点G，线段MG︰AG＝1︰2

m?1 ∴设M（m，?）， ············································ 9分

2m?1123?m?m?2 ∵点M在抛物线上 ∴?222解得m1?3，m2??1（舍去） ······························ 10分

则H（?根据中心对称图形性质知，MQ∥AF，MQ＝AF，NQ＝EF，

A D H O T C B x y=kx+1 图(9) -1 y E G A O F B x M N 图(9) -2

Q∴M点坐标为（3，?2） ····························································································· 11分 ∴N点坐标为（1，?3） ···························································································· 12分

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