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2008年广东省茂名市初中毕业生学业考试数学试题和答案


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广东省茂名市 2008 年初中毕业生学业考试 与高中阶段学校招生考试

数 学 试



1. 全卷分第一卷(选择题,满分 40 分,共 2 页)和第二卷(非选择题,满分 110 分, 考 共 8 页),全卷满分 150 分;考试时间 120 分钟. 生 2. 请认真填写答题卡和第二卷密封线内的有关内容,并在试卷右上角的座位号处填 须 上自己的座位号. 知 3. 第一卷用 2B 铅笔在答题卡上填涂;第二卷用黑色钢笔、签字笔作答. 4. 考试结束,将第一卷、第二卷和答题卡一并交回. 温馨提示:亲爱的同学,请你相信自己,仔细审题,沉着作答,就一定能考出好成绩, 温馨提示:亲爱的同学,请你相信自己,仔细审题,沉着作答,就一定能考出好成绩, 祝你成功 成功. 祝你成功.

第一卷(选择题,满分 40 分,共 2 页)
一、精心选一选(本大题共 10 小题,每小题 4 分,共 40 分. 每小题给出四个答案,其中只有一个是正确的). 1.- .-

1 的相反数是( 2

) B.2 D. ?

A.-2 C.

1 2

1 2


2.下列图形中,既是轴对称图形又是中心对称图形的是(

A 3.下列运算正确的是(
2 A.-2 =4

B ) B.2
?2





=-4

C. a · a

2

= a

2

D. a +2 a =3 a )

4.用平面去截下列几何体,截面的形状不可能是圆的几何体是( ... A.球 B.圆锥

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C.圆柱 D.正方体 5.任意给定一个非零数,按下列程序计算,最后输出的结果是(



m
A. m C. m +1

平方

-m

÷m

+2 B. m
2

结果

D. m -1

6.在数轴上表示不等式组 ?

?1 + x > 0 的解集,正确的是( ?2 x ? 4 ≤ 0



-2 -1

0

1 A

2 3

-2 -1

0 1 2 3 B

-2 -1

0 1 2 C

3

-2 -1

0 1 D

2 3 )

7.正方形内有一点 A,到各边的距离从小到大依次是 1、2、3、4,则正方形的周长是( A.10 C.24 B.20 D.25 )

8.一组数据 3、4、5、 a 、7 的平均数是 5,则它的方差是( A.10 C.5 9.已知反比例函数 y = B.6 D.2

a ( a ≠0)的图象,在每一象限内, y 的值随 x 值的增大而减少,则一次 x
) B.第二象限 D.第四象限

函数 y =- a x + a 的图象不经过( ... A.第一象限 C.第三象限

A E


10.如图,△ABC 是等边三角形,被一平行于 BC 的矩形所截, AB 被截成三等分,则图中阴影部分的面积是△ABC 的面积的 ( A.

H

1 9 1 C. 3

B.

2 9 4 D. 9

F B

G C

( (第 10 题图)

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广东省茂名市 2008 年初中毕业生学业考试 与高中阶段学校招生考试
数 学 试 卷

二 题 号



四 21

五 22 23 24

六 25

本卷得分

(11~15) (16~18) (19~20) 得 分 评卷人

第二卷(非选择题,共 8 页,满分 110 分) 第二卷 得 分 评卷人 二、耐心填一填(本大题共 5 小题,每小题 4 分,共 20 分.请你把答案 填在横线的上方). 11.据最新统计,茂名市户籍人口约为 7020000 人,用科学记数法表示是 12.分解因式:3 x 2 -27= . 人.

13.如图,点 A、B、C 在⊙O 上,AO∥BC,∴AOB = 50°, 则∴OAC 的度数是 .

O
C A B (第 13 题图)
全月应纳税所得税额 税率 5% 10% ……

14.依法纳税是每个公民应尽的义务,新的《中华人民共和 国个人所得税法》规定,从 2008 年 3 月 1 日起, 公民全月工薪不超过 2000 元的部分不必纳税,超过 2000 元的部分为全月应纳税所得税额,此项税款按右 表分段累进计算.黄先生 4 月份缴纳个人所得税税金 55 元,那么黄先生该月的工薪是 元.

不超过 500 元的部分 超过 500 元至 2000 元的部分 ……

15.有一个运算程序,可以使:a ⊕ b = n ( n 为常数)时, 得 ( a +1)⊕ b = n +1, a ⊕( b +1)= n -2 ⊕ 现在已知 1⊕1 = 2,那么 2008⊕2008 = ⊕ ⊕ 得 分 评卷人 .

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三、细心做一做 (本大题共 3 小题, 每小题 8 分,共 24 分) 16.(本题满分 8 分)计算:

座位号

下面解答题都应写 出文字说明、 出文字说明、证明过程 或演算步骤。 或演算步骤。请你一定 要注意噢! 要注意噢!

2a a a2 ?1 ( )· a ?1 a + 1 a
解:

y

17. (本题满分 8 分)如图,方格纸中有一条美丽可爱的小金鱼. (1)在同一方格纸中,画出将小金鱼图案绕原点 O 旋转 180°后得到的图案; 分) (4 (2)在同一方格纸中,并在 y 轴的右侧,将原小
0

金鱼图案以原点 O 为位似中心放大,使它们的位似 比为 1:2,画出放大后小金鱼的图案. 分) (4 (第 17 题图) 18. (本题满分 8 分)不透明的口袋里装有 3 个球,这 3 个球分别标有数字 1、2、3,这些球除了数 字以外都相同. (1)如果从袋中任意摸出一个球,那么摸到标有数字是 2 的球的概率是多少?(2 分) (2)小明和小东玩摸球游戏,游戏规则如下:先由小明随机摸出一个球,记下球的数字后放回, 搅匀后再由小东随机摸出一个球,记下球的数字.谁摸出的球的数字大,谁获胜.现请你 利用树状图或列表的方法分析游戏规则对双方是否公平?并说明理由. 分) (6 解:

x





评卷人

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四、沉着冷静,周密考虑(本大题共 2 小题,每小题 8 分,共 16 分) 沉着冷静,周密考虑 本大题共 小题,

19.(本题满分 8 分)2008 年 5 月 12 日 14 时 28 分我国四川汶川发生了 8.0 级大地震,地震发生后, 我市某中学全体师生踊跃捐款,支援灾区,其中九年级甲班学生共捐款 1800 元,乙班学生共 捐款 1560 元.已知甲班平均每人捐款金额是乙班平均每人捐款金额的 1.2 倍,乙班比甲班多 2 人,那么这两个班各有多少人? 解:

20. (本题满分 8 分)某文具店王经理统计了 2008 年 1 月至 5 月 A、B、C 这三种型号的钢笔平均每 月的销售量,并绘制图 1(不完整) ,销售这三种型号钢笔平均每月获得的总利润为 600 元,每 种型号钢笔获得的利润分布情况如图 2.已知 A、B、C 这三种型号钢笔每支的利润分别是 0.5 元、0.6 元、1.2 元,请你结合图中的信息,解答下列问题: (1)求出 C 种型号钢笔平均每月的销售量,并将图 1 补充完整; 分) (4 (2)王经理计划 6 月份购进 A、B、C 这三种型号钢笔共 900 支,请你结合 1 月至 5 月平均 每月的销售情况(不考虑其它因素) ,设计一个方案,使获得的利润最大,并说明理由. (4 分) 解:

平均每月 销售量/支
600

600
A 50% B 30% C 20%

300

300

A

B

C
(图 1)

型号
(图 2) (第 20 题图)

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五、开动脑筋,再接再厉 (本大题共 3 小题,每小题 10 分,共 30 分) 得 分 评卷人 21.(本题满分 10 分) 如图,某学习小组为了测量河对岸塔 AB 的高度,在塔底部 B 的正对岸点 C 处,测得 仰角∴ACB=30°. (1)若河宽 BC 是 60 米,求塔 AB 的高(结果精确到 0.1 米)(4 分) ; (参考数据: 2 ≈1.414, 3 ≈1.732) (2)若河宽 BC 的长度无法度量,如何测量塔 AB 的高度呢?小明想出了另外一种方法:从 点 C 出发,沿河岸 CD 的方向(点 B、C、D 在同一平面内,且 CD⊥BC)走 a 米,到达 D 处, 测得∴BDC=60°, 这样就可以求得塔 AB 的高度了. 请你用这种方法求出塔 AB 的高. (6 分) 解: A

B

D

C (第 21 题图)

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评卷人 22.(本题满分 10 分) 如图,⊙O 是△ABC 的外接圆,且 AB=AC,点 D 在弧 BC 上运动,过点 D 作 DE∥BC,DE

交 AB 的延长线于点 E,连结 AD、BD. (1)求证:∴ADB=∴E; 分) (3 (2)当点 D 运动到什么位置时,DE 是⊙O 的切线?请说明理由. 分) (3 (3)当 AB=5,BC=6 时,求⊙O 的半径. 分) (4 解:

A

B E

O

C D
(第 22 题图)





评卷人 23.(本题满分 10 分) 如图,在等腰梯形 ABCD 中,已知 AD∥BC,AB=DC,AD=2,BC=4,延长 BC 到 E,使 CE=AD.

(1)写出图中所有与△DCE 全等的三角形,并选择其中一对说明全等的理由; 分) (5 (2)探究当等腰梯形 ABCD 的高 DF 是多少时,对角线 AC 与 BD 互相垂直?请回答并说明 理由. 分) (5 解:

A

D

B

F C
(第 23 题图)

E

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六、充满信心,成功在望(本大题共 2 小题,每小题 10 分,共 20 分) 充满信心,成功在望 本大题共 小题, 得 分 评卷人 24.(本题满分 10 分) 我市某工艺厂为配合北京奥运,设计了一款成本为 20 元?件的工艺品投放市场进行试销. 经过调查,得到如下数据: 销售单价 x (元?件) 每天销售量 y (件) …… …… 30 500 40 400 50 300 60 200 …… ……

(1)把上表中 x 、 y 的各组对应值作为点的坐标,在下面的平面直角坐标系中描出相应的点, 猜想 y 与 x 的函数关系,并求出函数关系式; 分) (4 (2)当销售单价定为多少时,工艺厂试销该工艺品每天获得的利润最大?最大利润是多少? (利润=销售总价-成本总价) 分) (4 (3) 当地物价部门规定, 该工艺品销售单价最高不能超过 45 元/件, 那么销售单价定为多少时, .. 工艺厂试销该工艺品每天获得的利润最大?(2 分) 解: 800 700 600 500 400 300 200 100 0 10 20 30 40 50 60 70 80

y

x

(第 24 题图)





评卷人 25(本题满分 10 分) 相关链接 : 若 x1 , x2 是一元二次方程

如图,在平面直角坐标系中,抛物线 y =-

2 2 x +b x +c 3

经过 A(0,-4) 、B( x 1 ,0) C( x 2 ,0)三点,且 x 2 - x 1 =5. 、 (1)求 b 、 c 的值; 分) (4

ax 2 + bx + c = 0 (a ≠ 0) 的 两 b c 根, x1 + x2 = ? , x1 x2 = . 则 a a

(2)在抛物线上求一点 D,使得四边形 BDCE 是以 BC 为对角线的菱形; 分) (3

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(3)在抛物线上是否存在一点 P,使得四边形 BPOH 是以 OB 为对角线的菱形?若存在,求 出点 P 的坐标,并判断这个菱形是否为正方形?若不存在,请说明理由. 分) (3 解: y

B

C

x O

A

(第 25 题图)

茂名市 2008 年初中毕业生学业考试 与高中阶段学校招生考试

数学试题参考答案及评分标准 数学试题参考答案及评分标准
说明: 如果考生的解法和本解法不同,可根据试题的主要内容, 说明:1.如果考生的解法和本解法不同,可根据试题的主要内容,并参考评分标准制定相 应的评分细则后评卷. 应的评分细则后评卷 解答题右端所注的分数 表示考生正确做到这一步应得的累加分数. 题右端所注的分数, 2.解答题右端所注的分数,表示考生正确做到这一步应得的累加分数 一、选择题(本大题共 10 小题,每小题 4 分,共 40 分) 题 号 答 案 1 C 2 A 3 D 4 D 5 C 6 A 7 B 8 D 9 C 10 C

二、填空题(本大题共 5 小题,每小题 4 分,共 20 分). 11、7.02×10 6 12、3( x +3) x -3) 13、25° 14、2800 ( ) ( )
三、 (本大题共 3 小题,每小题 8 分,共 24 分) 16、解:解法一:原式=

15、-2005

2a a2 ?1 a a2 ?1 · · ···················································· 2 分 a ?1 a a +1 a
=

2a (a + 1)(a ? 1) a (a + 1)(a ? 1) · · ························· 4 分 a ?1 a a +1 a
································································ 6 分

=2· (a + 1) - ( a ? 1)

=2 a +2- a +1 ················································································ 7 分 ························································································ 8 分 = a +3

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解法二:原式=

2a (a + 1) ? a (a ? 1) a 2 ? 1 · a a2 ?1
2 a 2 + 3 a a ?1 · a a2 ?1

····························································· 3 分

=

·············································································· 5 分

=

a 2 + 3a a

························································································ 6 分

= a +3 17、解:

········································································································· 8 分

0

18、 解: 有数字是 2 的球的概率是

(说明:画图正确,每对一个给 4 分.) (1)从 3 个球中随机摸出一个,摸到标

1 ··································································································· 2 分 3 1 或 P(摸到标有数字是 2 的球)= ························································ 2 分 3
················································································ 3 分 或列表法: 1 (1,1)

(2)游戏规则对双方公平. 树状图法:

1

开始

2

3

2 3 1 2 3 1 2 3

(1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) (3,3)

小 明





1 (1,1) (2,1) (3,1)

2 (1,2) (2,2) (3,2)

3 (1,3) (2,3) (3,3)

1 2 3

(注:学生只用一种方法做即可) ················································································ 5 分 由图(或表)可知, P(小明获胜)=

1 1 , P(小东获胜)= , ····························· 7 分 3 3
········································ 8 分 ································· 1 分

∵P(小明获胜)= P(小东获胜) , ∴游戏规则对双方公平.

19、解:设甲班有 x 人,则乙班有( x +2)人,根据题意,得

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1800 1560 = ×1.2 x x+2
解这个方程,得

························································ 4 分 ················································ 6 分 ················································· 7 分 ··························································· 8 分

x =50

经检验, x =50 是所列方程的根. 所以,甲班有 50 人,乙班有 52 人. 20、解:

(1) 600×20%=120(元) ·················· 1 分 120÷1.2=100(支) ································ 2 分 600

平均每月 销售量/支 600 300 100 A B C 型号

作图如右图:

······································· 4 分 300

(2)A、B、C 这三种型号钢笔分别进 500 支、300 支、100 支. ······························ 7 分 理由是:利润大的应尽可能多进货,才可能获得最大利润. ······························· 8 分

21、解: (1)在 Rt△ABC 中,∵∴ACB=30°,BC=60, ∴AB=BC·tan∴ACB ···················································································· 1 分

=60×

3 =20 3 3

··············································································· 2 分

≈34.6(米) ······················································································· 3 分 . 所以,塔 AB 的高约是 34.6 米. ······························································· 4 分

(2)在 Rt△BCD 中,∵∴BDC=60°,CD= a , ··············································· 5 分 ∴BC=CD·tan∴BDC = 3 a ················································································· 6 分

. ························································································· 7 分

又在 Rt△ABC 中,AB=BC·tan∴ACB ························································ 8 分 = 3 a×

3 = a (米) . ········································ 9 分 3

所以,塔 AB 的高为 a 米. ·············································································· 10 分 22、解: (1)在△ABC 中,∵AB=AC, ∴∴ABC=∴C. ·············································· 1 分 ∵DE∥BC,∴∴ABC=∴E,

A

B E

O

C D

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∴∴E=∴C.

················································· 2 分

又∵∴ADB=∴C, ∴∴ADB=∴E. ·············································· 3 分 (2)当点 D 是弧 BC 的中点时,DE 是⊙O 的切线. ············································ 4 分 理由是:当点 D 是弧 BC 的中点时,则有 AD⊥BC,且 AD 过圆心 O. ························ 5 分 又∵DE∥BC,∴ AD⊥ED. ∴ DE 是⊙O 的切线. ················································· 6 分 (3)连结 BO、AO,并延长 AO 交 BC 于点 F, 则 AF⊥BC,且 BF=
O

A

B E

C D

1 BC=3.································ 7 分 2

又∵AB=5,∴AF=4. ················································· 8 分 设⊙O 的半径为 r ,在 Rt△OBF 中,OF=4- r ,OB= r ,BF=3,

A

B


O

r 2 =3 2 +(4- r ) 2

······························ 9 分

F

C

解得 r =

25 , 8

∴⊙O 的半径是

25 . ········································· 10 分 8

23、解: (1)△CDA≌△DCE,△BAD≌△DCE;·························································· 2 分 ① △CDA≌△DCE 的理由是: ∵AD∥BC, ∴∴CDA=∴DCE. 又∵DA=CE,CD=DC , ∴△CDA≌△DCE. ····························· 3 分 ······························· 4 分 ······································· 5 分

A

D

G
B F C E

或 ② △BAD≌△DCE 的理由是: ∵AD∥BC,

∴∴CDA=∴DCE. ········································································································· 3 分 又∵四边形 ABCD 是等腰梯形, ∴∴BAD=∴CDA, ∴∴BAD =∴DCE. ········································································································ 4 分 又∵AB=CD,AD=CE, ∴△BAD≌△DCE. ··································································································· 5 分 (2)当等腰梯形 ABCD 的高 DF=3 时,对角线 AC 与 BD 互相垂直. ··················· 6 分 理由是:设 AC 与 BD 的交点为点 G,∵四边形 ABCD 是等腰梯形,

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∴AC=DB. 又∵AD=CE,AD∥BC, ∴四边形 ACED 是平行四边形, ············································································ 7 分 ∴AC=DE,AC∥DE. ∴DB=DE. ············································································································ 8 分 则 BF=FE, 又∵BE=BC+CE=BC+AD=4+2=6, ∴BF=FE=3. ········································································································· 9 分 ∵DF=3, ∴∴BDF=∴DBF=45°,∴EDF=∴DEF=45°, ∴∴BDE=∴BDF+∴EDF=90°, 又∵AC∥DE ∴∴BGC=∴BDE=90°,即 AC⊥BD.·································································· 10 分 (说明:由 DF=BF=FE 得∴BDE=90°,同样给满分. ) 24. 解: (1)画图如右图; ····························· 1 分 800 700 600 500 400 300 200 100 0 10 20 30 40 50 60 70 80

y

由图可猜想 y 与 x 是一次函数关系, ·········· 2 分 设这个一次函数为 y = k x + b (k≠0) ∵这个一次函数的图象经过(30,500) (40,400)这两点, ∴?

x

?500 = 30k + b ?k = ?10 解得 ? ?400 = 40k + b ?b = 800

··········································································· 3 分 ············································································· 4 分

∴函数关系式是: y =-10 x +800

(2)设工艺厂试销该工艺品每天获得的利润是 W 元,依题意得 W=( x -20) (-10 x +800) ······················································································ 6 分 =-10 x 2 +1000 x -16000 =-10( x -50) 2 +9000 ·························································································· 7 分 ∴当 x =50 时,W 有最大值 9000. 所以,当销售单价定为 50 元?件时,工艺厂试销该工艺品每天获得的利润最大,最大利 润是 9000 元. ·············································································································· 8 分

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(3)对于函数 W=-10( x -50) +9000,当 x ≤45 时, W 的值随着 x 值的增大而增大, ················································································· 9 分 ∴销售单价定为 45 元?件时,工艺厂试销该工艺品每天获得的利润最大. ····························· 10 分 25. 解: (1)解法一:∵抛物线 y =-

2

2 2 x + b x + c 经过点 A(0,-4) , 3 2 2 x + b x + c =0 的两个根, 3

∴ c =-4 ·························································································· 1 分 又由题意可知, x 1 、 x 2 是方程- ∴ x 1+ x 2 =

3 b, 2

x 1 x 2 =-

3 c =6 ······························································ 2 分 2

由已知得( x 2 - x 1 ) 2 =25 又( x 2 - x
1

) 2 =( x 2 + x 1 ) 2 -4 x 1 x =

2

9 2 b -24 4

9 2 b -24=25 4 14 解得 b =± ··································································································· 3 分 3 14 当b = 时,抛物线与 x 轴的交点在 x 轴的正半轴上,不合题意,舍去. 3
∴ ∴ b =- 14 . ········································································································ 4 分 3 解法二:∵ x 1 、 x 2 是方程-

2 2 x + b x +c=0 的两个根, 3

即方程 2 x 2 -3 b x +12=0 的两个根.

∴x=

3b ±

9b 4

2

? 96

, ·················································································· 2 分

∴x 2-x

1

=

9b 2 ? 96 =5, 2

解得 b =±

14 ········································································································ 3 分 3

(以下与解法一相同. ) (2)∵四边形 BDCE 是以 BC 为对角线的菱形,根据菱形的性质,点 D 必在抛物线的 对称轴上, ··········································································································· 5 分

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又∵ y =-

2 2 14 2 7 25 x - x -4=- ( x + ) 2 + ····································· 6 分 3 3 3 2 6 7 25 )即为所求的点 D. ·········································· 7 分 ∴抛物线的顶点(- , 2 6

(3)∵四边形 BPOH 是以 OB 为对角线的菱形,点 B 的坐标为(-6,0) , 根据菱形的性质,点 P 必是直线 x =-3 与 抛物线 y =-

2 2 14 x - x -4 的交点, ·································································· 8 分 3 3 2 14 ∴当 x =-3 时, y =- ×(-3) 2 - ×(-3)-4=4, 3 3

∴在抛物线上存在一点 P(-3,4) ,使得四边形 BPOH 为菱形. ··················· 9 分 四边形 BPOH 不能成为正方形,因为如果四边形 BPOH 为正方形,点 P 的坐标 只能是(-3,3) ,但这一点不在抛物线上. ························································· 10 分



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