2006广东梅州市初中毕业生学业考试数学试题（含答案） - 试题 -

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2006年广东省梅州市初中毕业生学业考试

数学试卷

说明：全卷共8页，考试时间90分钟，满分120分． 题号 得分 一 二 三 16 17 18 19 20 21 22 23 24 25 总分 一、选择题：每小题3分，共15分；每小题给出四个答案，其中有一个是正确的，把所选答案的编号填写在题目后面的括号内． 1．?1等于（ ） 2

Ｂ．?2

Ｃ．?Ａ．2

1 2 Ｄ．

1 22．小明在镜中看到身后墙上的时钟，实际时间最接近8时的是下图中的（ ）

A． B． C． D．

3．我市大部分地区今年5月中、下旬的天气情况是：前5天小雨，后5天暴雨．那么能反映我市主要河流水位变化情况的图象大致是（ ）

水位 水位 水位 水位 （天） （天） （天） （天） 0 0 0 0 5 10 5 10 5 10 5 10 D． A． B． C． 4．如图1，把矩形ABCD沿EF对折，若?1?50，则?AEF等于（ ） Ａ．115 Ｃ．120

5．在同一平面直角坐标系中，直线y?x?3与双曲线y??

Ｂ．130 Ｄ．65

A B

E D 1 F

图1

C

1的交点个数为（ ） xＡ．0个 Ｂ．1个 Ｃ．2个 Ｄ．无法确定 二、填空题：每小题3分，共30分；答案填写在该题的横线上．

6．我市约有495万人口，用科学记数法表示为 人．

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7．如果一个几何体的主视图是等腰三角形，那么这个几何体可以是 ．（填上满足条件的一个几何体即可）

8．一个袋中装有6个红球、4个黑球、2个白球，每个球除颜色外完全相同，从袋中任意摸出一个球，那么摸出 球的可能性最大．

?1?9．计算：(sin60?1)????? ．

?2?0?110．计算：(a2b3?a2b2)?(ab)2? ．

11．将抛物y??(x?1)2向左平移1个单位后，得到的抛物线的解析式是 ．

x2?2x?312．当x? 时，分式的值为零．

x?313．能使平行四边形ABCD为正方形的条件是 ．（填上一个符合题目要求的条件即可）

B 14．如图2，两个半圆中，小圆的圆心O?在大O的直径CD上， A 长为4的弦AB与直径CD平行且与小半圆相切，那么圆中阴影部

C 分面积等于 ．

15．如图3，已知△ABC的周长为m，分别连接AB，BC，CA 的中点A1，B1，C1得△A1B1C1，再连接A，B1C1，C1A1的中点 1B1O? O

图2

A D A2，B2，C2得△A2B2C2，再连接A2B2，B2C2，C2A2的中点

A3，B3，C3得△A3B3C3，这样延续下去，最后得△AnBnCn． B A1 C B A2 A3 2B1 3C2 B3 C1 C

图3 ， △AnBnCn

设△A1B1C1的周长为l1，△A2B2C2的周长为l2，△A3B3C3的周长为l3的周长为ln，则ln? ． 三、解答题：本大题有10小题，共75分． 16．本小题满分6分．

因式分解：x(y?1)?2x(y?1)?(y?1)．

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17．本小题满分6分． 解不等式组：??5?3(x?4)?2，

?2x?3≥1.

18．本小题满分6分．

如图4是某文具店在2005年卖出供学生使用的甲、乙、丙三种品牌科学计算器个数的条形统计图，试解答下面问题：

（1）求卖出甲、乙、丙三种科学计算器的个数的频率；

（2）根据以上统计结果，请你为该文具店进货提出一条合理化建议．

个数 90 90 72 54 54 36 36

18 0 品牌

甲 乙 丙

19．本小题满分6分．

图4

，B，C的坐标分别是A(?1，?1)，B(?4，?3)，C(?4，?1)． 如图5，已知△ABC的顶点A（1）作出△ABC关于原点O中心对称的图形；

（2）将△ABC绕原点O按顺时针方向旋转90后得到△A1B1C1，画出△A1B1C1，并写出点A1的坐标．

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y C B A O x 图5

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20．本小题满分7分．

小明与小华在玩一个掷飞镖游戏，如图6?甲是一个把两个同心圆平均分成8份的靶，当飞镖掷中阴影部分时，小明胜，否则小华胜（没有掷中靶或掷到边界线时重掷）． （1）不考虑其他因素，你认为这个游戏公平吗？说明理由．

（2）请你在图6?乙中，设计一个不同于图6?甲的方案，使游戏双方公平．

图6?乙 图6 ?甲

21．本小题满分7分．

梅华中学九年级数学课外学习小组某下午实践活动课时，测量朝西教学楼前的旗杆AB的高度．如图7，当阳光从正西方向照射过来时，旗杆AB的顶端A的影子落在教学楼前的坪地

C处，测得影长CE?2m，DE?4m，BD?20m，DE与地面的夹角??30．在同一时

刻，测得一根长为1m的直立竹竿的影长恰为4m．根据这些数据求旗杆AB的高度．（可能用到的数据：2?1.414，3?1.732，结果保留两个有效数字）

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A B E C Ｄ ? 图7

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22．本小题满分8分．

某公司开发生产的1200件新产品需要精加工后才能投放市场，现有甲、乙两个工厂都想加工这批产品．公司派出相关人员分别到这两间工厂了解生产情况，获得如下信息： 信息一：甲工厂单独加工完成这批产品比乙工厂单独加工完成这批产品多用10天； 信息二：乙工厂每天比甲工厂多加工20件．

根据以上信息，求甲、乙两个工厂每天分别能加工多少件新产品？

23．本小题满分8分．

用两个全等的正方形ABCD和CDFE拼成一个矩形ABEF，把一个足够大的直角三角尺的直角顶点与这个矩形的边AF的中点D重合，且将直角三角尺绕点D按逆时针方向旋转． （1）当直角三角尺的两直角边分别与矩形ABEF的两边BE，EF相交于点G，H时，如图8?甲，通过观察或测量BG与EH的长度，你能得到什么结论？并证明你的结论． （2）当直角三角尺的两直角边分别与BE的延长线，EF的延长线相交于点G，H时（如

8?图乙），你在图8?甲中得到的结论还成立吗？简要说明理由．

H

A B

D F H A B

D F E

G

G C 图8?甲

E C

图8?乙

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24．本小题满分10分． 如图9，直线l的解析式为y?4x?4，l与x轴，y轴分别交于点A，B． 3（1）求原点O到直线l的距离；

（2）有一个半径为1的C从坐标原点出发，以每秒1个单位长的速度沿y轴正方向运动，设运动时间为t（秒）．当C与直线l相切时，求t的值．

25．本小题满分11分． 如图10，点A在抛物线y?y l B x A C O 图9

12x上，过点A作与x轴平行的直线交抛物线于点B，延长41AO，BO分别与抛物线y??x2相交于点C，D，连接AD，BC，设点A的横坐标为m，

8且m?0．

，B，D的坐标； （1）当m?1时，求点A（2）当m为何值时，四边形ABCD的两条对角线互相垂直； （3）猜想线段AB与CD之间的数量关系，并证明你的结论．

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y B O C 图10

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2006年广东省梅州市初中毕业生学业考试

数学试卷参考答案及评分意见

一、选择题：每小题3分，共15分 1．D 2．B 3．B 二、填空题：每小题3分，共30分

64．A 5．C

6．4.95?10； 7．圆锥或正三棱锥或正四棱锥；

8．红； 9．3

10．b?1；

11．y??x2； 12．x??1； 13．AC?BD且AC⊥BD或AB?BC且AB⊥BC等；

14．2?；

?1?15．??m．

?2?n三、解答下列各题：共75分

16．解：原式?(y2?1)(x2?2x?1) ···················································································· 2分 ?(y2?1)(x?1)2 ····························································································· 4分 ?(y?1)(y?1)(x?1)2 ···················································································· 6分 17．解：由5?3(x?4)?2，得x?5， ·············································································· 2分 由2x?3≥1，得x≥2， ····················································································· 4分 ?原不等式组的解集是：2≤x?5． ·································································· 6分

36?0.2 ············································ 2分

36?54?9054?0.3· 卖出乙计算器个数的频率：············································ 3分

36?54?9090?0.5． · 卖出丙计算器个数的频率：········································ 4分

36?54?90 （2）0.2:0.3:0.5?2:3:5， ····················································································· 5分 ?该文具店进甲、乙、丙三种科学计算器时，按2:3:5的比例进货． ·············· 6分

18．解：（1）卖出甲计算器个数的频率：

（或该文具店进货时，丙科学计算器进多一些，而甲、乙科学计算器进少一些．类

y 似这样的合理答案5分）

19．（1）正确画出图形 ················································································ 3分 B1 C1 B2 （2）正确画出图形 ················································································ 5分

A1 A2 C2 x A ······························································································· 6分 C ，1)1(?1A O 20．解：（1）这个游戏公平． ···································································· 2分 B 根据图6?甲的对称性，阴影部分的面积等于圆面积的一半， ?这个游戏公平． ··········································································································· 4分 （2）把图6?乙中的同心圆平均分成偶数等分，再把其中的一半作为阴影部分即可．（图略） ········································································································································· 7分

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21．解：如图，过点C，E分别作CF⊥AB于点F，EH⊥BD的延长线于H． ········ 1分 在Rt△DEH中，

DE?4m，?EDH?30，E C D H A ?EH?2m， ········································································ 2分 F ·············································· 3分 B DH?DE?EH?23m· 又

22AF1? ··················································································································· 5分 CF411 ?AF?CF?(EF?CE)

441 ?(BD?DH?CE)?6.4． ··········································································· 6分

48.4(m) · ?AB?EH?AF?．····················································································· 7分

22．解：设甲工厂每天能加工x件新产品， ········································································· 1分 则乙工厂每天能加工(x?20)件新产品． ····································································· 2分

12001200??10． ··········································································· 4分 xx?20 解得x?40或x??60（不合题意舍去）， ··································································· 6分 经检验x?40是所列方程的解， ?x?20?60． ·············································································································· 7分

依题意得方程

答：甲工厂每天能加工40件新产品，乙工厂每天能加工60件新产品． ··················· 8分 23．解：（1）BG?EH． ···································································································· 2分 四边形ABCD和CDFE都是正方形，

?DC?DF，?DCG??DFH??FDC?90，

?CDG??CDH??CDH??FDH?90，??CDG??FDH， ···················· 3分

△CDG≌△FD，H ? ?CG?FH， ················································································································ 4分

F?BG?EH BC?E，． ·························································································· 5分

（2）结论BG?EH仍然成立． ·························································································· 6分

同理可证△CDG≌△FDH，

，BC?，EF?B?G．E ?CG?FH ···································································· 8分

y 424．解：（1）在y?x?4中，令x?0，得y?4，得BO?4．

3l C? D? 令y?0，得x??3，得AO?3，

B ?AB?························································ 2分 AO2?BO2?5． ·

D C A 第8页 / 共10页

设点O到直线AB的距离为h，

O x

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11AOBO?ABh， 22AOBO?2.4 ?h?． ·································································································· 4分 AB

S△AOB? （其它解法参照给分）

（2）如图，设C与直线l相切于点D，连CD，则CD⊥AB， ·································· 5分

AO⊥BO，??BDC??BOA?90，

?ABO??CB DBCCD?， ·············································································· 6分 ABAO，BO?4，AB?5， 由（1）得AO?3BC1557?，?BC?，?OC?4??， ?533337 ?t?CO?（秒）． ···································································································· 8分

35 根据对称性得BC??BC?，

351717?t?O?C?（秒） ?OC??4??，． ······························································ 9分

333717 ?当C与直线l相切时，t?秒或秒． ···························································· 10分

33∽△CB，D? ?△ABO25．解：（1）

点A在抛物线y?12?1?x上，且x?m?1，?A?1························ 1分 ，?， ·4?4?

点B与点A关于y轴对称，?B??1····························································· 2分 ，?．·

??1?4? 设直线BD的解析式为y?kx，

?y?? ?k??，141x． ······························································································· 3分 41?y??x?1???4 解方程组?，得D?2，····································································· 4分 ??． ·

2???y??1x2?8?（2）当四边形ABCD的两对角线互相垂直时，由对称性得直线AO与x轴的夹角等于45所以点A的纵、横坐标相等， ····················································································· 5分

12x，得a?4，?A(4，，4)?m?4． 4 即当m?4时，四边形ABCD的两条对角线互相垂直．············································· 7分 （3）线段CD?2AB． ········································································································ 8分

这时，设A(a，a)，代入y?第9页 / 共10页

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点A在抛物线y?12?1?x，且x?m，?A?m，m2?， 4?4?mx， 4 得直线AO的解析式为y?m?y?x?1???4 解方程组?，得点C??2m··························································· 9分 ，?m2? ·

12???y??x2?8? 由对称性得点B??m，m2?，D?2m，???14????12?···················································· 10分 m?， ·

2?，CD?4m， ?AB?2m ?CD?2AB． ············································································································ 11分

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