2008年江苏省镇江市中考数学试卷及解析

江苏省镇江市2008年初中毕业升学考试数学试卷

本试卷共3大题,28小题,满分120分．考试用时120分钟．闭卷考试． 希望你沉着冷静,相信你一定能成功！

一、填空题:本大题共12小题,每小题2分,共24分．请把结果填在题中的横线上． 1．3-的相反数是 ；3-的绝对值是 ． 2．计算:23-+= ；(2)(3)-?-= ． 3．计算:2

a a a += ；3

4a

a -= ．

4．计算:(2)(1)x x +-= ；分解因式:2

1x -= ． 5．若代数式

1

2

x x -+的值为零,则x =

；函数y =,自变量x 的取值范围为 ． 6．一组数据13234，，，，,这一组数据的众数为 ；极差为 ． 7．如图(1),图中的1∠= ；如图(2),已知直线12l l ∥,135∠=,那么2∠=

．

8．如图,DE 是ABC △的中位线,2DE =cm,12AB AC +=cm,则BC = cm,梯形DBCE 的周长为 cm ． 9．如果1

1m m

-=-,则2m m += ；2221m m +-= ． 10．如图,

O 是等腰三角形ABC 的外接圆,AB AC =,45A ∠=,BD 为O 的直

径

,BD =连结CD ,则D ∠= ,BC = ．

11．圆柱的底面半径为1,母线长为2,则它的侧面积为 (结果保留π)．

12．如图所示,两个全等菱形的边长为1厘米,一只蚂蚁由A 点开始按ABCDEFCGA 的顺序

第7题图（1） 1 35

100 第7题图（2） 1 2 l 1 l 2 A E C B

D （第8题图）

（第10题图）

C A

D

B G （第12题图）

沿菱形的边循环运动,行走2008厘米后停下,则这只蚂蚁停在 点． 二、选择题:本大题共5小题,每小题3分,共15分．每小题都给出代号为A B C D ，，，的四个结论,其中只有一个结论是正确的,请将正确结论的代号填在题后的括号内． 13．两圆的半径分别为2和3,圆心距为5,则两圆的位置关系为( ) A ．外离 B ．外切 C ．相交 D ．内切

14．用代数式表示“a 的3倍与b 的差的平方”,正确的是( ) A ．2

(3)a b -

B ．2

3()a b -

C ．2

3a b -

D ．2

(3)a b -

15．下面几何体的正视图是( )

16．如下图,把矩形OABC 放在直角坐标系中,OC 在x 轴上,OA 在y 轴上,且

2OC =,4OA =,把矩形OABC 绕着原点顺时针旋转90得到矩形OA B C ''',则点B '的坐标

为( ) A ．(24)， B ．(24)-，

C ．(42)，

D ．(24)-，

17．福娃们在一起探讨研究下面的题目:

参考下面福娃们的讨论,请你解该题,你选择的答案是( )

贝贝:我注意到当0x =时,0y m =>． 晶晶:我发现图象的对称轴为12

x =． 欢欢:我判断出12x a x <<．

迎迎:我认为关键要判断1a -的符号． 妮妮:m 可以取一个特殊的值．

几何体

A ．

B ．

C ．

D ．

x

A

A '

B '

O

C B

C '

y 函数2y x x m =-+(m 为常数)的图象如左图,

如果x a =时,0y <；那么1x a =-时,函数值( ) A ．0y < B ．0y m <<

C ．y m >

D ．y m =

x y

O x 1 x 2

三、解答题:本大题共11小题,共81分．解答应写出必要的文字说明,证明步骤,推理过程． 18．(本小题满分10分)计算化简

(1)

)

1

112-??

- ???

；

(2)

241

42

x x +

-+．

19．(本小题满分10分)运算求解 解方程或不等式组 (1)2

20x x -=；

(2)921102

x x ->???-??≥，．

20．(本小题满分6分)实验探究

有A B ，两个黑布袋,A 布袋中有两个完全相同的小球,分别标有数字1和2．B 布袋中有三

个完全相同的小球,分别标有数字1-,2-和3-．小明从A 布袋中随机取出一个小球,记录其标有的数字为x ,再从B 布袋中随机取出一个小球,记录其标有的数字为y ,这样就确定点Q 的一个坐标为()x y ，．

(1)用列表或画树状图的方法写出点Q 的所有可能坐标； (2)求点Q 落在直线3y x =-上的概率．

21．(本小题满分6分)作图证明

如图,在ABC △中,作ABC ∠的平分线BD ,交AC 于D ,作线段BD 的垂直平分线EF ,分别交AB 于E ,BC 于F ,垂足为O ,连结DF ．在所作图中,寻找一对全等三角形,并加以证明．(不写作法,保留作图痕迹)

22．(本小题满分6分)推理运算

A B C

二次函数的图象经过点(03)A -，,(23)B -，,(10)C -，．

(1)求此二次函数的关系式；

(2)求此二次函数图象的顶点坐标；

(3)填空:把二次函数的图象沿坐标轴方向最少．．平移 个单位,使得该图象的顶点在原点．

23．(本小题满分6分)实际运用

512汶川大地震发生以后,全国人民众志成城．首长到帐篷厂视察,布置赈灾生产任务,下面是首长与厂长的一段对话:

首长:为了支援灾区人民,组织上要求你们完成12000顶帐篷的生产任务． 厂长:为了尽快支援灾区人民,我们准备每天的生产量比原来多一半． 首长:这样能提前几天完成任务？

厂长:请首长放心！保证提前4天完成任务！ 根据两人对话,问该厂原来每天生产多少顶帐篷？

24．(本小题满分6分)推理运算 如图,在直角坐标系xOy 中,直线1

22

y x =

+与x 轴,y 轴分别交于A B ，两点,以AB 为边在第二象限内作矩形ABCD ,

使AD =

(1)求点A ,点B 的坐标,并求边AB 的长；

(2)过点D 作DH x ⊥轴,垂足为H ,求证:ADH BAO △∽△；

(3)求点D 的坐标．

25．(本小题满分7分)实际运用

如图,奥运圣火抵达某市奥林匹克广场后,沿图中直角坐标系中的一段反比例函数图象传递．动点()T m n ，表示火炬位置,火炬从离北京路10米处的M 点开始传递,到离北京路1000米的N 点时传递活动结束．迎圣火临时指挥部设在坐标原点O (北京路与奥运路的十字路口),OATB 为少先队员鲜花方阵,方阵始终保持矩形形状且面积恒为10000平方米(路线宽度均不计)．

(1)求图中反比例函数的关系式(不需写出自变量的取值范围)；

(2)当鲜花方阵的周长为500米时,确定此时火炬的位置(用坐标表示)；

(3)设t m n =-,用含t 的代数式表示火炬到指挥部的距离；当火炬离指挥部最近时,

确定此时

x

火炬的位置(用坐标表示)．

26．(本小题满分7分)推理运算

如图,AB 为O 直径,CD 为弦,且CD AB ⊥,垂足为H ． (1)OCD ∠的平分线CE 交O 于E ,连结OE ．求证:E 为ADB 的中点；

(2)如果

O 的半径为1

,CD =①求O 到弦AC 的距离；

②填空:此时圆周上存在 个点到直线AC 的距离为12

．

27．(本小题满分9分)理解发现 阅读以下材料:

对于三个数a b c ，，,用{}M a b c ，，表示这三个数的平均数,用{}min a b c ，，表示这三个数中最小的数．例如:

{}1234

12333M -++-=

=，，；{}min 1231-=-，，；{}(1)min 121

(1).

a a a a -?-=?->-?≤；，，

解决下列问题:

(1)填空:{}min sin30cos 45tan30

=，， ；

如果{}min 222422x x +-=，，,则x 的取值范围为x ________≤≤_________． (2)①如果{}{}212min 212M x x x x +=+，，，，,求x ；

x

A

B

D

E

O C

H

②根据①,你发现了结论“如果{}{}min M a b c a b c =，，，，,那么 (填a b c ，，的大小关系)”．证明你发现的结论；

③运用②的结论,填空:

若{}{}2222min 2222M x y x y x y x y x y x y +++-=+++-，，，，, 则x y += ．

(3)在同一直角坐标系中作出函数1y x =+,2

(1)y x =-,2y x =-的图象(不需列表描点)．通过观察图象,

填空:{}

2min 1(1)2x x x +--，，的最大值为 ．

28．(本小题满分8分)探索研究

如图,在直角坐标系xOy 中,点P 为函数2

14

y x =

在第一象限内的图象上的任一点,点A 的坐标为(01)，,直线l 过(01)B -，且与x 轴平行,过P 作y 轴的平行线分别交x 轴,l 于C Q ，,连结AQ 交x 轴于H ,直线PH 交y 轴于R ． (1)求证:H 点为线段AQ 的中点； (2)求证:①四边形APQR 为平行四边形；

②平行四边形APQR 为菱形；

(3)除P 点外,直线PH 与抛物线2

14

y x =有无其它公共点？并说明理由．

镇江市2008年初中毕业升学考试数学试卷

参考答案及评分标准

x

x

一、填空题: 1．3,3

2．1,6

3．2

2a ,a

4．2

2x x +-,(1)(1)x x +-

5．1x =,2x ≥ 6．3,3 7．65,35 8．4,12 9．1,1 10．45,2 11．4π 12．A 二、填空题: 13．B 14．A 15．D 16．C 17．C 三、解答题:

18．(1)原式122=-+ ····························································· (3分,每对1个得1分)

1=． ········································································································· (5分) (2)原式41

(2)(2)2

x x x =

++-+ ······································································· (1分)

42

(2)(2)(2)(2)

x x x x x -=

++-+- ····································································· (2分)

2

(2)(2)

x x x +=

+-··························································································· (4分)

1

2

x =

-． ··································································································· (5分) 19．(1)(2)0x x -=． ··················································································· (3分)

10x ∴=,22x =． ························································································ (5分)

(2)由①,得4x <；························································································ (2分) 由②,得1x ≥． ··························································································· (4分) ∴原不等式组的解集为14x <≤． ·································································· (5分) 20．(1)用列表或画树状图的方法求点

Q 的坐标有

(11)-，,(12)-，,(13)-，,(21)-，,(22)-，,(23)-，．(4分,列表或树状图正确得2分,点坐标2分)

(2)“点Q 落在直线3y x =-上”记为事件A ,所以21

()63

P A ==, 即点Q 落在直线3y x =-上的概率为

13

． ························································· (6分) 21．(1)画角平分线,线段的垂直平分线． ··································· (3分,仅画出1条得2分) (2)BOE BOF DOF △≌△≌△ ··························· (4分,只要1对即可),证明全等．(6分) 22．(1)设2

3y ax bx =+-, ············································································· (1分) 把点(23)-，,(10)-，代入得423330.

a b a b +-=-??

--=?，

···················································· (2分)

解方程组得12.

a b =??

=-?， 2

23y x x ∴=--． ························································ (3分)

(也可设2

(1)y a x k =-+)

(2)2

2

23(1)4y x x x =--=--． ··································································· (4分)

∴函数的顶点坐标为(14)-，． ·

······································································· (5分) (3)5 ············································································································ (6分) 23．设该厂原来每天生产x 顶帐篷,根据题意得: ··················································· (1分)

1200012000

432

x x

-=． ·

·················································································· (3分) 解方程得:1000x =． ···················································································· (4分) 经检验:1000x =是原方程的根,且符合题意． ····················································· (5分)

答:该厂原来每天生产1000顶帐篷． ································································· (6分) 24．(1)(40)A -，,(02)B ，,

∴在Rt AOB △中

,AB =． ································· (2分)

(2)由90ADH DAH ∠+∠=,90BAO DAH ∠+∠=,

BAO ADH ∴∠=∠,又90AOB DHA ∠=∠=,

ADH BAO ∴△∽△． ················································································· (4分) (3)ADH BAO △∽△,

DH AH AD AO BO BA ∴

==,

即42DH AH ==

2DH ∴=,1AH =．

(52)D ∴-，． ······························································································· (6分) 25．(1)设反比例函数为(0)k

y k x

=

>． ·

··························································· (1分) 则10000OATB k xy mn S ====矩形, ································································· (2分)

10000

y x

∴=

．···························································································· (3分) (2)设鲜花方阵的长为m 米,则宽为(250)m -米,由题意得:

(250)10000m m -=． ················································································· (4分) 即:2

250100000m m -+=,

解得:50m =或200m =,满足题意．

∴此时火炬的坐标为(50200)，或(20050)，． ·

···················································· (5分) (3)

10000mn =,在Rt TAO △中

,TO ==

== ·

······························································ (6分) ∴当0t =时,TO 最小,

此时m n =,又10000mn =,0m >,0n >, 100m n ∴==,且101001000<<．

(100100)T ∴，． ··························································································· (7分) 26．(1)OC OE =,E OCE ∴∠=∠ ································································ (1分)

又OCE DCE ∠=∠,E DCE ∴∠=∠．

OE CD ∴∥． ·

··························································································· (2分) 又CD AB ⊥,90AOE BOE ∴∠=∠=．

E ∴为ADB 的中点． ··················································································· (3分) (2)①

CD AB ⊥,AB 为O 的直径

,CD =

,

122

CH CD ∴=

=． ················································································· (4分) 又1OC =

,2sin 12

CH COB OC ∴∠===． 60COB ∴∠=, ··························································································· (5分) 30BAC ∴∠=．

作OP AC ⊥于P ,则11

22

OP OA =

=． ·

··························································· (6分) ②3 ············································································································ (7分)

27．(1)1

2

·········································································· (1分,填sin30也得分)； 01x ≤≤ ·

································································································· (2分) (2)①{}21221

213

x x M x x x ++++==+，，． 法一:

2(1)1x x x -+=-．

当1x ≥时,则{}min 2122x x +=，，,则12x +=,1x ∴=．

当1x <时,则{}min 2122x x x +=，，,则12x x +=,1x ∴=(舍去)．

综上所述:1x =． ························································································· (4分) 法二:

{}{}2122121min 2123

x x

M x x x x x ++++=

=+=+，，，，,

212 1.x x x +?∴?

+?

≥，

≥ ···························································································· (3分) 11.

x x ?∴?

?≤，

≥ 1x ∴=． ··················································································· (4分) ②a b c == ································································································· (5分) 证明:

{}3

a b c

M a b c ++=

，，, 如果{}min a b c c =，，,则a c ≥,b c ≥． 则有

3

a b c

c ++=,即20a b c +-=． ()()0a c b c ∴-+-=．

又0a c -≥,0b c -≥．0a c ∴-=且0b c -=． a b c ∴==．

其他情况同理可证,故a b c ==． ····································································· (6分) ③4- ········································································································· (7分) (3)作出图象．

····································································· (8分)

1 ··············································································································· (9分) 28．(1)法一:由题可知1AO CQ ==．

90AOH QCH ∠=∠=,AHO QHC ∠=∠,

AOH QCH ∴△≌△．

················································································· (1分) OH CH ∴=,即H 为AQ 的中点． ·································································· (2分) 法二:

(01)A ，,(01)B -，,OA OB ∴=． ·

··························································· (1分) x

2x -

1x +

又BQ x ∥轴,HA HQ ∴=． ·········································································· (2分) (2)①由(1)可知AH QH =,AHR QHP ∠=∠,

AR PQ ∥,RAH PQH ∴∠=∠,

RAH PQH ∴△≌△． ·

················································································ (3分) AR PQ ∴=,

又AR PQ ∥,∴四边形APQR 为平行四边形． ··················································· (4分)

②设2

14

P m m ?

? ??

?

，,

PQ y ∥轴,则(1)Q m -，,则21

14

PQ m =+．

过P 作PG y ⊥轴,垂足为G ,在Rt APG △中

,

2114AP m PQ ====+=．

∴平行四边形APQR 为菱形． ········································································ (6分)

(3)设直线PR 为y kx b =+,由OH CH =,得22m H ??

???，,214P m m ??

???

，代入得: 2021.4m k b km b m ?+=???

?+=??， 22

1.

4

m k b m ?

=??∴??=-??，∴直线PR 为2124m y x m =-． ························· (7分) 设直线PR 与抛物线的公共点为2

14

x x ?

? ??

?

，,代入直线PR 关系式得:

22110424m x x m -+=,21()04x m -=,解得x m =．得公共点为214m m ?? ???

，． 所以直线PH 与抛物线2

14

y x =

只有一个公共点P ． (8分)