湖北鄂州

数 学

考生注意：

1．本卷共三道大题，27道小题，满分120分，考试时间120分钟． 2．考时不准使用计算器．

一、选择题（每小题3分，共42分） 1．下列计算正确的是（ ） A．

336

B．x6x3x2 D．a2(a)2a4

C．33

1aa2

2．已知1aa，则a的取值范围是（ ）

A．a≤0 B．a0 C．0a≤1 D．a0

1，6，1，x的众数为1，则这组数据的方差是（ ） 3．数据0，A．2

B．

345 C．2 D．

265

4．不等式组2x31x2≤4的解集在数轴上可表示为（ ）

A． B． C． D．

5．图1是由几个小立方块搭成的几何体的俯视图，小正方形中的数字表示在该位置的小立方块的个数，那么这个几何体的主视图是（ ）

A． B． C． D．

H是高AD6．如图2，已知△ABC中，ABC45，AC4，

2 1

3

图1

A E H

B D 图2

C

和BE的交点，则线段BH的长度为（ ） A．

6

B．4 C．23 D．5

7．在反比例函数y

4x的图象中，阴影部分的面积不等于4的是（ ）

A． B． C． D．

8．如图3，利用标杆BE测量建筑物DC的高度，如果标杆BE长为1.2米，测得AB1.6米，BC8.4米．则楼高CD是（ ） A．6.3米 B．7.5米 C．8米 D．6.5米 9．因为sin30D

E A B

C

图3

12，sin21012，

22所以sin210sin(18030)sin30；因为sin45，sin22522，所以

sin225sin(18045)sin45，由此猜想，推理知：一般地当为锐角时有sin(180)12，由此可知：sin240（ ） sin2232A． B． C． D．3

10．下列方程中，有两个不等实数根的是（ ） A．x3x8

22 B．x5x10

22

C．7x14x70 D．x7x5x3

11．如图4，直线y2x4与x轴，y轴分别相交于A，B两点，C为OB上一点，且

12，则S△ABC（ ）

A．1

C O y B 1 2 B．2 C．3 D．4

y 1x 3A1

2 1 0 1 2 x

H A

O

C B 图6

O1

H1 C1

A 图4

x 图5

12．△ABC是半径为15的圆内接三角形，以A为圆心，于D点，则ABAC的值为（ ） A．

3210

62为半径的A与边BC相切

B．4 C．

52 D．310 13．小明从图5所示的二次函数yax2bxc的图象中，观察得出了下面五条信息：①c0；②abc0；③abc0；④2a3b0；⑤c4b0，你认为其中正确信息的个数有（ ） A．2个

B．3个

C．4个

D．5个

Rt△ABC中，BC2，O，H分别为边AB，AC14．如图6，ACB90，CAB30，

的中点，将△ABC绕点B顺时针旋转120到△A1BC1的位置，则整个旋转过程中线段OH所扫过部分的面积（即阴影部分面积）为（ ） A．

73π783

B．

43π783 C．π D．

43π3

二、填空题（每小题3分，共18分）

15．在“a2□2ab□b2”方框中，任意填上“”或“”．能够构成完全平方式的概率是 ．

16．下列给出的一串数：2，5，10，17，26，？，50．仔细观察后回答：缺少的数？是 ．

17．如图7，正方体的棱长为2，O为边AD的中点，则以

O，A1，B三点为顶点的三角形面积为 ．

D1O A D B

C C118．已知在O中，半径r5，AB，CD是两条平行弦， 且AB8，CD6，则弦AC的长为 ． 19．已知，为方程x4x21450 ．

20．如图8，在△ABC中，BAC45，ADBC于D点，3A1图7 A B120的二实根，则

B

已知BD6，CD4，则高AD的长为 ．

三、解答题（21题6分，26题10分，27题12分，其余每题8分，总计60分）

21．设x1，x2是关于x的一元二次方程x2axa4a20的两实根，当a为何值时，

x1x2有最小值？最小值是多少？

22D

图8

C

22

22．如图9，教室窗户的高度AF为2.5米，遮阳蓬外端一点D到窗户上椽的距离为AD，某一时刻太阳光从教室窗户射入室内，与地面的夹角BPC为30，PE为窗户的一部分在教室地面所形成的影子且长为3米，试求AD的长度．（结果带根号）

图9

23．小王和小明用如图10所示的同一个转盘进行“配紫色”游戏，游戏规则如下：连续转动两次转盘．如果两次转出的颜色相同或配成紫色（若其中一次转盘转出蓝色，另一次转出红色，则配成紫色），则小王得1分，否则小明得1分（如果指针恰好指在分割线上，那么重转一次，直到指针指向一种颜色为止）

（1）请你通过列表法分别求出小王和小明获胜的概率．

（2）你认为这个游戏对双方公平吗？请说明理由；若不公平，请修改规则，使游戏对双方公平．

红 黄 蓝 绿

图10

24．甲乙两人同时登西山，甲、乙两人距地面的高度y（米）与登山时间x（分）之间的函数图象如图11所示，根据图象所提供的信息解答下列问题： （1）甲登山的速度是每分钟 米，

乙在A地提速时距地面的高度b为 米．

（2）若乙提速后，乙的速度是甲登山速度的3倍，请分别求出甲、乙二人登山全过程中，登山时距地面的高度y（米）与登山时间x（分）之间的函数关系式． （3）登山多长时间时，乙追上了甲？此时乙距A地的高度为多少米？

图11

25．如图12，已知：边长为1的圆内接正方形ABCD中，P为边CD的中点，直线AP交圆于E点．

（1）求弦DE的长．

（2）若Q是线段BC上一动点，当BQ长为何值时，三角形ADP与以Q，C，P为顶点的三角形相似．

B 图12

C A P E D

26．为了更好治理洋澜湖水质，保护环境，市治污公司决定购买10台污水处理设备．现有

A，B两种型号的设备，其中每台的价格，月处理污水量如下表： 价格（万元/台） 处理污水量（吨/月） A型 B型 a b 240 200 经调查：购买一台A型设备比购买一台B型设备多2万元，购买2台A型设备比购买3台B型设备少6万元．

（1）求a，b的值．

（2）经预算：市治污公司购买污水处理设备的资金不超过105万元，你认为该公司有哪几种购买方案．

（3）在（2）问的条件下，若每月要求处理洋澜湖的污水量不低于2040吨，为了节约资金，请你为治污公司设计一种最省钱的购买方案．

27．（1）如图13，A1，A2，A3是抛物线y14x图象上的三点，若A1，A2，A3三点的横坐

2标从左至右依次为1，2，3．求△A1A2A3的面积． （2）若将（1）问中的抛物线改为y14x212x2和yaxbxc(a0)，其他条件

2不变，请分别直接写出两种情况下△A1A2A3的面积． （3）现有一抛物线组：y112x213x；y216x2112x；y3112x2125x；

y4120x2142x；y5130x2163x；依据变化规律，请你写出抛物线组第n个式

子yn的函数解析式；现在x轴上有三点A(1，经过A，B，C向x轴作0)，B(2，0)，C(3，0)．垂线，分别交抛物线组y1，y2，y3，，yn于A1，B1，C1；A2，B2，C2；A3，B3，C3；

；An，Bn，Cn．记S△A1B1C1为S1，S△A2B2C2为S2，，S△AnBnCn为Sn，试求

S1S2S3S1的值．0（4）在（3）问条件下，当n10时有Sn10Sn9Sn8Sn的值不小于

11242，请

探求此条件下正整数n是否存在最大值，若存在，请求出此值；若不存在，请说明理由．

y

鄂州市2008年初中升学考试试卷

数学参及评分标准

说明：考生若写出其他正确答案，可参考本评分标准给分．

一、选择题（每小题3分，共42分） 1．D 2．C 3．B 4．D 5．A 8．B 9．C 10．D 11．C 12．D 二、填空题（每小题3分，共18分） 15．

12A3 A1 A2 O 1 2 3 A B C x

图13

6．B 13．C

7．B 14．C

16．37 17．6 18．2或52或72 19．2 20．12

三、解答题（第21题6分，第26题10分，第27题12分，其余每题8分，共60分） 21．解答：(2a)4(a4a2)≥0

a≤1222 ······················································································································· 1分

2又x1x22a，x1x2a4a2 ········································································ 2分

x1x2(x1x2)2x1x2

2222············································································································ 4分 2(a2)4·

a≤12

122时，x12x2的值最小 ·················································································· 5分

当a111此时xx224，即最小值为． ············6分

222D

21222A B F

G

P

22．解：过点E作EG∥AC交于PD于G点 ······· 1分

EGEPtan30333····················· 3分 1·

C

E

22题图

即ABAFBF2.511.5 ·················································································· 5分

BFEG1 ····················································· 4分

在Rt△ABD中，ADABtan301.53332················································· 7分 3（米） ·

AD的长为323米 ··································································································· 8分

23．解：（1） 第二次 第一次 红 黄 蓝 绿 红 （红红） （黄红） （蓝红） （绿红） 61610163858黄 （红黄） （黄黄） （蓝黄） （绿黄） 蓝 （红蓝） （黄蓝） （蓝蓝） （绿蓝） 绿 （红绿） （黄绿） （蓝绿） （绿绿） ····································································································································· 2分 从表中可知：P（小王获胜） P（小明获胜）（2）小王得分为18338········································································ 3分 ········································································ 4分

558，小明得分18

有：

3858

·············································································································· 6分 游戏不公平 ·

修改游戏规则：若两次出现颜色相同或配成紫色，小王得5分；否则小明得3分．

（注：答案不唯一，合理的修改规则均得分）······························································· 8分 24．（1）10，30 ············································································································ 2分

（2）由图知：

30030t2···································································· 3分 310 t11 ·

C(0，100)，D(20，300)

····················································· 4分 线段CD的解析式：y甲10x100(0≤t≤20) ·

A(2，30)，B(11，300)

15x (0≤t≤2) ············································· 6分 折线OAB的解析式为：y乙30x30 (2≤t≤11)（3）由y10x100y30x30解得x6.5y165 ·········································································· 7分

登山6.5分钟时乙追上甲．

此时乙距A地高度为16530135（米） ·································································· 8分

25．（1）如图1．过D点作DFAE于F点． 在Rt△ADP中，AP又S△ADP5512ADDPADDP122252································································ 1分

APDF

DF ················································································································· 2分

AD的度数为90

DEA45

105DE2DF································································································· 4分

A P D A P E

B （Q） 25题图2 ADQCDPCP A B

25题图1

F D D P E C C E

B Q 25题图3

C （2）如图2．当Rt△ADP∽Rt△QCP时有

得：QC1．

即点Q与点B重合，BQ0 ···················································································· 5分

ADPCPDQC如图3，当Rt△ADP∽Rt△PCQ时，有

1434

得QC，即BQBCCQ34 ············································································ 7分

当BQ0或BQab2时，三角形ADP与以点Q，C，P为顶点的三角形相似． ······· 8分

26．（1）a12·················································································· 2分 3b2a6b10（2）设购买污水处理设备A型设备X台，B型设备(10X)台，则：

····························································································· 3分 12X10(10X)≤105·

X≤2.5 ··················································································································· 4分 X取非负整数

X0，1，2·················································································································· 5分

有三种购买方案：①A型设备0台，B型设备10台；②A型设备1台，B型设备9台；

③A型设备2台，B型设备8台． ··············································································· 6分 （3）由题意：240X200(10X)≥2040 ································································ 7分

X≥1

又X≤2.5

············································································································· 8分 X为1，2．·

当X1时，购买资金为：121109102（万元）

当X2时，购买资金为：122108104（万元）

·············································10分 为了节约资金，应选购A型设备1台，B型设备9台 ·

27．（1）A11，，A2(2，，···································································· 1分 1)A33，·

4419S△A1A2A3S梯形AACAS梯形AABAS梯形ABCA

1312231912114442291114 ······························································· 3分 24（2）①S△AA12A314····································································································· 4分

②S△AA12A3 ············································································································· 5分

（3）由规律知：

yn1n(n1)x21(2n1)(n2)x或写成（yn1nn2x212n3n22··········· 6分 x） ·

由（1）（2）知：S1S2S3S10

1216112121313111014

11011111111101112

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

（4）存在

由上知：Sn10Sn9Sn8Sn 1(n10)(n9)1n101n101n91n11(n9)(n8)111(n8)(n7)11n71n1n(n1)1n1

n9n811n9n102n8

····················································································· 9分

11242Sn10Sn9Sn8Sn≥11n9n102

≥11242

n10

n9n100

2·································································································10分 n9n10≤242 ·解得12≤n≤21

又n10

10n≤21 ············································································································· 11分

··············································································12分 存在n的最大值，其值为n21·

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