2006年广西玉林市、防城港市(课改)

2006年玉林市、防城港市初中毕业升学考试
数学（课改卷）
亲爱的同学，展示才华的时候到了，相信自己，细心解答，遇到数字运算尽可能使用计算器，定会获得理想的成绩．祝你成功！
一、填空题：本大题共10小题，每小题2分，共20分．请你答案直接写在题中的横线上． 1．计算：(2)(1)-⨯-=
．
2．实数a b c ，，在数轴上的位置如图1所示，则最小的
数是 ．
3．写出一个主视图、左视图、俯视图都相同的几何体：
．
4．某人沿着一山坡向上走了400米，其铅直高度上升了200米，则山坡与水平面所成的锐角是 ． 5．若1003x y +=，2x y -=，则代数式22x y -的 值是 ．
6．如图2，火焰的光线穿过小孔O ，在竖直的屏幕上形成倒立的实像，像的高度为1.5c m ，48cm OA
=，
16cm OC =，那么火焰的高度是
cm ．
7．
1O 的半径3R =，2O 的半径为r ，且 125O O =，当r =
时，两圆外切．
8．商店里把塑料凳整齐地叠放在一起，据图3的信息，当有10张塑料凳整齐地叠放在一起时的高度是 cm ．
9．如图4，AB 为O 的直径，AB 经过弦CD 的中点E ，
150BOC ∠=，则ABD ∠=
．
10．某歌碟出租店有两种租碟方式：一种是用会员卡租碟， 办会员卡每月10元，租碟每张6角；另一种是零星租碟每
张1元．若小强经常来此店租碟，当每月租碟至少
张时，用会员卡租碟更合算．
二、选择题：本大题共8小题，每小题3分，共24分．在每小题给出的四个选项中，只有一项是符合题意的，请将你认为正确答案的序号填在题后的括号内．
b
图1
Ｂ
图4
11．截至2006年4月15日3时44分，我国神舟六号飞船轨道舱已环绕地球2920圈，用科学记数法表示这个数是（ ） Ａ．4
2.9210⨯圈 Ｂ．3
2.9210⨯圈
Ｃ．229.210⨯圈
Ｄ．4
0.29210⨯圈
12．老师将某班一次数学考试成绩分为A
B C D ，，，四个等级，绘制成图5的扇形统计图，则D 等级所占的百分数是（ ）
Ａ．5% Ｂ．8% Ｃ．10% Ｄ．20%
13．计算：111x x x +--，正确的结果是（ ） Ａ．1- Ｂ．0 Ｃ．2
Ｄ．1
14．不等式组230.52x x x >-+⎧⎨<⎩
，的解集是（ ）
Ａ．14x << Ｂ．4x <
Ｃ．1x >或4x < Ｄ．1x >
15．某厂前年缴税30万元，今年缴税36.3万元，若该厂缴税的年平均增长率为x ，则可列
方程是（ ）
Ａ．2
3036.3x =
Ｂ．230(1)36.3x -=
Ｃ．23030(1)30(1)36.3x x ++++= Ｄ．230(1)36.3x += 16．下列命题错误．．的是（ ） Ａ．五边形的外角和等于360
Ｂ．三角形，四边形和正六边形都可以密铺 Ｃ．经过三点可以作一个圆
Ｄ．圆锥的侧面展开图是一个扇形
17．如图6，四边形PAOB 是扇形OMN 的内接矩形，顶点P 在MN 上，
且不与M N ，重合，当P 点在MN 上移动时，矩形PAOB 的形状、大小随之变化，则AB 的长度（ ） Ａ．变大 Ｂ．变小
Ｃ．不变 Ｄ．不能确定
18．如图7，在五边形ABCDE 中，
A B ∠=∠，90C D E ∠=∠=∠=，4DE DC ==
，AB =，则五边形ABCDE 的周长是（ ）
Ａ．16
Ｂ．14
Ｃ．12
Ｄ．10
图5
Ｂ
Ｎ
Ｐ
Ｍ
Ｏ
图
6
图7
三八为解答题，满分共76分．解答应写出文字说明，证明过程或演算步骤． 三、本大题共2小题，满分共16分． 19．（本小题满分8分）
计算：0
8(1． 20．（本小题满分8分） 解方程：
65
1(1)
x x x x +=
++．
四、本大题共2小题，满分16分． 21．（本小题满分8分）
如图8，网络中每个小正方形的边长为1，点C 的坐标为(01)，．
（1）画出直角坐标系（要求标出x 轴，y 轴和原点）并写出
点A 的坐标； （2）以ABC △为基本图形，利用轴对称或旋转或平移设计一
个图案，说明你的创意．
解：（1）点A 的坐标是 ； （2）图案设计的创意是 ．
22．（本小题满分8分）
如图9，在四边形ABCD 中，AD BC <，AC 与BD 相交于O ，
现给出如下三个论断： ①AB DC =；②12∠=∠；③AD BC ∥．
请你选择其中两个论断为条件，另外一个论断为结论，构造一个
命题．
（1）在构成的所有命题中，是真命题的概率P = ； （2）在构成的真命题中，请选择一个加以证明．
你选择的真命题是：⎫
⇒⎬⎭
（用序号表示）
． 证明：
Ａ Ｂ
Ｃ
图
8
图9
五、本大题共1小题，满分10分． 23．（本小题满分10分）
某制衣厂近四年来关于销售额与总成本的统计图，如图10所示． （1）请你在图11画出四年利润（利润=销售额-总成本）的统计直方图（要求标出数字）； （2）根据图10，图11分别写出一条你发现的信息；
（3）若从2004年到2006年这两年间的利润年平均增长相同，请你预测2006年的利润是
多少万元？
六、本大题共1小题，满分10分．
24．（本小题满分10分）
为鼓励居民节约用水和保护水资源，A 市城区从2006年3月1日起，对居民生活用水采取按月按户实行阶梯式计量水价收费，其收费标准是：第一阶梯水价为1.28元／3
m ；第二阶梯水价为1.92元／3
m ．
（1）每户人口为4人（含4分）以内的，月用水量3
32m ≤执行第一阶梯水价，月用水量
332m >的部分．．
执行第二阶梯水价．如果某户人口4人，3月份用水量3
30m ，那么应交水费
元；4月份用水量3
35m ，那么应交水费
元．
（2）每户核定人数超过4人的，月用水量≤（3
8m ⨯核定人数）执行第一阶梯水价，月用
水量>（3
8m ⨯核定人数）的部分．．执行第二阶梯水价，若小江家人口有5人，设月用水量3
m x ，应交水费y 元． ①请你写出y 与x 的函数关系式；
②若小江家某月交水费60.8元，则该月用水量是多少3
m ？
图10 2002 2003 2004 2005 年份 利润（万元） 图11
七、本大题共1小题，满分12分． 25．（本小题满分12分）
在ABC △中，90ACB ∠=，O 为AC 上的动点． （1）当1
2
OA AC =
时，以O 为圆心，OA 的长为半径的 圆与AB 交于D ，连结CD （如图12），则图中相似的三
角形有：
．
（2）当OA 满足
1
2
AC OA AC <<时，以O 为圆心，OA 的长为半径的圆交AB 于D ，交AC 的延长线于E （如图
13）．
①请你在图中适当添加一条．．
辅助线，然后找出图中相似三角形（注：相似三角形只限于使用图中的六个字母），并加
以证明；
②若O 的半径为5，8AD =，求tan B ．
八、本大题共1小题，满分12分． 26．（本小题满分12分）
抛物线22(21)y x bx b =-+--（b 为常数）与x 轴相交于1(0)A x ，，2(0)B x ，（210x x >>）两点，设3OA OB =（O 为坐标系原点）． （1）求抛物线的解析式；
（2）设抛物线的顶点为C ，抛物线的对称轴交x 轴于点D ，求证：点D 是ABC △的外心； （3）在抛物线上是否存在点P ，使1ABP S =△？若存在，求出点P 的坐标；若不存在，请说明理由．
2006年玉林市、防城港市初中毕业升学考试
数学试题（课改）参考答案及评分标准
一、填空题（每小题2分，共20分） 1．2
2．b
3．球（或正方体）
4．
30
5．2006
Ｃ
图12
Ｅ
图13
y
x
Ｏ
6．4.5
7．2
8．50
9．15 10．26
二、选择题（每小题3分，共24分） 11．Ｂ 12．Ｃ 13．Ｄ 14．Ａ
15．Ｄ
16．Ｃ 17．Ｃ 18．Ｂ
三、19
．解：原式8=- ····················································································· 6分 8=． ········································································································ 8分 20．解：65x x =+． ············································································································ 2分 55x =． ·················································································································· 4分 1x =． ···················································································································· 6分 检验：当1x =时，120x +=≠，(1)1(11)20x x +=⨯+=≠．
∴原方程的解为1x =． ························································································· 8分 四、21．解：（1）正确画出直角坐标系，标出x 轴、y 轴和原点． ·································· 3分
(43)A -， ·························································································································· 5分
（2）答案略．正确画出设计图案． ······················································································ 7分 答案略．写出创意． ······································································································· 8分 22．解：（1）
2
3
······················································································································ 2分 （2）选择真命题一：⎫
⇒⎬⎭
①②③ ·
························································································· 3分 证明：AD BC ∵∥，AD BC <，AB DC =， ∴四边形ABCD 为等腰梯形． ·········································································· 4分 ABC DCB ∠=∠∴． ························································································· 5分 BC CB =∵，
ABC DCB ∴△≌△． ······················································································ 7分
12∠=∠∴． ······································································································ 8分
选择真命题二：
⎫
⇒⎬⎭
②①③ ·
··························································································· 3分 证明：12∠=∠∵， OB OC =∴． ···································································································· 4分 AD BC ∵∥，
2OAD ∠=∠∴，1ODA ∠=∠． ···································································· 5分 OAD ODA ∠=∠∴． OD OA =∴． ····································································································· 6分 AOB DOC ∠=∠∵， AOB DOC ∴△≌△． ····················································································· 7分 AB CD =∴． ····································································································· 8分 五、23．解：（1）正确画出统计直方图，并标出数字． ······················································ 4分
150
100 50
（2）答案不唯一．每写出一条正确的信息给1分． ······································ 6分 （3）从2004年到2005年的增长率120100
100%20%100
-=
⨯=． ·············· 8分
预测2006年的利润为：120(120%)144⨯+=（万元）． ·················· 10分 六、24．解：（1）38.4，46.72． ······················································································· 4分 （2）①当040x ≤≤时， 1.28y x =； ························································· 5分 当40x >时，
40 1.28(40) 1.92y x =⨯+-⨯
1.9225.6x =-． ············································································· 7分 ②40 1.2851.260.8⨯=<∵，可见用水量超过3
40m ．
∴当60.8y =时，1.9225.660.8x -=． ················································· 8分 解得45x =． ······························································································· 9分
∴小红家该月用水量为3
45m ． ······························································· 10分
七、25．解：（1）ACD ABC △∽△，ACD CBD △∽△，ABC CBD △∽△． ······ 3分 （2）解：①连结DE ，则ADE ACB △∽△，理由如下： ························· 5分
AE ∵是O 的直径，
90ADE ∠=∴． ············································································ 6分 90ACB ∠=∵，
ADE ACB ∠=∠∴． ······································································ 7分 A A ∠=∠∵，
ADE ACB ∴△∽△． ··································································· 8分
②6DE =
==． ·
···································· 9分 由①知ADE ACB △∽△，
AD DE
AC BC
=∴
． ············································································· 10分 8463
AC AD BC DE ===∴． ······························································· 11分
4
tan 3
AC B BC =
=∴. ······································································· 12分 八、26．（1）解：由题意，得1221x x b =-． ······················ 1分 3OA OB =∵，1OA x =2OB x =， 123x x =∴． ············································ 2分 213b -=∴． 2b =∴． ··················································· 3分
∴所求的抛物线解析式是：243y x x =-+-． ··································· 4分 （2）证明：2243(2)1y x x x =-+-=--+∵，
∴顶点(21)C ，，(20)D ，，1CD =． ················································· 5分
令0y =，得2
430x x -+-=．
解得11x =，23x =． ········································································· 6分
(10)A ，∴，(30)B ，，1AD DB ==． ··············································· 7分
AD DC DB ==∴．
D ∴为ABC △的外心． ····································································· 8分 （3）解法一：设抛物线存在点()P x y ，，使1ABP S =△． 由（2）可求得312AB =-=． 11
2122
ABP S AB y y =
=⨯=△∴． ············································· 9分 1y =±∴．
当1y =时，2
431x x -+-=，解得122x x ==． ····················· 10分 当1y =-时，
2
431x x -+-=-，解得2x =± ················ 11分
∴存在点P
，使1ABP S =△．点P 的坐标是(21)，
或(21)-或
(21)-． ··············································································· 12分 解法二：由（2）得11
21122
ABC S AB CD =
=⨯⨯=△． ······························ 9分 ∴顶点(21)C ，是符合题意的一个点． ··········································· 10分
另一方面，直线1y =-上任一点M ，能使1ABM S
=△，。