2010年内蒙古鄂尔多斯

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2010年鄂尔多斯市初中毕业升学考试

数 学

注意事项：

1． 本试题满分120分，考试用时120分钟． 答题前将密封线内的项目填写清楚．

2． 考试结束后将试卷按页码顺序排好，全部上交． 题号 得分 一 二 19 20 21 22 三 23 24 25 26 总分 1~10 11~18 一、选择题（本大题10个小题，每小题3分，共30分．每小题给出的四个选项中只有一个是正确的，请把正确选项填在下面的选项栏内） 题号 选项 1 2 3 4 5 6 7 8 9 10 1．如果a与1互为相反数，则a等于（ ）． A．2 B．?2 C．1 D．?1 2．如图，数轴上的点P表示的数可能是（ ）． A．5

B．-?5

C．?3.8

D．?10 3．下列计算正确的是（ ）． A．a?2a?3a C．(a3)2?a9

23第2题图

326

a?a B．a?D．a3?a4?a?1(a?0)

4．如图，形状相同、大小相等的两个小木块放在一起，其俯视图如图所示，则其主视图是

（ ）．

（俯视图） 第4题图

A． B．

C． D．

5．用折纸的方法，可以直接剪出一个正五边形．折纸过程如图所示，则??等于（ ）． A．108? B．90? C．72° D．60°

第5题图

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6．如图，小明从家走了10分钟后到达了一个离家900米的报亭，看了10分钟的报纸，然后用了15分钟返回到家，下列图象中能表示小明离家距离y（米）与时间x（分）关系的是（ ）．

A． B．

第6题图

C．

D．

7．如图，在ABCD中，E是BC的中点，且?AEC??DCE，则下列结论不正确的是（ ）． ．．．A．S△ADF?2S△EBF

?C．四边形AECD是等腰梯形

1DF 2D．?AEB??ADC

B．BF?第7题图

8．已知二次函数y??x2?bx?c中函数y与自变量x之间的部分对应值如右表所示，点A(x1，y1)，B(x2，y2)在函数

的图象上，当o?x1?1． ，2?x2?3时，y1与y2的大小关系正确的是（ ）A．y1≥y2

B．y1?y2

C．y1?y2

D．y1≤y2

?a?1(a≤b)?9．定义新运算：a?b??a，则函数y?3?x的图象大致是（ ）．

?(a?b且b?0)??bA．

B．

C．

第9题图

D．

10．某移动通讯公司提供了A、B两种方案的通讯费用y（元）与通话时间x（分）之间的关系，如图所示，则以下说法错误的．．是（ ）．

A．若通话时间少于120分，则A方案比B方案便宜20元 B．若通话时间超过200分，则B方案比A方案便宜

C．若通讯费用为60元，则B方案比A方案的通话时间多 D．若两种方案通讯费用相差10元，则通话时间是145分或185分

第10题图

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二、填空题（本大题8个小题，每小题3分，共24分） 11．在函数y?x?2中，自变量x的取值范围是__________.

12.把3??3a?2(a?1)?化简得_________.

13.“五一”期间，某服装商店举行促销活动，全部商品八折销售，小华购买一件原价为140元的运动服，打折后他比按原价购买节省了________元． 14．为参加“初中毕业升学体育考试”，小亮同学在练习掷实心球时，测得5次投掷的成绩分别为：8，8.2，8.5，8，8.6（单位：m），这组数据的众数、中位数依次是___________. 15.如图，用小棒摆下面的图形，图形（1）需要3根小棒，图形（2）需要7根小棒??照这样的规律继续摆下去，第n个图形需要__________根小棒（用含n的代数式表示）．

第15题图

2x?m?3的解是正数，则m的取值范围为________. x?217．如图，现有圆心角为90°的一个扇形纸片，该扇形的半径为50cm．小红同学为了在“圣

16．已知关于x的方程

诞”节联欢晚会上表演节目，她打算剪去部分扇形纸片后，利用剩下的纸片制作成一个底面半径为10cm的圆锥形纸帽（接缝处不重叠），那么被剪去的扇形纸片的圆心角应该是______度．

O1

第17题图

PO2

第18题图

18．如图，⊙O1和⊙O2的半径分别为1和2，连接O1O2，交⊙O2于点P，O1O2?5，若将⊙O1绕点P按顺时针方向旋转360°，则⊙O1与⊙O2共相切_________次．

三、解答题（本大题8个小题，共66分，解答时要写出必要的文字说明、演算步骤或推证过程） 19．（本小题满分8分）

?1?（1）计算：?2?3?27????(π?2)0；

?3?2?1

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a2?b2?2ab?b2?（2）先化简：再求值：2??a??，其中a?2?1，b?1．

a?ab?a?

20．（本小题满分7分）

近年来，随着经济的快速发展，我市城市环境不断改观，社会知名度越来越高，吸引了很多外地游客．某旅行社对5月份本社接待外地游客来我市观光的首选景点作了一次抽样调查，调查结果图表如下：

（1）此次共调查了多少人？并将上面的图表补充完整．

（2）如果将上表制成扇形统计图，那么“恩格贝”所对的圆心角是多少度？

（3）该旅行社预计6月份接待外地来我市的游客2 500人，请你估算一个首选去成陵观光的约有多少人？

景点 成陵 响沙湾 恩格贝 七星湖 巴图湾 频数 116 84 63 37 频率 29% 25% 21% 15.75% 9.25%

第20题图

21．（本小题满分6分）

如图，A信封中装有两张卡片，卡片上分别写着7cm、3cm；B信封中装有三张卡片，卡片上分别写着2cm、4cm、6cm；信封外有一张写着5cm的卡片．所有卡片的形状、大小都完全相同．现随机从两个信封中各取出一张卡片，与信封外的卡片放在一起，用卡片上标明的数量分别作三条线段的长度．

（1）求这三条线段能组成三角形的概率（画出树状图）； （2）求这三条线段能组成直角三角形的概率．

第21题图

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22．（本小题满分8分）

，E为CD的中点，EF∥AB交BC于如图，在梯形ABCD中，AD∥BC，?C?90°点F．

（1）求证：BF?AD?CF；

，BC?7，且BE平分?ABC时，求EF的长． （2）当AD?1 第22题图 23．（本小题满分7分）

某数学兴趣小组，利用树影测量树高，如图（1），已测出树AB的影长AC为12米，并测出此时太阳光线与地面成30°夹角．(2≈1.4，3≈1.7)

（1）求出树高AB；

（2）因水土流失，此时树AB沿太阳光线方向倒下，在倾倒过程中，树影长度发生了变化，假设太阳光线与地面夹角保持不变．（用图（2）解答） ①求树与地面成45°角时的影长； ②求树的最大影长．

第23题图

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24．（本小题满分9分）

??BE?，BD∥CE，连接AE并延长交BD于D． 如图，AB为⊙O的直径，劣弧BC求证：（1）BD是⊙O的切线；

·AD． （2）AB?AC

第24题图

25．（本小题满分10分）

在实施“中小学校舍安全工程”之际，某市计划对A、B两类学校的校舍进行改造，根据预算，改造一所A类学校和三所B类学校的校舍共需资金480万元，改造三所A类学校和一所B类学校的校舍共需资金400万元．

（1）改造一所A类学校的校舍和一所B类学校的校舍所需资金分别是多少万元？

（2）该市某县A、B两类学校共有8所需要改造．改造资金由国家财政和地方财政共同承担，若国家财政拨付的改造资金不超过770万元，地方财政投入的资金不少于210万元，其中地方财政投入到A、B两类学校的改造资金分别为每所20万元和30万元，请你通过计算求出有几种改造方案，每个方案中A、B两类学校各有几所．

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26．（本小题满分11分）

如图，四边形OABC是一张放在平面直角坐标系的矩形纸片，O为原点，点A在x轴上，点C在y轴上，OA?15，OC?9，在AB上取一点M，使得△CBM沿CM翻折后，点B落在x轴上，记作N点． （1）求N点、M点的坐标；

（2）将抛物线y?x2?36向右平移a(0?a?10)个单位后，得到抛物线l，l经过N点，求抛物线l的解析式；

（3）①抛物线l的对称轴上存在点P，使得P点到M，N两点的距离之差最大，求P点的坐标；②若点D是线段OC上的一个动点（不与O、C重合），过点D作DE∥OA交CN于E，设CD的长为m，△PDE的面积为S，求S与m之间的函数关系式，并说明S是否存在最大值．若存在，请求出最大值；若不存在，请说明理由．

第26题图

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2010年鄂尔多斯市初中毕业升学考试

数学试题参考答案及评分说明

（一）阅卷评分说明

1．正式阅卷前先进行试评，在试评中认真阅读参考答案，明确评分标准，不得随意拔高或降低评分标准．试评的试卷必须在阅卷后期予以复查，防止前后期评分标准宽严不一致． 2．评分方式为分步累计评分，解答过程的某一步骤发生笔误，只要不降低后继部分的难度，而后继部分再无新的错误，后继部分可评应得分数的50%；若是几个相对独立的得分点，其中一处错误不影响其它得分点的评分．

3．最小记分单位为1分，不得将评分标准细化至1分以下（即不得记小数分）．

4．解答题题头一律记该题的实际得分，不得用记负分的方式记分．对解题中的错误须用红笔标出，并继续评分，直至将解题过程评阅完毕，并在最后得分点处标上该题实际得分． 5．本参考答案只给出一至两种解法，凡有其它正确解法都应参照本评分说明分步确定得分点，并同样实行分步累计评分．

6．合理精简解题步骤者，其简化的解题过程不影响评分． （二）参考答案及评分标准

一、选择题（本大题10个小题，每小题3分，共30分） 题 号 选项 1 C 2 B 3 D 4 D 5 B 6 D 7 A

8 C 9 B 10 D 二、填空题（本大题8个小题，每小题3分，共24分） 11．x≥2 12．a?5 13.28 14.8，8.2 16．m??6且m??4

17．18(18°)

18．3

15．4n?1

三、解答题（本大题8个小题，共66分） 19．（本小题满分8分）

?1?（1）计算：?22?3?27????(π?2)0

?3?解：原式=?4?3?3 ············································································· 3分（一处正确给1分） ??10． ····································································································································4分

?1a2?b2?2ab?b2?（2）先化简：再求值：2??a??，其中a?2?1，b?1．

a?ab?a?(a?b)(a?b)(a?b)2解：原式= ····················································· 2分（一处正确给1分） ?a(a?b)a=

1 ·······································································································································3分 a?b?12 ···················································································································4分 ?2?1?1220．（本小题满分7分）

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景点 成陵 响沙湾 恩格贝 七星湖 巴图湾

频数 116 100 84 63 37 频率 29% 25% 21% 15.75% 9.25% 解：（1）84?21%?400（人）．答：共调查了400人． ····················································2分 400?25%?100（人），补充图表如下 ····························································· 4分（各1分） （2）360°?21%?75.6°．答：“恩格贝”所对的圆心角是75.6°． ·································6分 （3）2500?29%?725（人）．答：首选去成陵的人数约725人． ···································7分 21．（本小题满分6分） 解：（1）树状图：

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

P(组成三角形)?42··········································································································5分 ?．·

631（2）P(组成直角三角形)?． ································································································6分

622．（本小题满分8分） （1）证法一： 如图（1），延长AD交FE的延长线于N，

??NDE??FCE?90°，?DEN??FEC， DE?EC，图（1）

?△NDE≌△FCE． ············································································································3分 ?DN?CF． ·························································································································4分 ?AB∥FN，AN∥BF，?四边形ABFN是平行四边形． ·············································5分 ?BF?AD?DN?AD?FC． ···························································································6分

??1??BEF.??1??2，??2??BEF． （2）解：?AB∥EF，?EF?BF． ··························································································································7分

?EF?AD?CF?AD?BC1?7??4． ········································································8分 22（1）证法二：如图（2）

过D点作DN∥AB交BC于N，

?AD∥BN，AB∥DN，?AD?BN． ·························· 1分 ?EF∥AB，?DN∥EF． ··············································· 2分 ?△CEF∽△CDN． ························································ 3分

图（2）

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?CECF?． ························································································································4分 DCCNCE1CF1·······················································································5分 ?，??，即NF?CF． ·

DC2CN2??BF?BN?NF?AD?FC． ····························································································6分

23．（本小题满分7分） 解：（1）AB?ACtan30° ·····································································································1分

?12?3．（结果也可以保留一位小数，下同） ?43≈7（米）

3答：树高约7米．·····················································································································2分

（2）①如图（2），B1N?AN?AB1sin45°?43?2······························3分 ≈5（米） ·

2··········································································4分 NC1?NB1tan60°?26?3≈8（米） ·． AC1?AN?NC1?5?8?13（米）

答：树与地面成45°角时影长约13米． ················································································5分 ②如图（2）当树与地面成60°角时影长最大AC2（或树与光线垂直时影长最大或光线与半径为AB的⊙A相切时影长最大） ··························································································6分 ． AC2?2AB2≈14（米）

答：树的最大影长约14米． ···································································································7分

24．（本小题满分9分）

??BE?， 证明：（1）?CB?····································· 2分 ??1??2，AC??AE，AC?AE， ·

?AB?CE． ········································································· 3分 ?CE∥BD，?AB?BD． ··················································· 4分 ?BD是⊙O的切线． ···························································· 5分 （2）连接CB．

?AB是⊙O的直径，??ACB?90°． ···············································································6分 ??ABD?90°，??ACB??ABD． ···················································································7分 ??1??2，△?ACB∽△ABD． ························································································8分

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ACAB?，?AB2?AD·AC． ························································································9分 ABAD（证法二，连接BE，证明略） ?25．（本小题满分10分）

解：（1）设改造一所A类学校的校舍需资金x万元，改造一所B类学校的校舍需资金y万元， 则??x?3y?480 ············································································ 3分（正确一个方程组2分）

?3x?y?400?x?90． ···················································································································4分

y?130?解之得?答：改造一所A类学校的校舍需资金90万元，改造一所B类学校的校舍需资金130万元．

···················································································································································5分 （2）设A类学校应该有a所，则B类学校有(8?a)所，

则??20a?30(8?a)≥210 ································ 7分（正确一个不等式给1分）

(90?20)a?(130?30)(8?a)≤770??a≤3． ··························································································································8分

?a≥1解得??1≤a≤3，即a?1，2，3． ···································································································9分

答：有3种改造方案：

方案一：A类学校1所，B类学校7所； 方案二：A类学校2所，B类学校6所； 方案三：A类学校3所，B类学校5所． ··········································································· 10分 26．（本小题满分11分） 解：如图

（1）?CN?CB?15，OC?9，

······································· 1分 ?ON?152?92?12，?N(12，0)． ·又?AN?OA?ON?15?12?3，设AM?x，

································································· 2分 ?32?x2?(9?x)2， ·

?x?4，M(15，4)． ···············································································································3分

（2）解法一：设抛物线l为y?(x?a)?36，

则(12?a)?36. ·····················································································································4分 ． ··································································································5分 ?a1?6或a2?18（舍去）

22第 11 页 共 12 页

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······························································································6分 ?抛物线l:y?(x?6)2?36． ·

解法二：?x2?36?0，x1??6，x2?6，

··································································4分 0)和(6，0)． ·?y?x2?36与x轴的交点为(?6，0)向右平移6个单位到N点， ·由题意知，交点(6，·······························································5分

所以y?x2?36向右平移6个单位得到抛物线l:y?(x?6)2?36． ·································6分 （3）①由“三角形任意两边的差小于第三边”知，P点是直线MN与对称轴x?6的交点，

··············································7分

4??12k?b?0?k?设直线MN的解析式为y?kx?b，则?，解之得?3

?15k?b?4??b??16?y?4x?16.?P(6，?8)． ··································································································8分 3mDE4?ACB∽△ABD，??，DE?m． ·②?DE∥OA，△····································9分

912314234?S??m?(9?8?m)??m2?m． ·································································· 10分

233334234?3173?a???0，开口向下，又m?????9，?S有最大值， 32?2?3?42?????3?S最大

2?17?3417289． ··············································································· 11分 ????????3?2?3262第 12 页 共 12 页

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