南充中考数学试题及答案2008

市二○○八年四川省南充高中阶段学校招生统一考试

数学试卷

（满分100分，考试时间90分钟）

一、细心选一选（本大题共8个小题，每小题3分，共24分）每小题都有代号为A，B，C，D四个答案选项，其中只有一个是正确的，请把正确选项的代号填在相应的括号内．填写正确记3分，不填、填错或填出的代号超过一个记0分． 1．计算(?2)2?2的结果是（ ）

A．?6 B．2 C．?2 D．6 2．如图，下列选项中不是正六棱柱三视图的是（ ）

A．

3．某地区七、八月份天气较为炎热，小华对其中连续十天每天的最高气温进行统计，依次得到以下一组数据：34，35，36，34，36，37，37，36，37，37（单位：℃），则这组数据的中位数和众数分别是（ ） A．36，37 B．37，36 C．36.5，37 D．37，36.5 4．若?O1的半径为3cm，?O2的半径为4cm，且圆心距O1O2?1cm，则?O1与?O2的位置关系是（ ） A．外离 B．内切 5．已知数据

13B． C．

（第2题图）

D．

C．相交 D．内含

，?7，2.5，?，5其中分数出现的频率是（ ）

A．20% B．40% C．60% D．80%

6．“5·12”汶川大地震后，世界各国人民为抗震救灾，积极捐款捐物，截止2008年5月27日12时，共捐款人民币327.22亿元，用科学计数法（保留两位有效数字）表示为（ ） A．3.27?10

10 B．3.2?10

?10C．3.3?10

10D．3.3?10

117．如图，AB是?O直径，?AOC?130，则?D?（ ） A．65

?B．25

D B

?C．15

?D．35

y ?O A

O （第8题图）

x

C （第7题图）

- 1 -

8．二次函数y?ax2?bx?c的图像如图所示，则点Q?a，?在（ ）

?b??c?A．第一象限 B．第二象限 C．第三象限 D．第四象限

二、认真填一填（本大题共4个小题，每小题3分，共12分）请将答案直接写在题中横线上．

9．如图，四边形ABCD中，E，F，G，H分别是边AB，BC，CD，DA的中点．请你添加一个条件，使四边形EFGH为菱形，应添加的条件是 ．

H D G

C F （第9题图）

输入x y?x?2(x?0) A E B y?x?2x?1(0≤x?1) y?x?2x?1(x≥1) 22输出y （第10题图）

3时，输出的结果y? ．

10．根据下面的运算程序，若输入x?1?11．某商场为了解本商场的服务质量，随机调查了本商场的200名顾客，调查的结果如图所

示．根据图中给出的信息，这200名顾客中对该商场的服务质量表示不满意的有 人．

A A 48% D A：满意 C 9% B：基本满意 B C：说不清 36% D：不满意

D O P C E B （第12题图） （第11题图）

12．如图，从?O外一点P引?O的两条切线PA，PB，切点分别是A，B，若PA?8cm，

，过点C作?O的切线，分别交C是?AB上的一个动点（点C与A，B两点不重合）PA，PB于点D，E，则△PED的周长是 ．

三、（本大题共2个小题，每小题6分，共12分） 13．计算：

1?148?1?2．

14．化简?1???1??，并选择你最喜欢的数代入求值． ?2x?1?x?xx

- 2 -

四、（本大题共2个小题，每小题6分，共12分）

15．如图，?ABCD的对角线相交于点O，过点O任引直线交AD于E，交BC于F，

则OE OF（填“?”“?”“?”），说明理由．

A E D

O

B F C

（第15题图）

16．桌面上放有质地均匀、反面相同的3张卡片，正面分别标有数字1，2，3，这些卡片反面朝上洗匀后放在桌面上，甲从中任意抽出1张，记下卡片上的数字后仍反面朝上放回洗匀，乙再从中任意抽出1张，记下卡片上的数字，然后将这两数相加． （1）请用列表或画树形图的方法求两数和为4的概率；

（2）若甲与乙按上述方式做游戏，当两数之和为4时，甲胜，反之则乙胜；若甲胜一次得6分，那么乙胜一次得多少分，这个游戏才对双方公平？ 五、（本大题共2个小题，每小题8分，共16分）

17．在“5·12”汶川大地震的“抗震救灾”中，某部队接受了抢修映秀到汶川的“213”国道的任务．需要整修的路段长为4800m，为了加快抢修进度，获得抢救伤员的时间，该部队实际工作效率比原计划提高了20%，结果提前2小时完成任务，求原计划每小时抢修的路线长度．

- 3 -

?4)是一次函数y?kx?b的图像和反比例函数y?18．如图，已知A(?4，n)，B(2，mx的

图像的两个交点．

（1）求反比例函数和一次函数的解析式；

（2）求直线AB与x轴的交点C的坐标及三角形AOB的面积． y A O x C

B

（第18题图） 六、（本大题8分）

19．如图，已知?O的直径AB垂直于弦CD于点E，过C点作CG∥AD交AB的延长线于点G，连接CO并延长交AD于点F，且CF?AD． （1）试问：CG是?O的切线吗？说明理由； （2）请证明：E是OB的中点； （3）若AB?8，求CD的长．

O C E B G （第19题图）

A

F D - 4 -

七、（本大题8分）

20．某乒乓球训练馆准备购买10副某种品牌的乒乓球拍，每副球拍配x(x≥3)个乒乓球，已知A，B两家超市都有这个品牌的乒乓球拍和乒乓球出售，且每副球拍的标价都为20元，每个乒乓球的标价都为1元，现两家超市正在促销，A超市所有商品均打九折（按原价的90%付费）销售，而B超市买1副乒乓球拍送3个乒乓球，若仅考虑购买球拍和乒乓球的费用，请解答下列问题：

（1）如果只在某一家超市购买所需球拍和乒乓球，那么去A超市还是B超市买更合算？ （2）当x?12时，请设计最省钱的购买方案．

八、（本大题8分）

21．如图，已知平面直角坐标系中，有一矩形纸片OABC，O为坐标原点，AB∥x轴，

B(?3，3)，现将纸片按如图折叠，AD，DE为折痕，?OAD?30．折叠后，点O落

?在点O1，点C落在线段AB上的C1处，并且DO1与DC1在同一直线上． （1）求C1的坐标；

（2）求经过三点O，C1，C的抛物线的解析式；

B E C1 O1 D （第21题图） O x

y A （3）若?P的半径为R，圆心P在（2）的抛物线上运动， C ?P与两坐标轴都相切时，求?P半径R的值．

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参考答案及评分意见

一、细心选一选（本大题共8个小题，每小题3分，共24分）

1．B； 2．A； 3．A； 4．B； 5．B； 6．C； 7．B； 8．C． 二、认真填一填（本大题共4个小题，每小题3分，共12分）

9．AC?BD或EG?HF或EF?FG等（任填一个满足题意的均可）； 10．?1?3；

11．14； 12．16cm．

三、（本大题共2个小题，每小题6分，共12分） 13．解：原式?1?22?(2?1) ························································································· 4分

?1?22?2?1··················································································································· 5分

?2?322 ······························································································································ 6分

x?1?xx?11x(x?1)14．解：原式??1? ······················································································ 2分

x?11······································································································································· 5分 ??x·

??x(x?1) ··················································································································· 4分

选取除0与1以外的任何值，求代数式的值 ········································································· 6分 注：若选取的值为0与1，该步骤不得分． 四、（本大题共2个小题，每小题6分，共12分） 15．解：填“?”

理由：?四边形ABCD是平行四边形

······················································· 3分 ?OA?OC，AD∥BC ·

A E D 1 ??1??2，?3??4 ····························································· 4分 3 O 在△AOE和△COF中

??3??4???1??2 ?OA?OC?B 4 2 F C

（第15题图）

········································································································· 5分 ?△AOE≌△COF． ·

··························································································································· 6分 ?OE?OF ·

16．解：（1）

1 2 3 甲：

1 2 3 1 1 2 3 ······································ 2分 乙：2 3

- 6 -

P(两数之和为4)?39?13 ······································································································· 4分

13（2）由（1）P(两数之和为4)?1323，P(两数之和不为4)?23

设乙胜一次得x分，这个游戏才对双方公平，根据题意得

?6?x

x?3

答：乙胜一次得3分，这个游戏才对双方公平． ································································· 6分

五、（本大题共2个小题，每小题8分，共16分）

17．解：设原计划每小时抢修的路线长为xm，根据题意，得

4800x?4800x(1?20％)········································································································· 5分 ?2

解之得

································································································································· 7分 x?400 ·

检验：x?400是原方程的解，且符合题的实际意义．

答：原计划每小时抢修的路线长为400m． ·········································································· 8分

?4)在y?18．解：（1）?B(2，mx上

A ． ·································· 1分

C y O B （第18题图） ?m??8．

?反比例函数的解析式为：y???点A(?4，n)在y??8xx

8x上

?n?2

?A(?4，2) ······························································································································ 2分 ?y?kx?b经过A(?4，2)，B(2，?4)，

??4k?b?2 ??2k?b??4?解之得

?k??1 ??b??2?一次函数的解析式为：y??x?2 ···················································································· 4分

（2）?C是直线AB与x轴的交点

?当y?0时，x??2

?点C(?2，0) ·························································································································· 5分

- 7 -

································································································································ 6分 ?OC?2

?S△AOB?S△ACO?S△BCO ?12?2?2?12?2?4

········································································································································· 8分 ?6 ·

六、（本大题8分）

19．（1）解：CG是?O的切线 ··························································································· 1分 理由：?CG∥AD

??FCG??CFD?180

??CF?AD

??CFD?90 ??FCG?90

??即OC?CG．

·········································································································· 2分 ?CG是?O的切线． ·

（2）第一种方法： A 证明：连接AC，如图（第19题图1） ?CF?AD，AE?CD

F O 且CF，AE过圆心O

?D ??AC??AD，?AC?CC E B D ?AC?AD?CD

····················································· 3分 ?△ACD是等边三角形． ·

??D?60

??G （第19题图1）

??FCD?30······················································································································· 4分

在Rt△COE中，

OE?12OC 12OB

?OE??点E为OB的中点 ·············································································································· 5分

第二种方法：

证明：连接BD，如图（第19题图2） ?AB为?O的直径

??ADB?90

?A

F E B G

（第19题图2）

- 8 -

O C D 又??AFO?90 ??ADB??AFO

?

?CF∥BD

············································································································· 3分 ?△BDE∽△OCE ·

BEDE ??OECE?AE?CD

且AE过圆心O

··························································································································· 4分 ?CE?DE ·

?BE?OE

?点E为OB的中点．··········································································································· 5分 （3）解：?AB?8

?OC?12AB?4

又?BE?OE

································································································································ 6分 ?OE?2

································································································ 7分 ?CE?OE?cot30?23 ·?AB?CD

?············································································································· 8分 ?CD?2CE?43 ·七、（本大题8分）

20．解：（1）去A超市购买所需费用yA?0.9(20?10?10x)

即yA?9x?180····················································································································· 1分 去B超市购买所需费用yB?20?10?10(x?3)

即yB?10x?170 ··················································································································· 2分 当yA?yB时，即9x?180?10x?170 x?10

当yA?yB时，即9x?180?10x?170 x?10

当yA?yB时，即9x?180?10x?170

···································································································································· 4分 x?10 ·

综上所述：当x?10时，去A超市购买更合算；当x?10时，去A超市或B超市购买一样；

当3≤x?10时，去B超市购买更合算． ············································································ 5分 （2）当x?12时，即购买10副球拍应配120个乒乓球 若只去A超市购买的费用为：

··················································································· 6分 9x?180?9?12?180?288（元）·

若在B超市购买10副球拍，去A超市购买余下的乒乓球的费用为：

200?0.9(12?3)?10?281（元） ······················································································ 7分

- 9 -

?281?288

?最佳方案为：只在B超市购买10副球拍，同时获得送30个乒乓球，然后去A超市按九

折购买90个乒乓球． ············································································································· 8分 八、（本大题8分） 21．解：（1）过C1作C1F?x轴于点F，如图（第21题图） 在Rt△ADO中，?OAD?30，AO?BC?33?3

B E C1 O1 D F （第21题图） y A OD?tan30?OA???3?1 ···································· 1分

C O x

由对称性可知：

?ADO??ADO1?60 ??FDC1?60

???DF?C1F?cot60??3?33?1 ··················································································· 2分

?OF?DF?DO?1?1?2

?点C1的坐标为(?2，3) ······································································································ 3分

（2）设经过O，C1，C的抛物线的解析式为y?ax?bx?c，则

?c?0?2········································································································· 4分 ?(?2)a?2b?c?3 ·?2(?3)a?3b?c?0?2解之得

?3?a??2??33? ?b??2??c?0???32332?抛物线的解析式为：y??x?2x ······································································ 5分

（3）??P与两坐标轴相切

?圆心P应在第一、三象限或第二、四象限的角平分线上． 即在直线y?x或y??x上 ··································································································· 6分

若点P在直线y?x上，根据题意有

- 10 -

x??32x?2332x

解之得

x1?0，x2??3?233 ?R?0

?R?x?3?233··············································································································· 7分

若点P在直线y??x上，根据题意有

32332?x??x?2x

解之得x1?0，x2??R?0

?R?x?3?233233?3

??P的半径R为3?233或3?233． ············································································ 8分

- 11 -

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