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2013年全国初中数学联赛(初三组)初赛试卷及答案

发布时间:2014-03-28 15:12:51  

2013年全国初中数学联赛(初三组)初赛试卷

(3月7日下午4:00—6:00)

班级:: 姓名: 成绩:

考生注意:

1、本试卷共五道大题,全卷满分140分; 2、用圆珠笔、签字笔或钢笔作答; 3、解题书写不要超出装订线; 4、不能使用计算器。

一、选择题(本题满分42分,每小题7分)

本题共有6个小题,每题均给出了代号为A、B、C、D的四个答案,其中有且只有一个是正确的。将你选择的答案的代号填在题后的括号内。每小题选对得7分;不选、错选或选出的代号字母超过一个(不论是否写在括号内),一律得0分。

1、某件商品的标价为13200元,若以8折降价出售,仍可获利10%(相对于进货价),则该商品的进货价是( )

A、9504元 A、80?

B、9600元 C、9900元 D、10000元

B、100? C、140?

C

D

C

2、如图,在凸四边形ABCD中,AB?BC?BD,?ABC?80?,则?ADC等于( )

D、160?

D

A

B

第2题图

A

第4题图

B

3、如果方程?x?2??x2?4x?m??0的三根可以作为一个三角形的三边之长,那么,实数m的取值范围是( )

A、0?m?4 BD的长度是( )

A、7

B、4

B、m?3

C、m?4

D、3?m?4

4、如图,梯形ABCD中,AB//CD,?BAD?60?,?ABC?30?,AB?6且AD?CD,那么

C、2 D、42

5、如果?2014?a?0,那么|x?a|?|x?2014|?|x?a?2014|的最小值是( )

A、2014 B、a?2014 C、4028 D、a?4028

O

DF C B 6、方程x2?xy?y2?3?x?y?的整数解有( ) A、3组 C、5组 B、4组 D、6组 二、填空题(本大题满分28分,每小题7分)

1、如图,扇形AOB的圆心角?AOB?90?,半径为5,正方形CDEF内接于该扇形,则正方形CDEF的边长为 .

2、已知四个自然数两两的和依次从小到大的次序是:23,28,33,39,x,y,则x?y?____.

3、已知x?y?6,x2?xy?xy?y2?9,则x2?xy?xy?y2的值是.

4、有质地均匀的正方体形的红白骰子各一粒,每个骰子的六个面分别写有1、2、3、4、5、6的自然数,随机掷红、白两粒骰子各一次,红色骰子掷出向上面的点数比白色骰子掷出向上面的点数小的概率是 .

三、(本大题满分20分)

已知2a2?a?4?0,a?b?2,求

12?的值。 a?1b

,交BC于点E.

A

D

C E 四、(本大题满分25分) 在Rt?ABC中,?ACB?90?,AE垂直于AB边上的中线CD(1)求证:AC2?BC?CE (2)若CD?3,AE?4,求边AC与BC的长。

五、(本大题满分25分) 已知二次函数y?ax2?bx?c的图像经过点A(x1,0)、B(x2,0)、C(2,m),且0?x1?x2?2.

(1)求证:m?0;

(2)若b?1,求证:m?1

2013年全国初中数学联赛(初三组)初赛

评 分 细 则

一、选择题(本题满分42分,每小题7分)

1、B.

1

2、

3、

54、. 12

三、(本大题满分20分)

解:由已知得b?a?2, 所以12123a. ···················································· (10分) ????2a?1ba?1a?2a?a?22、C. 3、D. 4、C. 5、A. 6、D. 二、填空题(本大题满分28分,每小题7分) a显然a?0,由2a2?a?4?0得a2?2??. ·············································· (15分) 2

所以3a3a???2, a2?a?2?a?a2

12····················································································· (20分) ???2. ·a?1b所以

四、(本大题满分25分)

解:(1)因为CD是AB边上的中线,

所以CD=DB,

∠ABC=∠DCB=∠CAE,

∠ACB=∠ECA=90?,

ACCB, ?ECCA

B所以△ACB∽△ECA, ···················································································· (5分) 所以

所以AC2?BC?CE. ····················································································· (10分)

(2)因为CD是Rt△ABC的中线,

所以CD=AD=BD。

所以AB=6。

所以AC2?BC2?AB2?36. ········································································· (15分) 取BC中点F,连结DF,则DF//AC,∠DFC=∠ECA=90?,

所以△DFC∽△ECA, 所以DCFC.

?EACA

所以BC2CD3··················································································· (20分)

??, ·CAAE2

······················································· (25分) BC? ·故可解得AC?

法2:因为CD是Rt△ABC的中线,

所以CD=AD=BD。

所以AB=6。

所以AC2?BC2?AB2?36。 ········································································· (15分) 由(1)知△ACB∽△ECA, BCAB63所以··············································································· (20分)

???。 ·CAEA42

故可解得AC?

······················································· (25分) BC? ·五、(本大题满分25分)

证:(1)由已知可得方程x2?ax?b?0的两根为x1、x2,

a所以x1?x2??a,x1?x2,所以2?x2??, ············································ (5分) 2

a由已知可得,当x??时二次函数y?x2?ax?b的值随x的增大而增大. 2

所以二次函数在x?2的函数值大于在x?x2的函数值.

即m?0. ········································································································ (10分) 法2:又由已知可得x2?ax?b?(x?x1)(x?x2),

所以m?(2?x1)(2?x2)。 ··············································································· (5分)

又因为0?x1?x2?2,

所以2?x1?0,2?x1?0,

所以m?0. ···································································································· (10分)

(2)由已知得x2?ax?b?(x?x1)(x?x2)

令x?0得b?x1x2,

令x?2得m?(2?x1)(2?x2),

所以bm?x1x2(2?x1)(2?x2)?[1?(x1?1)2][1?(x2?1)2]。····························· (15分)

因为0?x1?x2?2,所以0?1?(x1?1)2?1,0?1?(x2?1)2?1,

并且1?(x1?1)2?1和1?(x2?1)2?1不能同时成立,

所以0?bm?1. ····························································································· (20分) 又b?1,所以m?1. ····················································································· (25分)

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