资阳市中考数学试题及答案

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资阳市2005年初中课改毕业年级学业考试

数  学

全卷分为第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.第Ⅰ卷1至2页,第Ⅱ卷3至8页.全卷满分120分,考试时间共120分钟.

答题前,请考生务必在答题卡上正确填涂自己的姓名、考号和考试科目,并将试卷密封线内的项目填写清楚;考试结束,将试卷和答题卡一并交回.

第Ⅰ卷(选择题 共40分)

注意事项:

每小题选出的答案不能答在试卷上,须用铅笔在答题卡上把对应题目的答案标号涂黑.如需改动,用橡皮擦擦净后,再选涂其它答案.

一、选择题:本大题共10个小题,每小题4分,共40分. 在每小题给出的四个选项中,只有一个选项符合题意要求.

1.的绝对值是

A. -2              B.             C. 2               D.

2. 图1所示的几何体的右视图是

3. 某服装销售商在进行市场占有率的调查时,他最应该关注的是

A. 服装型号的平均数                   B. 服装型号的众数

C. 服装型号的中位数                   D. 最小的服装型号

4. 下列命题中,正确的是

A. 同位角相等                        B. 平行四边形的对角线互相垂直平分

C. 等腰梯形的对角线互相垂直            D. 矩形的对角线互相平分且相等

5. 若“!”是一种数学运算符号,并且1!=1,2!=2×1=2,3!=3×2×1=6,4!=4×3×2×1,…,则的值为

A.              B. 99!             C. 9900             D. 2!

6. 如图2,若ABCPQ、甲、乙、丙、丁都是方格纸中的格点,为使△ABC∽△PQR,则点R应是甲、乙、丙、丁四点中的

A. 甲              B. 乙

C. 丙              D. 丁

7. 已知正比例函数y=k1x(k1≠0)与反比例函数y=(k2≠0)的图象有一个交点的坐标为(-2,-1),则它的另一个交点的坐标是

A. (2,1)              B. (-2,-1)         C. (-2,1)          D. (2,-1)

8. 若关于x的方程x2+2(k-1)x+k2=0有实数根,则k的取值范围是

A.            B.                C.               D. k

9. 若⊙O所在平面内一点P到⊙O上的点的最大距离为a,最小距离为b(a>b),则此圆的半径为

A.                B.            C.      D. a+ba-b

10. 已知二次函数y=ax2+bx+c(a≠0)的图象如图3所示,给出以下结论:① a+b+c<0;② a-b+c<0;③ b+2a<0;④ abc>0 . 其中所有正确结论的序号是

A. ③④            B. ②③             C. ①④         D. ①②③

第Ⅱ卷(非选择题 共80分)

注意事项:

本卷共6页,用黑色或蓝色钢笔或圆珠笔直接答在试卷上.

二、填空题:本大题共6个小题,每小题3分,共18分.把答案直接填在题中横线上.

11. 若1000张奖券中有200张可以中奖,则从中任抽1张能中奖的概率为______.

12. 若实数m,n满足条件m+n=3,且m-n=1,则m=________,n=___________.

13. 在△ABC中,若DE分别是边ABAC上的点,且DEBCAD=1,DB=2,则△ADE与△ABC的面积比为____________.

14. 函数的自变量x的取值范围是_______________.

15. 如图4,如果△APB绕点B按逆时针方向旋转30°后得到

APB,且BP=2,那么PP'的长为____________.

(不取近似值. 以下数据供解题使用:sin15°=,cos15°=)

16. 已知n(n≥2)个点P1P2P3,…,Pn在同一平面内,且其中没有任何三点在同一直线上. 设Sn表示过这n个点中的任意2个点所作的所有直线的条数,显然,S2=1,S3=3,S4=6,S5=10,…,由此推断,Sn=______________.
三. 解答题:本大题共8个小题,共62分. 解答应写出必要的文字说明,证明过程或演算步骤.

17. (本小题满分7分)

(1) 已知a=sin60°,b=cos45°,c=d=,从abcd这4个数中任意选取3个数求和;

(2) 计算: .

18. (本小题满分7分)

如图5,已知点MN分别是△ABC的边BCAC的中点,点P是点A关于点M的对称点,点Q是点B关于点N的对称点,求证:PCQ三点在同一条直线上.

19. (本小题满分7分)

甲、乙两同学开展“投球进筐”比赛,双方约定:① 比赛分6局进行,每局在指定区域内将球投向筐中,只要投进一次后该局便结束;② 若一次未进可再投第二次,以此类推,但每局最多只能投8次,若8次投球都未进,该局也结束;③ 计分规则如下:a. 得分为正数或0;b. 若8次都未投进,该局得分为0;c. 投球次数越多,得分越低;d. 6局比赛的总得分高者获胜 .

(1) 设某局比赛第n(n=1,2,3,4,5,6,7,8)次将球投进,请你按上述约定,用公式、表格或语言叙述等方式,为甲、乙两位同学制定一个把n换算为得分M的计分方案;

(2) 若两人6局比赛的投球情况如下(其中的数字表示该局比赛进球时的投球次数,“×”表示该局比赛8次投球都未进):

第一局

第二局

第三局

第四局

第五局

第六局

5

×

4

8

1

3

8

2

4

2

6

×

根据上述计分规则和你制定的计分方案,确定两人谁在这次比赛中获胜.

20. (本小题满分7分)

如图6,已知AB为⊙O的直径,弦CDAB,垂足为H.

(1) 求证:AHAB=AC2

(2) 若过A的直线与弦CD(不含端点)相交于点E,与⊙O相交于点F,求证:AEAF=AC2

(3) 若过A的直线与直线CD相交于点P,与⊙O相交于点Q,判断APAQ=AC2是否成立(不必证明).

21. (本小题满分8分)

已知某项工程由甲、乙两队合做12天可以完成,共需工程费用13800元,乙队单独完成这项工程所需时间是甲队单独完成这项工程所需时间的2倍少10天,且甲队每天的工程费用比乙队多150元.

(1) 甲、乙两队单独完成这项工程分别需要多少天?

(2) 若工程管理部门决定从这两个队中选一个队单独完成此项工程,从节约资金的角度考虑,应该选择哪个工程队?请说明理由.

22. (本小题满分8分)

甲骑自行车、乙骑摩托车沿相同路线由A地到B地,行驶过程中路程与时间的函数关系的图象如图7. 根据图象解决下列问题:

(1) 谁先出发?先出发多少时间?谁先到达终点?先到多少时间?

(2) 分别求出甲、乙两人的行驶速度;

(3) 在什么时间段内,两人均行驶在途中(不包括起点和终点)?在这一时间段内,请你根据下列情形,分别列出关于行驶时间x的方程或不等式(不化简,也不求解):① 甲在乙的前面;② 甲与乙相遇;③ 甲在乙后面.

23. (本小题满分9分)

阅读以下短文,然后解决下列问题:

如果一个三角形和一个矩形满足条件:三角形的一边与矩形的一边重合,且三角形的这边所对的顶点在矩形这边的对边上,则称这样的矩形为三角形的“友好矩形”. 如图8①所示,矩形ABEF即为△ABC的“友好矩形”. 显然,当△ABC是钝角三角形时,其“友好矩形”只有一个 .

(1) 仿照以上叙述,说明什么是一个三角形的“友好平行四边形”;

(2) 如图8②,若△ABC为直角三角形,且∠C=90°,在图8②中画出△ABC的所有“友好矩形”,并比较这些矩形面积的大小;

(3) 若△ABC是锐角三角形,且BC>AC>AB,在图8③中画出△ABC的所有“友好矩形”,指出其中周长最小的矩形并加以证明.


24. (本小题满分9分)

如图9,已知O为坐标原点,∠AOB=30°,∠ABO=90°,且点A的坐标为(2,0).

(1) 求点B的坐标;

(2) 若二次函数y=ax2+bx+c的图象经过ABO三点,求此二次函数的解析式;

(3) 在(2)中的二次函数图象的OB段(不包括点OB)上,是否存在一点C,使得四边形ABCO的面积最大?若存在,求出这个最大值及此时点C的坐标;若不存在,请说明理由.

资阳市2005年初中课改毕业年级学业考试暨高中阶段学校招生考试

数学试题参考答案及评分意见

说 明:

1. 解答题中各步骤所标记分数为考生解答到这一步的累计分数;

2. 给分和扣分都以1分为基本单位;

3. 参考答案都只给出一种解法,若考生的解答与参考答案不同,请根据解答情况参考评分意见给分 .

一、选择题:每小题4分,共10个小题,满分40分.

1-5. DABDC;6-10. CABCB.

二、填空题:每小题3分,共6个小题,满分18分.

11. ;12. m=2, n=1;13. 1:9;14. x,且x≠-1;15. ;16. .

(13题填为,16题填为2+3+…+n 或1+2+3+…+n-1均给分)

三、解答题:共8个小题,满分62分 .

17.(1) a+b+c=, a+b+d=, a+c+d=,

b+c+d=. ···························································································· 4分

(按考生的选择,得出正确结果都给分.正确写出所选a,b,c,d的值各1分,得出最后结果1分)

(2)原式= ······································································· 6分

=x2-y2 ···················································································· 7分

18.连结MNPCCQ. ······································································· 1分

∵点PA点关于点M的对称点,∴ MAP的中点, ························ 2分

MBC的中点,∴ MN是△APC的中位线,

CPMN .···························································································· 4分

同理可证,CQMN . ············································································· 5分

从而,CPCQ都经过点C且都平行于AB

PCQ三点在同一直线上. ····························································· 7分

(也可连结AQCQBPCP,由ABCQABPC为平行四边形证明,或根据全等三角形的性质证明)

19.(1)计分方案如下表:

n(次)

1

2

3

4

5

6

7

8

M(分)

8

7

6

5

4

3

2

1

 ············································································································· 4分

(用公式或语言表述正确,同样给分.)

(2) 根据以上方案计算得6局比赛,甲共得24分,乙共得分23分,········· 6分

所以甲在这次比赛中获胜 . ····································································· 7分

20.(1) 连结CB,∵AB是⊙O的直径,∴∠ACB=90°. ························· 1分

而∠CAH=∠BAC,∴△CAH∽△BAC . ················································· 2分

, 即AHAB=AC2 . ·························································· 3分

(2) 连结FB,易证△AHE∽△AFB, ······················································ 4分

AEAF=AHAB, ············································································· 5分

AEAF=AC2 . ···················································································· 6分

(也可连结CF,证△AEC∽△ACF)

(3) 结论APAQ=AC2成立 . ··································································· 7分

21.(1) 设甲队单独完成需x天,则乙队单独完成需要(2x-10)天.·············· 1分

根据题意有 =,·································································· 3分

解得x1=3(舍去),x2=20.············································································ 4分

∴ 乙队单独完成需要 2x-10=30 (天).

答:甲、乙两队单独完成这项工程分别需要20天、30天. ······················· 5分

(没有答的形式,但说明结论者,不扣分)

(2) 设甲队每天的费用为y元,则由题意有

12y+12(y-150)=138000,解得y=650 . ····················································· 7分

∴ 选甲队时需工程费用650×20=13000,选乙队时需工程费用500×30=15000.

∵ 13000 <15000,

∴ 从节约资金的角度考虑,应该选择甲工程队.······································· 8分

22.(1) 甲先出发;先出发10分钟;乙先到达终点;先到5分钟. ········· 2分

(2) 甲的速度为每分钟0.2公里, ···························································· 3分

乙的速度为每分钟0.4公里 . ··································································· 4分

(3) 在甲出发后10分钟到25分钟这段时间内,两人都行驶在途中. ········ 5分

设甲行驶的时间为x分钟(10<x<25),则根据题意可得:

甲在乙的前面:0.2x>0.4(x-10) ; ··························································· 6分

甲与乙相遇:0.2x=0.4(x-10) ;································································ 7分

甲在乙后面:0.2x<0.4(x-10) .··································································· 8分

(设甲行驶的时间x时,没有限定范围的,不扣分. 也可设乙行驶的时间列出相应的方程或不等式 .)

23. (1) 如果一个三角形和一个平行四边形满足条件:三角形的一边与平行四边形的一边重合,三角形这边所对的顶点在平行四边形这边的对边上,则称这样的平行四边形为三角形的“友好平行四边形”. 1分

(2) 此时共有2个友好矩形,如图的BCADABEF.

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

易知,矩形BCADABEF的面积都等于△ABC面积的2倍,∴ △ABC的“友好矩形”的面积相等.     4分

(3) 此时共有3个友好矩形,如图的BCDECAFGABHK,其中的矩形ABHK的周长最小 .  5分

证明如下:

易知,这三个矩形的面积相等,令其为S. 设矩形BCDECAFGABHK的周长分别为L1L2L3,△ABC的边长BC=aCA=bAB=c,则

L1=+2aL2=+2bL3=+2c .················· 6分

L1- L2=(+2a)-(+2b)=2(a-b),·· 7分

ab>Sa>b

L1- L2>0,即L1> L2 .······································ 8分

同理可得,L2> L3 .

L3最小,即矩形ABHK的周长最小.······················································ 9分

24.(1) 在Rt△OAB中,∵∠AOB=30°,∴ OB=. 过点BBD垂直于x轴,垂足为D,则 OD=BD=,∴ 点B的坐标为() . ············································································· 1分

(2) 将A(2,0)、B()、O(0,0)三点的坐标代入y=ax2+bx+c,得

 ················································································· 2分

解方程组,有 a=b=c=0. ·············································· 3分

∴ 所求二次函数解析式是 y=x2+x. ······································· 4分

(3) 设存在点C(x , x2+x) (其中0<x<),使四边形ABCO面积最大.

∵△OAB面积为定值,

∴只要△OBC面积最大,四边形ABCO面积就最大. ······························· 5分

过点Cx轴的垂线CE,垂足为E,交OB于点F,则

SOBC= SOCF +SBCF==

················································································································ 6分

CF=yC-yF=

SOBC= . ·································································· 7分

∴ 当x=时,△OBC面积最大,最大面积为. ······························· 8分

此时,点C坐标为(),四边形ABCO的面积为. ·················· 9分