2012年南平市初中毕业、升学考试数学试题及参考答案

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2012年福建省南平市初中毕业、升学考试

数 学 试 题

(满分:150分;考试时间:120分钟)

★友情提示:① 所有答案都必须填在答题卡相应的位置上,答在本试卷上一律无效;

② 可以携带使用科学计算器,并注意运用计算器进行估算和探究; ③ 未注明精确度、保留有效数字等的计算问题不得采取近似计算.

★参考公式:

b4ac b2 b

,抛物线y ax bx c(a≠0)的对称轴是x 顶点坐标是 4a 2a 2a

2

一、选择题(本大题共10小题,每小题4分,共40分.每小题只有一个正确的

选项,请在答题卡的相应位置填涂) ...

1.-3的相反数是 A.

1

3

B.

1 3

C.3 D.-3

2.计算: 2=

A.

B.5

C.

2

D.

2

3.若要对一射击运动员最近5次训练成绩进行统计分析,判断他的训练成绩是否稳定,

则需要知道他这5次训练成绩的 A.中位数 B.平均数 C.众数 D.方差 4.正多边形的一个外角等于30°,则这个多边形的边数为 A.6 B.9 C.12 D.15 5.下列计算正确的是 ..

325

A. a a a

44

B. a a a C. a a a

54

D.(ab) ab

236

6.为验证“掷一个质地均匀的骰子,向上的一面点数为偶数的概率是0.5”,下列模拟实验

中,不科学的是 ...

A.袋中装有1个红球1个绿球,它们除颜色外都相同,计算随机摸出红球的频率 B.用计算器随机地取不大于10的正整数,计算取得奇数的频率 C.随机掷一枚质地均匀的硬币,计算正面朝上的频率 D.如图,将一个可以自由转动的转盘分成甲、乙、丙3个相同的

扇形,转动转盘任其自由停止,计算指针指向甲的频率

数学试题 第1页(共4页)

7.一个三角形的周长是36,则以这个三角形各边中点为顶点的三角形的周长是 ..

A. 6

B. 12

C. 18

D. 36

8.已知反比例函数y

1

的图象上有两点A(1,m)、B(2,n), x

则m与n的大小关系为 A.m>n B.m<n C.m=n D.不能确定

9.如图所示,水平放置的长方体的底面是长为4和宽为2的矩形, 它的主视图的面积为12,则长方体的体积等于 .. A.16 B.24 C.32 D.48

10.如图,正方形纸片ABCD的边长为3,点E、F分别在边BC、

CD上,将AB、AD分别沿AE、AF折叠,点B、D恰好都落在点G处,已知BE=1,则EF的长为 A.

(第9题图)

D F

3 29 C.

4

B.

5 2

B E

(第10题图)

D.3

二、填空题(本大题共8小题,每小题3分,共24分.请将答案填入答题卡的...

相应位置)

11.计算:=.

12.样本数据2,8,3,5,6的极差是.

13.分解因式:2x 4x 2

14.如图,△ABC为⊙O的内接三角形,AB为⊙O的直径,

点D 在⊙O 上,∠ADC=68°,则∠BAC= °. 15.将直线y 2x向上平移1个单位长度后得到的直线 是 .

16.如图,在山坡AB上种树,已知∠C=90°,∠A=28°,AC=6米,

则相邻两树的坡面距离AB≈ 米.(精确到0.1米) 17.某校举行A、B两项趣味比赛,甲、乙两名学生各自随机

选择参加其中的一项,则他们恰好参加同一项比赛的概率 是 . 18.设 x 表示大于..x的最小整数,如 3 =4, 1.2 =-1,

则下列结论中正确的是 .(填写所有正确结论的序号) ..

) ① 0 0; f x ② [x) x的最小值是0;

2

B

(第14题图)

B

A

(第16题图)

C

fx③) x) [x) x=0.5成立.

[x) x的最大值是1; ④ 存在实数fx,使

数学试题 第2页(共4页)

三、解答题(本大题共8小题,共86分.请在答题卡的相应位置作答) ...

1 19.(1)(7分)计算:( 3) π 4 20120.

3

7,① 2x 1<

(2)(7分)解不等式组:

3x< 2x 8.②

2

1

6x x2

0. 20.(8分)解分式方程:x 3

x 3

21.(8分)如图,已知四边形ABCD是平行四边形,若点E、F分别在边BC、AD上,连接

AE、CF.请再从下列三个备选条件中,选择添加一个恰 ...........当的条件,使四边形AECF是平行四边形,并予以证明. .

备选条件:AE=CF,BE=DF,∠AEB=∠CFD. 我选择添加的条件是: . (注意:请根据所选择的条件在答题卡相应试题的图

B

C

D

(第21题图)

中,画出符合要求的示意图,并加以证明)

22.(10分)“六·一”前夕,质检部门从某超市经销的儿童玩具、童车和童装中共抽查了

300件儿童用品.以下是根据抽查结果绘制出的不完整的统计表和扇形图:

类 别

儿童玩具

90

童车

童装

抽查件数

%童车

25%

童装

%

(第22题图)

请根据上述统计表和扇形图提供的信息,完成下列问题: (1)分别补全上述统计表和扇形图;

(2)已知所抽查的儿童玩具、童车、童装的合格率分别为90%、88%、80%,若从该

超市的这三类儿童用品中随机购买一件,请估计能买到合格品的概率是多少?

23.(10分)如图,直线l与⊙O 交于C、D两点,且与

半径OA垂直,垂足为H,已知OD=2,∠O=60°. (1)求CD的长;

(2)在OD的延长线上取一点B,连接AB、AD,

若AD=BD,求证:AB是⊙O的切线.

A

(第23题图)

数学试题 第3页(共4页)

24.(10分)某乡镇决定对小学和初中学生按照每生每天3元的标准进行营养补助,其中家

庭困难寄宿生的补助标准为:小学生每生每天4元,初中生每生每天5元.已知该乡镇现有小学和初中学生共1 000人,且小学、初中均有2%的学生为家庭困难寄宿生. 设该乡镇现有小学生x人. (1)用含x的代数式表示:

该乡镇小学生每天共需营养补助费是 元; 该乡镇初中生每天共需营养补助费是 元; (2)设该乡镇小学和初中学生每天共需营养补助费为y元,求y与x之间的函数关系式; (3)若该乡镇小学和初中学生每天共需营养补助费为3029元,问小学生、初中生分别

有多少人?

25.(12分)在平面直角坐标系中,矩形OABC如图所示放置,

点A在x轴上,点B的坐标为(m,1)(m>0).将此矩形绕点O逆时针旋转90°,得到矩形OA B C . (1)写出点A、A 、C 的坐标; (2)设过点A、A 、C 的抛物线解析式为

y ax2 bx c,求此抛物线的解析式; (a、b、c可用含m的式子表示) (3)试探究:当m的值改变时,点B关于点O的对称点D

是否可能落在(2)中的抛物线上?若能,请求出此时

m的值.

26.(14分)如图,在△ABC中,点D、E分别在边BC、

AC上,连接AD、DE,且∠1=∠B=∠C. (1)由题设条件,请写出三个正确结论;(要求:不再....

添加其它字母和辅助线,找结论过程中添加的字母或辅助线不能出现在结论中,不必证明) 答:结论一: ;

结论二: ; 结论三: . (2)若∠B=45°,BC=2,当点D在BC上运动时

(点D不与点B、C重合), ① 求CE的最大值;

② 若△ADE是等腰三角形,求此时BD的长. (注意:在第(2)小题求解过程中,若有运用(1)

中得出的结论,须加以证明)

数学试题 第4页(共4页)

BB

(第25题图)

1(第26题图)

C

(备用图)

2012年福建省南平市初中毕业、升学考试

数学试题参考答案及评分说明

说明:

(1) 解答右端所注分数,表示考生正确作完该步应得的累计分数,全卷满分150分. (2) 对于解答题,评卷时要坚持每题评阅到底,勿因考生解答中出现错误而中断本

题的评阅.当考生的解答在某一步出现错误时,如果后续部分的解答未改变该题的考试要求,可酌情给分,但原则上不超过后面应得分数的一半,如果有较严重的错误,就不给分.

(3) 如果考生的解法与本参考答案不同,可参照本参考答案的评分标准相应评分. (4) 评分只给整数分.

一、选择题(本大题共10小题,每小题4分,共40分)

1.C; 2.A; 3.D; 4.C; 5.B; 6.D; 7.C; 8.A; 9.B; 10.B. 二、填空题(本大题共8小题,每小题3分,共24分) 11.2; 15.y 2x+1;

12.6; 16.6.8;

13.2(x 1)2; 17.

14.22;

1

(或0.5); 18.③ ④. 2

三、解答题(本大题共8小题,共86分) 19.(1)解:原式=9 3 ( 4) 1 ············································································· 4分 =27 4 1 ···················································································· 6分 =30 ······························································································· 7分 (2)由 ① 得 2x<7+1 ························································································· 2分 x<4 ································································································· 3分 由 ② 得 3x 2x 8 ······················································································ 5分 x<8 ································································································· 6分 ∴不等式组的解集为 x<4 ··············································································· 7分

20.解法一:原方程化为(x 3)(x 3) (6x x) 0 ··················································· 4分

∴x 9 6x x 0 ························································································· 6分

2

2

2

3

··········································································································· 7分 23

经检验,x=是原分式方程的解.

2

3

∴原方程的解是x= ························································································· 8分

2

2

解法二:原方程化为x(x 3) 3(x 3) (6x x) 0 ··········································· 4分

解得 x=

(以下与解法一相同)

数学试题 第5页(共4页)

21.情形一:选择添加的条件是 ······················· 2分

证法一:∵四边形ABCD是平行四边形

∴AD=BC ,AD∥BC ············································ 4分 ∵BE=DF,

∴AD-DF=BC-BE 即 AF=CE ·························· 6分 ∴四边形AECF是平行四边形 ································· 8分 证法二:∵四边形ABCD是平行四边形, ∴AB=CD,∠B=∠D 又∵BE=DF, ∴△ABE≌△CDF

∴AE=CF ······················································································································· 4分 又∵AD=BC,

∴AD-DF=BC-BE 即AF=CE ··············································································· 6分 ∴四边形AECF是平行四边形 ····················································································· 8分

情形二:选择添加的条件是∠AEB=∠CFD ································································· 2分 证法一:∵四边形ABCD是平行四边形,

∴AD∥BC ···················································································································· 4分 ∴∠AEB=∠EAF ············································································································ 5分 又∵∠AEB=∠CFD,

∴∠EAF=∠CFD ············································································································ 6分 ∴AE∥CF ······················································································································ 7分 又∵AF∥EC,

∴四边形AECF是平行四边形 ····················································································· 8分 证法二:∵∠AEB=∠CFD,

∴180°-∠AEB=180°-∠CFD,即∠AEC=∠CFA ······················································ 4分 又∵四边形ABCD是平行四边形, ∴AB=CD,∠B=∠D ∴△ABE≌△CDF

∴∠BAE=∠DCF ········································································································· 6分 又∵∠BAD=∠DCB,

∴∠BAD-∠BAE=∠DCB-∠DCF,即∠EAF=∠FCE ·········································· 7分 ∴四边形AECF是平行四边形 ················································································ 8分

数学试题 第6页(共4页)

BE

C

D

22.解:(1)

30%童车

25%

童装45 %

类 别

儿童玩具

抽查件数

7513590童车

童装

(补全统计图表每空2分,共8分) ································································· 8分 (2)

90 90% 75 88% 135 80%

0.85

300

答:从该超市这三大类儿童用品中随机购买一件能买到合格品的概率是0.85 ·························································································································· 10分

23.(1)解法一:∵OA⊥CD,∴∠OHD=90° ·································································· 1分

O 在Rt△OHD中,∠O=60°,OD=2,∴sinsin DOH

HD

····································· 2分 OD

DOHO ·∴HD OD sin ·································· 3分

=2sin60 3·········································· 4分 ∴CD 2HD ··································· 5分 解法二:∵OA⊥CD,∴∠OHD=90° ········· 1分 在Rt△OHD中,

∵∠O=60°,∴∠ODH=30° ·························· 2分 ∴OH=

1

OD=1 ·········································· 3分 2

∴HD OD2 OH2 ······················································· 4分 22 12 ·

∴CD 2HD ···························································································· 5分 (2)证法一:∵OA=OD,∠O=60°,

∴△AOD为等边三角形

∴∠OAD=∠ODA=60° ·························································································· 6分 ∵AD=BD,∴∠DAB=∠B ·················································································· 7分 ∵∠ODA=∠DAB+∠B, ∴∠DAB=

11

∠ODA=×60°=30° ········································································ 8分 22

数学试题 第7页(共4页)

∴∠OAB=∠OAD+∠DAB=60°+30°=90° ······························································ 9分 即OA⊥AB,∴AB是⊙O的切线 ······································································ 10分 证法二:∵OA=OD,∠O=60°,∴△AOD为等边三角形

∴OD= AD ············································································································ 6分 又∵AD=BD,∴OD=BD ····················································································· 7分 ∵DH⊥OA,∴OH=AH,∠OHD=90° ································································ 8分 ∴HD∥AB,∴∠OAB=∠OHD=90° ··································································· 9分 即OA⊥AB,∴AB是⊙O的切线 ······································································ 10分 证法三:∵OA=OD,∠O=60°,∴△AOD为等边三角形

∴∠O=∠DAO ······································································································· 6分 ∵AD=BD,∴∠B=∠DAB ·················································································· 7分 ∵∠OAB+∠O+∠B=180°,

∴∠OAB+∠DAO+∠DAB=180° ········································································ 8分 ∴2∠OAB=180°,∴∠OAB=90° ······································································· 9分 即OA⊥AB,∴AB是⊙O的切线 ···································································· 10分

数学试题 第8页(共4页)

24.解:(1)3.02x ·········································································································· 2分 2 y x

(

3x. 04 ····························································································· 4分

(2)依题意,得y 3.02x ( 3.04x 3040) ····························································· 6分

0.02x 3040 ··········································································· 7分

(3)由题意可知,当y 3029时,3029 0.02x 3040 ········································ 8分

解得 x 550 ···································································································· 9分

1000 x 1000 550 450

答:该乡镇有小学生550人,初中生450人 ·················································· 10分

25.解:(1)A(m,0),A′(0,m),C′(-1,0) ··················································· 3分 (2)依题意,得 c=m ······························································································· 4分

am2 bm m 0① ∴

a b m 0② 由①得 m(am b 1) 0,

∵m>0,∴am b 1 0 ③ ······································································ 5分 由②+③得 a(m 1) m 1 0,∴a 1 ····················································· 6分 ∴b m 1 ·········································································································· 7分 ∴这个抛物线的解析式为y x2 (m 1)x m ··············································· 8分 (3)答:可能 ············································································································· 9分

∵点D与点B(m,1)关于点O对称,

∴点D的坐标为( m, 1) ··························· 10假设点D( m, 1)在抛物线y x (m 1)x 则有 1 ( m) (m 1) ( m) m ∴2m2 2m 1

0 解得m

········································ 11分 2

2

∵m>0,故m 1 3舍去,

2

∴当m 1 时,点B关于点O的对称点D2

数学试题 第9页(共4页)

26.(1)答:如AB=AC;∠BAD=∠CDE;∠ADB=∠DEC;∠ADC=∠AED;

ABADBDAEADDE

△ABD∽△DCE;△ADE∽△ACD;;等 ;

DCDECEADACCD

(每写对一个结论得1分,共3分) ································································· 3分 (2)① ∵∠B=∠C=45°,∴AB=AC,∠BAC=90° ∵BC=2,∴AB=AC

··········· 4分 解法一:∵∠1+∠EDC=∠B+∠DAB,∠1=∠B ∴∠EDC=∠DAB,∴△ABD∽△DCE ··················· 6分

ABBDAD

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

E CEDCDC

即BD DC CE AB.设CE y,BD x,

BC D 22

有x(2 x),即y ······ 8分

x 2x ·

2

0,∴当x

1时,y最大值 ,∴CE

····················· 9分 解法二:∵∠1=∠C,∠DAE=∠CAD,∴△ADE∽△ACD ···································· 6分

ADAC ········································································································ 7分 AEAD

AD2 ∴AD2 AE AC (AC

CE) AC 2

∴CE ········· 8分 ∴当AD最小时,CE最大. 由垂线段最短,可知AD⊥BC,∵AB=AC,∴D为BC的中点

11

1 ∵∠BAC=90°,∴AD BC 2

1,∴CE 22

即CE

······································ 9分

② 分三种情形加以讨论:

1)如图,当AE=DE时,则∠DAE=∠1=45°

∵∠BAC=90°,∴AD平分∠BAC ······················ 10分

∵AB=AC,∴D为BC的中点 1∴BD=BC=1 ·················································· 11分 BC D

2

2)当AD=DE时, 解法一:∵∠1+∠EDC=∠B+∠DAB,∴∠EDC=∠DAB 又∵∠B=∠C,∴△ABD≌△DCE ········································································ 12分

∵∴AB=DC

BD=BC-DC=2

························································· 13分 解法二:∵∠1=∠C,∠DAE=∠CAD,∴△ADE∽△ACD···································· 12分 ∴当AD=DE时,DC=AC

BD=BC-DC=2

······························· 13分 3)当AD=AE时,则∠AED=∠1=45°,∠DAE=90° ∴此时点D与B重合,与题意不符,应舍去

综上所述,若△ADE是等腰三角形,则BD的长为1或2

························· 14分

数学试题 第10页(共4页)

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