2012年南平市初中毕业、升学考试数学试题及参考答案

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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