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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