湖北省荆门市2014年初中毕业生学业水平及升学考试数学试卷.pdf

文档介绍：该【湖北省荆门市2014年初中毕业生学业水平及升学考试数学试卷 】是由【mama】上传分享，文档一共【13】页，该文档可以免费在线阅读，需要了解更多关于【湖北省荆门市2014年初中毕业生学业水平及升学考试数学试卷 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。:.
湖北省荆门市2014年初中毕业生学业水平及升学考试试卷
数学
总分值120分考试时间120分钟
一、选择题(本大题共12小题,,共36分)
()×(-2)=1,则括号内填一个实数应该是()
11
.-2D.-
22
()
F
-1=-=±3C.(ab2)3=÷a2=a3
AB
,AB∥ED,AG平分∠BAC,∠ECF=70°,则∠FAG的度数是()
G
°°°°ECD
第3题图
=x2-6x+5向上平移2个单位长度,再向右平移1个单位长度后,得到的抛
物线解析式是()
=(x-4)2-=(x-4)2-=(x-2)2-=(x-1)2-3
-x-1=0较大的根,则下面对α的估计正确的选项是()
<α<<α<<α<<α<3
,AB是半圆O的直径,D,E是半圆上任意两点,连结AD,DE,AE与BD相交于
点C,要使△ADC与△ABD相似,
.....
是()
.
A.∠ACD=∠==BD··AB=AC·BD
y
yxb
1
E
P
DABC
-1Ox
C
D
-1
AOBykx1
2
第6题图第7题图第8题图第9题图
,直线y=x+b与y=kx-1相交于点P,点P的横坐标为-1,则关于x的不等
12
式x+b>kx-1的解集在数轴上表示正确的选项是()
-101-101-2-10-2-10
.
,电路图上有四个开关A、B、C、D和一个小灯泡,闭合开关D或同时闭合开关A、
B、C都可使小灯泡发光,则任意闭合其中两个开关,小灯泡发光的概率是():.
1111
.
2346
,在4×4的正方形网格中,每个小正方形的顶点称为格点,左上角阴影部分是一个
以格点为顶点的正方形(简称格点正方形).假设再作一个格点正方形,并涂上阴影,使这
两个格点正方形无重叠面积,且组成的图形是轴对称图形,又是中心对称图形,则这个格
点正方形的作法共有()

:点P(1-2a,a-2)关于原点的对称点在第一象限内,且a为整数,则关于x的分
x1
式方程=2的解是()
xa

,在第1个△ABC中,∠B=30°,AB=CB;在边AB上任取一点D,延长CA
1111
到A,使AA=AD,得到第2个△AAD;在边AD上任取一点E,延长AA到A,
2121122123
使AA=AE,得到第3个△AAE,…按此做法继续下去,则第n个三角形中以A为顶
23223n
点的内角度数是()
1111
A.()n·75°B.()n-1·65°C.()n-1·75°D.()n·85°
2222
B
A
D
E
F
BC
…AAAAC
31
第12题图
第11题图
,已知圆柱底面的周长为4dm,圆柱高为2dm,在圆柱的侧面上,过点A和点C
嵌有一圈金属丝,则这圈金属丝的周长最小为()

二、填空题(本大题共5小题,每题3分,共15分)
-2xm-ny2与3x4y2m+n是同类项,则m-3n的立方根是▲.
,正方形OABC与正方形ODEF是位似图形,点O为位似中心,相似比为1∶2,
点A的坐标为(0,1),则点E的坐标是▲.
yy
DEA
F
AB
E
AD
Ox
C
OCFxBC
B
第14题图
第16题图
第17题图:.
,:,=x,
111
则x=+x,解得x=,=.仿此方法,▲.
1033
,在□ABCD中,以点A为圆心,AB的长为半径的圆恰好与CD相切于点C,交

AD于点E,延长BA与⊙,则图中阴影部分的面积为
2
▲.
2
,已知:点A是双曲线y=在第一象限的分支上的一个动点,连结AO并延长交
x
另一分支于点B,以AB为边作等边△ABC,,点C的
k
位置也不断变化,但点C始终在双曲线y=(k>0)上运动,则k的值是▲.
x
三、解答题(本大题共7题,共69分)
18.(此题总分值8分)
11
(1)计算:24×-4××(1-2)0;
38
a2b2ab2
(2)先化简,再求值:(),
a22abb2baa2ab
其中a,b满足a1+|b-3|=0.
19.(此题总分值9分)如图①,正方形ABCD的边AB,AD分别在等腰直角△AEF的腰AE,
AF上,点C在△AEF内,则有DF=BE(不必证明).将正方形ABCD绕点A逆时针旋转
一定角度α(0°<α<90°)后,连结BE,
AA
在图②中用实线补全图形,这时DF=BE还成
DB
立吗?请说明理由.
C
FEFE
图①图②
第19题图
:.
20.(此题总分值10分),我国甲、乙两艘海监执法
船某天在钓鱼岛附近海域巡航,某一时刻这两艘船分别位于钓鱼岛正西方向的A处和正
东方向的B处,这时两船同时接到立即赶往C处海域巡查的任务,并测得C处位于A处
北偏东59°方向、位于B处北偏西44°、乙两船分别沿AC,BC方向航行,
其平均速度分别是20海里/小时,18海里/小时,试估算哪艘船先赶到C处.
(参考数据:cos59°≈,sin46°≈)C
北北
59°44°
AB
钓鱼岛
第20题图
21.(此题总分值10分)我市某中学七、八年级各选派10名选手参加学校举办的“爱我荆门”
知识竞赛,计分采用10分制,选手得分均为整数,成绩到达6分或6分以上为合格,到
,七、八年级两支代表队选手成绩分布的条形统计图
和成绩统计分析表如下所示,其中七年级代表队得6分、10分的选手人数分别为a,b.
选手/人数七年级队
a
八年级队
4
22
111111b
035678910成绩/分
队别平均分中位数方差合格率优秀率
%n
%10%
(1)请依据图表中的数据,求a,b的值;:.
(2)直接写出表中的m,n的值;
....
(3)有人说七年级的合格率、优秀率均高于八年级,所以七年级队成绩比八年级队好,但
.
22.(此题总分值10分)我国中东部地区雾霾天气趋于严重,
电器商场根据民众健康需要,代理销售某种家用空气净化器,其进价是200元/
市场销售后发现:在一个月内,当售价是400元/台时,可售出200台,且售价每降低10
元,,代理销
售商每月要完成不低于450台的销售任务.
(1)试确定月销售量y(台)与售价x(元/台)之间的函数关系式;
(2)求售价x的范围;
(3)当售价x(元/台)定为多少时,商场每月销售这种空气净化器所获得的利润w(元)最大?
最大利润是多少?
23.(此题总分值10分)已知:函数y=ax2-(3a+1)x+2a+1(a为常数).
(1)假设该函数图象与坐标轴只有两个交点,求a的值;
(2)假设该函数图象是开口向上的抛物线,与x轴相交于点A(x,0),B(x,0)两点,与y
12
轴相交于点C,且x-x=2.
21
①求抛物线的解析式;
②作点A关于y轴的对称点D,连结BC,DC,求sin∠DCB的值.
:.
24.(此题总分值12分)如图①,已知:在矩形ABCD的边AD上有一点O,OA=3,以O
为圆心,OA长为半径作圆,交AD于M,恰好与BD相切于H,过H作弦HP∥AB,弦
HP=(点E与C,D不重合),过E作直线EF∥BD交BC
于F,再把△CEF沿着动直线EF对折,=x,△EFG与矩形
ABCD重叠部分的面积为S.
(1)求证:四边形ABHP是菱形;
(2)问△EFG的直角顶点G能落在⊙O上吗?假设能,求出此时x的值;假设不能,请说
明理由;
(3)求S与x之间的函数关系式,并直接写出FG与⊙O相切时,S的值.
....
F
BCBC
HEH
G
ADAD
OMOM
P
图①图②(备用图)
第24题图
:.
湖北省荆门市2014年初中毕业生学业水平及升学考试试卷
数学***及评分说明
一、选择题(每题3分,共36分)
题号123456789101112
答案DCBBCDAACCCA
:当x>-1时,x+b>kx-1,即不等式x+b>kx-1的解集为x>-.
:
第一个开关
第二个开关
结果:任意闭合其中两个开关的情况共有12种,其中能使小灯泡发光的情况有6种,
1
.
2
:如图,组成的图形是轴对称图形,又是中心对称图形,
.
:根据题意,点P在第三象限内,
第9题图
12a0;1
x1
∵解得a2,∵a为整数,∴a=2得x.
a20,2x1

1
:∵AB=CB,∠B=30°,∴∠C=∠CAB=(180°-∠B)=75°.
11
2
1
又∵AA=AD,∴∠ADA=∠AAD=∠CAB.
12112121
2
1
也就是说:自A以后,这样得来的每一个角都等于前一个角的.
1
2
1A
∴∠A=()n-1·75°.故选C.
n
2
BC
第12题图:.
:过点A沿直径BC将圆柱纵向切开,得到半圆柱,并将半圆柱,
并将展开为矩形〔如图〕,由题意可知RT△ABC中,AB=BC=2,
∴AC=22,∴属丝周长的最小值为2AC=.
二、填空题(本大题共5小题,每题3分,共15分)
mn4;m2,
:∵-2xm-ny2与3x4y2m+n是同类项,∴解得
22mn,n2.
∴m-3n=8.∴382.
:根据题意,∵相似比为1∶2,∴OA:OD=1∶2,
∵点A的坐标为(0,1),即OA=1,∴OD=2,
∵四边形ODEF是正方形,∴DE=EF=OD=2.
∴E点的坐标为〔2,2〕.
:设x==……,那么100x=……,
……=45+……,
∴100x=45+x化简得99x=45,
45
解得x,
99
45
∴=.
99
:连接AC,∵DC是⊙A的切线,∴AC⊥CD.
又∵AB=AC=CD,∴△ACD是等腰直角三角形,∴∠CAD=45°.
又∵四边形ABCD是平行四边形,∴AD∥BC,∴∠CAD=∠ACB=45°.
又∵AB=AC,∴∠ACB=∠B=45°,∴∠FAD=45°.

∵EF的长为,∴错误!不能通过编辑域代码创建对象。,解得:r2.
2
14522
∴S=S=S=222.
阴影ACD扇形ACE23602
2第16题图
:设A(a,),∵点A与点B关于原点对称,∴OA=OB.
a
∵△ABC为等边三角形,∴AB⊥OC,OC=3AO,
212
∵AO=a2()2,∴CO=3a2.
aa2:.
过点C作CD⊥x轴于点D,则可得∠AOD=∠OCD〔都是∠COD的余角〕,
2
a2
设点C的坐标为(x,y),则tan∠AOD=tan∠OCD,即ax,解得:yx.

2
ay
12
在Rt△COD中,CD2OD2OC2,即y2x23a2.
a2
a21223
将yx代入,可得:x2,故x,
2a2a
a22323
∴y3a,则xy(3a)
2aa
三、解答题(此题包括7个小题,共69分)
32
:(1)原式=26×-4××1·······················································1分
34
=22-2······························································································2分
=2.····································································································3分
(ab)(ab)aa(ab)
(2)原式=[]
(ab)2abb2
ba(ab)a
==.················································································5分
abb2b
∵a1≥0,|b-3|≥0,a1+|b-3|=0,
∴a+1=0且b-3=0.∴a=-1,b=3.·················································7分
13
∴原式==-.················································································8分
33
:补全图形如下图.···········································································3分
DF=BE还成立,理由是:············································································4分
∵正方形ABCD和等腰△AEF,
∴AD=AB,AF=AE,∠FAE=∠DAB=90°.···················································6分
∴∠FAD=∠EAB.······················································································7分
ADAB,

在△ADF和△ABE中,FADEAB,

AFAE.
∴△ADF≌△ABE(SAS).∴DF=BE.·····························································9分:.
A
BC
α北北
59°44°
D
C
AB
FED
19题答案图20题答案图
:过C作CD⊥AB于D,设CD=h(海里),两船从A,B到C的时间分别是t、t(小
甲乙
时),
则∠ACD=59°,∠CBD=90°-44°=46°.
CDhh
在Rt△ACD中,cos59°===,则AC=.······························3分

CDhh
在Rt△BCD中,sin46°===,则BC=.

AChhBChh
∴t===,t===.

∵>,
∴t>t,即乙船先到达C处.···································································10分
甲乙
316a71819110b10,
:(1)依题意得:···············4分
a111b90%10或1a111b10.
a5,
解得································································································6分
b1.
(2)m=6,n=20%;·····················································································8分
(3)①八年级队平均分高于七年级队;②八年级队的成绩比七年级队稳定;③八年级队的成
绩集中在中上游,所以支持八年级队成绩好.(注:任说两条即可)·······················10分
:(1)依题意得:
400x
y=200+50×.················································································2分
10
化简得:y=-5x+2200.·············································································3分
(2)依题意有:
x≥300,
∵····················································································5分
5x2200≥450.
解得300≤x≤350.·····················································································6分
(3)由(1)得:w=(-5x+2200)(x-200)
=-5x2+3200x-440000=-5(x-320)2+72000.··············································8分
∵x=320在300≤x≤350内,∴当x=320时,w=72000.
最大
即售价定为320元/台时,可获得最大利润为72000元.····································10分
:(1)①当a=0时,y=-x+1,有两个交点(0,1),(1,0);·······················1分:.
1
②当a≠0且图象过原点时,2a+1=0,a=-,有两个交点(0,0),(1,0);········2分
2
③当a≠0且图象与x轴只有一个交点时,令y=0有:
△=(3a+1)2-4a(2a+1)==-1,有两个交点(0,-1),(1,0);
1
综上得:a=0或-或-1时,函数图象与坐标轴有两个交点.···························3分
2
3a12a1
(2)①依题意令y=0时,x+x=,xx=.······································4分
12a12a
3a14(2a1)
由x-x=2得:(x-x)2=4,则()2-=4.
2121aa
1
化简得:3a2-2a-1=:a=-,a=1.············································5分
132
∵△=(3a+1)2-4a(2a+1)=(a+1)2>0,且a>0,
1
∴a=-=1符合题意.
3
∴抛物线的解析式为y=x2-4x+3.(注:其它方法,请参照给分)························6分
②令y=0得:x2-4x+3=:x=1或3.
由x-x=2>0知x>x,∴A(1,0),B(3,0),D(-1,0),C(0,3).
2121
如图,过D作DE⊥BC于E,则有OB=OC=3,OD=1.
∴DE=BD·sin45°=22.
而CD=OC2OD2=10,
DE2225
∴在Rt△CDE中,sin∠DCB===.·······································10分
CD105
y
x=2
C
E
1
DOABx
-1
23题答案图
24.(1)连结OH,如图①.
∵AB∥HP,∠BAD=90°,∴AQ⊥,
13
∴HQ=HP=.
22
HQ313
在Rt△OHQ中,sin∠HOQ==×=,
OH232
∴∠HOQ=60°,则∠OHQ=30°,∠APH=60°.
又BD与⊙O相切,∴∠QHD=90°-∠OHQ=60°.∴∠APH=∠:.
∴AP∥BH.
又∵AB∥HP,∴四边形ABHP是平行四边形.·················································3分
由AB⊥AM,AM是直径知AB是⊙O的切线,而BD也是⊙O的切线,
∴AB=BH.
∴四边形ABHP是菱形.(注:其它方法,请参照给分)·······································4分
F
BC
HR
R
1E
AD
OQM(M)
1
P(P)
1
24题答案图①
(2)G点能落在⊙O上,如图①.
方法一:过C作射线CR⊥EF交EF于R,交AD于M,交BD于R,交AP于P,则C
111
关于EF对称点G在射线CR上.
当G点落在M上时,ME=CE=x,AB=CD=HP=3,AD=AB·tan60°=33,ED=CD
11
-CE=3-x.
ED3x1
在Rt△MDE中,cos60°===.解得x=2.··································6分
1MEx2
1
MDMD3
sin60°=1=1=,∴MD=3.
MEx21
1
而MD=AD-AM=3,∴M与M重合.·······················································7分
1
∴M在CP上,则MP⊥AP,而MP⊥AP,
11
∴P与P重合,这校射线CR与⊙O交于M,P.
1
由AP∥BD,CP⊥AP,CR=PR,知C与P关于BD对称.
11
由于点E不与点D重合,故点G不可能落在P点.
∴点G只能落在⊙O的M点上,此时x=2.····················································8分
CD3
方法二:连结CM,PM,如图①,由(1)知∠AMP=∠APH=60°,tan∠CMD===
MD3
3.∴∠CMD=∠AMP=60°.
∴C,M,P三点共线.
∵∠BDA=30°,∴CM⊥∥EF,
∴CM⊥EF,点C关于EF的对称点G落在CP上.
又∵点P到BD的距离等于点C到BD的距离(即点A到BD的距离),EF与BD不重合,∴
点G不能落在P点,可以落在⊙O上的M点.·················································6分
当点G落在⊙O上的M点时,ME=CE=x,:.
MD2
在Rt△MDE中,x==3×=2.
sin603
∴点G落在床⊙O上的M点,此时x=2.·······················································8分
方法三:证法略.
提示:过C作C′P⊥AP于P′,交BD于R′,可求CP′=2CR′=33,PM+CM=33,则
CP′=CM+MP,从而C,M,P三点共线,x的值求法同上.
(3)由(2)知:①当点G在CM上运动时,0<x≤2,
13
S=x·3x=x2.················································································9分
22
②当点G在PM上运动时,2<x<3,设FG交AD于T,EG交AD于N,如图②,
F
BC
H
E
M
AD
OTN
G
P
24题答案图②
13
则:EG=CE=x,ED=3-x,S=CE·CF=x2.
△EFG22
ED
NE==6-2x,GN=GE-NE=3x-6.
sin30
3
∵TG=GN·tan30°=(3x-6)×=3x-23.
3
333
S=S-S=x2-x2+63x-63
△EFG△TGN22
=-3x2+63x-63.·········································································11分
3
x2(0x≤2),
综上所述,S=2
3x263x63(2x3).

313
当FG与⊙O相切时,S=-6.····························································12分
6