中考数学试题含答案.docx
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中考数学试题含答案
2015年河南初中学业水平暨高级中等学校招生考试试题
数学
注意事项:
1.本试卷共6页,三个大题,满分120分,考试时间100分钟。
2.本试卷上不要答题,请按答题卡上注意事项的要求直接把答案填写在答题卡上。
答在试卷上的答案无效。
一、选择题(每小题3分,共24分)下列各小题均有四个答案,其中只有一个是正确的。
1.下列各数中最大的数是()
A.5 B.
C.π D.-8
2.如图所示的几何体的俯视图是()
3.据统计,2014年我国高新技术产品出口总额达40570亿元,将数据40570亿用科学记数法表示为()
A.4.0570×109B.0.40570×1010C.40.570×1011D.4.0570×1012HHH
4.如图,直线a,b被直线c,d所截,若∠1=∠2,∠3=125°,则∠4的度数为()
A.55° B.60° C.70° D.75°
5.不等式组
的解集在数轴上表示为()
6.小王参加某企业招聘测试,他的笔试,面试、技能操作得分分别为85分,80分,90分,若依次按照2:
3:
5的比例确定成绩,则小王的成绩是()
A.255分B.84分 C.84.5分 D.86分
7.如图,在□ABCD中,用直尺和圆规作∠BAD的平分线AG交BC于点E,若BF=6,AB=5,则AE的长为()
A.4 B.6 C.8 D.10
8.如图所示,在平面直角坐标系中,半径均为1个单位长度的半圆O1,O2,O3,…组成一条平滑的曲线,点P从原点O出发,沿这条曲线向右运动,速度为每秒
个单位长度,则第2015秒时,点P的坐标是()
A.(2014,0)B.(2015,-1)C.(2015,1)D.(2016,0)
二、填空题(每小题3分,共21分)
9.计算:
(-3)0+3-1=.
10.如图,△ABC中,点D、E分别在边AB,BC上,DE//AC,
若DB=4,DA=2,BE=3,则EC=.
11.如图,直线y=kx与双曲线
交于点
A(1,a),则k=.
12.已知点A(4,y1),B(
,y2),C(-2,y3)都在二次函数
y=(x-2)2-1的图象上,则y1,y2,y3的大小关系是.
13.现有四张分别标有数字1,2,3,4的卡片,它们除数字外完
全相同,把卡片背面朝上洗匀,从中随机抽取一张后放回,再
背面朝上洗匀,从中随机抽取一张,则两次抽出的卡片所标数
字不同的概率是.
14.如图,在扇形AOB中,∠AOB=90°,点C为OA的中点,
CE⊥OA交
于点E,以点O为圆心,OC的长为半径
作
交OB于点D,若OA=2,则阴影部分的面积为
.
15.如图,正方形ABCD的边长是16,点E在边AB上,AE=3,
点F是边BC上不与点B、C重合的一个动点,把△EBF沿
EF折叠,点B落在B′处,若△CDB′恰为等腰三角形,则DB′的长为.
三、解答题(本大题共8个小题,满分75分)
16.(8分)先化简,再求值:
,其中
,
.
17.(9分)如图,AB是半圆O的直径,点P是半圆上不与点A、B重合的一个动点,延长BP到点C,使PC=PB,D是AC的中点,连接PD,PO.
(1)求证:
△CDP≌△POB;
(2)填空:
①若AB=4,则四边形AOPD的最大面积为;
②连接OD,当∠PBA的度数为时,四边形BPDO是菱形.
18.(9分)为了了解市民“获取新闻的最主要途径”,某市记者开展了一次抽样调查,根据调查结果绘制了如下尚不完整的统计图。
根据以上信息解答下列问题:
(1)这次接受调查的市民总人数是;
(2)扇形统计图中,“电视”所对应的圆心角的度数是;
(3)请补全条形统计图;
(4)若该市约有80万人,请你估计其中将“电脑和手机上网”作为“获取新闻的最主要途径”的总人数.
19.(9分)已知关于x的一元二次方程(x-3)(x-2)=|m|.
(1)求证:
对于任意实数m,方程总有两个不相等的实数根;
(2)若方程的一个根是1,求m的值及方程的另一个根.
20.(9分)如图所示,某数学活动小组选定测量小河对岸大树BC的高度,他们在斜坡上D出测得大树顶端B的仰角是30°,朝大树方向走6米到达坡底A处,在坡底A处测得大树顶端B的仰角为48°.若坡角∠FAE=30°,求大树的高度.(结果保留整数,参考数据:
sin48°≈0.74,cos48°≈0.67,tan48°≈1.11,
≈1.73)
21.(10分)某游泳馆普通票价20元/张,暑假为了促销,新推出两种优惠卡:
①金卡售价600元/张,每次凭卡不再收费;
②银卡售价150元/张,每次凭卡另收10元.
暑期普通票正常出售,两种优惠卡仅限暑期使用,不限次数.设游泳x次时,所需总费用为y元.
(1)分别写出选择银卡、普通票消费时,y与x之间的函数关系式;
(2)在同一个坐标系中,若三种消费方式对应的函数图像如图所示,请求出点A、B、C的坐标;
(3)请根据函数图象,直接写出选择哪种消费方式更合算.
22.(10分)如图1,在Rt△ABC中,∠B=90°,BC=2AB=8,点D,E分别是边BC,AC的中点,连接DE.将△EDC绕点C按顺时针方向旋转,记旋转角为α.
(1)问题发现
①当
时,
;②当
时,
(2)拓展探究
试判断:
当0°≤α<360°时,
的大小有无变化?
请仅就图2的情况给出证明.
(3)问题解决
当△EDC旋转至A、D、E三点共线时,直接写出线段BD的长.
23.(11分)如图,边长为8的正方形OABC的两边在坐标轴上,以点C为顶点的抛物线经过点A,点P是抛物线上点A、C间的一个动点(含端点),过点P作PF⊥BC于点F.点D、E的坐标分别为(0,6),(-4,0),连接PD,PE,DE.
(1)请直接写出抛物线的解析式;
(2)小明探究点P的位置发现:
当点P与点A或点C重合时,PD与PF的差为定值.进而猜想:
对于任意一点P,PD与PF的差为定值.请你判断该猜想是否正确,并说明理由;
(3)小明进一步探究得出结论:
若将“使△PDE的面积为整数”的点P记作“好点”,则存在多个“好点”,且使△PDE的周长最小的点P也是一个“好点”.
请直接写出所有“好点”的个数,并求出△PDE的周长最小时“好点”的坐标.
参考答案
一、选择题:
题号
1
2
3
4
5
6
7
8
答案
A
B
D
A
C
D
C
B
二、填空题:
题号
9
10
11
12
13
14
15
答案
2
y3>y2>y1
和16
三、解答题
16.原式=
,原式=2.
原式=
÷
································································4分
=
·
=
···········································································6分
当a=
+1,b=
-1,原式=
=
=2··································8分
17.
(1)∵D是AC的中点,且PC=PB,
∴DP∥AB,DP=
AB,∴∠CPD=∠PBO·········································3分
∵OB=
AB,∴OB=DP,∴△CDP≌△POB·······································5分
(2)①4············································································7分
②60°(注:
若填60,不扣分)···················································9分
18.
(1)1000··········································································2分
(2)54°···········································································4分
(3)(按人数为100正确补全统计图)················································6分
(4)80×(26%+40%)=80×66%=52.8(万人)
所以估计该市将“电脑和手机上网”作为“获取新闻的最主要途径”的总人数大约为52.8万人
··················································································9分
19.
(1)原方程可化为x2-5x+6-|m|=0······················································1分
∴Δ=(-5)2-4×(6-|m|)=1+4|m|························································3分
∵|m|≥0,∴1+4|m|≥0
∴对于任意实数m,方程总有两个不相等的实数根·····································4分
(2)把x=1代入原方程,得|m|=2,∴m=±2··············································6分
|m|=2代入原方程,得x2-5x+4=0,解得x1=1,x2=4
∴m的值为±2,方程的另一个根是4················································9分
20.延长BD交AE于点G,过点D作DH⊥AE于点H
由题意得:
∠DAE=∠BGA=30°,DA=6,∴GD=DA=6
∴GH=AH=DA·cos30°=6×
=
,∴GA=
··································2分
设BC的长为x米,在Rt△GBC中,GC=
=
=
x···················4分
在Rt△ABC中,AC=
=
···········································6分
∵GC-AC=GA,∴
x-
=
···············································8分
∴x=13,即大树的高度约为13米····················································9分
21.
(1)银卡消费:
y=10x+150·························································1分
普通消费:
y=20x······························································2分
(2)把x=0代入y=10x+150,得y=150,∴A(0,150)········································3分
由题意得:
,∴
,∴B(15,300)································4分
(3)当0<x<15时,选择普通票消费更划算;(注:
若写成0≤x<15,不扣分)
当x=15时,选择购买普通票和银卡消费的总费用相同,均比金卡划算;
当15<x<45时,选择购买银卡消费更划算;
当x=45时,选择购买银卡和金卡消费的总费用相同,均比普通票划算;
当x>45时,选择购买金卡消费更划算.···············································10分
22.
(1)①
·······································································1分
②
;·······································································2分
(2)无变化(注:
若无判断,后续证明正确,不扣分)·····························3分
在图1中,∵DE是△ABC的中位线,
∴DE∥AB,∴
=
,∠EDC=∠B=90°,
如图2,∵△EDC在旋转过程中形状大小不变
∴
=
仍然成立·····························································4分
又∵∠ACE=∠BCD=α,
∴△ACE∽△BCD,∴
=
···················································6分
在Rt△ABC中,AC=
=
=
,
∴
=
=
,
=
,
∴
的大小不变································································8分
(3)
或
.······························································10分
【提示】当△EDC在BC上方,且A、D、E三点共线时,四边形ABCD是矩形,
∴BD=AC=
;当△EDC在BC下方,且A、D、E三点共线时,△ADC是直角三角形,
由勾股定理得AD=8,∴AE=6,根据
=
,得BD=
23.
(1)抛物线解析式为
;················································3分
(2)正确,理由:
设P(x,
),则PF=8-(
)=
·································4分
过点P作PM⊥y轴于点M,则
PD2=PM2+DM2=(-x)2+[6-(
)]2=
=
∴PD=
·······························································6分
∴PD-PF=
-
=2,∴猜想正确········································7分
(3)好点共有11个,·······························································9分
在点P运动时,DE的大小不变,∴PD与PE的和最小时,△PDE的周长最小
∵PD-PF=2,∴PD=PF+2,∴PD+PE=PE+PF+2,
当P、E、F三点共线,PE+PF的值最小,
此时点P、E的横坐标都为-4
将x=-4代入
,得y=6
∴P(-4,6),此时△PDE的周长最小,且△PDE的面积为12,点P恰为“好点”
∴当△PDE的周长最小的“好点”的坐标为(-4,6)································11分
【提示】△PDE的面积是S
,由-8≤x≤0,知4≤S≤13,
所以S的整数的整数值有10个,由函数图像得,当S=12时,“好点”有两个.
所以“好点”有11个.