【期中试卷】湖北省宜城市2018届九年级数学上期中试题含答案
湖北省宜城市2018届九年级数学上学期期中试题 一、选择题 (本大题有10个小题,每小题3分,共30分.)
1.下列方程是一元二次方程的一般形式的是( )
2222-+2A. 5x3x=0 B.3(x-2)=27 C. (x-1)=16 D.xx=8
222.已知方程的解是x=2,x=,3,则方程 ax,bx,c,0a(x,1),b(x,1),c,012
的解是( )
A. x=1,x=,4 B.x=,1,x=,4 C.x=,1,x=4 D. x=1,x=4njy.com 12121212
23.对于二次函数y=?3(x+1)-2的图象与性质,下列说法正确的是( )
A.对称轴是直线x=1,最小值是-2 B.对称轴是直线x=1,最大值是-2 C.对称轴是直线x=?1,最小值是-2 D.对称轴是直线x=?1,最大值是-2
24.菱形ABCD的一条对角线长为6,边AB的长是方程x,7x,10,0的一个根,则菱形ABCD的周长是( )
(20或8 (8 (20 (12 AB CD
5.下列图形中,既是轴对称图形又是中心对称图形的是( )
26.将二次函数y=2x的图象向左平移2个单位,再向上平移1个单位,所得图象的
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
达式是( )
22222(x,2)+1 B(y=2(x+2)+1 C(y=2(x,2),1 D(y=2(x+2),1 A(y=
,,7.如图,四边形PAOB是扇形OMN的内接矩形,顶点P在MN上,且不与M,N重合,当P点在MN上移动时,矩形PAOB的形状、大小随之变化,则AB的长度( )
A.变大 B(变小 C(不变 D(不能确定
8. 在平面直角坐标系中,O为坐标原点,点A的坐标为(2,0),将OA绕原点逆时针方向旋转60?得OB,则点B的坐标为( )
A.(1,3) B((1,-3) C((0,2) D((2,0)
9.如图,将?O沿弦AB折叠,圆弧恰好经过圆心O,点P是?O上的一点(点A,B除外),则?APB的度数为( )
A(45? B(60? C(120? D(60?或120?
第10题图 第9题图 第7题图
210. 已知抛物线=++的图象如图所示,则|++|+|,+|+|2+|=( ) yaxbxcabcabcab
A(2a+3 b B(2c,b C(2a,b D(b-2c
二、填空题 (本大题有6个小题,每小题3分,共18分.)
211.已知关于x的一元二次方程有实数根,则m的取值范围是 ( x,2x,m,0
212.若方程的两根是,,则的值为 ( x,3x,1,0xxx1++xx)(12212
213.公路上行驶的汽车急刹车时,刹车距离s(m)与时间t(s)的函数关系式为s,20t,5t,当遇到紧急情况时,司机急刹车,但由于惯性汽车要滑行__ __m才能停下来( 14.如图,在?ABC中,?BAC,33?,将?ABC绕点A按顺时针方向旋转50?,对应得到?AB′C′,则?′的度数为____( BAC
15.如图,?O的半径OD?弦AB于点C,连结AO并延长交?O于点E,连结EC.若AB,8,CD,2,则EC的长为 .
16.若点O是等腰?ABC的外心,且?BOC=60?,底边BC=6,则?ABC的面积为 .
第14题 第15题图
图 三、解答题(本大题共9个小题,计72分.)
322x,xx,12,,x17.(本题满分6分)先化简,再求值:,其中x+x-2017,0. 2x,1x,2x,1
18.(本题满分6分)如图,?ABC中,?B,10?,?ACB,20?,AB,4,?ABC逆时针旋转一定
角度后与?ADE重合,且点C恰好成为AD的中点( (1)指出旋转中心,并求出旋转的度数;
(2)求出?BAE的度数和AE的长(
19.(本题满分6分)如图,AC是?O的直径,点B在?O上,?ACB=30? (1)利用尺规作?ABC的平分线BD,交AC于点E,交?O于点D, 连接CD(保留作图痕迹,不写作法)
(2)在(1)所作的图形中,求AB与CD的比值(
20.(本题满分6分)为落实国务院房地产调控政策,使“居者有其屋”,某市加快了廉租房的建设
力度(2015年市政府共投资4亿元人民币建设了廉租房16万平方米,2017年
计划
项目进度计划表范例计划下载计划下载计划下载课程教学计划下载
投资9亿元人民
币建设廉租房,若在这两年内每年投资的增长率相同(
(1)求每年市政府投资的增长率;
(2)若这两年内的建设成本不变,问2017年建设了多少万平方米廉租房,
221.(本题满分7分)如图,二次函数的图象与y轴交 y,x,6x,n
于点C,点B在抛物线上,且与点C关于抛物线的对称轴对称,已知一 次函数y=kx+b的图象经过该二次函数图象上的点A(,2,0)及点B( (1)求二次函数与一次函数的解析式;
2x,6x,n(2)根据图象,写出满足?kx+b的x的取值范围(
22.(本题满分8分)如图,已知正方形ABCD的边长为6,E,F 分别是AB、BC边上的点,且?EDF=45?,将?DAE绕点D逆时针
旋转90?,得到?DCM(
(1)求证:EF=MF
(2)若AE=2,求FC的长(
23.(本题满分10分)某商家经销一种绿茶,用于装修门面已投资3000元(已知绿茶每千克成本
x(元/ kg) „ 70 75 80 85 90 „ 销售单价50元,经研究发现销量y(kg)随销售单价x(元/ kg)的变化而变化,具体变化规律如下表所示:
月销售量y(kg) „ 100 90 80 70 60 „
设该绿茶的月销售利润为w(元)(销售利润,单价×销售量,成本)
(1)请根据上表,求出y与x之间的函数关系式(不必写出自变量x的取值范围); (2)求w与x之间的函数关系式(不必写出自变量x的取值范围),并求出x为何值时,w的值最大,
(3)若在第一个月里,按使w获得最大值的销售单价进行销售后,在第二个月里受物价部门干预,
0元,要想在全部收回装修投资的基础上使第二个月的利润至少达到1700元,销售单价不得高于8
那么第二个月时里应该确定销售单价在什么范围内,
24.(本题满分10分)如图,已知:AB是?O的直径,点C在?O上,CD是?O的切线,AD?CD于点D.E是AB延长线上一点,CE交?O于点F,连结OC,AC.
(1)求证:AC平分?DAO. C
D(2)若?DAO=105?,?E=30?. F?求?OCE的度数. AEBO
22?若?O的半径为,求线段EF的长.
325.(本题满分13分)如图,抛物线经过A(,1,0),B(3,0),C(0,)三点( 2
(1)求抛物线的解析式; y
(2)在抛物线的对称轴上有一点P,使PA+PC的值最小,
C
AOBx
求点P的坐标;
(3)点M为x轴上一动点,在抛物线上是否存在一点N, 使以A,C,M,N四点构成的四边形为平行四边形,若存在, 求点N的坐标;若不存在,请说明理由(
2017-2018学年度上学期期中考试题
九 年 级 数 学参考答案
一、选择题 (本大题有10个小题,每小题3分,共30分.)
ABDCDBCADC
(本大题有6个小题,每小题3分,共18分.) 二、填空题
11.(m?-1); 12. 4; 13. 20; 14. 17?; 15. 213; 16. 6,33或6,33 三、解答题(本大题共9个小题,计72分.)
17.(本题满分6分)
2(1)(1)(1)xx,x,x,2解:原式,,,„„„„„„„„„3分 x,x,,x21(1)x,x,
22?x+x-2017,0,?x+x=2017. „„„„„„„„„„„„„„„„„„5分
?原式,2017. „„„„„„„„„„„„„„„„„„„„„„„„6分 18.(本题满分6分)
解:(1)旋转中心为点A („„„„„„„„„„„„„„„„„„„„„„„1分 由旋转可知,?DAE=?BAC=180?-10?-20?=150?. „„„„„„„„„„„2分 ?旋转角为150?. „„„„„„„„„„„„„„„„„„„„„„„„„„3分 (2)??DAE=?BAC=150?,
??BAE=360?-?DAE-?BAC=60?.„„„„„„„„„„„„„„„„„„„4分 由旋转可知,AD=AB,AE=AC
.?AB,4,点C为AD的中点
1?.?AE=2(„„„„„„„„„„„„„„„„„„„„„„„6分 AC,AD,22
19.(本题满分6分)
解:(1)如图所示;„„„„„„„„„„„„„„„„„„„„„„„„„„„3分
(2)如图2,连接OD,设?O的半径为r,
?AC是?O的直径,??ABC=90?.(
在R?ACB中,?ACB=30?, t
?AB=AC=r(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„4分 ?BD是?ABC的平分线,??ABD=?CBD=45? (
??DOC=2?CBD =90?
在R?ODC中,DC==r(„„„„„„„„„„„„„„„„„„5分 t
ABr2?(„„„„„„„„„„„„„„„„„„„„„„„„„„6分 ,,CD22r
20.(本题满分6分)
解:(1)设每年市政府投资的增长率为x,依题意得:
2 4(1+x)=9 „„„„„„„„„„„„„„„„„„„„„„„„„„„„„2分
解得x=0.5=50% x=-2.5(舍去) „„„„„„„„„„„„„„„„„„„3分 12
答:每年市政府投资的增长率为50% „„„„„„„„„„„„„„„„„„„„4分
2(2)16(1+50%)=24.„„„„„„„„„„„„„„„„„„„„„„„„„5分
答:2017年预计建设了24万平方米的廉租房.„„„„„„„„„„„„„„„„6分 21.(本题满分7分)
2解:(1)?抛物线经过点A(,2,0), y,x,6x,n
0,4,12,nn,8?. ?. „„„„„„„„„„„„„„„„„„„„„„„„1分
2?抛物线解析式为y=x+6x+8. „„„„„„„„„„„„„„„„„„„„„„„2分 ?点C坐标(0,8).
?对称轴x=,3,B、C关于对称轴对称,
?点B坐标(,6,8).„„„„„„„„„„„„„„„„„„„„„„„„„„3分 ?y=kx+b经过点A、B,
,6k,b,8,k,,2,,,?解得 ,,,2k,b,0.b,,4.,,
?一次函数解析式为y=,2x,4. „„„„„„„„„„„„„„„„„„„„„„5分
2(2)由图象可知,满足?kx+b的x的取值范围为,6?x?,2(„„„7分 x,6x,n
22.(本题满分8分)
解:(1)??DAE逆时针旋转90?得到?DCM,
??FCM=?FCD+?DCM=180?(
?F、C、M三点共线. „„„„„„„„„„„„„„„„„„„„„„„„„„1分 ?DE=DM,?EDM=90?(
??EDF+?FDM=90?,„„„„„„„„„„„„„„„„„„„„„„„„„2分
=45?,??=?=45?( ??EDFFDMEDF
??DEF??DMF(SAS),„„„„„„„„„„„„„„„„„„„„„„„3分 ?EF=MF(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„4分 (2)设EF=MF=x,
?AE=CM=2,且BC=6,?BM=BC+CM=6+2=8(„„„„„„„„„„„„„„5分 ?BF=BM,MF=BM,EF=8,x(„„„„„„„„„„„„„„„„„„„„„6分 ?EB=AB,AE=6,2=4(
222在Rt?EBF中,由勾股定理得EB+BF=EF(
222即4+(8,x)=x,„„„„„„„„„„„„„„„„„„„„„„„„„„7分 解得:=5,即=5( ?xFM
?FC=FM-CM=5-2=3(„„„„„„„„„„„„„„„„„„„„„„„„„„8分 23.(本题满分10分)
wkxb,,解:(1)设,将(70,100),(75,90)代入上式得: y,kx,b
70100kb,,k,,2,, 解得:,则,„„„„„„2分 y,,2x,240,,b,2407590kb,,,,
将表中其它对应值代入上式均成立,所以.„„„„„„3分 y,,2x,240
(2) w,(x,50)y
2,(x,50)(,2x,240),,2x,340x,9000„„„„„„„„„„„5分
22因此,w与x的关系式为 yxxx,,,,,,,,234090002(85)2450x,85w,2450当时,„„„„„„„„„„„„„„„„„„„„„„„6分 最大.
30002450550,, (3)由(2)知,第1个月还有元的投资成本没有收回(
w,2250 则要想在全部收投资的基础上使第二个月的利润达到1700元, 即才可以,
2xx,,75,95 可得方程,解得:„„„„„„7分根据题意,,,,2(85)24502250x12
x,85不合题意,应舍去.当,„„„„„„„„„8分?-2<0,?,当x,95x,80时,y,2400,2
时,随的增大而增大, wx
w,225075,x,80 当,且销售单价不高于80时,.„„„„„„„„„„„„9分
75,x,80答:当销售单价为元时,在全部收回投资的基础上使第二个月的利润不低于1700元.„„„„„„10分
24.(本题满分10分)
(1)
证明
住所证明下载场所使用证明下载诊断证明下载住所证明下载爱问住所证明下载爱问
:?直线与?O相切,?OC?CD. „„„„„„„„„„„„„„„„1分 又?AD?CD,?AD//OC. „„„„„„„„„„„„„„„„„„„„„„„„„2分 ??DAC=?OCA. „„„„„„„„„„„„„„„„„„„„„„„„„„„„3分 又?OC=OA,??OAC=?OCA.
??DAC=?OAC.
?AC平分?DAO. „„„„„„„„„„„„„„„„„„„„„„„„„„„„4分 (2)解:??AD//OC,?DAO=105?,??EOC=?DAO=105?.„„„„„„„„„5分
=30?,??=45?. „„„„„„„„„„„„„„„„„„„„„„„„„6分 ??EOCE
?作OG?CE于点G,可得FG=CG . „„„„„„„„„„„„„„„„„„„„7分
22?OC=,?OCE=45?.?CG=OG=2.
?FG=2. „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„8分
23?在Rt?OGE中,?E=30?,?GE=.„„„„„„„„„„„„„„„„„„9分
23?EF=GE-FG=-2. „„„„„„„„„„„„„„„„„„„„„„„„„10分 25.(本题满分13分)
2解:(1)设抛物线的解析式为y=ax+bx+c(a?0),
3?A(,1,0),B(5,0),C(0,)三点在抛物线上, 2
1,a,,,2,b,1?,解得(„„„„„„„„„„„„„„„„„„„2分 ,
,3,c,2,
132?抛物线的解析式为:.„„„„„„„„„„„„„„„„„3分 y,,x,x,22
132(2)?抛物线的解析式为, y,,x,x,22?其对称轴为直线:.„„„„„„„„„„„„„„4分
连接BC,设直线BC的解析式为, y,kx,b(k,0)
1,k,,,,3,2?B(3,0),C(0,),?解得„„„„„„„„„„5分 ,23,b,.,2,
13y,,x,?直线BC的解析式为.„„„„„„„„„„„„„„„„„„„„6分 22
13y,,,,1当x=1时,.?P(1,1);„„„„„„„„„„„„„„„„„„7分 22
3)存在(如图2所示,„„„„„„„„„„„„„„„„„„„„„„„„„8分 (
?当点N在x轴上方时,
33?抛物线的对称轴为直线x=1,C(0,),?N(2,);„„„„„„„„„„9分 122?当点N在x轴下方时,
如图,过点N作ND?x轴于点D,??AND??MCO. 2222
33,?ND=OC=,即N点的纵坐标为(„„„„„„„„„„„„„„„„„10分 2222
13321,71,7,,?x+x+=.解得x=或x=,„„„„„„„„„„„„„11分 222
331,71,7,,?N(,),N(,)(„„„„„„„„„„„„„„„12分 2322
3331,71,7,,综上所述,点N的坐标为(2,),(,),(,)(„„13分 222
y
NC1
MMD23MABx1O
N3N2图2