一、选择题(每小题<strong><span>3</span></strong><strong><span>分,共</span></strong><strong><span>30</span></strong><strong><span>分)</span></strong>
-
-
2.
要说明命题“若a2>b2 , 则a>b”是假命题,能举的一个反例是( )
A . a=1,b= -2
B . a=2,b= 1
C . a=4,b= - 1
D . a= -3,b= -2
-
3.
平面直角坐标系中,点P坐标是(﹣1,2),则点P关于y轴对称点的坐标是( )
A . (1,﹣2)
B . (1,2)
C . (﹣1,﹣2)
D . (﹣1,2)
-
-
5.
如图,AB∥CD,AD=CD,∠1=65°,则∠2的度数是( )
![](//tikupic.21cnjy.com/75/8e/758e8225313ee2bed70bb03ff25bfd7e.png)
A . 50°
B . 60°
C . 65°
D . 70°
-
6.
若点A(-3,y1),B(1,y2)都在直线y=x+5上,则y1与y2的大小关系是( )
A . y1>y2
B . y1=y2
C . y1<y2
D . 无法比较大小
-
7.
关于
x的不等式组
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmo%3E%7B%3C%2Fmo%3E%3Cmtable+columnalign%3D%22left%22%3E%3Cmtr%3E%3Cmtd%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%E2%88%92%3C%2Fmo%3E%3Cmn%3E6%3C%2Fmn%3E%3Cmo%3E%26lt%3B%3C%2Fmo%3E%3Cmn%3E1%3C%2Fmn%3E%3Cmn%3E5%3C%2Fmn%3E%3C%2Fmtd%3E%3C%2Fmtr%3E%3Cmtr%3E%3Cmtd%3E%3Cmn%3E2%3C%2Fmn%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmn%3E2%3C%2Fmn%3E%3Cmo%3E%26lt%3B%3C%2Fmo%3E%3Cmn%3E3%3C%2Fmn%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmn%3E3%3C%2Fmn%3E%3Cmi%3Ea%3C%2Fmi%3E%3C%2Fmtd%3E%3C%2Fmtr%3E%3C%2Fmtable%3E%3C%2Fmath%3E)
恰好有4个整数解,则
a满足( )
A . a=
B . -5≤a<
C . -5<a≤
D . -5≤a≤
-
8.
在同一直角坐标系内作一次函数y1=ax+b和y2=﹣bx+a图象,可能是( )
-
9.
已知:△
ABC纸片,将纸片分别按以下两种方法翻折:
如图1,沿着∠BAC的平分线AD翻折△ABD , 得到△AED , 设△CDE的周长为m .
如图2,沿着AB的垂直平分线翻折△BFG , 得到△AFG , 设△AGC的周长为n .
线段AB的长度用含m , n的代数式可表示为( )
![](//tikupic.21cnjy.com/2024/03/07/4f/e2/4fe2855a0982c157573caeb71687a9f0.png)
A . n-m
B .
C . m
D .
-
10.
甲、乙两辆汽车沿同一路线赶赴距出发地480千米的目的地,乙车比甲车晚出发2小时,图中折线
OABC、线段
DE分别表示甲、乙两车所行路程
y(千米)与时间
x(小时)之间的函数关系对应的图象(线段
AB表示甲因故障停车检修),下列说法正确的个数为( ).
①a=240;②b=60;③甲、乙两车第一次相遇的时间是离乙出发3小时;④甲、乙两车相距30km的时间分别为2.5小时、3.5小时、5.5小时、6.5小时.
![](//tikupic.21cnjy.com/2024/03/07/53/55/535553bbbe46bae34f743dabe2776993.jpeg)
A . 1个
B . 2个
C . 3个
D . 4个
二、填空题(每小题<strong><span>4</span></strong><strong><span>分,共</span></strong><strong><span>24</span></strong><strong><span>分)</span></strong>
-
11.
若函数
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmi%3Ey%3C%2Fmi%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmfrac%3E%3Cmrow%3E%3Cmsqrt%3E%3Cmrow%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmn%3E5%3C%2Fmn%3E%3C%2Fmrow%3E%3C%2Fmsqrt%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmrow%3E%3C%2Fmfrac%3E%3C%2Fmath%3E)
有意义,则自变量取值范围为
.
-
12.
若y与x﹣1成正比例,且x=2时y=6,则x=﹣2时y=.
-
13.
(2022八上·如皋月考)
如图,在
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmtext%3E%E2%96%B3%3C%2Fmtext%3E%3Cmi%3EA%3C%2Fmi%3E%3Cmi%3EB%3C%2Fmi%3E%3Cmi%3EC%3C%2Fmi%3E%3C%2Fmath%3E)
中,
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmi%3EA%3C%2Fmi%3E%3Cmi%3EB%3C%2Fmi%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmi%3EA%3C%2Fmi%3E%3Cmi%3EC%3C%2Fmi%3E%3C%2Fmath%3E)
,
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmi%3EB%3C%2Fmi%3E%3Cmi%3EC%3C%2Fmi%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmn%3E5%3C%2Fmn%3E%3Cmtext%3Ec%3C%2Fmtext%3E%3Cmtext%3Em%3C%2Fmtext%3E%3C%2Fmath%3E)
, AB的垂直平分线交AB于点D,交AC于点E,
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmtext%3E%E2%96%B3%3C%2Fmtext%3E%3Cmi%3EB%3C%2Fmi%3E%3Cmi%3EC%3C%2Fmi%3E%3Cmi%3EE%3C%2Fmi%3E%3C%2Fmath%3E)
的周长为12cm,则
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmtext%3E%E2%96%B3%3C%2Fmtext%3E%3Cmi%3EA%3C%2Fmi%3E%3Cmi%3EB%3C%2Fmi%3E%3Cmi%3EC%3C%2Fmi%3E%3C%2Fmath%3E)
的周长为
cm.
![](//tikupic.21cnjy.com/2022/11/03/94/50/945099d1d69f31f5bc3e241dccb08cb2_141x169.png)
-
14.
如图,函数
y=﹣5
x和
y=
mx+3图象相交于点
A(
n , 2),则不等式
mx+3≥﹣5
x>0的解集为
.
![](//tikupic.21cnjy.com/2024/03/07/95/e2/95e2c1c41827a1a023890b99d9e292e4.png)
-
15.
若等腰三角形一腰上的高线与另一条腰的夹角为20°,则该等腰三角形的底角为.
-
16.
如图是一个提供床底收纳支持的气压伸缩杆,除了
AB是完全固定的钢架外,
AD ,
BC ,
DE属于位置可变的定长钢架.如图1所示,
AD=13
cm ,
BC=20
cm , 伸缩杆
PQ的两端分别固定在
BC ,
CE两边上,其中
PB=13
cm ,
CQ=20
cm . 当伸缩杆完全收拢(即
CD∥
AB)时,如图2所示,床高(
CD与
AB之间的距离)为12
cm , 则此时伸缩杆
PQ的长度为
cm . 当∠
ADC成180°时,伸缩杆
PQ打开最大,此时
PQ的长度为
cm , 则固定钢架
AB的长度为
cm .
![](//tikupic.21cnjy.com/2024/03/07/83/06/8306e890521ccb366318abc3835542d2.png)
三、解答题<strong><span>(</span></strong><strong><span>本题</span></strong><strong><span>共</span></strong><strong><span>8</span></strong><strong><span>小题,共</span></strong><strong><span>66</span></strong><strong><span>分</span></strong><strong><span>)</span></strong>
-
17.
解一元一次不等式组
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmo%3E%7B%3C%2Fmo%3E%3Cmtable+columnalign%3D%22left%22%3E%3Cmtr%3E%3Cmtd%3E%3Cmrow%3E%3Cmn%3E2%3C%2Fmn%3E%3Cmtext%3Ex%3C%2Fmtext%3E%3Cmtext%3E-%3C%2Fmtext%3E%3Cmn%3E1%3C%2Fmn%3E%3Cmtext%3E%E2%89%A4%3C%2Fmtext%3E%3Cmo%3E%E2%88%92%3C%2Fmo%3E%3Cmtext%3Ex%3C%2Fmtext%3E%3Cmtext%3E%2B%3C%2Fmtext%3E%3Cmtext%3E2%3C%2Fmtext%3E%3C%2Fmrow%3E%3C%2Fmtd%3E%3C%2Fmtr%3E%3Cmtr%3E%3Cmtd%3E%3Cmrow%3E%3Cmfrac%3E%3Cmrow%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%E2%88%92%3C%2Fmo%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmrow%3E%3C%2Fmfrac%3E%3Cmtext%3E%EF%BC%9C%3C%2Fmtext%3E%3Cmfrac%3E%3Cmrow%3E%3Cmn%3E1%3C%2Fmn%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmn%3E2%3C%2Fmn%3E%3Cmtext%3Ex%3C%2Fmtext%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmn%3E3%3C%2Fmn%3E%3C%2Fmrow%3E%3C%2Fmfrac%3E%3C%2Fmrow%3E%3C%2Fmtd%3E%3C%2Fmtr%3E%3C%2Fmtable%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
, 并把它的解集表示在数轴上:
-
18.
如图,在 8×6 的网格中,每个小正方形的边长均为一个单位.
![](//tikupic.21cnjy.com/2024/03/07/20/ea/20ea11c9bab4bd7f87d047d36747ceae.png)
-
(1)
在图1中画出一个以BC为一边,面积为12的三角形;
-
-
(3)
在图 3中画出△
ABC的角平分线
BE(△
ABC 的三个顶点都在格点上).按要求完成作图:
①仅用无刻度直尺,且不能用直尺中的直角;②保留作图痕迹;③标注相关字母.
-
19.
如图,在Rt△
ABC中,∠
ACB=90°,
AC=
BC ,
AE⊥
CE ,
BF⊥
CE于点
F .
![](//tikupic.21cnjy.com/2024/03/07/3f/44/3f4454ff538b3fa6977a8181a4c92075.png)
-
-
-
20.
在平面直角坐标系中,一次函数
y=
kx+
b(
k≠0)的图象经过
A(0,3)和
B(2,2).
![](//tikupic.21cnjy.com/2024/03/07/c0/98/c0988aa049f4c2149b2c62ed42451a55.png)
-
-
(2)
若点D在x轴上,且△ACD为等腰三角形,求点D的坐标.
-
21.
某市组织20辆汽车装运食品、药品、生活用品三种救灾物资共100吨到灾区安置点,按计划20辆汽车都要装运,每辆汽车只能装运同一种救灾物资且必须装满,根据表提供的信息,解答下列问题:
物资种类 | 食品 | 药品 | 生活用品 |
每辆汽车运载量/吨 | 6 | 5 | 4 |
每吨所需运费/元 | 120 | 160 | 100 |
-
(1)
设装运食品的车辆数为x , 装运药品的车辆数为y , 求y与x的函数解析式;
-
(2)
若装运食品的车辆数不少于5,装运药品的车辆数不少于6,则车辆的安排有几种方案?并写出每种安排方案;
-
(3)
在(2)的条件下,若要求总运费最少,应采取哪种安排方案?并求出最少运费.
-
22.
定义:
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmi%3Ey%3C%2Fmi%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmfenced+open%3D%22%7B%22+close%3D%22%22%3E%3Cmtable+columnalign%3D%22left%22%3E%3Cmtr%3E%3Cmtd%3E%3Cmi%3Ek%3C%2Fmi%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmi%3Eb%3C%2Fmi%3E%3Cmo%3E%EF%BC%88%3C%2Fmo%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%EF%BC%9E%3C%2Fmo%3E%3Cmi%3Em%3C%2Fmi%3E%3Cmo%3E%EF%BC%89%3C%2Fmo%3E%3C%2Fmtd%3E%3C%2Fmtr%3E%3Cmtr%3E%3Cmtd%3E%3Cmo%3E-%3C%2Fmo%3E%3Cmi%3Ek%3C%2Fmi%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmi%3Eb%3C%2Fmi%3E%3Cmo%3E%EF%BC%88%3C%2Fmo%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%E2%89%A4%3C%2Fmo%3E%3Cmi%3Em%3C%2Fmi%3E%3Cmo%3E%EF%BC%89%3C%2Fmo%3E%3C%2Fmtd%3E%3C%2Fmtr%3E%3C%2Fmtable%3E%3C%2Fmfenced%3E%3C%2Fmath%3E)
叫做关于直线
x=m的“分边折叠函数”.
-
(1)
已知“分边折叠函数”
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmi%3Ey%3C%2Fmi%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmfenced+open%3D%22%7B%22+close%3D%22%22%3E%3Cmtable+columnalign%3D%22left%22%3E%3Cmtr%3E%3Cmtd%3E%3Cmn%3E3%3C%2Fmn%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E-%3C%2Fmo%3E%3Cmn%3E6%3C%2Fmn%3E%3Cmo%3E%EF%BC%88%3C%2Fmo%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%EF%BC%9E%3C%2Fmo%3E%3Cmn%3E4%3C%2Fmn%3E%3Cmo%3E%EF%BC%89%3C%2Fmo%3E%3C%2Fmtd%3E%3C%2Fmtr%3E%3Cmtr%3E%3Cmtd%3E%3Cmo%3E-%3C%2Fmo%3E%3Cmn%3E3%3C%2Fmn%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E-%3C%2Fmo%3E%3Cmn%3E6%3C%2Fmn%3E%3Cmo%3E%EF%BC%88%3C%2Fmo%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%E2%89%A4%3C%2Fmo%3E%3Cmn%3E4%3C%2Fmn%3E%3Cmo%3E%EF%BC%89%3C%2Fmo%3E%3C%2Fmtd%3E%3C%2Fmtr%3E%3C%2Fmtable%3E%3C%2Fmfenced%3E%3C%2Fmath%3E)
①直接写出该函数与y轴的交点坐标;
②若直线y=2x+t与该函数只有一个交点,求t的取值范围;
-
(2)
已知“分边折叠函数”
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmi%3Ey%3C%2Fmi%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmfenced+open%3D%22%7B%22+close%3D%22%22%3E%3Cmtable+columnalign%3D%22left%22%3E%3Cmtr%3E%3Cmtd%3E%3Cmi%3Ek%3C%2Fmi%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmi%3Ek%3C%2Fmi%3E%3Cmo%3E%EF%BC%88%3C%2Fmo%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%EF%BC%9E%3C%2Fmo%3E%3Cmi%3Em%3C%2Fmi%3E%3Cmo%3E%EF%BC%89%3C%2Fmo%3E%3C%2Fmtd%3E%3C%2Fmtr%3E%3Cmtr%3E%3Cmtd%3E%3Cmo%3E-%3C%2Fmo%3E%3Cmi%3Ek%3C%2Fmi%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmi%3Ek%3C%2Fmi%3E%3Cmo%3E%EF%BC%88%3C%2Fmo%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%E2%89%A4%3C%2Fmo%3E%3Cmi%3Em%3C%2Fmi%3E%3Cmo%3E%EF%BC%89%3C%2Fmo%3E%3C%2Fmtd%3E%3C%2Fmtr%3E%3C%2Fmtable%3E%3C%2Fmfenced%3E%3C%2Fmath%3E)
的图像被直线
x=m与
y轴所夹的线段长为
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmsqrt%3E%3Cmn%3E5%3C%2Fmn%3E%3C%2Fmsqrt%3E%3Cmo%3E%C2%B7%3C%2Fmo%3E%3Cmfenced+open%3D%22%7C%22+close%3D%22%7C%22%3E%3Cmi%3Em%3C%2Fmi%3E%3C%2Fmfenced%3E%3C%2Fmath%3E)
, 则
k的值为
.
-
23.
学完勾股定理后,小宇碰到了一道题:如图1,在四边形
ABCD中,
AC⊥
BD , 垂足为
O , 若
AB=5,
CD=4,
BC=6,则
AD的长为
▲ .
他不会做,去问同桌小轩,小轩通过思考后,耐心地对小宇讲道:“因为AC⊥BD , 垂足为O , 那么在四边形ABCD中有四个直角三角形,利用勾股定理可得AD2=OA2+OD2 , BC2=OB2+OC2 , AB2=OA2+OB2 , CD2=OC2+OD2...”小轩话没讲完,小宇就讲道:“我知道了,原来AD2+BC2与AB2+CD2之间有某种数量关系.”并对小轩表示感谢.
![](//tikupic.21cnjy.com/2024/03/07/29/5a/295aea9ace1b0cc9bb2df33e01d1dfb5.png)
-
-
(2)
如图2,分别在△
ABC的边
BC和边
AB上向外作等腰Rt△
BCQ和等腰Rt△
ABP , 连接
PC ,
PQ .
①若AC=4,BC=8,连接AQ , 交PC于点D , 当∠ACB=90°时,求PQ的长;
②如图3,若AB=10,BC=8,PC=
, 当∠ACB≠90°时,求△ABC的面积.
-
24.
如图,在平面直角坐标系中,直线
l1:
y=
kx-2交
x轴于点
A(2,0),且与直线
l2:
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmi%3Ey%3C%2Fmi%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmfrac%3E%3Cmrow%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmrow%3E%3C%2Fmfrac%3E%3Cmtext%3Ex%3C%2Fmtext%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmath%3E)
交于点
B , 点
C在(12,2),点
E在直线
l1上且位于点
B的右侧.
![](//tikupic.21cnjy.com/2024/03/07/ba/ee/baeec6918a5914f0ba6e38258c4d96a0.jpeg)
-
-
(2)
若在直线l2上存在点D , 使得S△BCD=15,求点D坐标;
-
(3)
若射线BE上存在点P , 直线l2上存在点Q , 使得点C、P、Q三点构成的△CPQ为等腰直角三角形,求出点P的坐标.