北京各区中考数学二模27题汇编及答案.docx
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北京各区中考数学二模27题汇编及答案
2015北京各区中考数学二模27题汇编及答案
2015北京各区中考数学二模27题汇编及答案
27.在平面直角坐标系xOy中,抛物线ymx22mxm4与y轴交于点A(0,3),与x轴交于点B,C(点B在点C左侧).
()
(1)求该抛物线的表达式及点B,C的坐标;
(2)抛物线的对称轴与x轴交于点D,若直线ykxb经过点D和点E(1,2),求直线DE的表达式;
(3)在
(2)的条件下,已知点P(t,0),过点P作垂直于x轴的直线交抛物线于点M,交直线DE于点N,若点M和点N中至少有一个点在x轴下方,直接写出t的取值范围.
27已知一次函数y1kxb(k≠0)的图象经过(2,0),(4,1)两点,二次函数
y2x22ax4(其中a>2).
(1)求一次函数的表达式及二次函数图象的顶点坐标(用含a的代数式表示);
(2)利用函数图象解决下列问题:
①若a
5
,求当y10且y2≤0时,自变量x的取值范围;2
②如果满足y10且y2≤0时的自变量x的取值范围内恰有一个整数,直接写出a
的取值范围.
27.在平面直角坐标系中,抛物线yax2bx+3a0与x轴交于点A(-3,0)、B(1,0)两点,D是抛物线顶点,E是对称轴与x轴的交点.
(1)求抛物线的解析式;
(2)若点F和点D关于x轴对称,点P是x轴上的一个动点,过点P作PQ∥OF交抛物线于点Q,是否存在以点O,F,P,Q为顶点的平行四边形?
若存在,求出点P坐标;若不存在,请说明理由.
27.在平面直角坐标系xOy中,抛物线yax2bx1经过A(1,3),B(2,1)两点.
(1)求抛物线及直线AB的解析式;
(2)点C在抛物线上,且点C的横坐标为3.将抛物线在点A,C之间的部分(包含点A,C)记为图象G,如果图象G沿y轴向上平移t(t0)个单位后与直线AB只有一个公共点,求t的取值范围.
27.已知关于x的方程mx3m1x2m20.2
(1)求证:
无论m取任何实数时,方程恒有实数根;
(2)若关于x的二次函数ymx3m1x2m2的图象经过坐标原点,得到抛2
物线C1.将抛物线C1向下平移后经过点A0,2进而得到
新的抛物线C2,直线l经过点A和点B2,0,求直线l和抛
物线C2的解析式;
(3)在直线l下方的抛物线C2上有一点C,求点C到直线l
的距离的最大值.
27.已知:
关于x的一元二次方程ax22(a1)xa20(a0).
(1)求证:
方程有两个不相等的实数根;
(2)设方程的两个实数根分别为x1,x2(其中x1>x2).若y是关于a的函数,且
yax2x1,求这个函数的表达式;(3)在
(2)的条件下,结合函数的图象回答:
若使y3a21,则自变量a的取值范围为.
27.已知抛物线yax2bxc经过原点O及点A(-4,0)和点B(-6,3).
(1)求抛物线的解析式以及顶点坐标;
(2)如图1,将直线y2x沿y轴向下平移后与
(1)中所求抛物线只有一个交点C,平移后的直线与y轴交于点D,求直线CD的解析式;
(3)如图2,将
(1)中所求抛物线向上平移4个单位得到新抛物线,请直接写出新抛物线上到直线CD距离最短的点的坐标及该最短距离.
图2
27.已知关于x的方程x2m2xm30.
(1)求证:
方程x2m2xm30总有两个实数根;
(2)求证:
抛物线yx2m2xm3总过x轴上的
一个定点;
(3)在平面直角坐标系xOy中,若
(2)中的“定点”记作A
抛物线yx2m2xm3与x轴的另一个交点为B与y轴交于点C,且△OBC的面积小于或等于8,求m的
取值范围.
27.在平面直角坐标系xOy中,抛物线yx2bxc经过点A(4,0)和B(0,2).
(1)求该抛物线的表达式;
(2)在
(1)的条件下,如果该抛物线的顶点为C,
点B关于抛物线对称轴对称的点为D,求直线
CD的表达式;
(3)在
(2)的条件下,记该抛物线在点A,B之
间的部分(含点A,B)为图象G,如果图象
G向上平移m(m>0)个单位后与直线CD只有一个公共点,请结合函数的图象,直接写出m的取值范围.
27.已知关于x的一元二次方程kx3k1x30(k≠0).214yOx
(1)求证:
无论k取何值,方程总有两个实数根;
(2)点Ax1,0、Bx2,0在抛物线ykx3k1x3上,其中x1<0<x2,且2
x1、x2和k均为整数,求A,B两点的坐标及k的值;
(3)设
(2)中所求抛物线与y轴交于点C,问该抛物线上是否存在点E,使得
S
ABESABC,若存在,求出E点坐标,若不存在,说明理由.y11
27.如图,在平面直角坐标系中,点A(5,0),B(3,2),点C在线段OA上,BC=BA,点Q是线段BC上一个动点,点P的坐标是(0,3),直线PQ的解析式为y=kx+b(k≠0),且与x轴交于点D.
(1)求点C的坐标及b的值;
(2)求k的取值范围;
(3)当k为取值范围内的最大整数时,过点B作BE∥x
﹣5ax(a≠0)的顶点在四边形ABED的内部,求a
27.已知关于x的方程mx2-(3m-1)x+2m-2=0
(1)求证:
无论m取任何实数时,方程恒有实数根.
(2)若关于x的二次函数y=mx2-(3m-1)x+2m-2的图象与x轴两交点间的距离为2时,
求二次函数的表达式.
答案
27.(本小题满分7分)
解:
(1)∵抛物线ymx22mxm4与y轴交于点A(0,3),
∴m43.
∴m1.
∴抛物线的表达式y22x3.…………………………………………………………………1x分
∵抛物线yx22x3与x轴交于点B,C,
∴令y0,即x22x30.
解得x11,x23.
又∵点B在点C左侧,
∴点B的坐标为(1,0),点C的坐标(3,0).…………………………………………………...……3分
(2)∵yx22x3(x1)24,
∴抛物线的对称轴为直线x1.
∵抛物线的对称轴与x轴交于点D,
∴点D的坐标(.…………………………………………………………………………...………4分
∵直线ykxb经过点D(1,0)和点E(1,2),
∴kb0,
kb2.为为为
k1,解得b1.
∴直线DE的表达式为yx1.………………………………………………………………………5分
(3)t1或
t3……………………………………………………………………………………………7分
27.解:
(1)∵一次函数y1kxb(k≠0)的图象经过(2,0),(4,1)两点,
2kb0,∴4kb1.
1k,解得2„„„„„„„„„„„„„„„„„„„„„„„„1分
b1.
∴y11x1.„„„„„„„„„„„„„„„„„„„„„„2分2
∵y2x22ax4(xa)24a2,
2∴二次函数图象的顶点坐标为(a,4a).
„„„„„„„„„„„„„3分
(2)①当a5时,y2x25x4.2
„„„„„„„„„„„„„4分
如图10,因为y10且y2≤0,由图象
得2<x≤4.„„„„„„„„„„6分
②
135≤a<.„„„„„„„„„„„7分62
27.解:
(1)据题意得
9a-3b+3=0,a1,
a+b+3=0.解得
b2.
∴解析式为y=-x2-2x+3„„3分
(2)当xb
2a1时,y=4
∴顶点D(-1,4)∴F(-1,-4)„4分若以点O、F、P、Q为顶点的平行四边形存在,则点Q(x,y)满足yEF4①当y=-4时,-x2-2x+3=-4
解得,x1
∴Q1(14),Q2(14)
∴P1(P2„„6分②当y=4时,-x2-2x+3=4
解得,x=-1
∴Q3(-1,4)
∴P3(-2,0)„„7分
综上所述,符合条件的点有三个即:
P1(P2P3(2,0)
27.解:
(1)∵抛物线yax2bx1过A(1,3),B(2,1)两点.
∴ab13
4a2b11.…….1分
解得,a2
b4.
∴抛物线的表达式是y2x24x+1.…….2分设直线AB的表达式是ymxn,
∴mn3m2,解得,.…….3分2mn1n5
∴直线AB的表达式是y2x5.…….4分
(2)∵点C在抛物线上,且点C的横坐标为3.
∴C(3,-5).…….5分
点C平移后的对应点为点C'(3,t5)
代入直线表达式y2x5,解得t4.…….6分
结合图象可知,符合题意的t的取值范围是0t4.…….7分
27.解:
(1)当m0时,x2
当m0时,3m14m2m22
9m26m18m28m
m22m1
m12
∵m10,∴0
综上所述:
无论m取任何实数时,方程恒有实数根;………………………3分
(2)∵二次函数ymx(3m1)x2m
222∴2m20
∴m1………………………4分
抛物线C1的解析式为:
yx2x
抛物线C2的解析式为:
yx2x2
设直线l所在函数解析式为:
ykxb
将A和点B2,0代入ykxb
∴直线l所在函数解析式为:
yx2………5分
(3)据题意:
过点C作CEx轴交AB于E,
可证DECOAB45,则CD222
2
设Ct,t2t2,Et,t2,0t3
∴ECyEyCt23t
39
t………………………6分
24
∵0∴当t
2
332
39时,ECmax24
∵CD随EC增大而增大,
∴CDmax
27.
(1)证明:
ax22(a1)xa20(a0)是关于x的一元二次方程,
.………………………7分[2(a1)]24a(a2)··················································································1分=4.即0.
·········································································2分方程有两个不相等的实数根.·
(2)解:
由求根公式,得x
∴x1或x1
2(a1)2
.
2a
2.······························································································3分a
a0,x1>x2,
x11,x21
2.·····························································································4分a
yax2x1a1.
即ya1(a0)为所求.………………………………………………………5分
(3)0<a≤
27.解:
(1)∵抛物线经过0,0,4,0,6,3三点,∴
2
.…………………………………………………………………………7分3
c0
„„„„„„„„„„„„„„„„„„„„„„„„„„1分16a4b0,
36a6b3.
解
得
1a,4
b1,„„„„„„„„„„„„„„„„„„„„„„„„„„„„2分c0.
1
∴抛物线的解析式为yx2x.
4
1112
∵yx2xx24x44x21
444
∴抛物线的顶点坐标为
2,1„„„„„„„„„„„„„„„„„„„„3分
(2)设直线CD的解析式为y2xm,
根
据
题
意
,
得
12
x4
2,x„„„„„„„„„„„„„„„„„„„„4分xm
化简整理,得x24x4m0,由
1616m0
,解得
m1,„„„„„„„„„„„„„„„„„„„5分
∴直线CD的解析式为y2x1.
(3)点的坐标为
2,7,„„„„„„„„„„„„„„„„„„„„„„„6分
最
短
距
离
为
.„„„„„„„„„„„„„„„„„„„„„„„„7分27.解:
(1)b24ac=m24m3........................................................1分=m24m44m12=m28m16=m4∵m40,
∴方程x2m2xm30总有两个实数根...............................................2分
22
2
(2)x1,2
2mm4
2
................................................3分
∴x11,x2m3,
∴抛物线yx2m2xm3总过x轴上的一个定点(-1,0).................4分(3)
∵抛物线yx2m2xm3与x轴的另一个交点为B,与y轴交于点C,∴B(3-m,0),C(0,m-3),...................................................................................5分∴△OBC为等腰直角三角形,∵△OBC的面积小于或等于8,∴OB,OC小于或等于4,
∴3-m4或m-34,.......................................................................................6分∴m-1或m7.
∴-1m7且m3.............................................................................................7分27.(本小题满分7分)
解:
(1)∵抛物线yx2bxc经过点A(4,0)和B(0,2).
1
4
12
44bc0,∴4„„„„„„„„„„„„„„„„„„1分
c2.
1
b,
2解得c2.
∴此抛物线的表达式为yx2
(2)∵yx2x2
∴C(1,
1
41
x2.„„„„„„„„„2分2
1412192
x1,44
9
).„„„„„„„„„„„„„„„„„„„„„„3分4
∵该抛物线的对称轴为直线x=1,B(0,2),
∴D(2,2).„„„„„„„„„„„„„„„„„„„„„„„4分设直线CD的表达式为y=kx+b.
9
kb,
由题意得4
2kb2.
1k,4解得
5b.2
∴直线CD的表达式为yx
1
45
.„„„„„„„„„„„„5分2
(3)0.5<m≤1.5.„„„„„„„„„„„„„„„„„„„„„„„7分
27.
(1)∵Δ=3k112k9k26k13k1≥0
∴方程总有两个实数根.„„„„„„„„„„„„„„„„„„„„2分
(2)由求根公式得:
x=
∴x=-3或x=-2
2
-(3k+1)?
(3k
2k
1)
1k
∵x1、x2和k均为整数
∴k=±1又∵x1<0<x2
∴k=-1„„„„„„„„„„„„„„„„„„„„„„„„„„„„3分∴A(-3,0),B(1,0)„„„„„„„„„„„„„„„„„„„„4分(3)(
-2,3),-1+
(
-3,-1-
)(
„„„„„„„„„„„„„„„„7分
)
27.解:
(1)直线y=kx+b(k≠0)经过P(0,3),
∴b=3.……………………………………………………1
过点B作BF⊥AC于F,∵A(5,0),B(3,2),BC=BA,∴点F的坐标是(3,0).∴点C的坐标是(1,0).…………………………………
(2)当直线PC经过点C时,k=﹣3.当直线PC经过点B时,k=
1
.………………………3
∴3k……………………………………………(3)3k且k为最大整数,∴k=﹣1.则直线PQ的解析式为y=﹣x+3.
1
3
13
∵抛物线y=ax2﹣5ax(a≠0)的顶点坐标是,
5
2525
a,对称轴为x.
24
5
xyx32
解方程组,得5
1xy22
即直线PQ与对称轴为x
551
的交点坐标为,…………………………………………6222
125
a2.24
82a.解得……………………………………………………………………………72525
∴
27.解:
(1)△=9m2-6m+1-8m2+8m=m2+2m+1,
=(m+1)2;
∴△=(m+1)2≥0,………………………………………….(1分)∴无论m取任何实数时,方程恒有实数根;
(2)设x1,x2为抛物线y=mx2-(3m-1)x+2m-2与x轴交点的横坐标.令y=0,则mx2-(3m-1)x+2m-2=0
由求根公式得,x1=2,,…………………………….(2分)
∴抛物线y=mx2-(3m-1)x+2m-2不论m为任何不为0的实数时恒过定点(2,0).∴x2=0或x2=4,∴m=1或)当m=1时,y=x2-2x,,∴抛物线解析式为y=x2-2x
18当时,yx22x
33
18
答:
抛物线解析式为y=x2-2x;或yx22x……….(3分)
33