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2014年中考数学压轴题精编--河南篇(试题及答案)

发布时间:2014-03-07 18:53:58  

2014年中考数学压轴题精编—河南篇

2014年中考数学压轴题精编—河南篇

1.(河南省)如图,直线y=k1x+b与反比例函数y=

(1)求k1、k2的值;

(2)直接写出k1x+b-k2>0时x的取值范围; xk2(x>0)的图象交于A(1,6),B(a,3)两点. x

(3)如图,等腰梯形OBCD中,BC∥OD,OB=CD,OD边在x轴上,过点C作CE⊥OD于E,CE和反比例函数的图象交于点P,当梯形OBCD的面积为12时,请判断PC和PE的大小关系,并说明理由.

1.解:

(1)由题意知:k2=1×6=6 ························································································· 1分

∴反比例函数的解析式为y=6

x

又B(a,3)在y=6

x的图象上,∴a=2,∴B(2,3)

∵直线y=k1x+b过A(1,6),B(2,3)两点

∴?k + b = 6 解得?

?2k?k1 = ?1-3 ·············································································

1+?···· 4分

b = 3b = 9

(2)x的取值范围为1<x <2 ······················································································ 6分

(3)当S梯形OBCD =12时,PC=PE ············································································· 7分

设点P的坐标为(m,n),∵BC∥OD,CE⊥OD,OB=CD,B(2,3)

∴C(m,3),CE=3,BC=m-2,OD=m+2

∴S梯形OBCD=11

2(BC+OD)·CE,即12=2×(m-2+m+2)×3

∴m=4,mn=6,∴n=3

2,即PE=1

2CE

∴PC=PE ···················································································································· 10分

2.(河南省)

(1)操作发现·

2014年中考数学压轴题精编—河南篇

1

2014年中考数学压轴题精编—河南篇

如图,矩形ABCD中,E是AD的中点,将△ABE沿BE折叠后得到△GBE,且点G在矩形ABCD内部.小明将BG延长交DC于点F,认为GF=DF,你同意吗?说明理由.

(2)问题解决

保持(1)中的条件不变,若DC=2DF,求(3)类比探究

保持(1)中的条件不变,若DC=n·DF,求 2.解:

(1)同意.连接EF,则∠EGF=∠D=90°,EG=AE=ED,EF=EF

∴Rt△EGF≌Rt△EDF,∴GF=DF ············································································· 3分 (2)由(1)知GF=DF,设DF=x,BC=y,则有GF=x,AD=y ∵DC=2DF,∴CF=x,DC=AB=BG=2x ∴BF=BG+GF=3x

在Rt△BCF中,BC +CF =BF ,即y +x =(3x) ∴y=22x,∴

2

2

2

2

2

2

AD

的值; AB

AD

的值. AB

yAD==2 ·············································· 6分 AB2x

(3)由(1)知GF=DF,设DF=x,BC=y,则有GF=x,AD=y ∵DC=n·DF,∴DC=AB=BG=nx ∴CF=(n-1)x,BF=BG+GF=(n+1)x

在Rt△BCF中,BC +CF =BF ,即y +[(n-1)x]=[(n+1)x] ∴y=2nx,∴

3.(河南省)在平面直角坐标系中,已知抛物线经过A(-4,0),B(0,-4),C(2,0)三点.

(1)求抛物线的解析式;

(2)若点M为第三象限内抛物线上一动点,点M的横坐标为m,△AMB的面积为S.求S关于m的函数关系式,并求出S的最大值.

(3)若点P是抛物线上的动点,点Q是直线y=-x上的动点,判断有几个位置能够使得点P、Q、B、O为顶点的四边形为平行四边形,直接写出相应的点Q的坐标.

2014年中考数学压轴题精编—河南篇

2

2

2

2

2

2

y22nAD

==(或) ·························································· 10分

nABnxn

2

2014年中考数学压轴题精编—河南篇

3.解:

2(1)设抛物线的解析式为y=ax+bx+c(a≠0),则有 1?a = ??16a -4b + c = 02?? 解得?b = 1 ?c = -4

?c = -4?4a + 2b + c = 0???

12∴抛物线的解析式为y=x+x-4 ············································· 3分 2 (2)过点M作MD⊥x轴于点D,设M点的坐标为(m,

12m-m+4 2

∴S=S△AMD+S梯形DMBO-S△ABO 12m+m-4

2 则AD=m+4,MD=-

=1121121(m+4)(-m-m+4)+(-m-m+4+4)(-m)-×4×4 22222

=-m-4m(-4<m<0) ····················································································· 6分 2

即S=-m-4m=-(m+2)+4 22

∴S 最大值=4 ···················································································································· 7分

(3)满足题意的Q点的坐标有四个,分别是:(-4,4),(4,-4) (-2+2,2-2),(-2-25,2+2) ··················································· 11分

2014年中考数学压轴题精编—河南篇

3

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