2014年中考数学压轴题精编--河南篇(试题及答案)

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12014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＝xk2（x＞0）的图象交于A（1，6），B（a，3）两点．（1）求k1、k2的值；（2）直接写出k1x＋b－xk2＞0时x的取值范围；（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＝x6又B（a，3）在y＝x6的图象上，∴a＝2，∴B（2，3）∵直线y＝k1x＋b过A（1，6），B（2，3）两点∴32611＝＋＝＋bkbk解得931＝＝－bk·······························································4分（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＝21(BC＋OD)·CE，即12＝21×(m－2＋m＋2)×3∴m＝4，mn＝6，∴n＝23，即PE＝21CE∴PC＝PE····························································································10分2．（河南省）（1）操作发现·xyOBCEPAD2014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2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ABAD的值；（3）类比探究保持（1）中的条件不变，若DC＝n·DF，求ABAD的值．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，BC2＋CF2＝BF2，即y2＋x2＝(3x)2∴y＝22x，∴ABAD＝xy2＝2····································6分（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，BC2＋CF2＝BF2，即y2＋[(n－1)x]2＝[(n＋1)x]2∴y＝n2x，∴ABAD＝nxy＝nn2（或n2）··············································10分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．GBCEFADxyOBCMAGBCEFAD2014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33．解：（1）设抛物线的解析式为y＝ax2＋bx＋c（a≠0），则有02440416＝＋＋＝＝＋－－cbaccba解得4121－＝＝＝cba∴抛物线的解析式为y＝21x2＋x－4···································3分（2）过点M作MD⊥x轴于点D，设M点的坐标为（m，21m2＋m－4）则AD＝m＋4，MD＝－21m2－m＋4∴S＝S△AMD＋S梯形DMBO－S△ABO＝21(m＋4)(－21m2－m＋4)＋21(－21m2－m＋4＋4)(－m)－21×4×4＝－m2－4m（－4＜m＜0）··································································6分即S＝－m2－4m＝－(m＋2)2＋4∴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＋52，2－52），（－2－52，2＋52）········································11分xyOBCMAD