轴交于点A、B，点A的坐标为（4，0）。 （1）求该抛物线的解析式；
（2）点Q是线段AB上的动点，过点Q作QE∥AC，交BC于点E，连接CQ。当△CQE的面积最大时，求点Q的坐标； （3）若平行于x轴的动直线l与该抛物线交于点P，与直线AC交于点F，点D的坐标为（2，0）。问：是否存在这样的直线l，使得△ODF是等腰三角形？若存在，请求出点P的坐标；若不存在，请说明理由。 Y
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?OBQDAX28题图
重庆市2008年初中毕业生学业暨高中招生考试
数学试题参考答案及评分意见
一、选择题：ABCCD，ADBAD 二、填空题：11．x?3； 14．点P在?O内；
12．a(x?y)；
?13．3.48?10； 17．x?1；
18．甲班；
615．60 16．18；
19．181； 20．①④⑤． 三、解答题： 21．解：（1）原式?2?3?1?1 ··················································································· （4分）
?5． ································································································ （5分）
?3?32?4?1?1（2）x? ························································································ （3分）
2?1??3?5． ······································································································· （5分） 2?3?5?3?5，x2?． 22所以原方程的解为：x1?22．解：作图如下：
（注：画正确一个给5分，共10分）
22题图
?a2?5a?2a?2?a2?423．解：原式?? ·················································· （2分） ???2a?2?a?4a?4?a?2
a2?4a?4a2?4a?4??2······································································ （4分）
a?2a?4(a?2)2(a?2)2 ······································································· （7分） ??a?2(a?2)(a?2)?a?2． ································································································· （8分）
当a?2?3时，原式?2?3?2?3． ····························································· （10分） 24．解：（1）设所求反比例函数的解析式为：y?k(k?0)． ································· （1分） x3)在此反比例函数的图象上， ?点A(1，k， ······················································································································· （3分） 1?k?3． ························································································································ （4分）
3
故所求反比例函数的解析式为：y?． ····································································· （5分）
x?3?（2）设直线BC的解析式为：y?k1x?b(k1?0)． ·················································· （6分）
?点B的反比例函数y??1?3，m?3． m31)， 的图象上，点B的纵坐标为1，设B(m，x，． ································································································· （7分） ?点B的坐标为(31)?1?3k1?b，由题意，得? ······························································································· （8分）
0?2k?b．?1解得：??k1?1， ·············································································································· （9分）
?b??2．········································································ （10分） ?直线BC的解析式为：y?x?2． ·
25．解：（1）画树状图如下：
被减数
减数
差
········································································································································· （4分） 或列表如下：
差 减 数 1 2 3 被减数减减减1 0 2 1 0 3 2 1 0 4 3 2 1 ?1 ?2 ?1 ········································································································································· （4分） 由图（表）知，所有可能出现的结果有12种，其中差为0的有3种， 所以这两数的差为0的概率为：p?31?．···························································· （6分） 124（2）不公平． ················································································································ （7分） 理由如下：
由（1）知，所有可能出现的结果有12种，这两数的差为非负数的有9种，其概率为：p1?这两数的差为负数的概率为：p2?因为
3， 41． ······································································· （9分） 431?，所以该游戏不公平． 44游戏规则修改为：
若这两数的差为正数，则小明赢；否则，小华赢． ··················································· （10分） 26．证明：（1）?CF平分?BCD，??BCF??DCF． ····································· （1分） 在△BFC和△DFC中，
?BC?DC，? ········································································································ （3分） ??BCF??DCF，?FC?FC．??△BFC≌△DFC． ·································································································· （4分） （2）连结BD． ············································································································· （5分） ?△BFC≌△DFC， ?BF?DF，
??FBD??FDB． ······································ （6分） ?DF∥AB，??ABD??FDB． ??ABD??FBD． ······································ （7分） ?AD∥BC，??BDA??DBC． ?BC?DC，??DBC??BDC． ??BDA??BDC． ········································ （8分） 26题图 又BD是公共边，?△BAD≌△BED． ······· （9分） ?AD?DE． ············································································································· （10分） 27．解：（1）设这批赈灾物资运往D县的数量为a吨，运往E县的数量为b吨． ··· （1分）
由题意，得??a?b?280， ······························································································ （2分）
?a?2b?20．
解得??a?180， ················································································································ （3分）
?b?100．?120?x?2x， ··················································································· （5分）
x?20≤25．?答：这批赈灾物资运往D县的数量为180吨，运往E县的数量为100吨． ·············· （4分） （2）由题意，得?解得??x?40，即40?x≤45．
?x≤45．····················································· （6分） ?x为整数，?x的取值为41，42，43，44，45． ·
则这批赈灾物资的运送方案有五种．
具体的运送方案是：
方案一：A地的赈灾物资运往D县41吨，运往E县59吨；
B地的赈灾物资运往D县79吨，运往E县21吨．
方案二：A地的赈灾物资运往D县42吨，运往E县58吨；
B地的赈灾物资运往D县78吨，运往E县22吨．
方案三：A地的赈灾物资运往D县43吨，运往E县57吨；
B地的赈灾物资运往D县77吨，运往E县23吨．
方案四：A地的赈灾物资运往D县44吨，运往E县56吨；
B地的赈灾物资运往D县76吨，运往E县24吨．
方案五：A地的赈灾物资运往D县45吨，运往E县55吨；
B地的赈灾物资运往D县75吨，运往E县25吨．
········································································································································· （7分） （3）设运送这批赈灾物资的总费用为w元．由题意，得
w?220x?250(100?x)?200(120?x)?220(x?20)?200?60?210?20
??10x?60800． ········································································································ （9分） 因为w随x的增大而减小，且40?x≤45，x为整数．
所以，当x?41时，w有最大值．则该公司承担运送这批赈灾物资的总费用最多为： w?60930（元）． ······································································································ （10分）
?0?16a?8a?c，28．解：（1）由题意，得? ································································ （1分）
4?c．?1??a??，解得?················································································································ （2分） 2
??c?4．1······················································· （3分） ?所求抛物线的解析式为：y??x2?x?4． ·
20)，过点E作EG?x轴于点G． （2）设点Q的坐标为(m，