七、解答题（本题满分7分）
23．已知：关于x的一元二次方程mx2?(3m?2)x?2m?2?0(m?0)． （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m的函数，且
y?x2?2x1，求这个函数的解析式；
（3）在（2）的条件下，结合函数的图象回答：当自变量m的取值范围满足什么条件时，
y≤2m．
（1）证明：
（2）解：
（3）解：
八、解答题（本题满分7分）
2y 4 3 2 1 -4 -3 -2 -1 O 1 2 3 4 -1 -2 -3 -4 x 24．在平面直角坐标系xOy中，抛物线y?x?bx?c与x轴交于A，B两点（点A在点B0)，将直线y?kx沿y轴向上平移3个单位的左侧），与y轴交于点C，点B的坐标为(3，长度后恰好经过B，C两点．
（1）求直线BC及抛物线的解析式；
（2）设抛物线的顶点为D，点P在抛物线的对称轴上，且?APD??ACB，求点P的坐标；
y （3）连结CD，求?OCA与?OCD两角和的度数． 解：（1）
（2）
4 3 2 1 -2 -1 O -1 -2 1 2 3 4 x
（3）
九、解答题（本题满分8分） 25．请阅读下列材料：
P是线段DF问题：如图1，在菱形ABCD和菱形BEFG中，点A，B，E在同一条直线上，的中点，连结PG，PC．若?ABC??BEF?60，探究PG与PC的位置关系及
?PG的PC值．
小聪同学的思路是：延长GP交DC于点H，构造全等三角形，经过推理使问题得到解决． C D C D
P G F P
G
F
B A E A
B
图1 图2 E
请你参考小聪同学的思路，探究并解决下列问题：
PG的值； PC（2）将图1中的菱形BEFG绕点B顺时针旋转，使菱形BEFG的对角线BF恰好与菱形ABCD的边AB在同一条直线上，原问题中的其他条件不变（如图2）．你在（1）中得到
（1）写出上面问题中线段PG与PC的位置关系及的两个结论是否发生变化？写出你的猜想并加以证明．
（3）若图1中?ABC??BEF?2?(0???90)，将菱形BEFG绕点B顺时针旋转任
??PG的值（用含?的式子表示）． PCPG? ． 解：（1）线段PG与PC的位置关系是 ；PC意角度，原问题中的其他条件不变，请你直接写出（2）
2008年北京市高级中等学校招生考试
数学试卷答案及评分参考
阅卷须知：
1．一律用红钢笔或红圆珠笔批阅，按要求签名．
2．第Ⅰ卷是选择题，机读阅卷．
3．第Ⅱ卷包括填空题和解答题．为了阅卷方便，解答题中的推导步骤写得较为详细，考生只要写明主要过程即可．若考生的解法与本解法不同，正确者可参照评分参考给分．解答右端所注分数，表示考生正确做到这一步应得的累加分数．
第Ⅰ卷 （机读卷 共32分）
一、选择题（共8道小题，每小题4分，共32分） 题号 答案
1 A 2 D 3 C 4 C 5 B 6 B 7 B 8 D 第Ⅱ卷 （非机读卷 共88分）
二、填空题（共4道小题，每小题4分，共16分） 题号 答案 9 10 11 4 12 1x? 2a(a?b)(a?b) b20?7 ab3n?1(?1) nan三、解答题（共5道小题，共25分） 13．（本小题满分5分）
?1?解：8?2sin45?(2?π)???
?3??0?1?22?2?2········································································································· 4分 ?1?3 ·
2?2?2． ····························································································································· 5分
14．（本小题满分5分）
解：去括号，得5x?12≤8x?6． ······················································································ 1分 移项，得5x?8x≤?6?12． ······························································································· 2分 合并，得?3x≤6． ··············································································································· 3分 系数化为1，得x≥?2． ······································································································ 4分 不等式的解集在数轴上表示如下：
?3 ?2 ?1 0 1 2 3
················································································································································· 5分 15．（本小题满分5分） 证明：?AB∥ED， ??B??E． ························································································································ 2分 在△ABC和△CED中，
?AB?CE，? ??B??E，?BC?ED，?
?△ABC≌△CED． ·········································································································· 4分 ?AC?CD． ························································································································ 5分
16．（本小题满分5分）
1)在直线y?kx?3上， ·解：由图象可知，点M(?2，························································· 1分 ??2k?3?1． 解得k??2． ························································································································· 2分
························································································· 3分 ?直线的解析式为y??2x?3． ·令y?0，可得x??3． 2?3?··············································································· 4分 0?． ·?直线与x轴的交点坐标为??，2??令x?0，可得y??3．
?3)． ·················································································· 5分 ?直线与y轴的交点坐标为(0，17．（本小题满分5分） 解：
2x?y?(x?y) 22x?2xy?y?2x?y?(x?y) ················································································································· 2分
(x?y)2?2x?y． ····························································································································· 3分 x?y当x?3y?0时，x?3y． ··································································································· 4分
原式?6y?y7y7??． ···································································································· 5分
3y?y2y2四、解答题（共2道小题，共10分） 18．（本小题满分5分）
解法一：如图1，分别过点A，D作AE?BC于点E， DF?BC于点F． ··············································· 1分 ?AE∥DF． 又AD∥BC，
?四边形AEFD是矩形．
A D
?EF?AD?2． ··············································· 2分
?AB?AC，?B?45?，BC?42，
B E F
图1
C
?AB?AC．
1?AE?EC?BC?22．
2?DF?AE?22，
··········································································································· 4分 CF?EC?EF?2 ·
在Rt△DFC中，?DFC?90，
??DC?DF2?CF2?(22)2?(2)2?10． ························································· 5分
解法二：如图2，过点D作DF∥AB，分别交AC，BC于点E，F． ·························· 1分
?AB?AC，
??AED??BAC?90?． ?AD∥BC，
A E B ?D
??DAE?180???B??BAC?45?．
在Rt△ABC中，?BAC?90，?B?45，BC?42，
?F
图2
C
?AC?BC?sin45??42?2··················································································· 2分 ?4 2??在Rt△ADE中，?AED?90，?DAE?45，AD?2，
?DE?AE?1．
?CE?AC?AE?3． ········································································································ 4分
在Rt△DEC中，?CED?90，
········································································ 5分 ?DC?DE2?CE2?12?32?10． ·
19． （本小题满分5分） 解：（1）直线BD与?O相切． ···························································································· 1分 证明：如图1，连结OD． ?OA?OD， ??A??ADO．
???C?90?， ??CBD??CDB?90?．
又??CBD??A，
D A
O E C ??ADO??CDB?90?． ??ODB?90?．
B
图1 ······································································································ 2分 ?直线BD与?O相切． ·