发布时间 : 星期二 文章广东省梅州市2009年初中毕业生学业考试数学试题（含答案）更新完毕开始阅读e4e6b2010740be1e650e9aee
y C A O B x 21．本题满分 8 分． 如图10，已知抛物线y??点C．
（1）求A，B，C三点的坐标； （2）求证：△ABC是直角三角形；
33x?2233图10 x?3与x轴的两个交点为A、B，与y轴交于（3）若坐标平面内的点M，使得以点M和三点A、B、C 为顶点的四边形是平行四边形，求点M的坐标．（直接写出点的坐标，不必写求解过程）
22．本题满分 10 分．
如图 11，矩形ABCD中，AB?5，AD?3．点E是CD上的动点，以AE为直径的⊙O与AB交于点F，过点F作FG⊥BE于点G． （1）当E是CD的中点时： ①tan?EAB的值为______________； ② 证明：FG是⊙O的切线；
（2）试探究：BE能否与⊙O相切?若能，求出此时DE的长；若不能，请说明理由．
A D O E G F 图11
B C
C x
y A O 图10 B
23．本题满分 11 分．
（提示：为了方便答题和评卷，建议在答题卡上画出你认为必须的图形）
如图 12，已知直线L过点A(0，1)和B(1，0)，P是x轴正半轴上的动点，OP的垂直平分线交L于点Q，交x轴于点M． （1）直接写出直线L的解析式；
t，△OPQ的面积为S，（2）设OP?求S关于t的函数关系式；并求出当0?t?2时，S的最大值；
（3）直线L1过点A且与x轴平行，问在L1上是否存在点C， 使得△CPQ是以Q为直角顶点的等腰直角三角形？若存在，求出点C的坐标，并证明；若不存在，请说明理由．
y
L
L1
A Q
x O M P B
图12
2009年梅州市初中毕业生学业考试数学
参考答案及评分意见
一、选择题：每小题 3分，共 15 分．每小题给出四个答案，其中只有一个是正确的． 1．C 2．B 3．D 4．B 5．C 二、填空题：每小题 3分，共 24 分．
6．a 7．3.6?106 8．40 9．4（1分），72（2分） 10．小张 11．?12 12．50 13．7（1分），2n?1（2分）
三、解答下列各题：本题有 10 小题，共 81 分．解答应写出文字说明、推理过程或演算步骤．
14．本题满分7分． （1）90 ························································································································ 2分 （2）30 ························································································································ 4分 43 ···················································································································· 7分 15．本题满分 7 分． （1）3 ·························································································································· 2分 （2）1 ·························································································································· 4分 （3）15 ························································································································ 7分 16．本题满分 7 分． ?1?解：(3?2)0????3?3?1?4cos30°?|?12|．
?1?3?4?2····························································································· 4分 ?1 ·2 ?4?23?2 ·3····································································································· 6分
?4 ··························································································································· 7分
17．本题满分 7 分．
解：由x?1≥1?x得x≥1， ····················································································· 2分 由x?8?4x?1，得x?3． ·······················································································4 分 所以不等式组的解为：1≤x?3， ·············································································6 分 所以不等式组的整数解为：1，2． ················································································7 分 18．本题满分 8 分． 解：
x?4x?4x?4x?2x?2?1
22?x?xx?12?x?(x?2)(x?2)(x?2)2?x(x?1)x?1········································ 3分 ?x ·
?
?2xx?232 ························································································································ 6分
2?32??6． ················································································ 8分
当x?时，原式?3219．本题满分8 分．
?2（1）证明：∵梯形ABCD，AB∥CD， ∴?CDF??FGB，?DCF??GBF， ···················· 2 分 ∴△CDF∽△BGF． ··························3分 （2） 由（1）△CDF∽△BGF， 又F是BC的中点，BF?FC
∴△CDF≌△BGF， ∴DF?FG，CD?BG ·············································6分 又∵EF∥CD，AB∥CD，
∴EF∥AG，得2EF?BG?AB?BG．
A E D C F B 19题图
G
∴BG?2EF?AB?2?4?6?2， ∴CD?BG?2cm． ·································································································· 8分 20．本题满分 8 分．
解：（1）30；20． ·····································································································2 分 （2）
12． ···················································································································4 分
（3）可能出现的所有结果列表如下： 小李抛到 的数字 小张抛到 的数字 1 2 3 4 或画树状图如下：
小张
1
2
3
4
（1，1） （2，1） （3，1） （4，1） （1，2） （2，2） （3，2） （4，2） 开始
（1，3） （2，3） （3，3） （4，3） （1，4） （2，4） （3，4） （4，4） 1 2 3 4
小李 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
共有 16 种可能的结果，且每种的可能性相同，其中小张获得车票的结果有6种： （2，1），（3，1），（3，2），（4，1），（4，2），（4，3）， ∴小张获得车票的概率为P?616?38；则小李获得车票的概率为1?38?58．
∴这个规则对小张、小李双方不公平． ····················································8 分