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?10 8．40 9．4（1分），72（2分） 10．小张 11．?61 12．50 13．7（1分），2n?1（2分） 2三、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)????4cos30°?|?12|．
?3?0?1 ?1?3?4?32?1 ·2······································································································ 4分
?4?23?2 ················································································································ 6分 3 ?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．
x2?4x2?x(x?2)(x?2)x(x?1)??x???x ·解：2············································ 3分
x?4x?4x?1(x?2)2x?1 ?x?2?1 x?2
2x ··································································································································· 6分 x?232?32??6． ·当x?时，原式?······················································································· 8分 32?22?19．本题满分8 分．
（1）证明：∵梯形ABCD，AB∥CD， ∴?CDF??FGB，?DCF??GBF， ······················ 2 分
D C ∴△CDF∽△BGF． ···························· 3分
（2） 由（1）△CDF∽△BGF，
F E 又F是BC的中点，BF?FC ∴△CDF≌△BGF， A G
B
∴DF?FG，CD?BG ················································ 6分
19题图 又∵EF∥CD，AB∥CD，
∴EF∥AG，得2EF?BG?AB?BG． ∴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）
1． ···························································································································· 4 分 2（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 1 2 3 4 （1，1） （2，1） （3，1） （4，1） （1，2） （2，2） （3，2） （4，2） （1，3） （2，3） （3，3） （4，3） （1，4） （2，4） （3，4） （4，4） 或画树状图如下： 开始
1 2 3 4 小张
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?6335?；则小李获得车票的概率为1??． 16888∴这个规则对小张、小李双方不公平． ························································ 8 分
21．本题满分 8 分．
（1）解：令x?0，得y?3，得点C(0，3)． ························································ 1分
令y?0，得?3223，x2?3， x?x?3?0，解得x1??133，，0)B(3，0)． ·∴A(?1································································································· 3分
（2）法一：证明：因为AC2?12?(3)2?4， ······················· 4分 BC?3?(3)?12，AB?16， ·
∴AB?AC?BC， ··············································· 5分 ∴△ABC是直角三角形． ·········································· 6分 法二：因为OC?3，OA?1，OB?3，
2222y M2 C N M1 B x
2222A O M3 21题图
OB， ·∴OC?OA?·············································································································· 4分 OCOB?，又?AOC??COB， OAOC∴Rt△AOC∽Rt△COB． ································································································ 5分 ∴?ACO??OBC，?OCB??OBC?90°， ∴?ACO??OCB?90°，
∴?ACB?90°， 即△ABC是直角三角形． ····················································· 6 分
∴
（3）M1(4，3)，M2(?4，3)，M3(2，（只写出一个给1分，写出2个，得1.5?3)．
分） ····································································· 8分
22．本题满分 10 分．
E 6D C ····································································· 2分
5O G ②法一：在矩形ABCD中，AD?BC，
?ADE??BCE，又CE?DE， B A F ∴△ADE≌△BCE， ················································ 3分
22题图 得AE?BE，?EAB??EBA，
连OF，则OF?OA， ∴?OAF??OFA， ?OFA??EBA， ∴OF∥EB， ·················································································· 4 分 ∵FG⊥BE， ∴FG⊥OF， ∴FG是⊙O的切线 ································································································· 6分 （法二：提示：连EF，DF，证四边形DFBE是平行四边形．参照法一给分．） （2）法一：若BE能与⊙O相切， ∵AE是⊙O的直径， ∴AE⊥BE，则?DEA??BEC?90°，
又?EBC??BEC?90°， ∴?DEA??EBC，
∴Rt△ADE∽Rt△ECB， （1）①
∴
ADDE3x??， ，设DE?x，则EC?5?x，AD?BC?3，得ECBC5?x32整理得x?5x?9?0． ······································································································· 8 分 ∵b?4ac?25?36??11?0， ∴该方程无实数根．
∴点E不存在，BE不能与⊙O相切． ·········································· 10分 法二： 若BE能与⊙O相切，因AE是⊙O的直径，则AE⊥BE，?AEB?90°，
222设DE?x，则EC?5?x，由勾股定理得：AE?EB?AB，
2即(9?x2)?[(5?x)2?9]?25， 整理得x?5x?9?0， ······································· 8分 ∵b?4ac?25?36??11?0， ∴该方程无实数根．
∴点E不存在，BE不能与⊙O相切． ·········································· 10分 （法三：本题可以通过判断以AB为直径的圆与DC是否有交点来求解，参照前一解法给分） 23．本题满分 11 分．
（1）y?1?x ························································································································ 2分 （2）∵OP?t，∴Q点的横坐标为①当0?221t， 211t?1，即0?t?2时，QM?1?t， 22∴S△OPQ?1?1?····································································································· 3分 t?1?t?． ·
2?2?11t?t?1， 22②当t≥2时，QM?1?∴S△OPQ?1?1?t?t?1?． 2?2??1?1?0?t?2，?2t?1?2t?，???∴S?? ······························································································ 4分
?1t?1t?1?，t≥2.???22???当0?11?1?11t?1，即0?t?2时，S?t?1?t???(t?1)2?， 22?2?441． ······························································································ 6分 4∴当t?1时，S有最大值
（3）由OA?OB?1，所以△OAB是等腰直角三角形，若在L1上存在点C，使得△CPQ
是以Q为直角顶点的等腰直角三角形，则PQ?QC，所以OQ?QC，又L1∥x轴，则C，
O两点关于直线L对称，所以AC?OA?1，得C(11)，． ················································ 7 分
下证?PQC?90°．连CB，则四边形OACB是正方形．
y 法一：（i）当点P在线段OB上，Q在线段AB上 （Q与B、C不重合）时，如图–1．
由对称性，得?BCQ??QOP，?QPO??QOP， ∴ ?QPB??QCB??QPB??QPO?180°，
∴ ?PQC?360°?(?QPB??QCB??PBC)?90°． ················································· 8分 （ii）当点P在线段OB的延长线上，Q在线段AB上时，如图–2，如图–3
∵?QPB??QCB，?1??2， ∴?PQC??PBC?90°． ·························· 9分 （iii）当点Q与点B重合时，显然?PQC?90°． 综合（i）（ii）（iii），?PQC?90°．
O P B 23题图-1 x L A Q C L1
，，使得△CPQ是以Q为直角顶点的等腰直角三角形． ·∴在L1上存在点C(11)··········· 11 分
y L A Q O 2 1 y L C L1 A 1 C L1 P B 23题图-2 x O B 2 Q P x 23题图-3
法二：由OA?OB?1，所以△OAB是等腰直角三角形，若在L1上存在点C，使得△CPQ是以Q为直角顶点的等腰直角三角形，则PQ?QC，所以OQ?QC，又L1∥x轴， 则C，
O两点关于直线L对称，所以AC?OA?1，得C(11)，． ·············································· 7 分
延长MQ与L1交于点N．
（i）如图–4，当点Q在线段AB上（Q与A、B不重合）时，