八、解答题（本题满分14分）
24．如图，已知抛物线y?ax2?bx?2(a?0)与x轴交A、B两点，与y轴交于C点，直线BD交抛物线于点D，且D（2，3），
tan?DBA?12于并
．
（1）求抛物线的解析式；
（2）已知点M为抛物线上一动点，且在第三象限，顺次连接点B、M、C、A，求四边形BMCA面积的最大值；
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（3）在（2）中四边形BMCA面积最大的条件下，过点M作直线平行于y轴，在这条直线上是否存在一个以Q点为圆心，OQ为半径且与直线 AC相切的圆，若存在，求出圆心Q的坐标，若不存在，请说明理由．
绝密★启用前
考试时间：2013年6月15日上午9∶00－11∶00]
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[四川省自贡市2013年初中毕业生学业考试
数学参考答案及评分标准
第Ⅰ卷(选择题 共40分)
一、选择题：(每小题4分，共40分) 1．B 2．A 3．C 4．D 5．C 6．D 7．B 8．A 9．B 10．A 绝密★启用前
四川省自贡市2013年初中毕业生学业考试
数学参考答案及评分标准
第Ⅱ卷(非选择题 共110分)
说明：
一、如果考生的解法与下面提供的参考解法不同，只要正确一律给满分，若某一步出现错误，可参照该题的评分意见进行评分。
二、评阅试卷时，不要因解答中出现错误而中断对该题的评阅，当解答中某一步出现错误，影响了后继部分，但该步以后的解答未改变这一道题的内容和难度，后来发生第二次错误前，出现错误的那一步不给分，后面部分只给应给分数之半；明显笔误，可酌情少扣；如有严重概念性错误，则不给分；在同一解答中，对发生第二次错误起的部分不给分。
三、涉及计算过程，允许合理省略非关键性步骤。 四、在几何题中，考生若使用符号“”进行推理，其每一步应得分数，可参照该题的评分意见进行评分。
二、填空题：(每小题4分，共计20分) 11．x-1
12．1
13．255 14．①② 15．4，
8n(n?1)
三、解答题：(每小题8分，共计16分)
x?4 2x…2 x…1 16．解：解不等式① 得 3x?6…··············································· (2′)
解不等式② 得 (4′) 2x?1?3x?3 ?x??4 x?4 ········································
1、2、3 ··························································· (8′)
?4a(6′) ≤x?4 ········································································· ?不等式组的解集是 1?不等式组的所有的整数解是
17．解：原式?a?1?a?12(a?1)(a?1)?2a?1a2 ·························································· (4′)
2?a?0且a??1 ?a? ?当a?时 ············································ (6′)
原式?42?22 ····························································································· (8′)
四、解答题：(每题8分，共计16分) 18．解：?a?0 ···················· (1′)
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?x?2bax?(b2a)??2ca?(b2a)2 ····················(3′) ?(x?2b2a)?2b?4ac4a22 ············· (4′)
当b2?4ac…0 ，x?b2a??b?4ac2a?b?2 ··········································································· (6′)
?x1??b?b?4ac2a2，x2?b?4ac2a ·································································· (7′)
当b2····················································································· (8′) ?4ac
19．解：（1）设：该校大寝室每间住x人，小寝室每间住y人 ······························· (0.5′) 可得方程组??55x?50y?740?50x?55y?730 ···················· (2.5′) 解方程组得??x?8?y?6 ··············· (3.5′)
答：该校大寝室每间住8人，小寝室每间住6人 ························································· (4′) （2）设应安排小寝室z间 ··················· (4.5′) 解不等式得
z≤5 ···············(6.5′) ?z(5.5′) 6z?8(80?z)…630 ······················
为自然数 ?z?0,1,2,3,4,5 ·················· (7.5′)
答：共有6种安排住宿方案 ·························································································· (8′) 五、解答题：(共2个题，每题10分，共20分) 20．解：(1)
x?4?20%?20（个） ········ (1′) 20?4?5?4?3?2?2（个） ············ (2′)
6?4?5?5?4?4?3?3?2?2?1?220(3′) ?4 ································································
答：该年级平均每班有4名文明行为劝导志愿者． ······················································ (4′) 补充条形图正确 ············································································································· (5′) （2）解法一 解法二
··········(9′)
?P（同一班级）?412?13 ······················· (9′)
412?13 ······················ (10′) ?P（同一班级）? ········· (10′)
21．（1）证明：连接CO，交DB于E，??CDB?30° ··········································· (1′) ∴∠O＝2∠D=60° ······················································· (2′) 又∵∠OBE＝30°∴∠BEO=180°-60°-30°=90° ··· (3′) ∵AC?BD ∴∠ACO=∠BEO=90° ························ (4′) ∴AC是?O的切线 ························································ (5′)
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(2)解：
?OE?DB ∴EB?12(6′) DB?33 ·································
EBOB在Rt△EOB中，cos30°? ∴OB?33?32(7′) ?6 ·············································
又∵∠D=∠DBO，DE=BE，∠CED=∠OEB ∴?CDE??OBE(ASA) ··················· (8′)
?S?CDE?S(9′) ?S阴影?OBE ····················
?S扇形OCB?60360(10′) ??6?6?(cm) ············
22六、解答题：(本题满分12分)
22．解：由题可得?BAC?90° ·········· (1′) ?BC?∴轮船航行速度为167?4340?(83)?16722 ··············· (3′)
································································ (4′) ?127(km/h)．
（2）解法一：作BD?l于D，CE?l于E，延长BC交l于F ·································· (5′) 在Rt?BDA中 AD?ABsin?BAD?20
DB?ABcos?BAD?203 ·························· (6′) 在Rt?ACE中
CE?ACsin?CAE?43
AE?ACcos?CAE?12 ···························· (7′)
?BD?lCE?l?CE?BD
·········································· (8′)
··································· (9′) ?x?8 ··················· (10′)
AN?20.5 ?19.5?AF?20.5
??FDB∽?FEC?CEBD?FEFD设EF?x?43203?xx?20?12?AF?AE?EF?20 ?AM?19.5?轮船不改变航向继续航行正好能与码头MN靠岸． ················································ (12′)
解法二：作BD?l于D，CE?l于E，延长BC交l于F ··········································· (5′) 在Rt?BDA中 AD?ABsin?BAD?20 DB?ABcos?BAD?203(6′) 在Rt?ACE中
?B(?20,20CE?ACsin?CAE?43
AE?ACcos?CAE?12 ···························· (7′)
3),C(12,43) ···································· (8′)
32,b?103
设直线BC的解析式为：y?kx?b，把B,C代入得k??BC的解析式为：y?? ?19.5?32x?10(9′)
3，令y?0,?x?20 (10′)
AF?20.5?轮船不改变航向继续航行正好能与码头MN靠岸． ············ (12′)
23．（1）证明：??B1CB?45°，?B1CA1(1′) ?90° ??B1CQ??BCP?45° ·········· 112又B1C?BC，?B1??B ??B1CQ??BCP（ASA） ······· (2′) ?CQ?CP ·············· (3′) 11（2）作PD?CA于D， ??A?30°，?PD?11??PCD?45° ?1(4′) AP?1 ······································ 1?2PD?12PD1CP1?sin45°?22 ················ (5′) ?CP1 ·········· (6′)
又CP1?CQ，?CQ?2 ······························································································ (7′)
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