大兴区2009~2010学年度第一学期期末检测试卷
初三数学参考答案
第Ⅰ卷(共32分,答在答题纸上)
一、选择题(本大题共8个小题，每小题4分，共32分）
下列各题的四个备选答案中，只有一个符合题意．请将符合题意选项前的字母填在答题纸．．中相应题号下的空格内．
题号 答案 1 B 2 A 3 B 4 C 5 A 6 D 7 A 8 D 第Ⅱ卷(共88分,答在答题纸上)
二、填空题: （共4道小题，每小题4分，共16分）
9. 16：25 10. 80? 11.y??2(x?3)2?1 12．27 三、解答题：（本大题共3个小题，共15分） 13.（本题满分5分）
解：原式?2?21?2??1?3?3.......................................................................4分 22?2?1....................................................................................................5分14.(本题满分5分)
k解：?反比例函数y?的图象经过点（?1，2），xk??2,...........................................................................................................................2分?1?k??2.............................................................................................................................3分
?2?反比例函数解析式为：y?..................................................................................4分x当x?2时，y??1...........................................................................................................5分15.(本题满分5分)
解：设这个二次函数的解析式为y?a(x?1)?2………………………………..2分 ∵二次函数的图象过坐标原点,
∴0?a(0?1)?2
22…………………………………………………….….…3分
a?2 ………………………………………………………………………4分
1
∴这个二次函数的解析式是y?2(x?1)2?2，
即y?2x2?4x………………………………………………………………………5分
四、解答题：（本大题共3个小题，共15分）
16.（本题满分5分） 证明： ∵ AC?3,BC?6?3,DC2 CE42 ………………………………..2分 ∴ AC?BC.
DCCE ………………………………………………3分又 ∠ACB=∠DCE=90°，……………………………………4分 ∴ △ACB∽△DCE．……………………………………….5分
17．（本题满分5分） 解：(1)∵AD是BC上的高，
∴∠ADB=∠ADC=90°．
∵sinB=
4，AD=12， 5∴AB=15. …………………………………………………………2分 BD?AB2?AD2
?152?122?9??????????????????????????3分
∵BC=14， ∴DC=BC-BD=14-9=5． …………………………………………………4分
5∴tan?DAC?……………………………………………………………5分
1218．（本题满分5分） 解：如图， 连结OA. ∵ CD=10cm，
∴ OA＝5cm，………………………………1分 ?AB?CD,
??AMO?90?.在Rt△AOM中， ?OM?3cm,
?AM?OA2?OM2?52?32?4??????????????????????3分
又∵ CD是直径,
AB是弦,
AB⊥CD于M ，
2
∴ AB＝2AM .
∴ AB＝8cm . ………………………5分 五、解答题：（本大题共2个小题，共10分）
19．(本小题满分5分)
解：如图，过C点作CD垂直于AB交BA的延长线于点D. ············································· 1分
??CAB?120?,
??CAD?180???CAB?180??120??60?.
在Rt△CDA中，AC=300,
··············································································· 2分 CD?AC?sin?CAD?300sin60??1503.
·················································································· 3分 AD?AC?cos?CAD?300?cos60??150 ·在Rt△CDB中，
C ?BC?700,BD2?BC2?CD2 ?BD?7002?(1503)2?650
D A B ·························································································· 4分 ?AB?BD?AD?650?150?500 ·
答：A、B两个凉亭之间的距离为500m. ··············································································· 5分
20． （本题满分5分）
开始 解：
6 7 2 6 2 7 6 2 7 6 2 7
···································································································································· 2分
1． ······································································································ 3分 34（2）P（两数字和大于10）=． ································································································ 5分
9（1）P（两数字相同）=
六、解答题：（本大题共2个小题，共11分） 21．（本题满分5分）
x?解：（1）根据题意，得y?(2400?2000?x)?8?4???，
50??22x?24x?3200．…………………………………………………3分 2522（2）对于y??x?24x?3200，
25即y??
3
当x??24?150时，……………………………………………….4分
?2?2?????25?150??y最大值?(2400?2000?150)?8?4???250?20?5000．
50??所以，每台彩电降价150元时，商场每天销售这种彩电的利润最大，最大利润是5000元．………………………………………………………………………………………………5分 22．（本题满分6分）（1）直线BD和⊙O相切． ··························································· 1分
证明：
∵?AEC??ODB，?AEC??ABC， ∴?ABC??ODB． ························································ 2分 ∵OD⊥BC，
∴?DBC??ODB?90°． ∴?DBC??ABC?90°． 即?DBO?90°． ∵AB是⊙O直径，
∴直线BD和⊙O相切． ··················································· 3分 （2）连接AC． ∵AB是⊙O直径， ∴?ACB?90°．
在Rt△ABC中，AB?10，BC?8， ∴AC?······························································································· 4分 AB2?BC2?6． ·
∵BD和⊙O相切，
∴?OBD?90°．
∴?ACB??OBD?90°． 由（1）得?ABC??ODB， ∴△ABC∽△ODB． ········································································································· 5分
ACBC?． OBBD68∴?， 5BD20∴BD?． ························································································································· 6分
3∴
六、解答题：（本大题共3个小题，共21分）
23. （本题满分6分）
解：以A为圆心，以a为半径作圆.延长BA交⊙A于E点，
连接ED…………………………………………………………..………..1分 ∵AB∥CD,
∴?CAB??DCA,
?DAE??CDA.
4 ∵AC=AD,
??DCA??CDA,
??DAE??CAB.…………………………………………………..…..…..2分
在△ABC和△DAE中，
?AD?AC,???DAE??CAB, ?AE?AB.?∴△CAB≌△DAE, ……………………………………………….…….……..3分 ∴ED=BC=b……………………………………………………………………..4分 ∵BE是直径， ∴?EDB?90? 在Rt△EDB中， ED=b, BE=2a, 由勾股定理得
ED2?BD2?BE2.
∴BD?BE2?ED2?(2a)2?b2?4a2?b2.……………………………….5分
BD?cos?DBA??BE4a2?b2………………………………………………………6分 2a24. （本题满分7分）（1）①由题意知：
对称轴为直线x?∴?1，b=1 2b1?, 2a2∴a??1. …………………………………………………………………..1分
12992②y??x?x?2??(x?)??…………………………………………2分
244∵y的值为正整数
∴y1?1,2y2?2 ………………………………………………………………3分
2∴?x?x?2?1或?x?x?2?2 ∴x?x?1?0或x?x?0 解，得 x1?221?51?5,x2? ，x3?0, x4?1……………………4分 222（2）∵当a?a1时，抛物线y?ax?x?2与x轴的正半轴相交于点M（m，0） ∴0?a1m?m?2 ①
∵当a?a2时，，抛物线y?ax?x?2与x轴的正半轴相交于点N（n，0） ∴0?a2n?n?2 ②
2225