2008年青海省初中毕业生学业考试数学试题（含答案）(2)

青海省2008年初中毕业升学考试数学试卷

参考答案及评分标准

一、填空题（本大题共12小题15空，每空2分，共30分） 1．

21；? 33?2．3；3y(x?1)2

3．2.58?10

54．某人以5千米/时的速度走了x小时，他走的路程是5x千米（答案不唯一） 5．45

6．13；4

7．50

8．5

9．(?11)，

10．四

11．正方形

12．49π2?400

二、选择题（本大题共8小题，每小题3分，共24分） 题号 选项 13 D 14 D 15 C 16 A 17 C 18 C 19 B 20 A 三、（本大题共3小题，每小题7分，共21分）

1?(x?2)(x?2)(x?2)? ···································································· （5分）

x?23?x3?x(x?2)(x?2)? ? ············································································ （6分） x?23?x ?x?2． ································································································· （7分） 22．解：解不等式①，得x≥2， ················································································· （2分）

3解不等式②，得x?． ································································································ （4分）

23··········································································· （5分） ?原不等式组的解集是?2≤x?． 2?1，0，1． ·则原不等式组的整数解是?2，········································································ （6分）

21．解：原式?························································· （7分） ?所有整数解的和是：?2?(?1)?0?1??2． ·

23．解：（1）2007年七月份至十二月份牛肉价格的极差：29?20?9（元/千克），

········································································································································· （2分） （2）七月份牛肉消费金额：180?40%?72（元）， ·················································· （3分） 十月份牛肉消费金额：180?(1?20%?25%?21%)?180?34%?61.2（元），······· （4分） 十二月份牛肉消费金额：180?25%?45（元）． ························································ （5分） （3）合理即可（答案不唯一） ······················································································ （7分） 四、（本大题共3小题，每小题8分，共24分） 24．解：在Rt△AMN中，

································· （2分） AN?MN?tan?AMN?MN?tan60??30?3?303． ·

在Rt△BMN中，

BN?MN?tan?BMN?MN?tan30??30?3································· （4分） ?103． ·

3···························································· （5分） ?AB?AN?BN?303?103?203． ·则A到B的平均速度为：

AB203． ··········································································· （6分） ??103?17（米/秒）

22?70千米/时?175米/秒?19米/秒?17米/秒， ························································· （7分） 9·································································································· （8分） ?此车没有超过限速． ·

25．（1）证明：?AF∥BC， ??AFE??DBE． ···································································································· （1分） ?E是AD的中点， ?AE?DE．

又??AEF??DEB， ?△AEF≌△DEB． ·································································································· （2分） ?AF?DB． ················································································································ （3分） ?AF?DC， ?DB?DC．

即D是BC的中点． ······································································································ （4分） （2）解：四边形ADCF是矩形， ················································································ （5分） 证明：?AF∥DC，AF?DC，

················································································ （6分） ?四边形ADCF是平行四边形． ·

?AB?AC，D是BC的中点， ?AD?BC．

即?ADC?90． ·········································································································· （7分） ···························································································· （8分） ?四边形ADCF是矩形． ·

26．解：（1）树状图如图：

开始

肠 枣 锦1 锦2

枣 锦1 锦2 肠 锦1 锦2 肠 枣 锦2 肠 枣 锦1

········································································································································· （2分）

??P（吃到两只粽子都是什锦馅）?21?． ···························································· （4分） 126（2）模拟试验的树状图为：

开始

肠 枣 锦1 锦2

肠 枣 锦1 锦2 肠 枣 锦1 锦2 肠 枣 锦1 锦2 肠 枣 锦1 锦2

········································································································································· （6分）

?P（吃到两只粽子都是什锦馅）?411?? ························································ （7分） 1646····································································································· （8分） ?这样模拟不正确． ·

五、（本大题共2小题，第27题10分，第28题11分，共21分） 27．（1）证明：连接OD． ?OA?OD，

??OAD??ODA． ·································· （1分） ??OAD??DAE， D ??ODA??DAE． ··································· （2分） ?DO∥MN．············································· （3分）

M E A ?DE?MN，

C O B N ??ODE??DEM?90?．

即OD?DE． ··············································································································· （4分）

?D在?O上，

?DC是?O的切线． ··································································································· （5分）

?（2）解：??AED?90，DE?6，AE?3，

······························································· （6分） ?AD?DE2?AE2?62?32?35． ·连接CD．

?AC是?O的直径，

??ADC??AED?90?． ·························································································· （7分） ??CAD??DAE， ?△ACD∽△ADE． ·································································································· （8分） ADAC??． AEAD?35AC． ?335则AC?15（cm）． ······································································································· （9分）

······························································································ （10分） ??O的半径是7.5cm．

28．解：（1）设y?kx，

4)代入，得k?2． 把(2，?y?2x． ····················································································································· （1分）

自变量x的取值范围是：0≤x≤30．········································································ （2分） （2）当0≤x≤5时，

设y?a(x?5)2?25， ·································································································· （3分）

0)代入，得25a?25?0，a??1． 把(0，············································································ （5分） ?y??(x?5)2?25??x2?10x． ·当5≤x≤15时，

y?25 ····························································································································· （6分）

??x2?10x(0≤x≤5)即y??．

?25(5≤x≤15)（3）设王亮用于回顾反思的时间为x(0≤x≤15)分钟，学习效益总量为Z， 则他用于解题的时间为(30?x)分钟． 当0≤x≤5时，

································ （7分） Z??x2?10x?2(30?x)??x2?8x?60??(x?4)2?76． ·

························································································ （8分） ?当x?4时，Z最大?76． ·当5≤x≤15时，

Z?25?2(30?x)??2x?85． ·················································································· （9分） ?Z随x的增大而减小，

?当x?5时，Z最大?75．

综合所述，当x?4时，Z最大?76，此时30?x?26． ········································· （10分） 即王亮用于解题的时间为26分钟，用于回顾反思的时间为4分钟时，学习收益总量最大．

······································································································································· （11分） 注：以上各题用不同于本参考答案的解法做正确的相应给分．