得 分 评卷人 (第28题15分)
28．已知等腰三角形ABC的两个顶点分别是A(0，1)、B(0，3)，第三个顶点
C在x轴的正半轴上．关于y轴对称的抛物线y＝ax2＋bx＋c经过A、D(3，－2)、P三点，且点P关于直线AC的对称点在x轴上． (1)求直线BC的解析式；
(2)求抛物线y＝ax2＋bx＋c的解析式及点P的坐标； (3)设M是y轴上的一个动点，求PM＋CM的取值范围．
y B
A x O D (第28题图) 2007年南通市初中毕业、升学考试 数学试题参考答案和评分标准
说明：本评分标准每题只提供一种解法，如有其他解法，请参照本标准的精神给分．
一、选择题：本大题共10小题，第1~8题每小题3分，第9~10题每小题4分，共32分． 1．D 2．B 3．C 4．C 5．A 6．B 7．C 8．A 9．D 10．B
龙门书局的初中数学北师大版的《三点一测》值得一看.欢迎大家评论.
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二、填空题：本大题共8小题，每小题3分，共24分．
11．x≥2 12．7.315?107 13．6 14．1800 15．x1?16．1 17．
4，x2??2 31 18．12 5三、解答题：本大题共10小题，共94分． 19．（1）解：原式?4?1?2 ································································································ 3分 ?5············································································································ 5分
x2?2xy?y2x?yx2?y（2）解： ??5x2?4xy5x?4yx(x?y)25x?4yx2?y ? ············································································ 3分 ??x(5x?4y)x?yxx?yx2?y? ?······································································································· 4分 xxx?x2 ?
x ?x?1． ················································································································· 6分 ?当x?2007，y?2008时，
x2?2xy?y2x?yx2?y??的值为2008． ································································ 7分 25x?4xy5x?4yx20．解：由2a?3x?1?0，3b?2x?16?0，
3x?12x?16可得a?． ··························································································· 2分 ，b?23?3x?1≤4， (1)??2 ················································································· 3分 ?a≤4?b，??2x?16??4． (2)??3由（1），得x≤3． ··············································································································· 4分 由（2），得x??2． ·············································································································· 5分
·························································································· 7分 ?x的取值范围是?2?x≤3． ·
21．解：（1）如图，菱形ABCD正确． ·············································································· 3分 （3）菱形ABCD的面积是15． ··························································································· 6分 y/米
2400
D 2000
1600 C
1200 800 400 22．解：（1）160； ··············································································································· 2分 A /分 ·xO 5 10 15 20 25 30 35 40 B （2）①图象正确； ················································································································ 4分
②2 ·········································································································································· 5分 （第22题） （第21题） ·
③根据题意，得40k?400?2400．求得k?50．所以y?50x?400． 由函数的图象可知，在出发后25分钟到40分钟之间最后一次相遇．
当25≤x≤40时，周华从体育场到家的函数关系式是y??160x?6400． ················· 7分
龙门书局的初中数学北师大版的《三点一测》值得一看.欢迎大家评论.
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?y?50x?400，200得x?． ···················································································· 8分
y??160x?64007?200所以，周华出发后经过分钟与刘明最后一次相遇． ···················································· 9分
723．解：过C作CF?AB，交AB的延长线于点F． C D 由? E A
B F
（第23题） ·由条件，得CF?80cm，BF?90cm． ··········································································· 1分 在Rt△CAF中，tanA?CF． ·························································································· 2分 AFCF80························································································· 4分 ≈?500． ·
tan9?0.16． ······································································ 5分 ?AB?AF?BF?500?90?410（cm）
答：从斜坡起点A到台阶前点B的距离为410cm． ····················· 6分
A E 24．（1）证明：连接OA．?DA平分?BDE，??BDA??EDA．
D ?OA?OD，??ODA??OAD．
·································· 3分 ??OAD??EDA，?OA∥CE． ·
O ?AE?DE，??AED?90?，??OAE??DEA?90?．
································· 5分 ?AE?OA．?AE是?O的切线． ·
C B ?（2）?BD是直径，??BCD??BAD?90． ??DBC?30?，??BDC?60?，
（第24题） ?··············································································································· 6分 ??BDE?120． ·
?DA平分?BDE，??BDA??EDA?60?．
································································································· 8分 ??ABD??EAD?30?． ·??在Rt△AED中，?AED?90，?EAD?30，?AD?2DE．
??在Rt△ABD中，?BAD?90，?ABD?30，?BD?2AD?4DE．
········································································· 10分 ?DE的长是1cm，?BD的长是4cm． ·
?AF?25．解：（1）800； ················································································································ 2分
（2）借阅自然科学类图书的频率是0.25，在扇形统计图中对应 的圆心角是90；借阅文学艺术类图书的频率是0. 30，在扇形统 计图中对应的圆心角是108；借阅生活百科类图书的频率是0.20， 在扇形统计图中对应的圆心角是72；借阅金融经济类图书的频率
????文学艺术 生活百科 自然科学 金融经济 是0.25，在扇形统计图中对应的圆心角是90．（扇形图正确） ········································· 6分
（3）因为10000?0.30?3000，所以如果该市图书馆添置这四类图书10000册，则“文学艺术”类图书应添置3000册较合适． ····································································（第·········25·····题）······· ···· 8分 26．解：（1）每台彩电的利润是(3900?100x?3000)元，每天销售(6?3x)台， ········· 1分 则y?(3900?100x?3000)(6?2x)??300x?2100x?5400． ··································· 4分 （2）y??300(x?3.5)?9075．当x?3或4时，y最大值?9000． ····························· 6分 当x?3时，彩电单价为3600元，每天销售15台，营业额为3600?15?54000元， 当x?4时，彩电单价为3500元，每天销售18台，营业额为3500?18?63000元，
所以销售该品牌彩电每天获得的最大利润是9000元，此时每台彩电的销售价是3500元时，能保证彩电的销售量和营业额较高． ······························································································· 9分
22龙门书局的初中数学北师大版的《三点一测》值得一看.欢迎大家评论.
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27．解：（1）如图1，??BAC?90?，AB?AC，??B??ACB?45?，
?DE∥AB，??DEC??DCE?45?，?EDC?90?，?DE?CD?22，
················································································································ 1分 ?CE?CE??4． ·
3?如图2，在Rt△ACE中，?E?AC?90，AC?23，CE??4，?cos?ACE??，
2················································································································· 3分 ??ACE??30?． ·
???（2）如图2，?D?CE???ACB?45，?ACE??30，??D?CA??E?CB?15， CD?AC2又．?△D?CA∽△E?CB． ································································· 5分 ??CE?BC2??D?AC??B?45?，
··············································································· 7分 ??ACB??D?AC，?AD?∥BC． ·
??B?45?，?D?CB?60?，??ABC与?D?CB不互补，?AB与D?C不平行．
··································································································· 8分 ?四边形ABCD?是梯形． ·
（3）在图2中，过点C作CF?AD?，垂足为F． ?AD?∥BC，?CF?BC．
??FCD???ACF??ACD??30?． D? A F 在Rt△ACF中，AF?CF?6，?S△ACF?3，
在Rt△D?CF中，CD??22，?FCD??30，
?E? M ?D?F?2，?S△D?CF?3．
B C 同理，SRt△AE?C?23，SRt△D?E?C?4． ·········································································· 10分
?AME?∽△D?MC． ??AME???D?MC，?E?AM??CD?M，△（第27题图2） 2?1?CE???2S△AME?AE??2??1． ·
??·················································································· 11分
S△D?MCCD?2CD?221?S△AE?M?S△CD?M． （1）
2?S△EMC?S△AE?M?S△AE?C?23， （2） S△E?MC?S△CD?M?S△D?EC?4． （3）
（3）－（2），得S△C?DM?S△AE?M?4?23，由（1），得S△CD?M?8?43，
? S△AD?M?S△ACF?S△DCF?S△CD?M?33?5．
························································································ 12分 ?△AD?M的面积是33?5． ·28．解：（1）?A(0，1)，B(0，3)，?AB?2，
?△ABC是等腰三角形，且点C在x轴的正半轴上，?AC?AB?2，
0)． ······································································ 2分 ?OC?AC2?OA2?3．?C(3，设直线BC的解析式为y?kx?3，?3k?3?0，?k??3．
················································································ 4分 ?直线BC的解析式为y??3x?3．
2（2）?抛物线y?ax?bx?c关于y轴对称，?b?0． ················································ 5分
y 2又抛物线y?ax?bx?c经过A(01)，，D(3，?2)两点．
B O M 龙门书局的初中数学北师大版的《三点一测》值得一看.欢迎大家评论. 9
C? A C Q D P x （第28题）
1??c?1，?a??，解得???3
9a?c??2．???c?1.1························· 7分 ?抛物线的解析式是y??x2?1． ·
3在Rt△AOC中，OA?1，AC?2，易得?ACO?30?．
?在Rt△BOC中，OB?3，OC?3，易得?BCO?60． ?CA是?BCO的角平分线．
?直线BC与x轴关于直线AC对称．
点P关于直线AC的对称点在x轴上，则符合条件的点P就是直线BC与抛物线y??12x?1的交3点． ········································································································································· 8分
?点P在直线BC：y??3x?3上，
?3x?3)． 故设点P的坐标是(x，1?3x?3)在抛物线y??x2?1上， 又点P(x，31??3?3??x2?1．解得x1?3，x2?23．
3，0)，P2(23，?3)． 故所求的点P的坐标是P··························································· 10分 1(3（3）要求PM?CM的取值范围，可先求PM?CM的最小值．
0)时，点P与点C重合，故PM?CM?2CM． I）当点P的坐标是(3，显然CM的最小值就是点C到y轴的距离为3，
·············· 13分 ?点M是y轴上的动点，?PM?CM无最大值，?PM?CM≥23． ·?3)时，由点C关于y轴的对称点C?(?3，0)，故只要求PM?MC?的II）当点P的坐标是(23，最小值，显然线段PC?最短．易求得PC??6． ?PM?CM的最小值是6．
同理PM?CM没有最大值，?PM?CM的取值范围是PM?CM≥6．
0)时，PM?CM≥23， 综上所述，当点P的坐标是(3，?3)时， PM?CM≥6． ·当点P的坐标是(23，························································ 15分
二00八年南通市初中毕业、升学考试
数 学
（满分150分，考试时间120分钟）
题号 得分 一 二 三 19~20 21~22 23~24 25~26 27 28 总分 结分人 核分人 龙门书局的初中数学北师大版的《三点一测》值得一看.欢迎大家评论.
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