2008年江西省南昌市中考数学试题（含答案）

江西省南昌市2008年初中毕业暨中等学校招生考试

数学试题参考答案及评分意见

说明：

1．如果考生的解答与本答案不同，可根据试题的主要考查内容参考评分标准制定相应的评分细则后评卷．

2．每题都要评阅到底，不要因为考生的解答中出现错误而中断对该题的评阅，当考生的解答在某一步出现错误，影响了后续部分时，如果该步以后的解答未改变这一题的内容和难度，则可视影响的程度决定后面部分的给分，但不得超过后面部分应给分数的一半，如果这一步以后的解答有较严重的错误，就不给分．

3．解答右端所注分数，表示考生正确做到这一步应得的累加分数． 4．只给整数分数．

一、选择题（本大题共8小题，每小题3分，共24分） 1．D 2．C 3．D 4．D 5．A 6．A 7．B 8．C 二、填空题（本大题共8小题，每小题3分，共24分） 9．1.514?10 13．125

9

10．x(x?2)(x?2) 11．y??3x?1

16．①②③

212．

1 414．x1?0，x2?2

15．4

说明：第16题，填了④的，不得分；未填④的，①，②，③中每填一个得1分． 三、（本大题共4小题，每小题6分，共24分）

17．解：原式?x?2x?(x?1) ······································································ 2分

22?x2?2x?x2?1··························································································· 3分

··································································································· 4分 ?2x?1． ·当x??1?1?时，原式?2?????1?0． ···························································· 6分 2?2?18．解：（1）符合条件的点D的坐标分别是

D1(2，1)，D2(?2，?1)． ·1)，D3(0，··································································· 3分 1)时，设直线BD1的解析式为y?kx?b， （2）①选择点D1(2，1?k?，???k?b?0，?3由题意得? 解得? ······························································· 5分

1?2k?b?1?b??3??直线BD1的解析式为y?11x?． ································································· 6分 331)时，类似①的求法，可得 ②选择点D2(?2，直线BD2的解析式为y??x?1． ······································································ 6分

?1)时，类似①的求法，可得直线BD3的解析式为y??x?1． ·③选择点D3(0，········· 6分

说明：第（1）问中，每写对一个得1分． 19．解：（1）从计算器中随机抽取一个，再从保护盖中随机取一个，有Aa，Ab，Ba，Bb四种情况．

恰好匹配的有Aa，Bb两种情况，

?P(恰好匹配)?21··············································································· 2分 ?． ·

42B

b

A

a

b

A

a B

b

A

b B

a

（2）用树形图法表示：

A

B

a

所有可能的结果AB Aa Ab BA Ba Bb aA aB ab bA bB ba ·················· 4分 可见，从计算器和保护盖中随机取两个，共有12种不同的情况． 其中恰好匹配的有4种，分别是Aa，Bb，aA，bB，

?P(恰好匹配)?41·············································································· 6分 ?． ·

123 A B a b A BA aA bA B AB aB bB a Aa Ba ba b Ab Bb ab 或用列表法表示：

······························································· 6分 可见，从计算器和保护盖中随机取两个，共有12种不同的情况． 其中恰好匹配的有4种，分别是Aa，Bb，aA，bB，

?P(恰好匹配)?41·············································································· 6分 ?． ·

12320．（1）证：由题意得B?F?BF，?B?FE??BFE， ········································ 1分 在矩形ABCD中，AD∥BC，

A?

??B?EF??BFE，

B? E A

················································ 2分 D ??B?FE??B?EF． ·

?B?F?B?E．

························································· 3分 ?B?E?BF． ·

B C

（2）答：a，b，c三者关系不唯一，有两种可能情况： F

（ⅰ）a，b，c三者存在的关系是a?b?c． ················································· 4分 证：连结BE，则BE?B?E．

由（1）知B?E?BF?c，?BE?c． ······························································ 5分 在△ABE中，?A?90，?AE?AB?BE．

222222····························································· 6分 AE?a，AB?b，?a2?b2?c2． ·

（ⅱ）a，b，c三者存在的关系是a?b?c． ················· 4分

证：连结BE，则BE?B?E．

由（1）知B?E?BF?c，?BE?c． ·························· 5分 在△ABE中，AE?AB?BE，

··························································· 6分 ?a?b?c． ·

说明：1．第（1）问选用其它证法参照给分；

2．第（2）问a?b?c与a?b?c只证1种情况均得满分； 3．a，b，c三者关系写成a?c?b或b?c?a参照给分． 四、（本大题共3小题，每小题8分，共24分） 21．解：（1）答案不唯一，只要合理均可．例如：

①BC?BD；②OF∥BC；③?BCD??A；④△BCE∽△OAF；⑤BC?BEAB；⑥BC?CE?BE；⑦△ABC是直角三角形；⑧△BCD是等腰三角形． ············ 3分 （2）连结OC，则OC?OA?OB．

C F 2222222A?

D B? E A

C

F

B

?D?30，??A??D?30，??AOC?120． ······ 4分

AB为O的直径，??ACB?90．

在Rt△ABC中，BC?1，?AB?2，AC?3． ········ 5分

A

O E D B

OF?AC，?AF?CF．

OA?OB，?OF是△ABC的中位线．

11?OF?BC?．

22?S△AOC?1113ACOF??3??． ························································· 6分 22241?············································································· 7分 S扇形AOC???OA2?． ·

33?S阴影?S扇形AOC?S△AOC??3?． ······························································· 8分 34说明：第（1）问每写对一条得1分，共3分．

22．解一：设乙同学的速度为x米/秒，则甲同学的速度为1.2x米/秒， ······················ 1分 根据题意，得??60?60?6???50， ································································ 3分 1.2xx??解得x?2.5． ······························································································· 4分

经检验，x?2.5是方程的解，且符合题意． ························································ 5分

?甲同学所用的时间为：

60， ···················································· 6分 ?6?26（秒）

1.2x60乙同学所用的时间为：． ······························································ 7分 ?24（秒）

x··········································································· 8分 26?24，?乙同学获胜． ·

解二：设甲同学所用的时间为x秒，乙同学所用的时间为y秒， ······························ 1分 ?x?y?50，?根据题意，得?6060

?x?6?1.2?y?解得? ····························································· 3分

?x?26， ································································································ 6分

y?24.?经检验，x?26，y?24是方程组的解，且符合题意．

·············································································· 8分 x?y，?乙同学获胜． ·

23．（1）可从不同角度分析．例如：

①甲同学的平均偏差率是16％，乙同学的平均偏差率是11％； ②甲同学的偏差率的极差是7％，乙同学的偏差率的极差是16％； ③甲同学的偏差率最小值是13％，乙同学的偏差率最小值是4％； ④甲、乙两同学的偏差率最大值都是20％；

⑤甲同学对字数的估计能力没有明显的提高，乙同学对字数的估计能力有明显提高． ························································· 4分 （2）可从不同角度分析．例如： ①从平均偏差率预测：

甲同学的平均偏差率是16％，估计的字数所在范围是84~116； ································ 6分 乙同学的平均偏差率是11······························· 8分 ％，估计的字数所在范围是89~111； ·②从偏差率的中位数预测：

甲同学偏差率的中位数是15％，估计的字数所在范围是85~115； ····························· 6分 乙同学偏差率的中位数是10％，估计的字数所在范围是90~110； ····························· 8分 ③从偏差率的变化情况预测：

甲同学的偏差率没有明显的趋势特征，可有多种预测方法，如偏差率的最大值与最小值的平均值是16.5··································· 6分 ％，估计的字数所在范围是84~116或83~117． ·

乙同学的偏差率是0％~4％，估计的字数所在的范围是96~104或其它． ··················· 8分 说明：1．第（1）问每写对一条结论得1分；

2．每写对一条偏差率及估计字数范围的各得1分； 3．答案不唯一，只要合理均参照给分． 五、（本大题共2小题，每小题12分，共24分） 24．解：（1）

2点P??，?在抛物线y1??ax?ax?1上，

?19??28?119??a?a?1?， ··················································································· 2分

428