梦想不会辜负每一个努力的人 2008年天津市初中毕业生学业考试
数学参考答案及评分标准
评分说明：
1．各题均按参考答案及评分标准评分．
2．若考生的非选择题答案与参考答案不完全相同但言之有理，可酌情评分，但不得超过该题所分配的分数．
一、选择题：本大题共10小题，每小题3分，共30分． 1．A 2．D 3．C 4．B 5．A 6．C 7．A 二、填空题：本大题共8小题，每小题3分，共24分． 11．?4?x?3 15．6
12．5
13．（4，5）
14．112.6；25.9，93?
8．B
9．B
10．D
16．3
17．y?x?2 （提示：答案不惟一，如y??x2?5x?6等）
18．O1，O3，如图① （提示：答案不惟一，过O1O3与O2O4交点O的任意直线都能将四个圆分成面积相等的两部分）；
． O5，O，如图② （提示：答案不惟一，如AO4，DO3，EO2，CO1等均可）
C E o3 C o4 o3 o D o4 o B o5 D o1 B o1 o2 A
第（18）题图①
三、解答题：本大题共8小题，共66分． 19．本小题满分6分．
A o2 第（18）题图②
?3x?5y?8，①解 ∵?
2x?y?1.②?由②得y?2x?1，③ ··········································································································· 2分
将③代入①，得3x?5(2x?1)?8．解得x?1．代入③，得y?1．
?x?1，
∴原方程组的解为? ········································································································ 6分
y?1.?
20．本小题满分8分．
解 （Ⅰ）∵点P（2，2）在反比例函数y?∴2?k
的图象上， x
k．即k?4． ················································································································· 2分 2 1
梦想不会辜负每一个努力的人 ∴反比例函数的解析式为y?∴当x??3时，y??4． x
4． ········································································································· 4分 34， ····························································· 6分 3（Ⅱ）∵当x?1时，y?4；当x?3时，y?又反比例函数y?
4
在x?0时y值随x值的增大而减小， ··················································· 7分 x
4········································································ 8分 ?y?4． ·
3∴当1?x?3时，y的取值范围为
21．本小题满分8分． 解（Ⅰ）∵AB∥CD，
∴?BAD??ADC?180?． ··································································································· 1分 ∵⊙O内切于梯形ABCD，
D C 1∴AO平分?BAD，有?DAO??BAD，
E 2O 1DO平分?ADC，有?ADO??ADC．
2A B 1∴?DAO??ADO?(?BAD??ADC)?90?．
2∴?AOD?180??(?DAO??ADO)?90?． ············································································· 4分 （Ⅱ）∵在Rt△AOD中，AO?8cm，DO?6cm，
∴由勾股定理，得AD?AO2?DO2?10cm． ··································································· 5分 ∵E为切点，∴OE?AD．有?AEO?90?． ········································································· 6分 ∴?AEO??AOD．
又?OAD为公共角，∴△AEO∽△AOD． ······································································ 7分 ∴
OEAOAO?OD，∴OE?············································································· 8分 ??4.8cm． ·
ODADAD22．本小题满分8分． 解 观察直方图，可得
车速为50千米/时的有2辆，车速为51千米/时的有5辆， 车速为52千米/时的有8辆，车速为53千米/时的有6辆， 车速为54千米/时的有4辆，车速为55千米/时的有2辆，
车辆总数为27， ······················································································································· 2分 ∴这些车辆行驶速度的平均数为
1···················································· 4分 (50?2?51?5?52?8?53?6?54?4?55?2)?52.4． ·
27∵将这27个数据按从小到大的顺序排列，其中第14个数是52，
2
梦想不会辜负每一个努力的人 ∴这些车辆行驶速度的中位数是52． ················································································· 6分 ∵在这27个数据中，52出现了8次，出现的次数最多，
∴这些车辆行驶速度的众数是52． ··························································································· 8分 23．本小题满分8分．
解 如图，过点A作AD?BC，垂足为D，
根据题意，可得?BAD?30?，?CAD?60?，AD?66． ··················································· 2分 在Rt△ADB中，由tan?BAD?BD， AD3?223． 3B A
D
得BD?AD?tan?BAD?66?tan30??66?在Rt△ADC中，由tan?CAD?CD， ADC 得CD?AD?tan?CAD?66?tan60??66?3?663． ······················································ 6分 ∴BC?BD?CD?223?663?883?152.2．
答：这栋楼高约为152.2 m． ························································································ 8分 24．本小题满分8分． 解 （Ⅰ） 速度（千米／时） 所用时间（时） 所走的路程（千米） 10 10 骑自行车 乘汽车 x 2x 10 x10 2x ···································································· 3分 （Ⅱ）根据题意，列方程得
10101········································································· 5分 ??． ·
x2x3解这个方程，得x?15． ··································································································· 7分 经检验，x?15是原方程的根． 所以，x?15．
答：骑车同学的速度为每小时15千米． ················································································ 8分 25．本小题满分10分．
（Ⅰ）证明 将△ACM沿直线CE对折，得△DCM，连DN，
则△DCM≌△ACM． ······································································································ 1分
3
梦想不会辜负每一个努力的人 有CD?CA，DM?AM，?DCM??ACM，?CDM??A． 又由CA?CB，得 CD?CB． ·············································· 2分 由?DCN??ECF??DCM?45???DCM，
A
B
C
?BCN??ACB??ECF??ACM ?90??45???ACM?45???ACM，
M E
N D F
得?DCN??BCN． ················································································································· 3分 又CN?CN，
∴△CDN≌△CBN． ········································································································ 4分 有DN?BN，?CDN??B．
∴?MDN??CDM??CDN??A??B?90?． ····································································· 5分 ∴在Rt△MDN中，由勾股定理，
得MN2?DM2?DN2．即MN2?AM2?BN2． ································································ 6分 （Ⅱ）关系式MN2?AM2?BN2仍然成立． ····································································· 7分 证明 将△ACM沿直线CE对折，得△GCM，连GN， 则△GCM≌△ACM． ···························································· 8分 有CG?CA，GM?AM，
G E M
A N F
B
C
?GCM??ACM，?CGM??CAM．
又由CA?CB，得 CG?CB．
由?GCN??GCM??ECF??GCM?45?，
?BCN??ACB??ACN?90??(?ECF??ACM)?45???ACM．
得?GCN??BCN． ············································································································· 9分 又CN?CN， ∴△CGN≌△CBN．
有GN?BN，?CGN??B?45?，?CGM??CAM?180???CAB?135?， ∴?MGN??CGM??CGN?135??45??90?． ∴在Rt△MGN中，由勾股定理，
得MN2?GM2?GN2．即MN2?AM2?BN2． ································································ 10分 26．本小题满分10分．
解（Ⅰ）当a?b?1，c??1时，抛物线为y?3x2?2x?1， 方程3x2?2x?1?0的两个根为x1??1，x2?1． 3 4
梦想不会辜负每一个努力的人 0?． ·∴该抛物线与x轴公共点的坐标是??1····················································· 2分 ，0?和?，
（Ⅱ）当a?b?1时，抛物线为y?3x2?2x?c，且与x轴有公共点．
?1?3??
1对于方程3x2?2x?c?0，判别式??4?12c≥0，有c≤． ·············································· 3分
3①当c?111时，由方程3x2?2x??0，解得x1?x2??． 333此时抛物线为y?3x2?2x??1?10?． ·与x轴只有一个公共点??，····································· 4分
3?3?②当c?1时， 3x1??1时，y1?3?2?c?1?c， x2?1时，y2?3?2?c?5?c．
1由已知?1?x?1时，该抛物线与x轴有且只有一个公共点，考虑其对称轴为x??，
3应有??y1≤0，?1?c≤0， 即?
?5?c?0.?y2?0.1或?5?c≤?1． ··························································································· 6分 3解得?5?c≤?1． 综上，c?（Ⅲ）对于二次函数y?3ax2?2bx?c，
由已知x1?0时，y1?c?0；x2?1时，y2?3a?2b?c?0， 又a?b?c?0，∴3a?2b?c?(a?b?c)?2a?b?2a?b． 于是2a?b?0．而b??a?c，∴2a?a?c?0，即a?c?0．
∴a?c?0． ·························································································································· 7分 ∵关于x的一元二次方程3ax2?2bx?c?0的判别式 ??4b2?12ac?4(a?c)2?12ac?4[(a?c)2?ac]?0，
∴抛物线y?3ax2?2bx?c与x轴有两个公共点，顶点在x轴下方． ·································· 8分 又该抛物线的对称轴x??b， 3ay
O 1 x 5
梦想不会辜负每一个努力的人 由a?b?c?0，c?0，2a?b?0， 得?2a?b??a， ∴
1b2???． 33a3又由已知x1?0时，y1?0；x2?1时，y2?0，观察图象，
可知在0?x?1范围内，该抛物线与x轴有两个公共点． ·················································· 10分
6
2008年天津中考数学试题答案
