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，
∴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，
M． ·则△DCM≌△AC··································································································· 1分
有CD?CA，DM?AM，?DCM??ACM，?CDM??A． 又由CA?CB，得 CD?CB． ············································ 2分 由?DCN??ECF??DCM?45???DCM， ?BCN??ACB??ECF??ACM ?90??45???ACM?45???ACM，
C A
M E
N
D
F
B
得?DCN??BCN． ············································································································ 3分 又CN?CN，
DN≌BN． ·∴△C△C······································································································ 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， M． ·则△GCM≌△AC·························································· 8分
C 有CG?CA，GM?AM，
?GCM??ACM，?CGM??CAM．
G E M
A N F
B
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， GN≌BN． ∴△C△C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∴该抛物线与x轴公共点的坐标是??1···················································· 2分 0?． ·，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?1与x轴只有一个公共点??，··································· 4分 0?． ·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?y1≤0，?1?c≤0，应有? 即?
y?0.5?c?0.?2?解得?5?c≤?1．
1综上，c?或?5?c≤?1． ······················································································· 6分
3（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 由a?b?c?0，c?0，2a?b?0， 得?2a?b??a，
O 1 x 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