2 7．（本题满分 14 分）

1 1

?如图①，一次函数 y＝x ? 2 的图像交 x 轴于点 A，交 y 轴于点 B，二次函数 y＝x2 ? bx ? c

22的图像经过 A、B 两点，与 x 轴交于另一点 C．

（1）求二次函数的关系式及点 C 的坐标；

（2）如图②，若点 P 是直线 AB 上方的抛物线上一点，过点 P 作 PD∥x 轴交 AB 于点 D，PE

∥y 轴交 AB 于点 E，求 PD＋PE 的最大值； （3）如图③，若点 M 在抛物线的对称轴上，且∠AMB＝∠ACB，求出所有满足条件的点 M

的坐标．

y

C O

A

B

y

P C

D

y

C

A

E B

B

x

O

x

O A

x

①

②

（第 27 题图）

③

九年级数学试卷 第 6页（共 6页）

九年级数学参考答案及评分标准

（阅卷前请认真校对，以防答案有误！）

一、选择题（每小题3分，共24分）

题号 答案 1 B 2 C 3 D 4 C 5 A 6 C 7 D 8 B 二、填空题（每小题3分，共24分） 9． x≤3． 13．

25． 410．1.3×105． 14．36°．

11．甲． 12．3．

15．3或25?2． 16．22?1．

三、解答题

17．解：原式＝5－1＋2－3 ······························································································· 4分

＝3． ············································································································· 6分

说明：每算对一个给1分． 18．解：原式＝

46a?3?? ········································································ 3分 a?3(a?3)(a?3)3＝＝

42 ······························································································ 5分 ?a?3a?32． ······································································································ 6分 a?319．解：??x?5?5(x?2)?3x?6，①?1?4x．??2②

由不等式①，得x≤8． ····························································································· 3分 由不等式②，得x＞－1． ························································································· 6分 ∴不等式组的解集为－1＜x≤8．·············································································· 8分 120．解：（1）抽到数字恰好为3的概率为． ······························································ 3分

3（2）画树状图（或列表）如下： ··············································································· 6分

九年级数学试卷 第 7页（共 6

页）

十位数 个位数 31开始

355153由树状图可知，所有等可能的结果共有6种，其中恰好是51有1种． ∴P（两位数恰好是“51”）＝

1． ········································································· 8分 621．解：（1）10． ············································································································· 2分

（2）72°； ······················································································································ 4分 （3）根据题意得：1200×(1－5%)＝1 140（人）， ··············································· 7分 答：估计测试成绩合格以上（含合格）的人数有1 140人． ································ 8分 22．解：（1）如图所示． ································································································· 4分

A BDCEM

说明：作出点C给1分；作出BD给2分，作出点E给1分．

（2）BD＝DE． ·············································································································· 5分 理由如下：

A∵BD平分∠ABC，∴∠1＝

1∠ABC． 21BD1∵AB＝AC，∴∠ABC＝∠4．∴∠1＝∠4．

242C3EM∵CE＝CD，∴∠2＝∠3． ∵∠4＝∠2＋∠3，∴∠3＝

1∠4． 2∴∠1＝∠3．∴BD＝DE． ························································································ 10分

九年级数学试卷 第 8页（共 6

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23．解：（1）5． ··············································································································· 3分

（2）①3÷0.2＝15，即运动员第1次到过点P用时15 min， ······························· 5分 ∵该运动员从第一次过P点到第二次过P点所用的时间为24 min，

∴该运动员从甲地出发到第二次经过P点所用的时间是15＋24＝39（min）， ∴直线AB经过点（25，5），（39，3）．

设AB所在直线的函数表达式为s＝kt＋b，将（25，5），（39，3）代入，得：

1?k?，＝??25k?b＝5，160?7∴?解得?∴AB所在直线的函数表达式为s＝?t?． ··· 7分

6039k?b3．77＝??b＝．?7?160160②∵s＝?t?，∴当s＝0时，?t?＝0，解得t＝60．

7777答：该运动员跑完赛程用时60 min． ····································································· 10分 24．解：（1）设该商场购进LED灯泡x个，普通白炽灯泡的数量为y个．根据题意，得

?x?y＝300， ·········································································· 2分 ?(60?45)x?(0.9?30?25)y3200．＝??x＝200，解得? ·············································································································· 4分

y100．＝?答：该商场购进LED灯泡与普通白炽灯泡的数量分别为200个和100个． ······· 5分 （2）设该商场再次购进LED灯泡a个，这批灯泡的总利润为W元．则购进普通白炽灯泡(120－a)个．根据题意得

W＝(60－45)a＋(30－25)(120－a)＝10a＋600． ················································· 7分 ∵10a＋600≤[45a＋25(120－a)]×30%，解得a≤75， ······································· 9分 ∵k＝10＞0，∴W随a的增大而增大，

∴a＝75时，W最大，最大值为1350，此时购进普通白炽灯泡(120－75)＝45个． 答：该商场再次购进LED灯泡75个，购进普通白炽灯泡45个，这批灯泡的总利润为1 350元． ······················································································································ 10分 25．解：（1）证明：∵AE＝EC，BE＝ED，

九年级数学试卷 第 9页（共 6

页）

∴四边形ABCD是平行四边形． ················································································ 2分 ∵AD为直径，∴∠AED＝90°，即AC⊥BD． ··························································· 3分 ∴四边形ABCD 是菱形． ···························································································· 5分 （2）由（1）知，四边形ABCD是菱形． ∴AD＝DC，DE⊥AC．∴∠ADE＝∠CDE． 如图，过点C作CG⊥AD，垂足为G，连接FO． ∵BF切圆O于点F，∴OF⊥AD，且OF＝

1AD＝3． 2D

FCEBOGA∵BC∥AD，OF⊥BC，CG⊥AD，∴CG＝OF＝3． 在Rt△CDG中，sin∠ADC＝

CG31＝＝，∴∠ADC＝30°． ································ 7分 CD62连接OE，∵菱形ABCD中，AE＝EC，AO＝OD， ∴OE∥DC，∴∠AOE＝∠ADC＝30°．

AE的长＝∴?30?π?3π＝． ··················································································· 10分 218026．解：（1）45°． ············································································································ 3分

（2）如图1，过点C作CD⊥AB于点D． yyCCCFB

AD

BBOGAExOMANx图

1 图2 图3

在Rt△ACD中，∠A＝45°，∴AC＝2DC． ···························································· 3分 在Rt△BCD中，∠B＝30°，∴BC＝2DC． ································································· 4分 ∴

BC＝2．∴△ABC是智慧三角形． ·································································· 7分 AC（3）由题意可知∠ABC＝90°或∠BAC＝90°．

九年级数学试卷 第 10页（共 6页）