湖北省孝感市2014年中考数学试卷（含答案）

参考答案及评分说明

一、选择题

题号 答案 二、填空题

1 A 2 D 3 C 4 A 5 D 6 B 7 C 8 A 9 C 10 B 11 D 12 C 113．x≠1； 14．①③； 15．1； 16．； 17．6； 18．(63，32) ．

3三、解答题

19．解：原式＝

1＋2－?2 ···································································· 2分 12(?)2＝4＋2－2 ··················································································· 4分 ＝4 ···················································································· 6分

20．解：（1）如图:

················································· 4分

（2）AB与⊙O相切． ·········································································· 6分

证明：作OD⊥AB于D，如图．

∵BO平分∠ABC，∠ACB＝90°，OD⊥AB， ∴OD＝OC，

∴AB与⊙O相切． ································································ 8分

A

D

O C （第20题答案图）

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21．（1）40； …………………………………2分

（2）54°，如图：…………………………………4分 （3）700； …………………………………7分 （4）画树形图如下：

（第21题答案图）

EFGFH6E1GH12?2GEFHEHFG ················ 8分

∴P（选中小明）＝ ． ····················································· 10分

22．解：（1）由题意可知：

··································· 1分 ???(2k?3)?2?4(k2?1)?0， ·

即?12k?5?0 ································· 2分

∴k?5． ····································· 3分 12（2）∵???x1?x2?2k?3?0， ····································· 5分 2??x1x2?k?1?0∴x1?0,x2?0． ····································· 6分

（3）依题意，不妨设A（x1,0），B（x2,0）.

∴OA?OB?x1?x2??(x1?x2)??(2k?3)，

······························ 8分 OAOB??x1?x2?(?x1)(?x2)?x1x2?k2?1，

∵OA?OB?2OAOB?3， ∴?(2k?3)?2(k2?1)?3，

解得k1＝1，k2＝－2． ······································· 9分 ∵k?5，∴k＝－2． ···································· 10分 1223．解：（1）依题意可知零售量为（25－x）吨，则

y=12 x +22(25－x) +30×15 ····································································· 2分 ∴y =－10 x+1000 ············································································ 4分

（2）依题意有：

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?x?0?·································· 6分 ?25?x?0 ， 解得：5≤x≤25． ·

?25?x?4x?∵－10＜0，∴y随x的增大而减小． ······································· 7分

∴当x=5时，y有最大值，且y最大＝950（百元）．

∴最大利润为950百元. ····································· 10分

24． 解：（1）∵PD切⊙O于点C，∴OC⊥PD． ······························································ 1分

又AD⊥PD，∴OC∥AD．∴∠ACO＝∠DAC． 又OC＝OA，∴∠ACO＝∠CAO，

∴∠DAC＝∠CAO，即AC平分∠DAB． ························································ 3分

（2）∵AD⊥PD，∴∠DAC＋∠ACD＝90°．

又AB为⊙O的直径，∴∠ACB＝90°． ∴∠PCB＋∠ACD＝90°，∴∠DAC＝∠PCB． 又∠DAC＝∠CAO，∴∠CAO＝∠PCB．…… 4分 A ∵CE平分∠ACB，∴∠ACF＝∠BCF， ∴∠CAO＋∠ACF＝∠PCB＋∠BCF， ∴∠PFC＝∠PCF， …………… 5分

∴PC＝PF，∴△PCF是等腰三角形．…………… 6分

D C O F B P E

（第24题答案图）

（3）连接AE．∵CE平分∠ACB，∴AE?BE，∴AE?BE?72．

∵AB为⊙O的直径，∴∠AEB＝90°． 在Rt△ABE中，AB?AE2?BE2?14． ···································· 7分

∵∠PAC＝∠PCB，∠P＝∠P，∴△PAC∽△PCB， ·········································· 8分 ∴

PCAC4AC4PC4．又tan∠ABC＝，∴??，∴?．

PBBC3BC3PB3设PC?4k，PB?3k，则在Rt△POC中，PO?3k?7，OC?7， ∵PC2?OC2?OP2，∴(4k)2?72?(3k?7)2，

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∴k＝6 （k＝0不合题意，舍去）．

∴PC?4k?4?6?24． ····················································· 10分

25．（1）A（0，3），B（4，3），C（4，－1），D（0，－1）． ······························· 4分

（2）①设直线BD的解析式为y?kx?b(k?0)，由于直线BD经过D（0，－1），B（4，3），

∴???1?b?k?1，解得?，∴直线BD的解析式为y?x?1． ············· 5分

3?4k?bb??1??设点P的坐标为(x,x2?4x?3)，则点H(x,x?1)，点G(x,3)． 1°当x?1且x≠4时，点G在PH的延长线上，如图①．

∵PH＝2GH，∴(x?1)?(x2?4x?3)?2?3?(x?1)?， ∴x2?7x?12?0，解得x1?3，x2?4． 当x2?4时，点P，H，G重合于点B，舍去．

∴x?3．∴此时点P的坐标为(3,0)． ······································· 6分 2°当0?x?1时，点G在PH的反向延长线上，如图②,PH＝2GH不成立．………7

分

3°当x?0时，点G在线段PH上，如图③．

∵PH＝2GH，∴(x2?4x?3)?(x?1)?2?3?(x?1)?， ∴x2?3x?4?0，解得x1??1，x2?4（舍去）， ∴x??1．此时点P的坐标为(?1,8)．

综上所述可知，点P的坐标为(3,0)或(?1,8)． ··································· 8分

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②如图④，令x2?4x?3?0，得x1?1，x2?3，∴E(1,0)，F(3,0)，∴E F＝2．

∴s?AEF?1EFOA?3． ……………………9分 22s?PH?∵?KPH∽?AEF，∴?KPH??，

s?AEF?EF??∴s?KPH?33PH2?(?x2?5x?4)2 ． …………11分 44∵1?x?4, ∴当x?

5243时，s?KPH的最大值为 ． …………12分 264注意：1．按照评分标准分步评分，不得随意变更给分点；

2．上述各题的其它解法，只要思路清晰，解法正确，均应参照上述标准给予相应分数．

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