A F D EBC 数学试题答案及评分参考评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分参考制定相应的评分细则．2．对于计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分．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．选择题和填空题不给中间分．一、选择题：共10小题，每小题4分，满分40分；在每小题给出的四个选项中，只有一项是符合题目要求的，请在答题卡的相应位置填涂．1．A 2．C 3．A 4．B 5．B6．A 7．B 8．C 9．C 10．D二、填空题：共6小题，每小题4分，满分24分，请在答题卡的相应位置作答．11．112．14 13．15 14．4 15．18 16．94三、解答题：共9小题，满分86分，请在答题卡的相应位置作答．17．（本小题满分8分）解：解不等式①，得x ≤3． ······························································································ 3分解不等式②，得x ＞1 ． ···························································································· 5分∴原不等式组的解集是1 ＜x ≤3， ··············································································· 6分 将该不等式组解集在数轴上表示如下：······························································· 8分18．（本小题满分8分）证明：∵点E ，F 在BC 上，BE CF ，∴BE EF CF EF ，即BF CE ． ········································································································· 3分在△ABF 和△DCE 中，AB DC B C BF CE ，，， ∴△ABF ≌△DCE ， ······························································································· 6分∴∠A ∠D ． ······································································································· 8分12345-1-2-3 -4-5019．（本小题满分8分） 解：原式221(1)(1)(1)x x x x ······················································································· 3分 2(1)(1)111x x x x x ·························································································· 4分 221111x x x x ·································································································· 5分 21x ． ··········································································································· 6分当1x时，原式 ················································································· 7分． ····················································································· 8分20．（本小题满分8分）解：画法一：画法二：······························································· 4分如图，点C ，D 分别为（1），（2）所求作的点． ························································ 5分（2）证明如下：由（1）得BC ∥OA ，BC 12OA ， ∴∠DBC ∠DAO ，∠DCB ∠DOA ，∴△DBC ∽△DAO ， ············································································ 7分∴12DC BC DO AO ， ∴OD 2CD ． ····················································································· 8分21．（本小题满分8分）解：（1）由图1可得甲的速度是1202=60 m/min ． ································································ 2分由图2可知，当43x 时，甲，乙两人相遇， 故4(60)2003v 乙， 解得90v 乙m/min ． ···························································································· 4分答：甲的速度是60 m/min ，乙的速度是90 m/min ．（2）由图2可知：乙走完全程用了b min ，甲走完全程用了a min ，∴20020909b ， ······························································································· 6分 20010603a ． ································································································ 8分 ∴a 的值为103，b 的值为209．22．（本小题满分10分）解：（1）依题意得100a ． ······························································································ 2分这1000户家庭月均用水量的平均数为：2406100101801428018220221002660302014.721000x ， ········· 6分 ∴估计这1000户家庭月均用水量的平均数是14.72．（2）解法一：不合理．理由如下： ··············································································· 7分由（1）可得14.72在12≤x ＜16内，∴这1000户家庭中月均用水量小于16 t 的户数有40100180280600 （户）， ···························································· 8分∴这1000户家庭中月均用水量小于16 t 的家庭所占的百分比是600100%60%1000， ∴月均用水量不超过14.72 t 的户数小于60%． ············································· 9分∵该市政府希望70%的家庭的月均用水量不超过标准m ，而60%＜70%，∴用14.72作为标准m 不合理． ······························································· 10分解法二：不合理．理由如下： ··············································································· 7分∵该市政府希望70%的家庭的月均用水量不超过标准m ，∴数据中不超过m 的频数应为700， ·························································· 8分即有300户家庭的月均用水量超过m ．又2060100160300 ，2060100220380300 ，∴m 应在16≤x ＜20内． ·········································································· 9分而14.72＜16，∴用14.72作为标准m 不合理． ······························································· 10分23．（本小题满分10分）（1）证明：连接OD ，AD ．∵AB 为⊙O 直径，点D 在⊙O 上，∴∠ADB 90°，分∴∠ADC 90°． ∵E 是AC 的中点，∴DE ＝AE ，∴∠EAD ∠EDA ． ·分 ∵OA OD ，∴∠OAD ∠ODA ． ······················································································· 3分 ∵∠OAD ∠EAD ∠BAC 90°，∴∠ODA ∠EDA 90°，即∠ODE 90°， ···························································································· 4分 ∴OD ⊥DE ．∵D 是半径OD 的外端点，∴DE 是⊙O 的切线． ····················································································· 5分（2）解法一：过点F 作FH ⊥AB 于点H ，连接OF ， ∴∠AHF 90°．∵AB 为⊙O 直径，点F 在⊙O 上， ∴∠AFB 90°， ∴∠BAF ∠ABF 90°．∵∠BAC 90°，∴∠G ∠ABF 90°， ∴∠G ∠BAF ． ························································································· 6分 又∠AHF ∠GAB 90°，∴△AFH ∽△GBA ， ···················································································· 7分 ∴AF FH GB BA． ··························································································· 8分 由垂线段最短可得FH ≤OF ， ········································································ 9分 当且仅当点H ，O 重合时等号成立．∵AC ＜AB ， ∴ BD上存在点F 使得FO ⊥AB ，此时点H ，O 重合， ∴AF FH GB BA ≤12OF BA ， ············································································ 10分即AF GB 的最大值为12． 解法二：取GB 中点M ，连接AM ．∵∠BAG 90°， ∴AM 12GB ． ·分 ∵AB 为⊙O 直径，点F 在⊙O 上， ∴∠AFB 90°，∴∠AFG 90°，∴AF ⊥GB ．分 由垂线段最短可得AF ≤AM ， ········································································ 8分 当且仅当点F ，M 重合时等号成立，此时AF 垂直平分GB ，即AG ＝AB ．∵AC ＜AB ， ∴ BD上存在点F 使得F 为GB 中点， ∴AF ≤12GB ， ··························································································· 9分 ∴AF GB ≤12， ···························································································· 10分 即AF GB 的最大值为12．24．（本小题满分12分）（1）①证明：∵∠AED 45°，AE DE ，∴∠EDA 18045267.5°． ······································································· 1分 ∵AB AC ，∠BAC 90°，∴∠ACB ∠ABC 45°，∠DCA 22.5°， ························································· 2分 ∴∠DCB 22.5°，即∠DCA ∠DCB ，∴CD 平分∠ACB ． ····················································································· 3分②解：过点D 作DF ⊥BC 于点F ，∴∠DFB 90°． ∵∠BAC 90°，∴DA ⊥CA ． 又CD 平分∠ACB ， ∴AD FD ， ································································································· 4分 ∴AD FD DB DB． 在Rt △BFD 中，∠ABC 45°，∴sin ∠DBF FD DB ················································································ 5分 ∴AD DB ． ······························································································· 6分 （2）证法一：过点A 作AG ⊥AE 交CD 的延长线于点G ，连接BG ，∴∠GAE 90°．又∠BAC 90°，∠AED 45°，∴∠BAG ∠CAE ，∠AGE 45°，∠AEC 135°， ·············································· 7分 ∴∠AGE ∠AEG ，∴AG AE ． ······························································································· 8分 ∵AB AC ，∴△AGB ≌△AEC ， ···················································································· 9分 ∴∠AGB ∠AEC 135°，CE BG ，∴∠BGE 90°． ························································································ 10分 ∵AE ⊥BE ，F B A C D E。