2012年福建省南平市初中毕业、升学考试
数学试题参考答案及评分说明
说明：
（1） 解答右端所注分数，表示考生正确作完该步应得的累计分数，全卷满分150分． （2） 对于解答题，评卷时要坚持每题评阅到底，勿因考生解答中出现错误而中断本
题的评阅．当考生的解答在某一步出现错误时，如果后续部分的解答未改变该题的考试要求，可酌情给分，但原则上不超过后面应得分数的一半，如果有较严重的错误，就不给分．
（3） 如果考生的解法与本参考答案不同，可参照本参考答案的评分标准相应评分． （4） 评分只给整数分．
一、选择题（本大题共10小题，每小题4分，共40分）
1．C； 2．A； 3．D； 4．C； 5．B； 6．D； 7．C； 8．A； 9．B； 10．B． 二、填空题（本大题共8小题，每小题3分，共24分） 11．2； 15．y?2x＋1；
12．6； 16．6.8；
13．2(x?1)2； 17．
14．22；
1（或0.5）； 18．③ ④． 2三、解答题（本大题共8小题，共86分） 19．（1）解：原式＝9?3?(??4)?1 ············································································· 4分 ＝27?4???1 ···················································································· 6分 ＝30?? ······························································································· 7分 （2）由 ① 得 2x＜7＋1 ························································································· 2分 x＜4 ································································································ 3分 由 ② 得 3x?2x?8 ······················································································ 5分 x＜8 ································································································ 6分 ∴不等式组的解集为 x＜4 ··············································································· 7分
20．解法一：原方程化为(x?3)(x?3)?(6x?x)?0 ··················································· 4分
∴x?9?6x?x?0 ························································································ 6分
2223 ··········································································································· 7分 23经检验，x＝是原分式方程的解．
23∴原方程的解是x＝ ························································································· 8分
22解法二：原方程化为x(x?3)?3(x?3)?(6x?x)?0 ·········································· 4分
解得 x＝
（以下与解法一相同）
数学试题参考答案及评分说明 第1页 (共6页)
21．情形一：选择添加的条件是BE=DF ······················· 2分
证法一：∵四边形ABCD是平行四边形
∴AD=BC ，AD∥BC ············································ 4分 ∵BE=DF，
∴AD－DF=BC－BE 即 AF=CE ·························· 6分 ∴四边形AECF是平行四边形 ································· 8分 证法二：∵四边形ABCD是平行四边形， ∴AB=CD，∠B=∠D 又∵BE=DF， ∴△ABE≌△CDF
∴AE=CF ······················································································································· 4分 又∵AD=BC，
∴AD－DF=BC－BE 即AF=CE ··············································································· 6分 ∴四边形AECF是平行四边形 ····················································································· 8分
情形二：选择添加的条件是∠AEB=∠CFD ································································· 2分 证法一：∵四边形ABCD是平行四边形，
∴AD∥BC ···················································································································· 4分 ∴∠AEB=∠EAF············································································································ 5分 又∵∠AEB=∠CFD，
∴∠EAF=∠CFD ············································································································ 6分 ∴AE∥CF ······················································································································ 7分 又∵AF∥EC，
∴四边形AECF是平行四边形 ····················································································· 8分 证法二：∵∠AEB=∠CFD，
∴180°－∠AEB=180°－∠CFD，即∠AEC=∠CFA ······················································ 4分 又∵四边形ABCD是平行四边形， ∴AB=CD，∠B=∠D ∴△ABE≌△CDF
∴∠BAE=∠DCF ········································································································· 6分 又∵∠BAD=∠DCB，
∴∠BAD－∠BAE=∠DCB－∠DCF，即∠EAF=∠FCE ·········································· 7分 ∴四边形AECF是平行四边形 ················································································ 8分
B E
C
A F D
数学试题参考答案及评分说明 第2页 (共6页)
22．解：（1）
儿童玩具30%童车
25%
童装45 %
类 别
儿童玩具
90
抽查件数
75 135 童车
童装
（补全统计图表每空2分，共8分） ································································· 8分 （2）
90?90%?75?88%?135?80%?0.85
300答：从该超市这三大类儿童用品中随机购买一件能买到合格品的概率是0.85 ·························································································································· 10分
数学试题参考答案及评分说明 第3页 (共6页)
23．（1）解法一：∵OA⊥CD，∴∠OHD=90° ································································ 1分
?O?在Rt△OHD中，∠O=60°，OD=2，∴sinsin?DOHHD ····································· 2分 OD?DOHO ·∴HD?OD?sin?···································3分
=2sin60??3 ··········································4分 ∴CD?2HD?23 ····································5分 解法二：∵OA⊥CD，∴∠OHD=90° ··········1分 在Rt△OHD中，
∵∠O=60°，∴∠ODH=30° ···························2分 ∴OH=
C H A D l B O 1OD＝1 ···········································3分 2∴HD?OD2?OH2?······················································· 4分 22?12?3 ·
∴CD?2HD?23 ···························································································· 5分 （2）证法一：∵OA=OD，∠O=60°，
∴△AOD为等边三角形
∴∠OAD=∠ODA=60° ·························································································· 6分 ∵AD=BD，∴∠DAB=∠B ·················································································· 7分 ∵∠ODA=∠DAB+∠B，
11∠ODA=×60°=30° ········································································ 8分 22∴∠OAB=∠OAD+∠DAB=60°+30°=90° ······························································ 9分 ∴∠DAB=
即OA⊥AB，∴AB是⊙O的切线 ······································································ 10分 证法二：∵OA=OD，∠O=60°，∴△AOD为等边三角形
∴OD= AD ············································································································ 6分 又∵AD=BD，∴OD=BD ····················································································· 7分 ∵DH⊥OA，∴OH=AH，∠OHD=90° ································································ 8分 ∴HD∥AB，∴∠OAB=∠OHD=90° ··································································· 9分 即OA⊥AB，∴AB是⊙O的切线 ······································································ 10分 证法三：∵OA=OD，∠O=60°，∴△AOD为等边三角形
∴∠O=∠DAO ······································································································· 6分 ∵AD=BD，∴∠B=∠DAB ·················································································· 7分 ∵∠OAB+∠O+∠B=180°，
∴∠OAB+∠DAO+∠DAB=180° ········································································ 8分 ∴2∠OAB=180°，∴∠OAB=90° ······································································· 9分 即OA⊥AB，∴AB是⊙O的切线 ··································································· 10分
数学试题参考答案及评分说明 第4页 (共6页)
24．解：（1）3.02x ·········································································································· 2分 2? y ? x
?(3x.?04 ····························································································· 4分
（2）依题意，得y?3.02x?(?3.04x?3040) ····························································· 6分
??0.02x?3040 ··········································································· 7分
（3）由题意可知，当y?3029时，3029??0.02x?3040 ········································ 8分
解得 x?550 ···································································································· 9分
1000?x?1000?550?450
答：该乡镇有小学生550人，初中生450人 ·················································· 10分
25．解：（1）A（m，0），A′（0，m），C′（－1，0） ·················································· 3分 （2）依题意，得 c＝m ······························································································· 4分
?am2?bm?m?0① ∴?
?a?b?m?0② 由①得 m(am?b?1)?0，
∵m＞0，∴am?b?1?0 ③ ······································································ 5分 由②＋③得 a(m?1)?m?1?0，∴a??1 ···················································· 6分 ∴b?m?1 ·········································································································· 7分 ∴这个抛物线的解析式为y??x2?(m?1)x?m ··············································· 8分 （3）答：可能 ············································································································· 9分
∵点D与点B(m,1)关于点O对称，
∴点D的坐标为(?m,?1) ··························· 10分 假设点D(?m,?1)在抛物线y??x2?(m?1)x?m上， 则有?1??(?m)2?(m?1)?(?m)?m ∴2m2?2m?1?0
1?3解得m? ········································ 11分 2B' y A' C B C' D O A x ∵m＞0，故m?1?3舍去，
2∴当m?1?3时，点B关于点O的对称点D落在抛物线上 ······················· 12分 2
数学试题参考答案及评分说明 第5页 (共6页)
26．（1）答：如AB=AC；∠BAD=∠CDE；∠ADB=∠DEC；∠ADC=∠AED；
ABADBDAEADDE△ABD∽△DCE；△ADE∽△ACD；；等 ??;??DCDECEADACCD（每写对一个结论得1分，共3分） ································································· 3分 （2）① ∵∠B=∠C=45°，∴AB=AC，∠BAC=90° ∵BC=2，∴AB=AC=2 ··········· 4分 解法一：∵∠1+∠EDC=∠B+∠DAB，∠1=∠B A ∴∠EDC=∠DAB，∴△ABD∽△DCE ····················6分
ABBDAD∴ ·······················································7分 ?E CEDCDC1 即BD?DC?CE?AB．设CE?y,BD?x，
B C D 22有x(2?x)?2y，即y??·······8分 x?2x ·
2222＜0，∴当x?1时，y最大值?，∴CE的最大值为 ····················· 9分 222 解法二：∵∠1=∠C，∠DAE=∠CAD，∴△ADE∽△ACD ···································· 6分
ADAC∴ ······································································································· 7分 ?AEAD2AD2 ∴AD2?AE?AC?(AC?CE)?AC?2?2CE ∴CE?2?········· 8分 2∴当AD最小时，CE最大． 由垂线段最短，可知AD⊥BC，∵AB=AC，∴D为BC的中点
2211?1?∵∠BAC=90°，∴AD?BC??2?1，∴CE?2? 22222即CE的最大值为 ······································ 9分
2A
② 分三种情形加以讨论：
1）如图，当AE=DE时，则∠DAE=∠1=45°
E ∵∠BAC=90°，∴AD平分∠BAC ······················ 10分
∵AB=AC，∴D为BC的中点 1 1∴BD=BC=1 ·················································· 11分 B C D
22）当AD=DE时， 解法一：∵∠1+∠EDC=∠B+∠DAB，∴∠EDC=∠DAB 又∵∠B=∠C，∴△ABD≌△DCE ······································································· 12分
∵?∴AB=DC=2，∴BD=BC－DC=2－2 ························································· 13分 解法二：∵∠1=∠C，∠DAE=∠CAD，∴△ADE∽△ACD ··································· 12分 ∴当AD=DE时，DC=AC＝2，∴BD=BC－DC=2－2 ······························· 13分 3）当AD=AE时，则∠AED=∠1=45°，∠DAE=90° ∴此时点D与B重合，与题意不符，应舍去
综上所述，若△ADE是等腰三角形，则BD的长为1或2－2 ·························· 14分
数学试题参考答案及评分说明 第6页 (共6页)