根据对称性可得，在BD下方还存在一个点P'也满足条件．
15(5?1). ······································ 14分 2(注：只求出一个点P并计算正确的扣1分.)
∴存在两个点P，到OD的距离都是
2013年龙岩市初中毕业、升学考试 参 考 答 案 及 评 分 标 准
数 学
说明：评分最小单位为1分，若学生解答与本参考答案不同，参照给分．
（第25题图）
一、选择题（本大题共10题，每题4分，共40分）
题号 答案 1 A 2 C 3 D 4 D 5 B 6 C 7 A 8 C 9 B 10 B 二、填空题（本大题共7题，每题3分，共21分．注：答案不正确、不完整均不给分） 11．a(a?2)
12．9 13．8
14．3
15．70? 16．①④
a2?b2 17．．
ab三、解答题（本大题共8题，共89分）
18．(10分，第（1）小题5分，第（2）小题5分)
（1）解：原式=2?1?(?1)?2?3 ····························································· 4分 = 2?3 ············································································· 5分 （2）解：方程两边同乘（2x+1），得 4=x+2x+1 ················································ 2分 3=3x
x=1 ························································································· 3分 检验：把x=1代入2x+1=3≠0 ······················································· 4分 ∴原分式方程的解为x=1. ···························································· 5分 19．（8分）解：原式=
x(2x?3)(2x?3)1?? ········································ 4分 2x?332x?3
x ·············································································· 6分 32当x=2时，原式=. ·································································· 8分
3=
20．(10分)
(1)证明：（法一）如图：∵四边形ABCD是平行四边形
∴AD=BC，AD∥BC，∠3=∠4 ····················· 1分
∵∠1=∠3+∠5, ∠2=∠4+∠6 ························································ 2分 ∠1=∠2
∴∠5=∠6 ················································································· 3分 ∴△ADE≌△CBF ···································································· 5分 ∴AE=CF ················································································ 6分 （法二）如图：连接BD交AC于点O ··············· 1分 在平行四边形ABCD中
OA=OC，OB=OD ·································· 2分 ∵∠1=∠2,∠7=∠8
∴△BOF≌△DOE ······································································· 4分 ∴OE=OF ················································································ 5分 ∴OA－OE=OC－OF
即AE=CF. ················································································· 6分
(2) )证明：（法一）∵∠1=∠2,
∴DE∥BF ············································································· 7分
∵△ADE≌△CBF
∴DE=BF ·············································································· 9分 ∴四边形EBFD是平行四边形. ················································ 10分
（法二）∵OE=OF，OB=OD ······························································· 9分
∴四边形EBFD是平行四边形. ··············································· 10分
其他证法，请参照标准给分.
21．（10分，第（1）小题4分，第（2）小题3分，第（3）小题3分） （1） 0.11 ， 540 ； （注：每空2分） （2）36?；
（3）9000． 22．（12分，每小题4分）
（1）6 ······························································································ 4分 （2）3?1··························································································· 8分 2（3）∵∠C＝90?，BC=3，EC＝1 ∴tan∠BEC＝
BC=3 CE∴∠BEC＝60? ······················································································ 9分 由翻折可知：∠DEA＝45? ······································································· 10分 ∴?AEA??75?＝?D?ED?? ····································································· 11分 ∴l ?755?3 ································································· 12分 ?2??3?3601223．（12分，第（1）小题5分，第（2）小题7分）
解：（1）设需租赁甲、乙两种设备分别为x、y天． ········································ 1分
x?7y?80?12则依题意得 ? ··················································· 3分
10x?1y0?100??x?2解得 ? ·········································································· 4分
y?8?答：需租赁甲种设备2天、乙种设备8天. ··················································· 5分 （2）设租赁甲种设备a天、乙种设备(10－a)天，总费用为?元． ······················ 6分
依题意得
?a?5?10?a?7???12a?7(10?a)?80??10a?10(10?a)?100
∴3≤a≤5． ∵a为整数，
∴a＝3、4、5． ·················································································· 8分 方法一：
∴共有三种方案．
方案（1）甲3天、乙7天，总费用400×3+300×7＝3300; ·············· 9分
[来源学科网ZXXK]
方案（2）甲4天、乙6天，总费用400×4+300×6＝3400; ················· 10分 方案（3）甲5天、乙5天，总费用400×5+300×5＝3500. ················ 11分 ∵3300<3400<3500 ∴方案（1）最省，最省费用为3300元． ····· 12分
方法二：
则?＝400a+300(10－a)＝100a+3000 ··········································· 10分 ∵100＞0，
∴?随a的增大而增大．
∴当a＝3时，?最小＝3300． ······················································ 11分
答：共有3种租赁方案：①甲3天、乙7天；②甲4天、乙6天；③甲5天、乙5天.最少租赁费用3300元． ················································································ 12分
方法三：能用穷举法把各种方案枚举出来，并得出三种符合条件的方案，求出最省费用的，参照标准酌情给分．
24.（1）设F（x，y），(x＞0，y＞0) ．
则OC=x, CF=y ···················································································· 1分
1∴S?OCF?xy?3． ··········································································· 2分
2∴xy=23．
∴k=23． ························································································ 3分 ∴反比例函数解析式为y=23(x＞0) ． ············································· 4分x[来源学科网ZXXK]
（2）该圆与y轴相离． ·········································································· 5分
理由：过点E作EH⊥x轴，垂足为H，过点E作EG⊥y轴，垂足为G．