五、解答题(本题共22分，第23题7分，第24题7分，第25题8分)
23. 抛物线y?x2?kx?3与x轴交于点A，B，与y轴交于点C，其中点B坐标为(1+k，0)． （1）求抛物线对应的函数表达式；
（2）将（1）中的抛物线沿对称轴向上平移，使其顶点M落在线
段BC上，记该抛物线为G，求抛物线G对应的函数表达式；
（3） 将线段BC平移得到线段B'C'（B的对应点记作B'，C的
对应点记作C'），使其经过（2）中所得使得抛物线G的顶点M，求点B'到直线OC'的距离h的取值范围．
24. 四边形ABCD是正方形，△BEF是等腰直角三角形，∠BEF=90°，BE=EF．G为DF的中点，连接EG，CG ，EC．
（1）如图1，若点E在CB边的延长线上，直接写出EG与GC的位置关系及
y321-3-2-1O-1-2-3-5-61234xEC的值； GC（2）将图1中的△BEF绕点B顺时针方向旋转至图2所示位置，在（1）中所得的结论是否仍然成立？若成立，请写出证明过程；若不成立，请说明理由；
（3）将图1中的△BEF，绕点B顺时针旋转?（0°
AGFDAGFD
ADEE
B图1
CB图2
C
B备用图
C
6
25. 定义1：在 △ABC中，若顶点A，B，C按逆时针方向排列，则规定它的面积为△ABC的“有向面积”；若△ABC的顶点A，B，C按顺时针方向排列，规定它的面积的相反数为
△ABC的“有向面积”. “有向面积”用S表示.例如图1中，S?ABC?S?ABC；图2中，S?ABC??S?ABC.
定义2：在平面内任取一个△ABC和点P(点P不在△ABC三边所在直线上),称有序数
组(S?PBC,S?PCA,S?PAB)为点P关于△ABC的“面积坐标”，记作P(S?PBC,S?PCA,S?PAB). 例如图３中，菱形ABCD的边长为2，∠ABC=60°，则S?ABC?3，点D关于
AD△ABC的“面积坐标”D(S?DBC,S?DCA,S?DAB)为 D(3,?3,3).
在图３中，我们知道，S?ABC?S?DBC?S?DAB?S?DCA，利用“有向面积”我们也可以把上式表示为S?ABC?S?DBC?S?DAB?S?DCA.
应用新知：
（1）如图4，正方形ABCD的边长为1，则S?ABC?_____，点D关于△ABC的“面积坐标”是D ( ____________)；
探究发现：
（2）如图4，在平面直角坐标系xOy中，点A(0,2)，B(?1,0).
①若点P是第二象限内任意点(不在直线AB上)，点P关于△ABO的“面积坐标”为
B图4
CP(m,n,k)，试探究m?n?k的值是否为定值？如果是，求出这个定值；如果不是，说明理
由；
②若点P(x,y)是第四象限内任意点，直接写出点P关于△ABO的“面积坐标”（用含x，y表示）； 解决问题：
（3）在（2）的条件下，点C(1,0)，D(0,1)，点Q在抛物线y?x2?2x?4上，求当
S?QAB?S?QCD 的值最小时，求点Q的横坐标.
yyAB-121AB121Ox-1O1x备用图 备用图
7
北京市西城区2014年初三一模试卷
数学试卷参考答案及评分标准
一、选择题（本题共32分，每小题4分） 题号 答案 1 A 2 B 3 B 4 C 5 C 6 B 7 D 2014. 4
8 D 二、填空题（本题共16分，每小题4分）
9 10 答案不唯一，如：11 12 （4，0） 2 B或F 2(a－1)2 1 x?215 三、解答题（本题共30分，每小题5分）
1-12-1）-27+2cos30+() 13．解：（
20o=1?33?2?3········································································································· 4分 ?2 ·
2=3?2······················································································································ 5分 3. ·
14. 证明：（1）∵BF＝CE，
∴ BF+ FC＝CE+CF，
即BC=EF. …………1分 在△ABC和△DEF中,
AD
?AB?DE,???B??E, ?BC?EF,?BFCE∴△ABC≌△DEF ························································································· 4分 ∴∠ACB＝∠DFE． ························································································· 5分
?3(x?1)?x?7,?15．解：?2x?1
≤x?1.??3由①得x?5． ········································································································ 2分
x?5． ············································································ 5分
由②得x??4． ················································································································· 4分 ∴ 原不等式组的解集是?4?16. 解：(x?1)(3x?1)?(x?2)=3x=3x22?4．
?3x?x?1?(x2?4x?4)?4． ············································································ 2分
?3x?x?1?x2?4x?4?4． ·············································································· 3分
2=2x
2?6x?9． ··················································································································· 4分
8
当x2?3x?1时，原式=2?1?9??7． ············································································ 5分
17．解：设乙班有学生x人，则甲班有学生（x+8）人．
依题意得 解得x12001200． ·
·················································································· 5分 ?1.2?xx?8?40． ·························································································································· 3分
?40是原方程的解，并且符合题意． ······························································· 4分
经检验，x （x+8）=48．
答: 甲班有学生48人，乙班有学生40人． ········································································ 5分
18. 解：（1）∵一次函数y?x?n和反比例函数
y??6的图象都经过点A(3,m)． xy∴m??6??2． ································ 1分 31∴点A的坐标为A（3,?2），. ························ 2分 ∴?2?3?n． ∴nO1xA??5．
y?x?5． ······················ 3分 ∴一次函数表达式为
（2）点B的坐标为(1,?6)或(6,?1)． ··········································································· 5分
四、解答题（本题共20分，每小题5分） 19．解：（1）∵CE∥AD且CE=AD，
∴四边形ADCE是平行四边形． ······················································································ 1分
又在△ABC中，AB=AC，AD平分∠BAC， ∴AD⊥BC． ∴∠ADC=90°．.
∴四边形ADCE是矩形． ································································································ 2分 （2）作 OH⊥CE于点H，
∵△ABC是边长为4的等边三角形， ∴∠ACB =60°,
11?DAC??BAC?30?,CD?BC?2．
221AC?2． 2由（1）知四边形ADCE是矩形， ∴AC与DE互相平分，AO=OC?BDOCHFA∴FC=OC=2． ························································ 3分 ∵在矩形ADCE中．∠AED=∠DCE=90°． ∴∠ACE=∠DCA=30°． 在Rt△COH中， ∴CHE1·············································································· 4分 OH?OC?1. ·
2?EH?3．
9
∴S四边形AOFE11··············· 5分 ?S?ACE?S?FOC?AE?CE?CF?OH?23?1． ·
2220．解：（1）1.6；补全条形统计图如图；…………2分
（2）2013； …………3分 （3）①C.②4.4． …………5分
（1）证明：连接AD．
∵AB是直径， ∴∠ACB=90°．
∴AD⊥BC． ∵AB=AC，
∴∠1=∠2． ∴∠1=∠ODA． ∴∠2=∠ODA．
∴OD∥AC． ···························································································································· 2分
(2)解: ∵EF是⊙O的切线，
可得∠ODE=90°．
由（1）知OD∥AC，∠1=∠2，∠ADB=90°． ∴∠AFE=∠ODE=90°，∠ADF=∠ABC．
A12OFBEDC5在Rt△ADB中，AB=10，cos?ABC?，
5∴
AD?45,BD?25 ,OD?5．
cos?ADF?cos?ABC?5， 5在Rt△AFD中，
可得
AF?8． ···························································································································· 4分
∵OD∥AC，
∴△EDO∽△EFA．
ODDE5DE?5?，可得?． AFDA8DE?1010∴DE=．
310∴DE长为 ································································································································· 5分
3由
10