2014年北京市西城区初三数学一模试题及答案评分标准 doc

北京市西城区2014年初三一模

数 学 试 卷 2014. 4

学校 姓名 准考证号 考生须知 1．本试卷共6页，共五道大题，25道小题，满分120分。考试时间120分钟。 2．在试卷和答题卡上准确填写学校名称、姓名和准考证号。 3．试题答案一律填涂或书写在答题卡上，在试卷上作答无效。 4．在答题卡上，选择题、作图题用2B铅笔作答，其他试题用黑色字迹签字笔作答。 5．考试结束，将本试卷、答题卡和草稿纸一并交回。 一、选择题(本题共32分，每小题4分)

下面各题均有四个选项，其中只有一个是符合题意的． 1．?2的绝对值是 A．2

B．?2

C．

1 2 D．?

122．2014年3月5日，李克强总理在政府工作报告中指出：2013年全国城镇新增就业人数 约13 100 000人，创历史新高．将数字13 100 000用科学记数法表示为 A．13.1×106

B．1.31×107

B．

C．1.31×108

D．0.131×108

3．由5个相同的正方体搭成的几何体如图所示，则它的主视图是 A．

C．

4．从1到9这九个自然数中任取一个，是奇数的概率是 A．

D．

2 9 B．

4 9 C．

5 9 D．

2 3OACDB5．如图，表示一圆柱形输水管的横截面，阴影部分为有水部分，如果输水管的半径为5cm，水面宽AB为8cm，则水的最大深度CD为 A．4cm

B．3cm

C．2cm

D． 1cm

6．为了解某小区家庭使用垃圾袋的情况，小亮随机调查了该小区10户家庭

一周垃圾袋的使用量，结果如下：7，9，11，8，7，14，10，8，9，7（单位：个）．关于这组数据，下列结论正确的是 A．极差是8 A．m??1

B．众数是7

C．中位数是8 B．m?1

D．m??1且m?0

D．平均数是9

7．已知关于x的一元二次方程mx2?2x?1?0有两个不相等的实数根，则m的取值范围是 C．m?1且m?0

1

8．如图，在平面直角坐标系xOy中，点A（2，3）为顶点作一直角 ∠PAQ，使其两边分别与x轴、y轴的正半轴交于点P，Q．连接PQ，过点A作AH⊥PQ于点H．若点P的横坐标为x，AH的长为y，则下列图象中，能表示y与x的函数关系的图象大致是 yyyyyAQHOPx3 2O6.5x3 2O6.5x3 2O6.5x3 2O6.5x A B C D 二、填空题(本题共16分，每小题4分) 9．分解因式：2a2?4a?2=______．

10．写出一个只含字母x的分式，满足x的取值范围是x≠2，所

写的分式是： ．

11．如图，菱形ABCD中，∠DAB=60°，DF⊥AB于点E，且DF=DC，

连接FC，则∠ACF的度数为 度．

12. 如图，在平面直角坐标系xOy中，点A(1，0)，B(2，0)，将正六边形ABCDEF沿x轴正

方向无滑动滚动，当点D第一次落在x轴上时，D的横坐标为 ；在运动的过程中，点A的纵坐标的最大值是 ；保持上述运动过程，经过(2014,3)的正六边形的顶点 ．

yED1DAEFOBCyFC2B1EDCFA12BOA1xOx

三、解答题(本题共30分，每小题5分) 13．计算：(2?1)0?27?2cos30??()?1．

14．如图，点C，F在BE上，BF＝CE，AB＝DE，∠B＝∠E．

求证：∠ACB＝∠DFE．

AD12?3(x?1)?x?7,?15．解不等式组?2x?1

≤x?1.??316．已知x2?3x?1，求代数式(x?1)(3x?1)?(x?2)2?4的值．

2

BFCE17．列方程（组）解应用题：

某校甲、乙两班给贫困地区捐款购买图书，每班捐款总数均为1200元，已知甲比乙班多8

人，乙班人均捐款是甲班人均捐款多1.2倍，求甲、乙两班各有多少名学生．

18．在平面直角坐标系xOy中，一次函数y?x?n和反比例函数y??y6的图象都经过点A(3,m)． x1（1）求m的值和一次函数的表达式；

6（2）点B在双曲线y??上，且位于直线y?x?n的x下方，若点B的横、纵坐标都是整数，直接写出点B的坐标．

四、解答题(本题共20分，每小题5分)

O1x19．如图，在△ABC中，AB=AC，AD平分∠BAC，CE∥AD且CE=AD． B（1）求证：四边形ADCE是矩形；

（2）若△ABC是边长为4的等边三角形，对角线AC，DE相交于

点O，在CE上截取CF=CO，连接OF，求FC的长及四边形AOFE的面积．

20．以下是根据北京市统计局公布的2010—2013年北京市城镇居民人均可支配收入和农民

人均现金收入的数据绘制的统计图的一部分：

CFEDOA

根据以上信息，解答下列问题：

（1）2012年农民人均现金收入比2011年城镇居民人均可支配收入的一半少0.05万元，则

2012年农民人均现金收入是______万元，请根据以上信息补全条形统计图，并标明相应的数据（结果精确到0.1）；

（２）在2010—2013年这四年中，北京市城镇居民人均可支配收入和农民人均现金收入相

差数额最大的年份是______年；

（３）①2011—2013年城镇居民人均可支配收入的年平均增长率最接近______；

A．14%

3

B．11% C．10% D．9%

②若2014年城镇居民人均可支配收入按①中的平均增长率增长，请预测2014年的城镇居民人均可支配收入为_____万元（结果精确到0.1）．

21.如图，在△ABC中，AB=AC，以AB为直径作半圆⊙O，交BC于点D，连接OD，过点D作⊙O的切线，交AB延长线于点E，交AC于点F． A（1）求证：OD∥AC； （2）当AB=10，cos?ABC?

22.阅读下列材料：

问题：在平面直角坐标系xOy中，一张矩形纸片OBCD

5时，求AF及BE的长． 5BEOFDC按如图1所示放置，已知OB=10，BC=6．将这张纸片折叠，使点O落在边CD上，记作点A，折痕与边OD（含端点）交于点E，与边OB（含端点）或其延长线交于点F，求点A的坐标．

小明在解决这个问题时发现：要求点A的坐标，只要求出线段AD的长即可．连接

OA，设折痕EF所在直线对应的函数表达式为y?kx?b（k?0,n?0），于是有 E（0，n），F（?n，0）．所以在Rt△EOF中， 得到tan∠OFE=?k，在Rt△AOD中，利用等角的三k 1角函数值相等，就可以求出线段DA的长（如图1）． yDE1ACyDACyDC111O FBxOBxO1Bx 图1 请回答：

图2 备用图

（1）如图1，若点E的坐标为（0，4），直接写出点A的坐标；

（2）在图2中，已知点O落在边CD上的点A处，请画出折痕所在的直线EF（要求：尺

规作图，保留作图痕迹，不写作法）； 参考小明的做法，解决以下问题：

（3）将矩形沿直线y??x?n折叠，求点A的坐标；

（4）将矩形沿直线y?kx?n折叠，点F落在OB边上（含端点），直接写出k的取值范围．

tan∠OFE=?k= tan∠AOD， 12?k?OD6，DA??6k， ?DADA2?DA?6，2??6k?6

4

5

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

22．解：（1）点A的坐标（23，0）；……………… 1分

（2）如图； ………………2分

（3）EF垂直平分OA，

则∠AOD=∠OFE． ∴tan∠AOD =tan∠OFE=

yDE1AC1． 2O1BFx在Rt△AOD中，DA= OD tan∠AOD?3．

∴点A的坐标为?3，···························································································· 3分 6?； ·（4）?1?1····················································································································· 5分 k?? ·3五、解答题（本题共22分，第23题7分，第24题7分，第25题8分） 23．（1）解：∵(k?1)2∴ k∴

··············································································· 1分 ?k(k?1)?3?0， ·

?2.

····································································································· 2分 y?x2?2x?3． ·

（2）由（1），得B (3,0)，C (0,?3)，

∴经过B，C的直线：y·························································································· 3分 ?x?3． ·

设平移后的抛物线G的的顶点M的坐标为：(1,t)． ∵M在线段BC上， ∴t?1?3??2． ···················································································································· 4分

y?(x?1)2?2．

∴平移后的抛物线G为：即

·················································································································· 5分 y?x2?2x?1． ·

（3）作OE⊥BC于点E．

∴B?C??BC?32，OE∴S?OB?C?∴

132BC?． 22119?OC??h?B?C??OE?． 222?5?OC??17．

11

∴

91795?h?．··············································································································· 7分 175

24．（1）解：EG?GC，

EC····················································································· 2分 ?2； ·

GC （2）(1)中的结论仍然成立．

证明：取线段BF的中点M，连接EM，MG，

∵△BEF是等腰直角三角形， ∴EM?MB?FM?1． FB，且∠FME=90°2连接BD，取线段BD的中点N，连接GN，CN， ∵ABCD是正方形， ∴GN?BN?DN?1． BD，且∠CND=90°2∵G是DF的中点， ∴GN?1FB?EM，GN∥FB． 21BD?CN，MG∥BD． 2∴∠1=∠2． 同理MG?∴∠2=∠3． ∴GN?1FB?EM，GN∥FB． 2∴∠1=∠3．

∴∠EMG=∠EMF+∠1=∠CND+∠2=∠GNC．

∴△EMG≌△GNC． ·························································································· 4分 ∴EG=GC． ∴∠EGM=∠GCN．

在△CNG中，∠GNC+∠GCN+∠CGN=180°． ∴∠3+∠GCN+∠CGN=90°． ∴∠2+∠EGM+∠CGN=90°． 即EG⊥GC,

EC···························································································· 5分 ?2． ·

GCA12

（3）当E，F，D三点共线时，连接BD.

FGDE∵BE=1，AB=∴BF2，

?2，BD=2．

?BE1?． BD2在Rt△BED中，sin?EDB∴?EDB?30?． ∴DE?3．

···································································································· 6分 DF?DE?EF?3?1． ·∴?FBD??EFB??EDB∴?ABF?45??30??15?．

??ABD??FBD?45??15??30?．

∴tan?ABF?3． ················································································································· 7分 325．解：（1）

1111, (,?,) ········································································································· 2分 2222（2）答：m?n?k分两种情况：

?S?ABO．

(ⅰ) 当点P在△ABO的内部时，

m?n?k?S?PBO?S?POA?S?PAB?S?PBO?S?POA?S?PAB?S?ABO?S?ABO．

·········································································································································· 3分

(ⅱ) 当点P在△ABO的外部时， yPABP-1OyA2121B1x-1O1xm?n?k?S?PBO?S?POA?S?PAB?S?PBO?S?POA?S?PAB?S?ABO?S?ABO．

综上，m?n?k② P(·························································································· 4分 ?S?ABO. ·

11························································································ 5分 y,?x,1?x?y). ·

22（3）当x>0或x<0时，均有：

13

111y，S?QOA??x，S?QDO?x，S?QOC?y.

2221由S?QBO?S?QOA?S?QAB?S?ABO，得y?x?S?QAB?1.

2111由S?QDO?S?QOC?S?QCD?S?CDO，得x?y?S?QCD?.

222111113S?QAB?S?QCD?(y?x?1)?(x?y?)??x?y?

22222213??x?(x2?2x?4)?

2235?x2?x?

22331?(x?)2?.

41635可以验证当x=0时，y=4，S?QAB?S?QCD?1??，上式仍然成立.

2233∴当S?QAB?S?QCD的值最小时，x??，即点Q的横坐标为?. ······························ 5分

44S?ABO?

QADB-112yyQADB1C21

Ox-1O1Cx 14

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