2016-2017学年北京市平谷区九年级一模数学试卷及答案

平谷区2017年初三统一练习（一）

一、选择题(本题共30分，每小题3分)

1．为解决“最后一公里”的交通接驳问题，平谷区投放了大量公租自行车供市民使用．据统计，目前我区共有公租自行车3 500辆．将3 500用科学记数法表示应为 A．0.35×104 B． 3.5×103

C．3.5×102 D． 35×102

2．把一个边长为1的正方形如图所示放在数轴上，以正方形的对角线为半径画弧交数轴于点A，则点A对应的数是 AA．1 B．2 C．3 D．2

3．右图是某几何体从不同角度看到的图形，这个几何体是

A．圆锥

B．圆柱 D．三棱锥

主视图 左视图 俯视图

-10123C．正三棱柱

2x2y?4．如果x+y=4，那么代数式2的值是

x?y2x2?y2A．﹣2 B．2 C．

11 D．? 225．下列图形中，既是轴对称图形，又是中心对称图形的是

A． B． C． D．

6．某商场一楼与二楼之间的手扶电梯如图所示．其中AB、CD分别表示一楼、二楼地面的水平线，∠ABC=150°，BC的长是8 m，则乘电梯从点B到点C上升的高度h是

A．43m B．8 m

D．4 m

A 150° B C h D 83m C．3

7．在我国古代数学著作《九章算术》中记载了一道有趣的数学问题：“今有凫（凫：野鸭）起南海，七日至北海；雁起北海，九日至南海.今凫雁俱起，问何日相逢？”意思是：野鸭从南海起飞，7天飞到北海；大雁从北海起飞，9天飞到南海．野鸭与大雁从南海和北海同时起飞，经过几天相遇．设野鸭与大雁从南海和北海同时起飞，经过x天相遇，根据题意，下面所列方程正确的是

A．(9?7)x?1 D. (B.(9?7)x?1

C.

11?)x?1 7911(?)x?179

8．如图，是利用平面直角坐标系画出的天安门广场的平面示意图，若

这个坐标系分别以正东、 正北方向为x轴、y轴的正方向， 表示国旗杆的点的坐标为(0，2.5)， 表示中国国家博物馆的点的坐标为(4，1)， 则表示下列建筑的点的坐标正确的是 A．天安门(0， 4)

B．人民大会堂(﹣4，1) C．毛主席纪念堂(﹣1，﹣3) D．正阳门(0，﹣5)

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9．1-7月份，某种蔬菜每斤的进价与每斤的售价的信息如图所示，则出售该种蔬菜每斤利润最大的月份是

A．3月份 B．4月份 C．5月份 D．6月份

10．AQI是空气质量指数（Air Quality Index）的简称，是描述空气质量状况的指数．其数值越大说明空气污染状况越严重，对人体的健康危害也就越大．AQI共分六级，空气污染指数为0－50一级优，51－100二级良，101－150三级轻度污染，151－200四级中度污染，201－300五级重度污染，大于300六级严重污染．小明查阅了2015年和2016年某市全年的AQI指数，并绘制了如下统计图，并得出以下结论：①2016年重度污染的天数比2015年有所减少；②2016年空气质量优良的天数比2015年有所增加；③ 2015年和2016年AQI指数的中位数都集中在51－100这一档中；④2016年中度污染的天数比2015年多13天．以上结论正确的是

A． ①③ B． ①④ C．②③ D．②④ 二、填空题(本题共18分，每小题3分)

bx?311．如果分式的值为0，那么x的值是 ．

x?112．如图，一个正方形被分成两个正方形和两个一模一样的矩形，请根据图形，写出一个含有a，b的正确的等式 ．

13．请写出一个在各自象限内，y的值随x值的增大而增大的反比例函数表达式 ．

14．一个猜想是否正确，科学家们要经过反复的论证．下表是几位科学家“掷硬币”的实验数据： 实验者 掷币次数 出现“正面朝上”的次数 频率 德·摩根 6 140 3 109 0.506 蒲丰 4 040 2 048 0.507 费勒 10 000 4 979 0.498 皮尔逊 36 000 18 031 0.501 80 640 39 699 0.492 aab罗曼诺夫斯基 请根据以上数据，估计硬币出现“正面朝上”的概率为 （精确到0.01）．

15．如图，圆桌面正上方的灯泡发出的光线照射桌面后，在地面上形成阴影（圆形）．已知灯泡距离地面2.4m，桌面距离地面0.8m（桌面厚度不计算），若桌面的面积是1.2m2，则地面上的阴影面积是 m2．

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16．小米是一个爱动脑筋的孩子，他用如下方法作∠AOB的角平分线： 作法：如图，

（1）在射线OA上任取一点C，过点C作CD∥OB； A（2）以点C为圆心，CO的长为半径作弧，交CD于点E； （3）作射线OE．

所以射线OE就是∠AOB的角平分线． CED请回答：小米的作图依据是____________________________ ____________________________________________________．

三、解答题（本题共72分，第17-26题，每小题5分，第27

OB分，第28题7分，第29题8分）解答应写出文字说明、演算步骤或证明过程. 17．计算：1?3?12?2cos30??20170．

?3x?2?x, 18．解不等式组???2x?1x?1并写出它的所有非负整数解．．．．．． ?5?2,

19．如图，在矩形ABCD中，点E是BC上一点，且DE=DA，AF⊥DE于F，求证：AF=CD． AD F BEC

20．已知关于x的一元二次方程x2-(m+2)x+2m=0．

（1）求证：方程总有两个实数根； （2）当m=2时，求方程的两个根．

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题7

21．在平面直角坐标xOy中，直线y?kx?1?k?0?与双曲线y?与x轴交于点B．

(1) 求m的值和点B的坐标；

m?m?0?的一个交点为A（﹣2，3），x(2) 点P在y轴上，点P到直线y?kx?1?k?0?的距离为2，直接写出点P的坐标．

y A

B

xO

22．随着人们“节能环保，绿色出行”意识的增强，越来越多的人喜欢骑自行车出行．某自行车厂生产的某型号自行车去年销售总额为8万元．今年该型号自行车每辆售价预计比去年降低200元．若该型号车的销售数量与去年相同，那么今年的销售总额将比去年减少10%，求该型号自行车去年每辆售价多少元？

23．如图，在△ABC中，BD平分∠ABC交AC于D，EF垂直平分BD，分别交AB，BC，BD于E，F，G，连接DE，DF．

（1）求证：DE=DF； （2）若∠ABC=30°，∠C=45°，DE=4，求CF的长．

A

DE

G

BC F

24．阅读以下材料：

2017年1月28日至2月1日农历正月初一至初五，平谷区政府在占地面积6万平方米的琴湖公园举办主题为“逛平谷庙会乐百姓生活”的平谷区首届春节庙会．

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本次庙会共设置了文艺展演区、非遗展示互动区、特色商品区、儿童娱乐游艺区、特色美食区等五个不同主题的展区．展区总面积1720平方米．文艺展演区占地面积600平方米，占展区总面积的34.9％；非遗展示区占地190平方米，占展区总面积的11.0％；特色商品区占地面积是文艺展演区的一半，占展区总面积的17.4％；特色美食区占地200平方米，占展区总面积的11.6％；还有孩子们喜爱的儿童娱乐游艺区．

此次庙会本着弘扬、挖掘、展示平谷春节及民俗文化，以京津冀不同地域的特色文化为出发点，全面展示平谷风土人情及津冀人文特色．大年初一，来自全国各地的约3.2万人踏着新春的脚步，揭开了首届平谷庙会的帷幕．大年初二尽管天气寒冷，市民逛庙会热情不减，又约有4.3万人次参观了庙会，品尝特色美食，观看绿都古韵、秧歌表演、天桥绝活，一路猜灯谜、赏图片展，场面火爆．琳琅满目的泥塑、木版画、剪纸、年画等民俗作品也让游客爱不释手，纷纷购买．大年初三，单日接待游客约4万人次，大年初四风和日丽的天气让庙会进入游园高峰，单日接待量较前日增长了约50%．大年初五，活动进入尾声，但庙会现场仍然人头攒动，仍约有5.5万人次来园参观． （1）直接写出扇形统计图中m的值；

（2）初四这天，庙会接待游客量约_______万人次；

（3）请用统计图或统计表，将庙会期间每日接待游客的人数表示出来．

25．如图，⊙O为等腰三角形ABC的外接圆，AB=AC，AD是⊙O的直径，切线DE与AC的延长线相交于点E．

（1）求证：DE∥BC；

CE（2）若DF=n，∠BAC=2α，写出求CE长的思路．

DAF O B

26．有这样一个问题：探究函数y??x+2?x的图象与性质．?

小军根据学习函数的经验， 对函数y??x+2?x的图象与性质进行了探究．? 下面是小军的探究过程， 请补充完整：

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（1）函数y??x+2?x的自变量x的取值范围是 ；

（2）下表是 y与x的几组对应值 ? x ﹣2﹣1.9 ﹣1.5﹣10 1 2 3 4 … ﹣0.5 y 2 0 ﹣0.72﹣1.41 ﹣0.37 0 0.76 1.55 … 1.60 0.80 在平面直角坐标系xOy中， 描出了以上表中各对对应值为坐标的点，根据描出的点， 画出该函数的图象；

y4321O–2–1–1–2–312345x（3）观察图象，函数的最小值是 ； （4）进一步探究，结合函数的图象， 写出该函数的一条性质（函数最小值除．．外）： ．

27．直线y??3x?3与x轴，y轴分别交于A，B两点，点A关于直线x??1的对称点为点C． （1）求点C的坐标；

（2）若抛物线y?mx?nx?3m?m?0?经过A，B，C三点，

2

54y求该抛物线的表达式；

（3）若抛物线y?ax?bx?3?a?0? 经过A，B两点，且顶点

232在第二象限，抛物线与线段AC有两个公共点，求a的取值范围． 1 –5–4–3–2–1O12x–1

–2

28．在△ABC中，AB=AC，∠A=60°，点D是BC边的中点，作射线DE，与边AB交于点E，射线DE绕点D顺时针旋转120°，与直线AC交于点F． （1）依题意将图1补全；

（2）小华通过观察、实验提出猜想：在点E运动的过程中，始终有DE=DF．小华把这个猜想与同学们

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进行交流，通过讨论，形成了证明该猜想的几种想法：

想法1：由点D是BC边的中点，通过构造一边的平行线，利用全等三角形，可证DE=DF； 想法2：利用等边三角形的对称性，作点E关于线段AD的对称点P，由∠BAC与∠EDF互补，可得∠AED与∠AFD互补，由等角对等边，可证DE=DF；

想法3：由等腰三角形三线合一，可得AD是∠BAC的角平分线，由角平分线定理，构造点D到AB，AC的高，利用全等三角形，可证DE=DF…….

请你参考上面的想法，帮助小华证明DE=DF（选一种方法即可）； （3）在点E运动的过程中，直接写出BE，CF，AB之间的数量关系．

A E BDC 图1

AEBDC备用图 7

29．在平面直角坐标系中，点Q为坐标系上任意一点，某图形上的所有点在∠Q的内部（含角的边），这时我们把∠Q的最小角叫做该图形的视角．如图1，矩形ABCD，作射线OA，OB，则称∠AOB为矩形ABCD的视角．

图1

图2

备用图

（1）如图1，矩形ABCD，A（﹣3，1），B（3，1），C（3，3），D（﹣3，3），直接写出视角∠AOB的度数；

（2）在（1）的条件下，在射线CB上有一点Q，使得矩形ABCD的视角∠AQB=60°，求点Q的坐标； （3）如图2，⊙P的半径为1，点P（1，3），点Q在x轴上，且⊙P的视角∠EQF的度数大于60°，若Q（a，0），求a的取值范围．

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平谷区2016—2017学年度初三统练（一）

数学答案2017.4

一、选择题(本题共30分，每小题3分) 题号 答案 1 B 2 B 3 A 4 C 5 D 6 D 7 C 8 B 9 A 10 C 二、填空题（本题共18分，每小题3分） 11．3；12．?a?b??a2?2ab?b2； 13．答案不唯一，如y??21；14．0.50；15．2.7； x16．两直线平行，内错角相等； ············································································· 1

等腰三角形两底角相等； ················································································ 3 （其他正确依据也可以）．

三、解答题（本题共72分，第17-26题，每小题5分，第27题7分，第28题7分，第29题8分）解答应写出文字说明、演算步骤或证明过程. 17．解：1?3?12?2cos30??2017

0=3?1?23?2?3········································································ 4 ?1 ·

2=﹣2． ···································································································· 5

① ?3x?2?x ?18．解：?2x?1x?1，

②?5?2 ?············································································· 1 解不等式①得x≤1， ·········································································· 2 解不等式②得x＞﹣3， ·

····························································· 3 ∴不等式组的解集是：﹣3＜x≤1． ······························································ 5 ∴不等式组的非负整数解为0，1． ·

19．证明：∵矩形ABCD，

AD∴AD∥BC．

···································· 1 ∴∠ADE=∠DEC． ·

∵AF⊥DE于F，

F·································· 2 ∴∠AFD=∠C=90°． ·

BEC·············································· 3 ∵DE=DA， ·

···································· 4 ∴△ADF≌△DEC．

·············································· 5 ∴AF=CD． ·

20．（1）证明: ∵ Δ=[-(m+2)]2-4×2m ······································································ 1

2

=(m-2) ? ∵ (m-2)2≥0，

∴方程总有两个实数根． ································································ 2

2

························································· 3 （2）当m=2时，原方程变为x-4x+4=0． ·

······················································································· 5 解得x1=x2=2． ·

21．解：（1）∵双曲线y?m?m?0?经过点，A（﹣2，3）， x9

∴m??6．···················································································· 1 ∵直线y?kx?1?k?0?经过点A（﹣2，3），

∴k??1． ···················································································· 2 ∴此直线与x轴交点B的坐标为（1，0）. ··········································· 3 （2）(0，3)，(0，-1)． ············································································· 5

22．解：设去年该型号自行车每辆售价x元，则今年每辆售价为（x﹣200）元． ·············· 1

由题意，得

8000080000?1?10%?···························································· 2 ， ?xx?200···················································································· 3 解得：x=2000． ·

······························································ 4 经检验，x=2000是原方程的根． ·

·············································· 5 答：去年该型号自行车每辆售价为2000元． ·

23．（1）证明：∵EF垂直平分BD，

∴EB=ED，FB=FD． ··············································································· 1 ∵BD平分∠ABC交AC于D， A∴∠ABD=∠CBD．

DE∵∠ABD+∠BEG=90°，∠CBD+∠BFG=90°，

G∴∠BEG=∠BFG．

∴BE=BF． BCFH∴四边形BFDE是菱形． ∴DE=DF． ···························································································· 2 （2）解：过D作DH⊥CF于H． ∵四边形BFDE是菱形， ∴DF∥AB，DE=DF=4．

在Rt△DFH中，∠DFC=∠ABC=30°， ∴DH=2．

∴FH=23． ························································································· 3 在Rt△CDH中，∠C=45°， ∴DH=HC=2． ························································································ 4 ∴CF=2+23． ······················································································ 5

24．（1）扇形统计图中m的值是25.1%； ······························································· 1

（2）6； ····································································································· 2 （3）如图． ································································································ 5

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25．（1）证明：∵AB=AC，AD是⊙O的直径，

∴AD⊥BC于F． ···················································································· 1 ∵DE是⊙O的切线， ∴DE⊥AD于D．2 ∴DE∥BC． ··························································································· 2 （2）连结CD．

由AB=AC，∠BAC=2α，可知∠BAD=α． ··················································· 3 由同弧所对的圆周角，可知∠BCD=∠BAD=α． 由AD⊥BC，∠BCD =α，DF=n， CE根据sinα=

DF，可知CD的长． ················ 4 CDA由勾股定理，可知CF的长

由DE∥BC，可知∠CDE=∠BCD． 由AD是⊙O的直径，可知∠ACD=90°． 由∠CDE=∠BCD，∠ECD=∠CFD， 可知△CDF∽△DEC，可知

OFBDDFCF=，可求CE的长． ····························· 5 CECD26．（1）x??2； ······················································································· ····· 1

（2）该函数的图象如图所示； ········································································ 3?

y4321O–2–1–1–212345x （3）-2； ·························································································· ····· 4 （4）该函数的其它性质:当?2?x?0时，y随x的增大而减小； ····················· 5

（答案不唯一，符合函数性质即可写出一条即可）

27．解：（1）令y=0，得x=1．

∴点A的坐标为（1,0）． ·································································· 1

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–3∵点A关于直线x=﹣1对称点为点C， ∴点C的坐标为（﹣3,0）． ·················· 2 （2）令x=0，得y=3．

∴点B的坐标为（0,3）． ∵抛物线经过点B，

∴﹣3m=3，解得m=﹣1． ····················· 3 ∵抛物线经过点A，

∴m+n﹣3m=0，解得n=﹣2．

∴抛物线表达式为y??x?2x?3． ········ 4

254321yBC–4–3–2–1AO–1–212x（3）由题意可知，a<0．

根据抛物线的对称性，当抛物线经过（﹣1,0）时，开口最小，a=﹣3， ········· 5 此时抛物线顶点在y轴上，不符合题意.

当抛物线经过（﹣3,0）时，开口最大，a=﹣1. ········································· 6

结合函数图像可知，a的取值范围为?3?a??1. ····································· 7

28．解：（1）如图1， ························································································ 1

AEBDFC图1

（2）

A

AAGEBDFCBEDPFCEMBDNFC图2 图3

图4

想法1证明：如图2，过D作DG∥AB，交AC于G， ···································· 2

∵点D是BC边的中点， ∴DG=

1AB． 2∴△CDG是等边三角形． ∴∠EDB+∠EDG=120°． ∵∠FDG+∠EDG=120°， ∴∠EDB =∠FDG． ················································································· 3 ∵BD=DG，∠B=∠FGD=60°， ∴△BDE≌△GDF．················································································· 4 ∴DE=DF． ···························································································· 5 想法2证明：如图3，连接AD， ∵点D是BC边的中点， ∴AD是△ABC的对称轴．

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作点E关于线段AD的对称点P，点P在边AC上， ······································· 2 ∴△ADE≌△ADP．

∴DE=DP，∠AED=∠APD． ∵∠BAC+∠EDF=180°， ∴∠AED+∠AFD=180°． ∵∠APD+∠DPF=180°， ∴∠AFD=∠DPF． ·················································································· 3 ∴DP=DF． ···························································································· 4 ∴DE=DF． ···························································································· 5 想法3证明：如图4，连接AD，过D作DM⊥AB于M，DN⊥AB于N， ············ 2 ∵点D是BC边的中点， ∴AD平分∠BAC．

∵DM⊥AB于M，DN⊥AB于N， ∴DM=DN． ··························································································· 3 ∵∠A=60°，

∴∠MDE+∠EDN=120°． ∵∠FDN+∠EDN=120°， ∴∠MDE=∠FDN．

∴Rt△MDE≌Rt△NDF． ············································································ 4 ∴DE=DF． ···························································································· 5 （3）当点F在AC边上时，BE?CF?1AB； ··········································· 6 21当点F在AC延长线上时，BE?CF?AB． ········································ 7

2

29．解：（1）120°； ··························································································· 1

（2）连结AC，在射线CB上截取CQ=CA，连结AQ． ··································· 2

∵AB=23，BC=2，

∴AC=4．··························· 3

∴∠ACQ=60°．

∴△ACQ为等边三角形， 即∠AQC=60°． ················ 4 ∵CQ=AC=4，

∴Q（3，﹣1）． ············· 5

（3）

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图1 图2

如图1，当点Q与点O重合时，∠EQF=60°， ∴Q（0，0）． ··············································································· 6 如图2，当FQ⊥x轴时，∠EQF=60°， ∴Q（2，0）． ··············································································· 7 ∴a的取值范围是0＜a＜2． ······························································ 8

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