昌平区2008—2009学年第二学期初三年级第一次统一练习

昌平区2008—2009学年第二学期初三年级第一次统一练习

数

考生须知

学 试 卷 2009．5

1．本试卷共5页，共九道大题，25个小题，满分120分。考试时间120分钟。 2．在试卷和答题卡上认真填写学校名称、姓名和考试编号。 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．考试结束，请将本试卷和答题卡一并交回。

一、选择题（共8道小题，每小题4分，共32分） 下列各题均有四个选项，其中只有一个是符合题意的． ．．1．?3的相反数是 A．?1 3B．

1 3C．?3 D．3

2．今年两会期间，新华网、人民网、央视网等各大网站都推出了“向总理提问”的网上互动话题，上百万网民给总理提出了内容广泛的问题．在新华网推出的“总理，请听我说”栏目中，网民所提出的问题就达200 000多条. 将200 000用科学记数法表示应为 A．0.2?10

6B．20?10

4C．2?10

4D．2?10

53．如图，在Rt?ABC中, ?C?90?，D是AC上一点，直线DE∥CB交AB于点E，若?A?30?，则?AED的度数为 DCAA．30? B．60? C．120? D．150?

E

4．把代数式a?2ab?b分解因式，下列结果中正确的是 A．?a?b?

222BB．?a?b?

2C．?a?b??a?b?

D．a?b

225．在下列所表示的不等式的解集中，不包括．．．?5的是

A．x??4 B．x??5 C．x??6 D．x??7

6．某校初三学生为备战5月份中考体育测试，分小组进行训练. 其中一个小组7名同学的一次训练的成绩（单位：分）为：18，27，30，27，24，28，25. 这组数据的众数和中位数分别是 A．27，30 B．27，25 C．27，27 D．25，30 7．把点A?1,2?、B??1,2?、C?1,?2?、D??1,?2?分别写在四张卡片上，随机抽取一张，该点在函数y??2x的图象上的概率是 A．

1 3 B．

1 2C．

2 3D．

3 4

8．将左图中的正方体纸盒沿所示的粗线剪开，其平面展开图的示意图为 ．．

纸 盒裁剪线A B C D

二、填空题（共4道小题，每小题4分，共16分） 9．在函数y?1中，自变量x的取值范围是 ． 1?x210．若?x?4??y?2?0，则x?y的值为 ．

B顺时针旋转，使得点

11．如图所示，把一个直角三角尺ACB绕着30°角的顶点A落在CB的延长线上的点E处，则∠BDC的度数为 ．

12．一组按规律排列的式子：

x3,?yx5,2yx7?,3yx9（,xy?0）， 其中第6个式子是 ，第n个式子是 （n为正整数）． ?4y

三、解答题（共5道小题，每小题5分，共25分） 13．计算：27?(1??)0?2sin60???2．

2214．已知x?1?0，求代数式x(x?x)?x(3x?1)?4的值 ．

3

15．解分式方程：

x6?2?1． x?1x?1

16．已知：如图，在矩形ABCD中，点E、F在 AD上，AE?DF，连接BE、CF． 求证：BE?CF．

AEFDBC

?2x?y?4,?x?m,k17．已知方程组?的解为? 又知点A?m,n?在双曲线y?上，求该双曲线的解析式． ?k?0?xx?y?5y?n.??

四、解答题（共2道小题，每小题5分，共10分）

18．如图，在梯形ABCD中，AD∥BC，?A?90?，AB?2AD?4，求BE的长．

19．如图，点A、B、F在?O上，?AFB?30?，OB的

AD?C?45?，E是CD的中点，

EBC延长线交直线AD于点接AB.

D，过点B作BC?AD于C，?CBD?60?，连（1）求证：AD是?O的切线； （2）若AB?6，求阴影部分的面积.

五、解答题（本题满分6分）

20．某校欲从甲、乙、丙三名候选人中挑选一名作为学生会主席，根据设定的录用程序，首先，随机抽取校内200名学生对三名候选人进行投票选举，要求每名学生最多推荐一人. 投票结果统计如下：

200名学生投票结果统计图 三名候选人得票情况统计图 8070 甲60 丙 25P % 40得票数30 乙弃权28 100甲乙丙 图1

图2

其次，对三名候选人进行了笔试和面试两项测试，成绩如下表所示：

测试项目 笔试 面试

甲 75 93

测试成绩（分）

乙 80 70

丙 90 68

请你根据以上信息解答下列问题： （1）补全图1和图2；

（2）若每名候选人得一票记1分，根据投票、笔试、面试三项得分按3:4:3的比例确定个人综合成绩，综合成绩高的被录用，请你分析谁将被录用.

六、解答题（共2道小题，21题5分，22题4分，共9分） 21．列方程或方程组解应用题：

为保证学生有足够的睡眠，政协委员于今年两会向大会提出一个议案，即“推迟中小学生早晨上课时间”，这个议案当即得到不少人大代表的支持. 根据北京市教委的要求，学生小强所在学校将学生到校时间推迟半小时. 小强原来7点从家出发乘坐公共汽车，7点20分到校；现在小强若由父母开车送其上学，7点45分出发，7点50分就到学校了. 已知小强乘自家车比乘公交车平均每小时快36千米，求从小强家到学校的路程是多少千米？

22．请阅读下列材料：

问题：如图1，点A,B在直线l的同侧，在直线l上找一点P，使得AP?BP的值最小.

小明的思路是：如图2，作点A关于直线l的对称点A?，连接A?B，则A?B与直线l的交点P即为所求.

BAlBAlPA'图1图2

请你参考小明同学的思路，探究并解决下列问题：

（1）如图3，在图2的基础上，设AA?与直线l的交点为C，过点B作BD?l，垂足为D. 若CP?1，PD?2，AC?1，写出AP?BP的值；

（2）将（1）中的条件“AC?1”去掉，换成“BD?4?AC”，其它条件不变，写出此时AP?BP的值； （3）请结合图形，直接写出

七、解答题（本题满分7分）

23．已知：关于x的一元二次方程kx?2x?2?k?0． （1）若原方程有实数根，求k的取值范围； （2）设原方程的两个实数根分别为x1，x2． ①当k取哪些整数时，x1，x2均为整数；

②利用图象，估算关于k的方程x1?x2?k?1?0的解．

八、解答题（本题满分7分）

22?2m?3?2?1??8?2m?2?4的最小值.

BAlCA'PD图3y43211234K-4-3-2-1O-1-2y-34-4324．在平面直角坐标系xOy中，抛物线y??x?bx?c与x轴21交于A、B两点（点A在点B的左侧），过点A12345x-3-2-1O-1-2-3

的直线y?kx?1交抛物线于点C?2,3?． （1）求直线AC及抛物线的解析式； （2）若直线y?kx?1与抛物线的对称轴交于 点E，以点E为中心将直线y?kx?1顺时针 旋转90?得到直线l，设直线l与y轴的交点

为P，求?APE的面积；

（3）若G为抛物线上一点，是否存在x轴上的

点F，使以B、E、F、G为顶点的四边形为平行四边形，若存在，直接写出点F的坐标；若不存在，请说明理由．

九、解答题（本题满分8分）

25.已知?AOB?90?，OM是?AOB的平分线．将一个直角RPS的直角顶点P在射线OM上移动，点P不与点O重合.

（1）如图，当直角RPS的两边分别与射线OA、OB交于点C、D时，请判断PC与PD的数量关系，并证明你的结论；

GD3的值； PD，求OD2（3）若直角RPS的一边与射线OB交于点D，另一边与直线OA、直线OB分别交于点C、E，且以P、D、E为顶点的三角形与?OCD相似，请画出示意图；当OD?1时，直接写出OP的长.

（2）如图，在（1）的条件下，设CD与OP的交点为点G，且PG?APRCGODMBS

昌平区2008—2009学年第二学期初三年级第一次统一练习

数学试卷答案及评分参考

一、选择题（共8道小题，每小题4分，共32分）

题号 答案

1 D

2 D

3 B

4 A

5 C

6 C

7 B

8 C

二、填空题（共4道小题，每小题4分，共16分）

题号

9

10

11

12

答案

x?1 2

15°

x13?6y,

(?1)n?1x2n?1 ny三、解答题（共5道小题，每小题5分，共25分） 13．解：27?(1??)0?2sin60???2

?33?1?2?3··················································································· 4分 ?2 ·

2································································································ 5分 ?43?1． ·

14．解：x(x?x)?x(3x?1)?4

=x?x?3x?x?4 ··················································································· 2分 =4x?4． ··································································································· 3分 当x?1?0时，x=1． ················································································ 4分

333323222

原式?4?1?4?8． ······················································································ 5分 15．解：分母因式分解，得

x6············································ 1分 ??1 ·

x?1?x?1??x?1?方程两边同乘?x?1??x?1?，得x?x?1??6??x?1??x?1? ·································· 3分 解得 x??5． ······························································································ 4分 经检验，x??5是原分式方程的解． ································································· 5分 16．证明：∵四边形ABCD是矩形， ??A??D?90?，

EFDAAB?DC．???????????????2分

在△AEB和△DFC中，

?AB?DC，? ??A??D，?AE?DF,?BC?△AEB≌△DFC．···················································································· 4分 ?BE?CF． ······························································································· 5分

17．解：解方程组??2x?y?4,?x?3,得? ????????????????2分

?x?y?5?y??2.············································································· 3分 ?点A的坐标为?3,?2?． ·∵点A(3,?2)在双曲线y?k上， x??2?k. 3解得k??6． ······························································································· 4分

6····································································· 5分 ?该双曲线的解析式为y??． ·x四、解答题（共2道小题，每小题5分，共10分）

18．解：如图，分别过点D、E作DF?BC于点F，EH?BC于点H． ?EH∥DF，

?DFB??DFC??EHB??EHC?90?．

又?A?90?，AD∥BC， ??ABC?90? .

?四边形ABFD是矩形． ∵AB?2AD?4,

ADE?AD?2．

?BF?AD?2，DF?AB?4． ·················· 1分

在Rt△DFC中，?C?45，

?BFHC?FC?DF?4． ························································································· 2分 又∵E是CD的中点，

1?EH?DF?2． ······················································································ 3分

2?HC?EH?2． ?FH?2．

?BH?4． ································································································· 4分

在Rt△EBH中，

······················································ 5分 ?BE?BH2?EH2?42?22?25． ·

19．（1）证明：如图，连结OA． ??AFB?30?, 点F在⊙O上， ??AOB?60?． ??CBD?60?, ??CBD??AOB．

?OA∥BC ．???????1分 又?BC?AD, ?OA?AD ． ∵点A在?O上，

··················································································· 2分 ?AD是?O的切线．·

（2）解：∵?AOB?60?，OA?OB，

??OAB是等边三角形． ∵AB?6，

?OA?AB?6．

在Rt△OAD中，?OAD?90，

??tan?AOD?AD， OA?AD?6?tan60??63．

1?S?OAD??63?6?183． ······································································ 3分

2?S扇形AOB60???62??6?， ········································································ 4分

360··················································································· 5分 ?S阴影?183?6?． ·

五、解答题（本题满分6分）

20．解：（1）图1中，丙得票所占的百分比为35%． ············································ 1分 补全图2见下图． ························································································ 2分 三名候选人得票情况统计图

80706050403020100得票数甲乙丙

（2）∵x甲?50?30%?75?40%?93?30%?72.9，

30%?40%?30v?30%?80?40%?70?30%x乙??75.8，

30%?40%?30p?30%?90?40%?68?30%x丙??77.4． ·················································· 5分

30%?40%?30%∴丙被录用． ································································································ 6分

六、解答题（共2道小题，21题5分，22题4分，共9分）

21．解：设小强乘公交车的平均速度是每小时x千米，则小强乘自家车的平均速度是每小时(x?36)千米． 1分 依题意，得

205x?(x?36)． ······································································ 2分 6060解得x?12． ································································································ 3分

?20?12?4． ···························································································· 4分 60答：从小强家到学校的路程是4千米． ······························································· 5分 22．解：（1）AP?BP的值为32． ································································ 2分 （2）AP?BP的值为5． ··············································································· 3分 （3）?2m?3??1?2···································· 4分 ?8?2m??4的最小值为34． ·

2七、解答题（本题满分7分）

23．解：（1）∵一元二次方程kx?2x?2?k?0有实数根，

2??k?0,??2 ·············································································· 1分

2?4?k?2?k?0.??????k?0,?? 2??4?k?1??0.2∴当k?0时，一元二次方程kx?2x?2?k?0有实数根． ·································· 2分

（2）①由求根公式，得x??1?(k?1)．

k?x1?k?22?1?，x2??1．???????3分 kk2要使x1，x2均为整数，必为整数，

ky24y =k321-4-3-2-1O-1-2-3-4y = k - 11234K?2时，x1，x2均为整数． 所以，当k取?1、??????????????5分

2，x2??1代入方程 k2x1?x2?k?1?0中，得?k?1．

k22设y1?，y2?k?1，并在同一平面直角坐标系中分别画出y1?与y2?k?1的图象（如图所示）． 6分

kk②将x1?1?由图象可得，关于k的方程x1?x2?k?1?0的解为k1??1，k2?2．??????7分 八、解答题（本题满分7分）

24．解：（1）∵点C?2,3?在直线y?kx?1上，

?2k?1?3．

解得k?1．

··································································· 1分 ?直线AC的解析式为y?x?1． ·∵点A在x轴上，

?A(?1，0)．

?抛物线y??x2?bx?c过点A、C， ??1?b?c?0， ????4?2b?c?3．?b?2，解得?

c?3．?···························································· 2分 ?抛物线的解析式为y??x2?2x?3． ·（2）由y??x?2x?3???x?1??4，

22ly4P21EC，B(3，0)． 可得抛物线的对称轴为x?1?E?1,2?．???????3分

-3-2AOD-1-2-32B45x

根据题意，知点A旋转到点B处，直线l过点B、E． 设直线l的解析式为y?mx?n．

将B、E的坐标代入y?mx?n中，联立可得m??1,n?3．

∴直线l的解析式为y??x?3． ······································································ 4分

?P?0,3?．

过点E作ED?x轴于点D.

?S1?APE?S?APB?S?2?AB?PO?12?AB?ED?1EAB?2?4??3?2??2． ················ 5分 （3）存在，点F的坐标分别为?3?2,0?、?3?2,0?、??1?6,0?、??1?6,0?．九、解答题（本题满分8分）

25．解：（1）PC与PD的数量关系是相等 ． ··················································· 1分 证明：过点P作PH?OA，PN?OB，垂足分别为点H、N． ∵?AOB?90?，易得?HPN?90?．

??1??CPN?90?，

A而?2??CPN?90?，

PM??1??2．

H1∵OM是?AOB的平分线，

RC2?PH?PN， G又??PHC??PND?90?，

3?△PCH≌△PDN．

ONDB?PC?PD． ····················································································S·········· 2分 （2）?PC?PD，?CPD?90?， ??3?45?， ??POD?45?， ??3??POD．

又??GPD??DPO， ?△POD∽△PDG． ··················································································· 3分

?GDOD?PGPD． 7分

∵PG?3PD， 2?GDPG3． ····················································································· 4分 ??ODPD2（3）如图1所示，若PR与射线OA相交，则OP?1； ········································ 6分 如图2所示，若PR与直线OA的交点C与点A在点O的两侧，则OP?2?1．

·················································································································· 8分

AMAPCGREO图1DMPOEDBSSBC

R图2

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