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2009年梅州市初中毕业生学业考试
数 学 试 卷
说明:本试卷共 4 页,23 小题,满分 120 分.考试用时 90 分钟.
参考公式: 抛物线2y ax bx c =++的对称轴是直线2b x a
=-
, 顶点坐标是424b ac b a a 2??-- ???
,. 一、选择题:每小题 3分,共 15 分.每小题给出四个答案,其中只有一个是正确的. 1.1
2
-的倒数为( ) A .
12
B .2
C .2-
D .1-
2.下列图案是我国几家银行的标志,其中不是..轴对称图形的是( )
根据表中数据可知,全班同学答对的题数所组成的样本的中位数和众数分别是( ) A .8、8 B . 8、9 C .9、9 D .9、8 4.下列函数:①y x =-;②2y x =;③1y x
=-
;④2
y x =.当0x <时,y 随x 的增大而减小的函数有( )
A .1 个
B .2 个
C .3 个
D .4 个 5.一个正方体的表面展开图可以是下列图形中的( ) 二、填空题:每小题 3分,共 24 分. 6.计算:2
(
)
a
a
-÷= .
7.梅州是中国著名侨乡,
祖籍在梅州的华侨华人及港澳台同胞超过360万人,360万用科学计数法表示为
.
8.如图1,在O ⊙中,20ACB ∠=
°,则AOB ∠=_______度.
A .
B .
C .
D .
A .
B .
C .
D . 图1 图2
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9.如图2 所示,五角星的顶点是一个正五边形的五个顶点.这个五角星可以由一个基本图形(图中的阴影部分)绕中心O 至少经过____________次旋转而得到, 每一次旋转_______度.
10.小张和小李去练习射击,第一轮10发子弹打完后,两人的成绩如图3所示.根据图中的信息,小张和小李两人中成绩较稳定的是 .
11.已知一元二次方程2
2310x x --=的两根为12x x ,,则12x x = ___________.
12.如图4,把一个长方形纸片沿EF 折叠后,点D C 、分别落在11 D C 、的位置.若65EFB ∠=°,则1AED ∠等于_______度.
13. 如图5,每一幅图中有若干个大小不同的菱形,第1幅图中有1个,第2幅图中有3个,第3幅图中有5个,则第4幅图中有 个,第n 幅图中共有 个. 三、解答下列各题:本题有 10 小题,共 81 分.解答应写出文字说明、推理过程或演算步骤. 14.本题满分 7 分.
如图 6,已知线段AB ,分别以A B 、为圆心,大于1
2
AB 长为半径画弧,两弧相交
于点C 、Q ,连结CQ 与AB 相交于点D ,连结AC ,BC .那么: (1)∠ ADC =________度; (2)当线段460AB ACB =∠=,°时,ACD ∠= ______度, ABC △的面积等于_________(面积单位). 15.本题满分 7 分.
星期天,小明从家里出发到图书馆去看书,再回到家.他离家 的距离y (千米)与时间t (分钟)的关系如图7所示. 根据图象回答下列问题: (1)小明家离图书馆的距离是____________千米;
(2)小明在图书馆看书的时间为___________小时;
(3)小明去图书馆时的速度是______________千米/
小时.
16.本题满分
7 分. 计算:1
012)4cos30|3-?
?++- ???
°.
图
3 A
E D
C F B
D 1 C 1 图4
… … 第1幅 第2幅 第3幅 第n 幅 图5
C
B
D
A 图6
Q
(分)
图7
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17.本题满分 7 分.
求不等式组1184 1.
x x x x --??
+>-?≥,
的整数解.
18.本题满分 8 分.
先化简,再求值:2224441
x x x x x x x --+÷-+-,其中32x =.
19.本题满分 8 分.
如图 8,梯形ABCD 中,AB CD ∥,点F 在BC 上,连DF 与AB 的延长线交于点G . (1)求证:CDF BGF △∽△;
(2)当点F 是BC 的中点时,过F 作EF CD ∥交AD 于点E ,若6cm 4cm AB EF ==,,求CD 的长.
20.本题满分 8 分. “五·一”假期,梅河公司组织部分员工到A 、B 、C 三地旅游,公司购买前往各地的车票种类、数量绘制成条形统计图,如图9.根据统计图回答下列问题:
(1)前往 A 地的车票有_____张,前往C 地的车票占全部车票的________%;
(2)若公司决定采用随机抽取的方式把车票分配给 100 名员工,在看不到车票的条件下,每人抽取一张(所有车票的形状、大小、质地完全相同且充分洗匀),那么员工小王抽到去 B 地车票的概率为______; (3)若最后剩下一张车票时,员工小张、小李都想要,决定采用抛掷一枚各面分别标有数字1,2,3,4的正四面体骰子的方法来确定,具体规则是:“每人各抛掷一次,若小张掷得着地一面的数字比小李掷得着地一面的数字大,车票给小张,否则给小李.”试用“列表法或画树状图”的方法分析,这个规则对双方是否公平?
D C
F E A
G 图8
图9
地点
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21.本题满分 8 分.
如图10
,已知抛物线2y =x 轴的两个交点为A B 、,与y 轴交于点C . (1)求A B C ,,三点的坐标; (2)求证:ABC △是直角三角形;
(3)若坐标平面内的点M ,使得以点M 和三点 A B C 、、为顶点的四边形是平行四边形,求点M 的坐标.(直接写出点的坐标,不必写求解过程)
22.本题满分 10 分.
如图 11,矩形ABCD 中,53AB AD ==,.点E 是CD 上的动点,以AE 为直径的O ⊙与AB 交于点F ,过点F 作FG BE ⊥于点G .
(1)当E 是CD 的中点时:
①tan EAB ∠的值为______________;
② 证明:FG 是O ⊙的切线;
(2)试探究:BE 能否与O ⊙相切?若能,求出此时DE
23.本题满分 11 分.
如图 12,已知直线L 过点(01)A ,
和(10)B ,,P 是x 轴正半轴上的动点,OP 的垂直平分线交L 于点Q ,交x 轴于点M .
(1)直接写出直线L 的解析式;
(2)设OP t =,OPQ △的面积为S ,求S 关于t 的函数关系式;并求出当02t <<时,S 的最大值; (3)直线1L 过点A 且与x 轴平行,问在1L 上是否存在点C , 使得CPQ △是以Q 为直角顶点的等腰直角三角形?若存在,求出点C
L 1
x C B 图11
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2009年梅州市初中毕业生学业考试数学
参考答案及评分意见
一、选择题:每小题 3分,共 15 分.每小题给出四个答案,其中只有一个是正确的. 1.C 2.B 3.D 4.B 5.C 二、填空题:每小题 3分,共 24 分.
6.a 7.6
3.610? 8.40 9.4(1分),72(2分) 10.小张 11.1
2
-
12.50 13.7(1分),21n -(2分) 三、解答下列各题:本题有 10 小题,共 81 分.解答应写出文字说明、推理过程或演算步骤. 14.本题满分7分. (1)90 ···································································································································· 2分 (2)30 ··································································································································
·· 4分 ······························································································································ 7分 15.本题满分 7 分. (1)3 ····································································································································· 2分 (2)1 ····································································································································· 4分 (3)15 ····································································································································
7分 16.本题满分
7 分.
解:1
12)4cos30|3-??
++-
???
°.
134=++·································································································
·····
4分 43=+-················································································································ 6分 4= ······································································································································ 7分 17.本题满分 7 分.
解:由11x x --≥得1x ≥, ······························································································ 2分 由841x x +>-,得3x <. ······························································································ 4 分 所以不等式组的解为:13x <≤, ···················································································· 6 分 所以不等式组的整数解为:1,2. ······················································································· 7 分 18.本题满分 8 分.
解:2224441x x x x x x x --+÷-+-2
(2)(2)(1)
(2)1
x x x x x x x -+-=+÷-- ············································· 3分 2
12x x +=
+- 22
x x =- ··································································································································· 6分
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当32
x =时,原式3
226322
?
==--. ························································································ 8分
19.本题满分8 分.
(1)证明:∵梯形ABCD ,AB CD ∥, ∴CDF FGB DCF GBF ∠=∠∠=∠,, ······················ 2 分
∴CDF BGF △∽△. ···························· 3分
(2) 由(1)CDF BGF △∽△,
又F 是BC 的中点,BF FC = ∴CDF BGF △≌△, ∴DF FG CD BG ==, ················································ 6分
又∵EF CD ∥,AB CD ∥,
∴EF AG ∥,得2EF BG AB BG ==+. ∴22462BG EF AB =-=?-=, ∴2cm CD BG ==. ··········································································································· 8分 20.本题满分 8 分. 解:(1)30;20. ·············································································································· 2 分 (2)
1
2
. ···························································································································· 4 分
或画树状图如下:
共有 16 种可能的结果,且每种的可能性相同,其中小张获得车票的结果有6种: (2,1),(3,1),(3,2),(4,1),(4,2),(4,3), ∴小张获得车票的概率为63168P =
=;则小李获得车票的概率为35
188
-=. ∴这个规则对小张、小李双方不公平. ························································ 8 分
21.本题满分 8 分.
(1)解:令0x =,得y =(0C . ························································ 1分
D C F E
A B
G
19题图 1 2 3 4 1 1 2 3 4 2 1 2 3 4 3 1 2 3 4 4
开始
小张 小李
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令
0y =
,得20x =,解得121
3x x =-=,, ∴(10)(30)A B -,,,. ································································································
·· 3分 (2)法一:证明:因为22
214AC =+=, 2222
31216BC AB =+==,, ·
······················· 4分 ∴2
2
2
AB AC BC =+, ··············································· 5分 ∴ABC △是直角三角形. ·····································
····· 6分 法二:因为13OC OA OB ==,,
∴2OC OA OB = , ··············································································································· 4分 ∴
OC OB OA OC
=,又AOC COB ∠=∠, ∴Rt Rt AOC COB △∽△. ································································································ 5分 ∴90ACO OBC OCB OBC ∠=∠∠+∠=,°, ∴90ACO OCB ∠+∠=°,
∴90ACB ∠=°, 即ABC △是直角三角形. ····················································
· 6 分
(3
)1(4M
,2(4M -,3(2M .(只写出一个给1分,写出2个,得1.5分) 8分 22.本题满分 10 分.
(1)①
65
····································································· 2分
②法一:在矩形ABCD 中,AD BC =,
ADE BCE ∠=∠,又CE DE =, ∴ADE BCE △≌△, ················································ 3分
得AE BE EAB EBA =∠=∠,,
连OF ,则OF OA =, ∴OAF OFA ∠=∠, OFA EBA ∠=∠, ∴OF EB ∥, ·················································································· 4 分 ∵FG BE ⊥, ∴FG OF ⊥, ∴FG 是O ⊙的切线 ································································································· 6分 (法二:提示:连EF DF ,,证四边形DFBE 是平行四边形.参照法一给分.) (2)法一:若BE 能与O ⊙相切, ∵AE 是O ⊙的直径, ∴AE BE ⊥,则90DEA BEC ∠+∠=°,
又90EBC BEC ∠+∠=°, ∴DEA EBC ∠=∠,
∴Rt Rt ADE ECB △∽△, ∴AD DE EC BC =,设DE x =,则53EC x AD BC =-==,,得353x
x =-, 整理得2
590x x -+=. ······································································································· 8 分 ∵2
42536110b ac -=-=-<, ∴该方程无实数根.
22题图
x
21题图
M 1 3
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∴点E 不存在,BE 不能与O ⊙相切. ·········································· 10分 法二: 若BE 能与O ⊙相切,因AE 是O ⊙的直径,则90AE BE AEB ∠=⊥,°,
设DE x =,则5EC x =-,由勾股定理得:222
AE EB AB +=,
即22(9)[(5)9]25x x ++-+=, 整理得2
590x x -+=, ······································· 8分 ∵2
42536110b ac -=-=-<, ∴该方程无实数根.
∴点E 不存在,BE 不能与O ⊙相切. ·········································· 10分 (法三:本题可以通过判断以AB 为直径的圆与DC 是否有交点来求解,参照前一解法给分) 23.本题满分 11 分.
(1)1y x =- ························································································································ 2分 (2)∵OP t =,∴Q 点的横坐标为1
2
t , ①当1012t <
<,即02t <<时,112
QM t =-, ∴11122OPQ S t t ??
=
- ???
△. ·
····································································································· 3分 ②当2t ≥时,11
1122
QM t t =-
=-, ∴11122OPQ S t t ??
=
- ???
△. ∴1110222111 2.
22
t t t S t t t ???
-<< ?????=????- ?????,
,,≥ ······························································································ 4分
当1012t <
<,即02t <<时,211111(1)2244
S t t t ??=-=--+ ???, ∴当1t =时,S 有最大值
1
4
. ······························································································ 6分 (3)由1OA OB ==,所以OAB △是等腰直角三角形,若在1L 上存在点C ,使得CPQ △是以Q 为直角顶点的等腰直角三角形,则PQ QC =,所以OQ QC =,又1L x ∥轴,则C ,O 两点关于直线L
对称,所以1AC OA ==,得(11)C ,. ··············································································· 7 分
下证90PQC ∠=°.连CB ,则四边形OACB 是正方形.
法一:(i )当点P 在线段OB 上,Q 在线段AB 上
L 1
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(Q 与B C 、不重合)时,如图–1.
由对称性,得BCQ QOP QPO QOP ∠=∠∠=∠,, ∴ 180QPB QCB QPB QPO ∠+∠=∠+∠=°,
∴ 360()90PQC QPB QCB PBC ∠=-∠+∠+∠=°°. ················································· 8分 (ii )当点P 在线段OB 的延长线上,Q 在线段AB 上时,如图–2,如图–3
∵12QPB QCB ∠=∠∠=∠,, ∴90PQC PBC ∠=∠=°. ·························· 9分 (iii )当点Q 与点B 重合时,显然90PQC ∠=°. 综合(i )(ii )(iii ),90PQC ∠=°.
∴在1L 上存在点(11)C ,,使得CPQ △是以Q 为直角顶点的等腰直角三角形. ············ 11 分
法二:由1OA OB ==,所以OAB △是等腰直角三角形,若在1L 上存在点C ,使得CPQ △是以Q 为直角顶点的等腰直角三角形,则PQ QC =,所以OQ QC =,又1L x ∥轴, 则C ,O 两点关于直线L
对称,所以1AC OA ==,得(11)C ,. ············································································· 7 分
延长MQ 与1L 交于点N .
(i )如图–4,当点Q 在线段AB 上(Q 与A B 、不重合)时, ∵四边形OACB 是正方形,
∴四边形OMNA 和四边形MNCB 都是矩形,AQN △和QBM △都是等腰直角三角形. ∴90NC MB MQ NQ AN OM QNC QMB ====∠=∠=,,°. 又∵OM MP =, ∴MP QN =, ∴QNC QMP △≌△,
23题图-3
L 1
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∴MPQ NQC ∠=∠, 又∵90MQP MPQ ∠+∠=°, ∴90MQP NQC ∠+∠=°.
∴90CQP ∠=°. ············································································································ 8分 (ii )当点Q 与点B 重合时,显然90PQC ∠=°. ·············································· 9分 (iii )Q 在线段AB 的延长线上时,如图–5, ∵BCQ MPQ ∠=∠,∠1=∠2 ∴90CQP CBM ∠=∠=°
综合(i )(ii )(iii ),90PQC ∠=°.
∴在1L 上存在点(11)C ,,使得CPQ △是以Q 为直角顶点的等腰直角三角形. ········ 11分
法三:由1OA OB ==,所以OAB △是等腰直角三角形,若在1L 上存在点C ,使得CPQ △是以Q 为直角顶点的等腰直角三角形,则PQ QC =,所以OQ QC =,又1L x ∥轴,
则C ,O 两点关于直线L 对称,所以1AC OA ==,得(11)C ,. ························ 9分
连PC ,∵|1|PB t =-,12OM t =
,12
t
MQ =-, ∴2222
2
(1)122PC PB BC t t t =+=-+=-+,
2
2
2
22222
11222t t t OQ OP CQ OM MQ t ????===+=+-=-+ ? ?????
.
∴222
PC OP QC =+,∴90CQP ∠=°. ······································································· 10分
∴在1L 上存在点(11)C ,,使得CPQ △是以Q 为直角顶点的等腰直角三角形. ··········· 11分
L 1
23题图-5
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