茂名市 2009 年初中毕业生学业考试 与高中阶段学校招生考试 数学试题参考答案及评分标准
说明:1.如果考生的解与本解法不同,可根据试题的主要内容,并参照评分标准制定相应 的评分细则后评卷. 2.解答题右端所注的分数,表示考生正确做到这一步应得的累加分数. 一、选择题(本大题共 10 小题,每小题 4 分,共 40 分. ) 题号 答案 1 B 2 C 3 D 4 D 5 B 6 A 7 A 8 C 9 B 10 D
二、填空题(本大题共 5 小题,每小题 4 分,共 20 分. ) 11.1 12.
1 2
13.2
14.60
15.110
三、 (本大题共 3 小题,每小题 8 分,共 24 分. ) 16. (1)解:原式 1 2 8 ······························ 2 分 ··········· ·········· ········· ·········· ··········· ·········
4 . ··········· ··········· ·········· ··· 分 ··········· ·········· ··········· ·· 4 ·········· ··········· ··········· ··
(2)解:由① -② 得: y 3 , ····························· 分 ··········· ·········· ········ ·········· ··········· ······· 2 ∴ y 3 代入① 把 得: x 2 , ································ 分 ··········· ·········· ·········· 3 ·········· ··········· ··········
∴ 方程组的解为
x 2, ··········· ··········· ·········· ··· 分 ··········· ·········· ··········· ·· 4 ·········· ··········· ··········· ·· y 3.
17.解: (1)列表(或树状图)得:
a
b
1 2 3 4
1 (1,1) (1,2) (1,3) (1,4)
2 (2,1) (2,2) (2,3) (2,4)
3 (3,1) (3,2) (3,3) (3,4)
4 (4,1) (4,2) (4,3) (4,4)
因此,点 A(a,b) 的个数共有 16 个; ···························· 分 ··········· ·········· ······ 4 ·········· ··········· ······ (2)若点 A 在 y x 上,则 a b , 由(1)得 P(a b )
4 1 , 16 4 1 . ···················8 分 ··········· ········ ·········· ········ 4
因此,点 A(a,b) 在函数 y x 图象上的概率为 18.解:如图所示:每画对一个 3 分,共 6 分.
△ ABC 与 △A1B1C1 不一定全等. ····························· 8 分 ··········· ·········· ········ ·········· ··········· ·······
·
B1
B
A1
C1 B1
A
C A1 C1
四、 (本大题共 2 小题,每小题 8 分,共 16 分. ) 19、解: (1)九年级捐书数为:1000× 30%× 4=1200(本) ····················· 1 分 ··········· ·········· ·········· ·········· · 八年级捐书数为:1000× 35%× = 2100(本)························ 分 6 ··········· ·········· ·· 2 ·········· ··········· ·· 七年级捐书数为:1000× 35%× =700(本)························· 分 2 ··········· ·········· ··· 3 ·········· ··········· ··· ∴ 捐书总本数为:1200+2100+700=4000(本)······················ 分 ··········· ·········· · ·········· ··········· 5 因此,该校学生捐图书的总本数为 4000 本. ························ 分 ··········· ·········· ·· 6 ·········· ··········· ·· (2)4000÷ 1000=4(本) ··································· 分 ··········· ·········· ··········· ·· 7 ·········· ··········· ··········· ·· 因此,该校平均每人捐图书 4 本. ······························ 分 ··········· ·········· ········ 8 ·········· ··········· ········ 20.解:∵ 方程有实数根,∴b 4ac ≥ 0 ,∴(4)2 4(k 1) ≥ 0 ,即 k ≤ 3 . ···· 分 ··· 2 ···
2
解法一:又∵x
4 (4)2 4(k 1) ··········· ······· ·········· ········ 2 3 k , ··········· ······· 3 分 2
∴x1 x2 (2 3 k ) (2 3 k ) 4 , ························ 分 ··········· ·········· ·· 4 ·········· ··········· ·· ··········· ·········· ··· 5 ·········· ··········· ··· x1 x2 (2 3 k ) 3 k ) k 1 ························· 分 (2 若 x1 x2 x1 x2 ,即 k 1 4 ,∴k 3 .························ 7 分 ··········· ·········· ··· ·········· ··········· ··· 而这与 k ≤ 3 相矛盾,因此,不存在实数 k ,使得 x1 x2 x1 x2 成立. ········· 分 ········ 8 ········ 解法二:又∵x1 x2
x1 x2
c k 1 k 1 , ··········· ··········· ·········· · 5 分 ··········· ·········· ·······
···· · ·········· ··········· ··········· · a 1
b 4 4 , ··········· ··········· ···· 分 ··········· ·········· ····· ·········· ··········· ···· 4 a 1
(以下同解法一) 五、 (本大题共 3 小题,每小题 10 分,共 30 分. ) 21.解: (1)依题意得: y1 (2100 800 200)x 1100 x , ··················3 分 ··········· ······· ·········· ········
y2 ( 2 4 0 0 1 1 0 0 1 0 ) x 0
·· 0 0 0 ····· ···· ···· 0 2 0 0 0 0 2 0 , ··2············6 分 x 1 0 ·· ········ ····
(2)设该月生产甲种塑料 x 吨,则乙种塑料 (700 x) 吨,总利润为 W 元,依题意得:
W 1 1 0 0 1 2 0 0 ( 7 0 0 ) x x
∵
2 0 0 0 0x 1 0 0 . ···0········· 分 8 ·· 0 0 0 ···· · 7 ··· ······· ·· · 2 ····
x ≤ 400, 解得: 300 ≤ x ≤ 400 . ······················· 8 分 ··········· ·········· ·· ·········· ··········· ·· 700 x ≤ 400,
∵100 0 ,∴ 随着 x 的增大而减小,∴ x 300 时,W 最大=790000(元) ···· 9 分 W 当 . ···· ···· 此时, 700 x 400 (吨) . 因此,生产甲、乙塑料分别为 300 吨和 400 吨时总利润最大,最大利润为 790000 元. ···········10 分 ··········· ·········· 22.证明:
(1)连接 BM ,∵B、C 把 OA 三等分,∴1 5 60° , ·············· 1 分 ··········· ··· ·········· ····
1 5 30° , ························2 分 ··········· ·········· ··· ·········· ··········· ··· 2 1 又∵ 为 ⊙M 直径,∴ABO 90° ,∴AB OA OM , 3 60° ,····· 分 OA ···· 3 ···· 2 ∴1 3 , DOM ABO 90° , ·························· 分 ··········· ·········· ···· 4 ·········· ··········· ····
又∵OM BM ,∴2
1 3, 在 △OMD 和 △BAO 中, OM AB, ··········· ··········· 分 ··········· ·········· · ·········· ··········· 5 DOM ABO.
∴△OMD ≌△BAO (ASA) ······························· 6 分 ··········· ·········· ·········· ·········· ··········· ·········· (2)若直线 l 把 ⊙M 的面积分为二等份, y 则直线 l 必过圆心 M , ··············· 7
分 ··········· ···· ·········· ·····
3) ∵D(0, , 1 60°,
D 0, 3
4 C
OD 3 3, ∴OM tan 60° 3
∴M ( 3, , ···················· 8 分 ·········· ·········· 0) ···················· 把 M ( 3, 代入 y kx b 得: 0) ··········· ········ ·········· ········· 3k b 0 . ··················· 10 分 O
B
2 1 M
5
3 A x
23.解: (1)设 2006 年底至 2008 年底手机用户的数量年平均增长率为 x ,依题意得: ····· 分 ···· 1 ···· ··········· ·········· ··········· ······ ·········· ··········· ··········· ······ 50(1 x)2 72 , ······································ 3 分 ∴1 x 1.2 ,∴x1 0.2 , x2 2.2 (不合题意,舍去) ·············· 分 , ··········· ··· ·········· ··· 4 ∴ 2006 年底至 2008 年底手机用户的数量年平均增长率为 20%. ············· 分 ··········· ·· ·········· ·· 5 (2)设每年新增手机用户的数量为 y 万部,依题意得:·················· 分 ··········· ······ 6 ·········· ·······
[72(1 5%) y](1 5%) y ≥103.98 , ·························8 分 ··········· ·········· ··· ·········· ··········· ···
0.95 y ≥103.98, 0.95 0.95 y y ≥103.98 , 68.4 即 (68.4 y) 64.98 1.95 y ≥103.98 , 1.95 y ≥ 39 ,∴ y ≥ 20 (万部) ·············· 分 .··············9 ·········· ····
∴ 每年新增手机用户数量至少要 20 万部. ························ 分 ··········· ·········· ··· ······················· 10 六、 (本大题共 2 小题,每小题 10 分,共 20 分. ) 24、解: (1)当△ABC 与△DAP 相似时,∠ APD 的度数是 60° 30° ········· 2 分 或 . ········· ········· (2)设 PC x ,∵PD ∥ BA , BAC 90° ,∴PDC 90° ,··········· 分 ·········· 3 ··········
cos cos 又∵C 60° ,∴AC 24 60° 12 , CD x 60°
∴ AD 12
1 x, 2
1 3 x ,而 PD x 60° ··········· ·········· 4 ·········· ··········· sin x , ··········· ··········· 分 2 2
∴S△ APD
1 1 3 1 ··········· ·········· ··· ·········· ··········· ·· PDAD x 12 x
························5 分 2 2 2 2
3 2 3 ( x 24 x) ( x 12) 2 18 3 . 8 8
∴ PC 等于 12 时, △ APD 的面积最大,最大面积是 18 3 . ··············· 分 ··········· ··· 6 ·········· ···· (3)设以 BP 和 AC 为直径的圆心分别为 O1 、 O2 ,过 O2 作 O2 E ⊥ BC 于点 E ,
cos 设 ⊙O1 的半径为 x ,则 BP 2 x .显然, AC 12 ,∴O2C 6 ,∴CE 6 60° 3 ,
∴O2 E
62 32 3 3 ,
A
··········· ·········· O1E 24 3 x 21 x , ···········7 分 又∵⊙O1 和 ⊙O2 外切, ∴O1O2 x 6 . ················· 8 分 ··········· ······ ·········· ······· 在 Rt△O1O2 E 中,有 O1O O2 E O1E ,
2 2 2 2
D O2 60° C E
B
O1
P
∴( x 6)2 (21 x)2 (3 3)2 , ········9 分 ········ ······· 解得: x 8 , ∴BP 2 x 16 . ···························· 10 分 ··········· ·········· ······· ·········· ··········· ·······
25.解: (1)∵M 0, 在 y (2)由(1)得: y
1 4
1 1 1 1 x b 上,∴ 0 b ,∴b .········· 分 ········ 2 ········ 3 4 3 4
1 1 x , ∵B1 (1 y1 ) 在 l 上, , 3 4
∴ x 1 时, y1 当
1 1 7 7 1 ,∴B1 1, .················ 分 ··········· ····· ·········· ····· 3 3 4 12 12
2
解法一:∴ 设抛物线表达式为: y a (x 1)
2
7 (a 0) ,················ 分 ··········· ···· 4 ·········· ····· 12
7 7 ,∴a , ······· 分 ······· ······ 5 12 12(d 1) 2
又∵x1 d , ∴A (d, ,∴0 a (d 1) 0) 1
∴ 经过点 A、B1、A2 的抛物线的解析式为: y 1 解法二:∵x1 d ,∴ A (d, , A2 (2 d, , 0) 0) 1
7 7 ····· ···· ( x 1) 2 . ·····6 分 2 12(d 1) 12
( ∴ y a(x d ) x 2 d )(a 0) , ··························· 分 设 ··········· ·········· ······ ·········· ··········· ····· 4
把 B1 1 , 代入:
7 12
7 7 a(1 d )(1 2 d ) ,得 a , ·········· 5 分 ·········· ·········· 12 12(d 1) 2
∴ 抛物线的解析式为 y
7 ( x d ) x 2 d ) . ················6 分 ( ··········· ····· ·········· ····· 12(d 1)2
(3)存在美丽抛物线. ·········
·························· 分 ··········· ·········· ··········· ··· ·········· ··········· ··········· ·· 7 由抛物线的对称性可知, 所构成的直角三角形必是以抛物线顶点为直角顶点的等腰直角三角 形,∴ 此等腰直角三角形斜边上的高等于斜边的一半,又∵0 d 1 ,∴ 等腰直角三角形斜 边的长小于 2,∴ 等腰直角三角形斜边上的高必小于 1,即抛物线的顶点的纵坐标必小于 1.
1 1 7 1 1, 3 4 12 1 1 11 1, 当 x 2 时, y2 2 3 4 12 1 1 1 当 x 3 时, y3 3 1 1 , 3 4 4
∵ x 1 时, y1 当 y
Bn
l
B3
B2
B1
M O
…
A1
1
A2 2 A3
3
A4
An
n
An1
x
∴ 美丽抛物线的顶点只有 B1、B2 . ····························· 8 分 ··········· ·········· ········ ·········· ··········· ········ ①若 B1 为顶点,由 B1 1 , ,则 d 1
7 12
7 5 ; ··········· ········· 分 ··········· ········ 9 ·········· ········· 12 12
②若 B2 为顶点,由 B2 2, ,则 d 1 2 综上所述, d 的值为
11 12
11 11 1 , 12 12
5 11 或 时,存在美丽抛物线. ··················· 分 ·················· 10 ·········· ········ 12 12