2013年九年级学业水平考试试卷数学参考
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2013年九年级学业水平测试
数学参考答案及评分标准
(注意:以下答案仅供参考,解答题过程有些提供的是做题思路,有些提供的完整过
程,大题所给分数为累计得分,如有其它做法,可参照给分)
一、
1 2 3 4 5 6 7 8
B C B C C C D C
二、
9 10 11 12 13 14 15
5
73
x=2 50°23 (4,-1)或(-1,3)或
(-1,-1)
n(n+2)= n2+2n
三、16、解:原式=21x ……………………7分
当x=-1 时,原式 =31 ……………………9分
17、答案不唯一每图3分,共6分,
18、(1)300;……………………2分
(2)
…………………7分(每处1分,共5分)
(3)258或0.32. ……………………8分
4000×258=1280 ……………………9分
∴该中学大约有1280个学生喜欢文学类图书. ……………………10分
19、解:过点O作半径OC⊥AB交AB于D,连接OB, ……………………2分
则CD=4cm,BD=21AB=8……………………4分
设圆形截面的半径为xcm
在Rt⊿BOD中,由勾股定理得
222
8)4(xx
……………………6分
∴x=10 …………………8分
即圆形截面的半径为10cm……………………9分
O
B
A
C
D
20、 解:如图,过点C作CD⊥AB交AB于点D. ……………………1分
∵探测线与地面的夹角为30°和 60°
∴∠CAD=30°,∠CBD=60° ……………………2分
法一:可证AB=BC=3, ……………………4分
在Rt⊿BCD中,由sin60°=CBCD……………………6分
可求出)(6.2273.13233米CD……………………8分
古物所在点C的深度大约为2.6米. ……………………9分
法二;在Rt△BDC中,
BD
CD
60tan
∴360tanCDCDBD ------------------3分
在Rt△ADC中,ADCD30tan
∴3330tanCDCDAD ---------------4分
∵3BDADAB
∴3333CDCD --------------6分
∴)(6.2273.13233米CD ---------------8分
答:古物所在点C的深度大约为2.6米. ------------9分
21、(1)由SAS证明△ACD ≌△DFB……………………5分
(2)四边形CDBF是菱形,理由如下:……………………6分
由平移知CF∥AD, CF=AD……………………7分
∵AD=BD
∴
CF∥BD, CF=BD
∴
四边形CDBF是平行四边形……………………8分
∵△ABC是直角三角形,点D是AB的中点
∴CD
=21AB=BD……………………9分
∴
□
CDBF是菱形……………………10分
(也可再证DF⊥BC得□CDBF是菱形)
D
22、(1)解法一:设甲种消毒液购买x瓶,则乙种消毒液购买(100)x瓶. 1分
依题意,得69(100)780xx.
解得:40x. ····················································································································· 3分
1001004060x
(瓶). ······················································································· 4分
答:甲种消毒液购买40瓶,乙种消毒液购买60瓶. ························································· 5分
解法二:设甲种消毒液购买x瓶,乙种消毒液购买y瓶. ·················································· 1分
依题意,得10069780xyxy,. ·································································································· 3分
解得:4060xy,. ······················································································································ 4分
答:甲种消毒液购买40瓶,乙种消毒液购买60瓶. ························································· 5分
(2)设再次购买甲种消毒液y瓶,刚购买乙种消毒液2y瓶.·········································· 6分
依题意,得6921200yy≤. ························································································ 8分
解得:50y≤. ···················································································································· 9分
答:甲种消毒液最多再购买50瓶. ···················································································· 10分
23、解:(1)令y=0,解得11x或23x
∴A(-1,0)B(3,0); ……………………2分
将C点的横坐标x=2代入223yxx得y=-3,
∴C(2,-3)…………3分
求出直线AC的表达式是y=-x-1…………………… 5分
(2)设P点的横坐标为x(-1≤x≤2)(注:x的范围不写不扣分)
则P、E的坐标分别为:P(x,-x-1), E(2(,23)xxx
∵PE=EPyy=22(1)(23)2xxxxx ……………7分
∴当12x时,PE的最大值=94 ……………8分
(3) 存在4个这样的点F,分别是 F1(1,0) F2(-3,0)
F3(7+4 ,0) F4(-7+4 ,0)………12分(分别过点C、点G作x轴的垂线,
通过三角形全等求) (此问共4分,对1个得1分)
思路:分别以点F在点A的左右侧,AC为边、为对角线讨论