建邺区2015年九年级学情调研测试(1)
数 学
注意事项:
1.本试卷共6页.全卷满分120分.考试时间为120分钟.考生答题全部答在答题卡上,答在本试卷上无效.
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一、选择题(本大题共6小题,每小题2分,共12分.在每小题所给出的四个选项中,恰有一项是符合题目要求的,请将正确选项前的字母代号填涂在答题卡...相应位置....上) 1.下列计算结果是负数的是
A .3-2
B .3 (-2)
C .3-2
D .3
2.计算a 3·(1
a
)2的结果是
A .a
B .a 5
C .a 6
D .a 8
3.面积为10m 2的正方形地毯,它的边长介于
A .2 m 与3 m 之间
B .3 m 与4 m 之间
C .4 m 与5 m 之间
D .5 m 与6 m 之间
4.如图是一几何体的三视图,这个几何体可能是
A .四棱柱
B .圆锥
C .四棱锥
D .圆柱
5.如图,将一张矩形纸片和一张直角三角形纸片叠放在一起,∠1+∠2的值是 A .180°
B .240°
C .270°
D .300°
6.“一般的,如果二次函数y =ax 2+bx +c 的图像与x 轴有两个公共点,那么一元二次方程
ax 2+bx +c =0有两个不相等的实数根.”——苏科版《数学》九年级(下册)P 25.参考上述教材中的话,判断方程 x 2-2x =1
x -2实数根的情况是
A .有三个实数根
B .有两个实数根
C .有一个实数根
D .无实数根
(第4题)
2
(第
5题)
1
左视图
俯视图 主视图
二、填空题(本大题共10小题,每小题2分,共计20分.不需写出解答过程,请把答案直接填写在答题卡...相应位置....
上) 7.函数y =1
1-x 中,自变量x 的取值范围是 ▲ .
8.若a -b =3,ab =2,则a 2b -ab 2= ▲ .
9.若关于x 的一元二次方程x 2-kx -2=0有一个根是1,则另一个根是 ▲ .
10.由我国倡议筹建的亚洲基础设施投资银行(简称亚投行),法定资本100 000 000 000美
元.若1美元兑换6.254元人民币,则亚投行法定资本换算成人民币为 ▲ 元人民币(用科学记数法表示).
11.在同一平面直角坐标系中,反比例函数y 1=k
x
(k 为常数,k ≠0)的图像与正比例函数
y 2=ax (a 为常数,a ≠0)的图像相交于A 、B 两点.若点A 的坐标为(2,3),则点B 的坐标为 ▲ .
12.如图,四边形ABCD 是⊙O 的内接四边形,若⊙O 的半径为3 cm ,∠A =110°,则劣弧
BD ⌒的长为 ▲ cm .
13.如图,正五边形FGHIJ 的顶点在正五边形ABCDE 的边上,若∠1=20°,则∠2= ▲ °.
14.一组数据4、5、6、7、8的方差为S 12,另一组数据3、5、6、7、9的方差为S 22,那么
S 12 ▲ S 22(填“>”、“=”或“<”).
15.如图,△ABC 与△DEF 均为等边三角形,⊙O 是△ABC 的内切圆,同时也是△DEF 的
外接圆.若AB =1cm ,则DE = ▲ cm .
16.如图,在矩形ABCD 中,AB =8.将矩形的一角折叠,使点B 落在边AD 上的B ′点处,
若AB ′=4,则折痕EF 的长度为 ▲ .
D
C
B
A F
E
(第16题)
B ′
(第15题)
D
C
B A
J
E (第13题)
I H
G F 2
1
(第12题)
三、解答题(本大题共有11小题,共计88分.请在答题卡指定区域内作答,解答时应写出
必要的文字说明、证明过程或演算步骤) 17.(6分)计算 (212-1
3
)?6.
18.(6分)解不等式组?????2-x >0,
5x +12
+1≥2x -13.
19.(7分)已知:如图,在□ABCD 中,线段EF 分别交AD 、AC 、BC 于点E 、O 、F ,
EF ⊥AC ,AO =CO . (1)求证:△ABF ≌△CDE ;
(2)在本题的已知条件中,有一个条件如果去掉,并不影响(1)的证明,你认为这个
多余的条件是 ▲ (直接写出这个条件).
20.(8分)如图①,南京中山陵的台阶拾级而上被分成坡度不等的两部分.图②是台阶的
侧面图,若斜坡BC 长为120 m ,在C 处看B 处的仰角为25°;斜坡AB 长70 m ,在A 处看B 处的俯角为50°,试求出陵墓的垂直高度AE 的长.
(参考数据:sin50°≈0.77,cos50°≈0.64,tan50°≈1.19,sin25°≈0.42,cos25°≈0.91,tan25°≈0.47)
图①
图②
(第20题)
D C
B
A
E
F
D
C
B
A
E
(第19题) O
F
21.(7分)A、B、C三把外观一样的电子钥匙对应打开a、b、c三把电子锁.(1)任意取出一把钥匙,恰好可以打开a锁的概率是▲;
(2)求随机取出A、B、C三把钥匙,一次性对应打开a、b、c三把电子锁的概率.22.(8分)如图,是某卖场国产大米牌手机的宣传广告.
(第22题)
(1)你认为大米手机5月份的销售量必定是三个品牌手机中最高的吗?通过计算说明你的理由.
(2)若各品牌手机2015年4月的销售量如下:
求该卖场5月份三个品牌手机销售量的平均增长率.
23.(9分)某日,小敏、小君两人约好去奥体中心打球.小敏13∶00从家出发,匀速骑自
行车前往奥体中心,小君13∶05从离奥体中心6 000m 的家中匀速骑自行车出发.已知小君骑车的速度是小敏骑车速度的1.5倍.设小敏出发x min 后,到达离奥体中心y m 的地方,图中线段AB 表示y 与x 之间的函数关系.
(1)小敏家离奥体中心的距离为 ▲ m ;她骑自行车的速度为 ▲ m/min ; (2)求线段AB 所在直线的函数表达式;
(3)小敏与小君谁先到奥体中心,要等另一人多久?
24.(8分)某商品的进价为每件40元,售价为每件50元,每个月可卖出210件.如果每
件商品的售价每上涨1元,则每个月少卖10件.当每件商品的售价定为多少元时,每个月的利润恰为2200元?
25.(10分)在正方形ABCD 中,AD =2,l 是过AD 中点P 的一条直线.O 是l 上一点,以
O 为圆心的圆经过点A 、D ,直线l 与⊙O 交于点E 、F (E 、F 不与A 、D 重合,E 在F 的上面).
(1)如图,若点F 在BC 上,求证:BC 与⊙O 相切.并求出此时⊙O 的半径.
(2)若⊙O 半径为23
3,请直接写出∠AED 的度数.
(第25题)
(第23题)
26.(9分)已知函数y =x 2+(2m +1)x +m 2-1. (1)m 为何值时,y 有最小值0;
(2)求证:不论m 取何值,函数图像的顶点都在同一直线上.
27.(10分) 问题提出
如图①,已知直线l 与线段AB 平行,试只用直尺作出AB 的中点. 初步探索
如图②,在直线l 的上方取一个点E ,连接EA 、EB ,分别与l 交于点M 、N ,连接MB 、NA ,交于点D ,再连接ED 并延长交AB 于点C ,则C 就是线段AB 的中点.
推理验证
利用图形相似的知识,我们可以推理验证AC =CB .
(1)若线段a 、b 、c 、d 长度均不为0,则由下列比例式中,一定可以得出b =d 的是( ▲ )
A .a b =c d
B .a b =a
d
C .a b =d a
D .a c =d b
(2)由MN ∥AB ,可以推出△EFN ∽△ECB ,△EMN ∽△EAB ,△MND ∽△BAD ,
△FND ∽△CAD .
所以,有FN CB =( ▲ )( ▲ )=MN AB =( ▲ )( ▲ )=FN
AC ,
所以,AC =CB .
拓展研究
如图③,△ABC 中,D 是BC 的中点,点P 在AB 上. (3)在图③中只用直尺....作直线l ∥BC .
(4)求证:l ∥BC .
C B A
D
E
l
M
N
F 图②
B
A
l
图①
(第27题)
图③
建邺区2015年九年级学情分析卷
数学参考答案及评分标准
说明:本评分标准每题给出了一种或几种解法供参考,如果考生的解法与本解答不同,参照本评分标准的精神给分.
一、选择题(每小题2分,共计12分)
二、填空题(每小题2分,共计20分)
7.x ≠1 8.6 9.-2 10.6.254×1011 11.(-2,-3) 12.7π3 13.52 14.< 15.1
2 16.55
三、解答题(本大题共11小题,共计88分) 17.(本题6分)
解: (212-
1
3
)× 6 =212×6-
1
3
× 6 ····································································· 2分 =122- 2 ······················································································ 4分 =112. ·························································································· 6分 18.(本题6分)
解:解不等式①,得x <2. ········································································ 2分
解不等式②,得x ≥-1. ····································································· 4分 所以不等式组的解集是-1≤x <2. ························································ 6分
19.(本题7分)
解:(1)∵四边形ABCD 是平行四边形,
∴AB =CD ,∠B =∠D ,AD =BC ,AD ∥BC .
∵AD ∥BC .
∴∠EAO =∠FCO , 又∵∠AOE =∠COF ,
∴△AOE ≌△COF . ······································································ 3分 ∴CF =AE ,
∴AD -AE =BC -CF , 即DE =BF .
∵AB =CD ,∠B =∠D ,BF =DE ,
∴△ABF ≌△CDE . ······································································ 5分 (2)EF ⊥AC . ················································································· 7分
20.(本题8分)
解:在Rt △BDC 中,sin C =
BD
BC
,································································· 1分 ∴BD =BC ·sin C =BC ·sin25°=120×0.42=50.4 m . ······························ 3分 在Rt △AFB 中,sin ∠ABF =
AF
AB
, ··························································· 4分 ∴AF =AB ·sin ∠ABF =AB ·sin50°=70×0.77=53.9 m . ·························· 6分 ∴AE =AF +FE =AF +BD =50.4+53.9=104.3 m .
答:陵墓的垂直高度AE 的长为104.3 m .················································ 8分
21.(本题7分)
解:(1)13 .··························································································· 2分
(2)用树状图列出所有可能出现的结果:
一共有6种可能的结果,它们是等可能的,其中符合要求的有1种. P (一次性对应打开a 、b 、c 三把电子锁)=1
6
.
答:一次性对应打开a 、b 、c 三把电子锁的概率为 1
6
. ························· 7分
22.(本题8分)
解:(1)不一定.
例如:芒果手机4月份销售200台,则5月份销售量为240台;四星手机4月份销售200台,则5月份销售量为100台;大米手机4月份销售50台,则5月份销售量为100台,从而可知大米手机5月份的销售量不是三个品牌手机中最高的. ················································································· 4分
(2)
200×20%-80×50%+120×100%
200+80+120
=30%.
答:该卖场5月份三个品牌手机销售量的平均增长率是30%. ··············· 8分
23.(本题9分)
解:(1)6000,200; ················································································ 2分 (2)设AB 所在直线的函数表达式为y =kx +b ,
将点A (0,6 000),B (30,0)代入y =kx +b 得:
开 始
第一次
第二次
第三次
所有可能出现的结果
A
A A
B B
B
C C C A
B
C C
B A
(A ,B ,C ) (B ,A ,C ) (A ,C ,B ) (B ,C ,A ) (C ,A ,B ) (C ,B ,A )
?????b =6 000,30k +b =6, 解得 ?
????b =6 000,k =-200, ∴AB 所在直线的函数表达式为y =-200x +6 000. ····························· 5分 (3)设小君骑公共自行车时与奥体中心的距离为y 1 m ,
则y 1=-300(x -5)+6 000, 当y 1=0时,x =25. 30-25=5.
∴小君先到达奥体中心,小君要等小敏5分钟. ································ 9分
24.(本题8分)
解法一:设每件商品的售价上涨x 元,
(210-10x )(50+x -40)=2200 ·························································· 4分 解得x 1 =1,x 2=10, ··································································· 6分 ∴当x =1时,50+x =51,当x =10时,50+x =60; ·························· 7分
解法二:设每件商品的售价为x 元,
[210-10(x -50)] (x -40)=2200 ······················································· 4分 解得x 1 =51,x 2=60, ·································································· 7分 答:当每件商品的售价定为51或60元时,每个月的利润恰为2200元. · 8分
25.(本题10分)
解:(1)连接OA 、OD ,
在⊙O 中,OA =OD ,
∵在△AOD 中,点P 是AD 的中点, ∴OP ⊥AD .
∵四边形ABCD 是正方形, ∴AD ∥BC , ∴OP ⊥BC ,
且OF 是⊙O 半径,
∴BC 与⊙O 相切. ······································································· 3分
∵在△AOD 中,点P 是AD 的中点, ∴AP =DP =1.
在Rt △AOP 中,∠APO =90°, ∵AP 2+OP 2=AO 2,
∴12+(2-r )2=r 2,求得r =5
4 . ······················································ 6分
(2)120°或60°. ·········································································· 10分
26.(本题9分) 解:(1)当y =0时,
4ac -b 24a =4(m 2-1)-(2m +1)24=-4m -5
4=0, ···································· 3分 ∴m =-54
. ················································································· 5分
(2)函数y =x 2+(2m +1)x +m 2-1的顶点坐标为(-2m +12,-4m -5
4
)
设顶点在直线y 1=kx +b 上,则-2m +12k +b =-4m -5
4
求得k =1,b =3
4
,
不论m 取何值,该函数图像的顶点都在直线y 1=x -3
4
上. ··················· 9分
27.(本题10分)
解:(1)B ;···························································································· 1分 (2)EN EB ,ND
AD ; ·················································································· 3分
(3)如图①,画图正确; ······································································ 6分 (4)如图②,过点Q 作MN ∥BC ,交AB 、AC 分别于点M 、N ,
∵MN ∥BC ,
∴△AMQ ∽△ABD ,△AQN ∽△ADC ,
∴MQ BD =AQ AD ,AQ AD =QN DC
, ∴MQ BD =QN DC . ············································································· 7分 ∵点D 是BC 的中点, ∴BD =CD , ∴MQ =NQ . ∵MN ∥BC ,
∴△PMQ ∽△PBC ,△EQN ∽△EBC , ∴MQ BC =PQ PC ,NQ CB =EQ EB , ∴PQ PC =EQ EB , ∴
PQ QC =EQ
QB
, ··············································································· 8分 又∵∠PQE =∠CQB ,
∴△PQE ∽△CQB , ······································································ 9分 ∴∠EPQ =∠BCQ , ∴PE ∥BC ,
即l ∥BC . ················································································ 10分
(图①)
(图②)