人教版初二下册数学期
末试卷及答案
TTA standardization office【TTA 5AB- TTAK 08- TTA 2C】
附中2010-2011学年度下期期末考试
初二数学试题
(总分:150分 考试时间:120分钟)
一、选择题(每小题4分,共40分)
1.在22
923 3.1407
π-,,,,,这六个实数中,无理数的个数是( ) A .4个
B .3个
C .2个
D .1个
2.下列计算正确的是( ) A .532-= B .824+=
C .2733=
D .(12)(12)1+-=
3.已知Rt △ABC 中,90C ∠=?,BC = 8,4sin 5
A =,则AC = ( ) A .6
B .8
C .10
D .
323
4.如图,小正方形的边长均为1,则下列图中三角形(阴影部分)与△ABC 相似的是( )
5.如图,设M 、N 分别为直角梯形ABCD 两腰AD 、CB 的
中点,DE ⊥AB 于点E ,将△ADE 沿DE 翻折,M 与N 恰好重合,则AE ∶BE 等于( ) A .2∶1 B .1∶2 C .3∶2
D .2∶3
6.关于x 的方程22(21)10k x k x +-+=有实数根,则下列结论正确的是( ) A .当12
k =时方程两根互为相反数 B .当14
k ≤时方程有实数根
C D
M
N
(第5题图)
A
B C D
(第4题图)
(第9题图)
(第10题图)
(第14题图)
C .当1k =±时方程两根互为倒数
D .当k = 0时方程的根是x = – 1
7.已知实数x 、y 满足222222()(1)2x y x y x y +++=+,则的值为( ) A .1 B .2
C .– 2或1
D .2或 – 1
8.
已知22y x =的图象是抛物线,若抛物线不动,将x 轴、y 轴分别
向上、向右平移2个单位,在新坐标系下,所得抛物线解析式为( ) A .22(2)2y x =-+ B .22(2)2y x =+- C .22(2)2y x =--
D .22(2)2y x =++
9.如图,△OAP 、△ABQ 均是等腰直角三角形,点P 、Q 在函数4
(0)y x x
=
>的图象上,直角顶点A 、B 均在x 轴上,则点B 的坐标为( ) A 1,0) B 1,0) C .(3,0)
D 1,0)
10 2
D .1
3
二、填空题(每小题3分,共30分)
11.方程220x x +=的解为_________________.
12.已知α为锐角,若1sin cos 3
αα==,则_________________. 13.已知2
21(2)m m y m m x +-=+是反比例函数,则m = _________________.
14.在实数范围内定义一种运算“*”,其规则为22a b a b *=-,根据这个规则,方程
(2)50x +*=的解为_________________.
15.如图是一个二次函数当40x -≤≤的图象,则此时函数y 的取值范围是
_________________.
16.小亮同学想利用影长测量学校旗杆的高度,如图,他在某一时刻立1m 长的标杆
测得其影长为2 m ,同时旗杆的投影一部分在地面上,另一部分在某建筑物的墙
(第16题图)
上,分别测得其长度为9.6 m 和2 m ,则学校旗杆的高度为_________________m .
17.一个三角形两边长为3和4角形的形状为_________________18.已知开口向下的抛物线过A (– 1,0),B (3,0)两点,与y 轴交于C
,且
BC =_________________.
19.如图,□ABCD 中,E 为CD 上一点,DE ∶CE = 2∶3,连结AE 、BE 、BD ,且
AE 、BD 交于点F ,则S △DEF ∶S △EBF ∶S △ABF = _________________.
20.如图,二次函数2y ax bx c =++的图象经过点(– 1,2)和(1,0),且与y 轴交于
负半轴,给出以下四个结论:① abc < 0;② 2a + b > 0;③ a + c = 1;④ a > 1.其中正确结论的序号是_________________.
(第19
题
(第20题
三、解答题(共80分) 21.(6分) (1)
计算:0111()()|tan302
3
--+?
(6分) (2) 解方程:21302
x x ++=
22.(10分) 已知抛物线2y ax bx c =++的图象如图所示.
(1) 抛物线的解析式为____________________. (2) 抛物线的顶点坐标为______________,且y 有最________(填“大”或“小”)值.
(3) 当x _____________时,y 随x 的增大而减小. (4) 根据图象可知,使不等式2
0ax bx c ++<成立的x 的
取值范围是______________________.
23.(8分) 如图,在△ABC 中,4
30sin 105
B C AC ∠=?==,,,求AB 的长.
24.(10分) 如图,反比例函数k y x
=的图象与一次函数y
1,3),B
(n ,– 1
)两点.
(1)
求反比例函数与一次函数的解析式;
(2) 根据图象,直接写出使反比例函数的值大于一次函数的值的x 的取值范围.
(第23题图)
(第24题
(第22题
25.(10分) 西瓜经营户以2元/千克的价格购进一批小型西瓜,以3元/千克的价格出售,每天可售出200千克.为了促销,该经营户决定降价销售,经调查发现,这种小型西瓜每降价元/千克,每天可多售出40千克,另外,每天的房租等固定成本共24元,该经营户要想每天盈利200元,应将每千克小型西瓜的售价降低多少元?
26.
27.(10分) 如图,在等腰直角三角形ABC中,904
,,点D在线段BC上
∠=?==
BAC AB AC
运动(不与B、C重合),过D作45
∠=?,交AC于E.
ADE
(1)求证:△ABD∽△DCE;
(2)设BD = x,AE = y,求y与x之间的函数关系式,并写出自变量的取值范围.
(第26题图)
(10分) 如图,甲、乙两辆大型货车同时从A 地出发驶往P 市.甲车沿一条公路向北偏东53?方向行驶,直达P 市,其速度为30千米/小时.乙车先沿一条公路向正东方向行驶1小时到达B 地,卸下部分货物(卸货的时间不计),再沿一条北偏东37?方向的公路驶往P 市,其速度始终为35千米/小时. (3) 求AP 间的距离.
(4) 已知在P 市新建的移动通信接收发射塔,其信号覆盖面积只可达P 市周围方圆50千米的区域(包括边缘地带),除此以外,该地区无其他发射塔,问甲、乙两司机至少经过多少小时可以互相正常通话? (5)
(3
434sin37cos53cos37sin53tan37tan535
5
4
3
?=?=?=?=?=?=,,,)
53?
37?
(第27题图)
28.(10分) 如图,抛物线2y ax bx c =++与x 轴交于A 、B 两点(点A 在点B 的左侧),与y
轴交于点C ,这条抛物线的顶点是M (1,– 4),且过点(4,5). (1) 求这条抛物线的解析式;
(2) P 为线段BM 上的一点,过点P 向x 轴引垂线,垂足为Q ,若点P 在线段BM 上运动(点P 不与点B 、M 重合),四边形PQAC 的面积能否等于7?如果能,求出点P 的坐标;如果不能,请说明理由.
(3) 设直线m 是抛物线的对称轴,是否存在直线m 上的点N ,使以N 、B 、C 为顶点的三角形是直角三角形?若存在,请求出点N 的坐标;若不存在,请说明理由.
附中2010—2011学年度下期期末考试
初二数学试题参考答案
一、选择题(每小题4分,共40分)
题号 1 2 3 4 5 6 7
8 9 10 选项
C
C
A
B
A
B
A B
D
C
二、填空题(每小题3分,共30分) 11.x 1 = 0,x 2 = – 2 12.
22
13.– 1 14.x 1 = 3,x 2 = – 7
15.24y -≤≤
16.
17.等腰或直角三角形 18.223y x x =-++ 19.4∶10∶25 20.②③④
三、解答题(共80分) 21.(1) 解:原式233
13|3|=+?
--
····················································· 3分 23
123=+- ································································· 5分 43
1=+
········································································ 6分 (2) 解:a = 1,b = 3,c =1
2
O x
y
A
Q
B P
M
C
(第28题图)
∵ 21494172
b a
c -=-??= ∴
x =···································································· 5分 ∴
12x x =
=
··················································· 6分 22.(1) 213
222
y x x =-++
(2) (325
28
,) 大 (3) 32
>
(4) 14x x <->或(每空2分)
23.解:过A 作AD ⊥BC 于D ································································ 1分
∵ AD ⊥BC ,4
sin 5
C =,AC = 10
∴ sin 8AD AC C == ··························· 5分 ∵ 30B ∠=?
∴ AB = 2AD = 16 ······························ 8分
24.解:(1) ∵ A (1,3),B (n ,– 1)在反比例函数k y x
=的图象上
∴ 31k
k n =???-=
??
········································································· 2分 ∴ 33k n ==-, ···································································· 3分 ∵ A (1,3),B (n ,– 1)在一次函数y mx b =+的图象上 ∴313m b
m b
=+???
-=-+?? ·································································· 4分
1
2m b =???=??
解得 ········································································· 5分 ∴ 反比例函数与一次函数的解析式分别为3
2y y x x
==+, ············· 6分
(2) 301x x <-<<或 ·································································· 10分
25.解:设每千克小型西瓜的售价降低x 元,由题意 ·································· 1分
(32)(20040)242000.1
x
x --+?
-= ·
··················································· 5分
(第23题图)
∴ x 1 = ,x 2 = ········································································· 9分 答:应将每千克小型西瓜的售价降低元或元. ··································· 10分 26.(1) 证明:∵ 90BAC ∠=?,AB = AC
∴ 45B C ∠=∠=? ································································ 1分 ∵ 23180C ∠+∠+∠=?
∴ 23135∠+∠=? ················2分 ∵ 1218045ADE ADE ∠+∠+∠=?∠=?, ∴ 12135∠+∠=? ················3分 ∴ 13∠=∠ ························4分 ∴ △ABD ∽△DCE ············5分
(2) 解:∵ AB = AC = 4
∴ 42BC =········································································· 6分 ∵ BD = x ,AE = y
∴ 424CD x CE y =-=-, ······················································· 7分 ∵ △ABD ∽△DCE , ∴
42AB BD x
DC CE y
x ==
--4即4 ·················································· 9分 ∴ 21
24(042)4
y x x x =-+<< ·············································· 10分
27.解:(1) 过P 作PD ⊥AB 延长线于D ·················································· 1分
由题意知:AB = 35千米 设PD = x 千米
∵ 在Rt △PAD 中,tan37PD
AD
?= ∴ 4
tan373
x AD x =
=?
··················· 2分 C
A
D
E (第26题图)
1
2
3
53?
37?
(第27题图)
D
∵在Rt △PBD 中,tan53PD
BD
?= ∴ 3
tan534
PD BD x =
=?
································································ 3分 ∵ AD – BD = AB 即43353
4
x x -=
∴ x = 60,即 AD = 60千米 ····················································· 4分 ∴在Rt △PAD 中,sin37PD
AP
?= ∴ 60
100sin370.6
PD AP =
==?千米·
··················································· 5分 (2) ∵在Rt △PBD 中,sin53PD
BD
?=
∴ 60
754
sin 533
PD PB =
==?千米 ···················································· 6分 设甲、乙两司机分别出发t 1、t 2小时后手机有信号 ∴ 150100505
30303
PA t --===小时 ·
·············································· 7分 2503575506012
3535357PB AB t +-+-=
===小时
································ 8分 ∵ 125351236
3
21721
t t ==
<==
······················································· 9分 ∴ 甲、乙两司机至少经过
12
7
小时可以正常通话. ···················· 10分 28.解:(1) 设2(1)4y a x =--代入(4,5)得a = 1,∴ 223y x x =-- ·············· 2分
······················································································ 8分 ③若∠C = Rt ∠,则222NB BC NC =+,∴22418(3)1n n +=+++ ∴ 4n =- ······················································································ 9分
综上,存在这样的点N
,其坐标为(1
或(1或(1
2),或(1,4-) ······················································································ 10分