2007 年黑龙江省中考数学真题及答案(非课改区)
(本试卷满分 120 分,考试时间 120 分钟)
一、填空题(每小题 3 分,满分 30 分)
1.函数
y
x
中,自变量 x 的取值范围是
3
.
2.2008 年北京奥运会火炬接力,火炬手达到 21 780 人,把这个数用科学记数法表示约为
3.在 ABC△
y
人(保留两个有效数字).
BC ,
6
x
,当 0
4.对于函数
中,
3
AC ,
4
x 时, y 随 x 的增大而
5
AB ,则 tan A 的值为
.
5.在直径为 1 000mm 的圆柱形油槽内装入一些油后,截面如图所示,若
AB
800mm
,则油的最大深度为
油面宽
6.“红星”商场对商品进行清仓处理,全场商品一律八折,小亮在该商
场购买了一双运动鞋,比按原价购买该鞋节省了 16 元,他购买该鞋实际
用
mm.
元.
.
O
A
B
(第 5 题)
7.抛物线
y
8. ABC△
∠
BOD
2
x
3
bx
经过点 (3 0), ,则b 的值为
是 O 的内接三角形, OD BC
40
,则 BAC∠
的度数为
,垂足为 D ,若
.
E
A
.
.
9.一组数据 3,7,8, x ,4 的平均数是 5,这组数据的中位数
为
10.如图,矩形纸片 ABCD ,
边上,且
EF ,则 AE 的长为
二、单项选择题(将正确答案的代号填在题后括号内,每小题 3 分,满分 30 分)
11.下列运算正确的是(
BC ,点 M 在 BC
CM ,将矩形纸片折叠使点 D 落在点 M 处,折痕为
AB ,
12
M
B
.
)
8
4
(第 10 题)
A. 6
x x
x
6
B.
3
(2 )
a
6
a
3
C. 5
x
2
x
3
x
D. 2
x
3
2
2
x
2
6
x
D
F
C
12.在下列图形中,既是轴对称图形,又是中心对称图形的是(
)
A.
B.
13.下列二次根式是最简二次根式的为(
C.
)
D.
A. 2 3a
B.
28x
C. 3y
D.
b
4
A
B
O
E
F
14.如图, AB CD∥ , :
的中点,则 :EF AF 的值为(
A.1
B.2
1: 4
BO OC
)
C.3
,点 E F, 分别是OC OD,
D.4
C
D
(第 14 题)
15.点
P ac
2 b
, 在第二象限,点 (
a
Q a b, 关于原点对称的点在(
)
)
B.第二象限
A.第一象限
16.已知 O 的半径为 5,点 P 在直线l 上,且
A.相切
B.相交
C.相离
C.第三象限
D.相切或相交
D.第四象限
OP ,直线l 与 O 的位置关系是(
5
)
17.若
x
y
2
x
,则 x
y
y 的值为(
)
3
B.1
0
C.3
)
D. 3
B. AB CD
A. 1
18.不能判定四边形 ABCD 是平行四边形的题设是(
A. AB CD∥ , AB CD
C. AD BC , A
C∠ ∠
19.国家实施惠农政策后,某镇农民人均收入经过两年提高 44%,这两年该镇农民人均收
入平均年增长率是(
A. 22% B. 20% C.10%
20.如图, ABC△
CE
是等边三角形,点 D E, 分别在 BC AC, 上,
, BE AD, 相交于点 F ,连接 DE ,
, AD BC
D. AB CD∥ , B
D∠ ∠
D.11%
BD
BC
A
且
,
)
1
3
AC
1
3
AFE
,正确的结论有(
;② DE
60
则下列结论:①
∠
④ AF BE AE AC
A.4 个
B.3 个
三、解答题(满分 60 分)
21.(本题 6 分)
C.2 个
D.1 个
AC
;③ 2CE
DF DA
;
B
)
E
C
F
D
(第 20 题)
先化简,再求值:
2
x
1
2
x
1
x
x
1
1
x
1
x
2
x
,其中
x
2 1
.
22.(本题 6 分)
x, 是关于 x 的一元二次方程 2
kx
x
已知 1
(1)求 k 的取值范围;
2
4
x
的两个不相等的实数根.
3 0
(2)是否存在这样的实数 k ,使 1
x
2
2
x
2
3
x x
1
2
2
成立?若存在,求 k 的值;若不存在,
请说明理由.
23.(本题 6 分)
为了解“宏亮”中学初四男生身高情况,抽测了该校初四 20 名男生身高,结果如下(单位:
厘米):
165 172
176 175
结合所列出的样本频率分布表回答下列问题:
175 171
182 173
170
176
174 175
179 172
181
177
183
169
179
188
分组
频数
164.5~169.5
169.5~174.5
174.5~179.5
179.5~184.5
184.5~189.5
2
8
3
频率
0.10
0.40
0.15
合计
20
1.00
(1)在这个问题中,样本的容量是
(2)填写表中未完成的部分;
(3)如果该校初四男生共有 400 人,那么该校初四男生身高不低于 175 厘米的约有多少人?
;
24.(本题 6 分)
在数学活动课上,小明做了一梯形纸板,测得一底为 10cm,高为 12cm,两腰长分别为 15cm
和 20cm,求该梯形纸板另一底的长.
25.(本题 8 分)
甲、乙二人骑自行车同时从张庄出发,沿同一路线去李庄.甲行驶 20 分钟因事耽误一会儿,
事后继续按原速行驶.下图表示甲、乙二人骑自行车行驶的路程 y (千米)随时间 x (分)
变化的图像(全程),根据图像回答下列问题:
(1)乙比甲晚多长时间到达李庄?
(2)甲因事耽误了多长时间?
(3) x 为何值时,乙行驶的路程比甲行驶的路程多 1 千米?
y (千米)
15
C D
甲
乙
10
E
A
B
26.(本题 8 分)
O
20
60
80
x (分)
中, AB AC
在 ABC△
AB 于点 E , PF
若点 P 在 BC 边上(如图 1),此时
请直接应用上述信息解决下列问题:
当点 P 分别在 ABC△
请给予证明;若不成立,PD PE PF
不需要证明.
A
,点 P 为 ABC△
所在平面内一点,过点 P 分别作 PE
AC∥ 交
AB∥ 交 BC 于点 D ,交 AC 于点 F .
PD ,可得结论: PD PE PF AB
0
.
内(如图 2), ABC△
外(如图 3)时,上述结论是否成立?若成立,
, , 与 AB 之间又有怎样的数量关系,请写出你的猜想,
E
B
F
C
(
)P D
图 1
A
E
B
F
C
P
D
图 2
A
E
F
C
B
D
P
图 3
27.(本题 10 分)
天宇便利店老板到厂家购进 A B, 两种香油, A 种香油每瓶进价 6.5 元, B 种香油每瓶进
价 8 元,购进 140 瓶,共花了 1 000 元,且该店销售 A 种香油每瓶 8 元, B 种香油每瓶 10
元.
(1)该店购进 A B, 两种香油各多少瓶?
(2)将购进 140 瓶香油全部销售完可获利多少元?
(3)老板打算再以原来的进价购进 A B, 两种香油共 200 瓶,计划投资不超过 1 420 元,
且按原来的售价将这 200 瓶香油销售完成获利不低于 339 元,请问有哪几中购货方案?
28.(本题 10 分)
如图,点 A 为 x 轴负半轴上一点,点 B 为 x 轴正半轴上一点,
OA OB OA OB
,
(
)
的长分
2
4
mx m
别是关于 x 的一元二次方程 2
x
(1)求 ABC∠
的度数;
(2)过点C 作CD AC
(3)在第(2)问的条件下, y 轴上是否存在点 P ,使 PBA
交 x 轴于点 D ,求点 D 的坐标;
2 0
∠
的两根, (0 3)
C , ,且
S
△
ABC
6
.
∠
ACB
?若存在,请直接
写出直线 PD 的解析式;若不存在,请说明理由.
y
C
A
O
B
x
黑龙江省 2007 年初中升学统一考试
数学试题参考答案
一、填空题(每小题 3 分,满分 30 分)
4
3
2.2 10
≥
2.
3.
1.
3
x
4
4.增大
5.200
6.64
7. 4
8. 40 或140
9.4
10.2
12.D
二、单项选择题(将正确答案的代号填在题后括号内,每小题 3 分,满分 30 分)
11.C
三、解答题(满分 60 分)
21.(本题 6 分)
15.A
16.D
17.A
18.C
19.B
20.A
13.A
14.B
解:原式
(
x
1)
1)(
x
(
1)
x
2
x
x
1
1
x
1
x
································································· 2 分
1
1x
········································································································· 2 分
当
x
2 1
时,原式
1
2 1 1
2
2
···························································· 2 分
······················································································· 2 分
22.(本题 6 分)
解:(1)由题意知, 0
24
k 且
4 ( 3) 0
k
4
3
k 且 0
k .························································································ 1 分
(2)存在.
x
1
x
2
4
k
x x
,
1
2
3
k
·············································································1 分
又
2
x
1
2
x
2
3
x x
1 2
2
,
.
k
2
8
k
2
4
k , 2
k (不符合题意,舍去).·······················································1 分
k .···································································· 1 分
解得 1
存在满足条件的 k 值,即 4
23.(本题 6 分)
解:(1)20;··································································································2 分
(2)在表中所填数据由左至右,由上至下依次为 6,0.30,1,0.05;······················ 2 分
(3) 400 (0.4 0.15 0.05)
(人),
240
于是可以估计,该校初四男生身高不低于 175 厘米的约有 240 人.····························2 分
24.(本题 6 分)
AD
于点 F .
于点 E ,
.································1 分
,分别过点 A D, 作 AE BC
20cm
EF AD
解:不妨设
BC
DF
AB
AE DF
15cm
,
12cm
CD
,
10cm
10cm
,
在 Rt ABE△
中,
BE
2
AB
2
AE
9(cm)
····················································· 1 分
同理可求
CF
16cm
.···················································································· 1 分
A
D
A
D
A
B
E
F
图1
C
E
F
B
图 2
C
B
C
E
图 3
D
F
分三种情况:
(1)如图 1,
(2)如图 2,
(3)如图 3,
BC BE EF CF
BC EF BE CF
BC BE EF CF
35(cm)
······················································ 1 分
17(cm)
······················································ 1 分
3(cm)
························································1 分
综上所述,该梯形纸板另一底的长为 35cm 或 17cm 或 3cm.
25.(本题 8 分)
解:(1)设直线OD 解析式为
y
k x ,
1
由题意可得 160
y
x
···························································· 1 分
k , 1
k ,
1
6
x , 90
10
1
6
1
6
当 15
y 时,
15
x ,90 80 10
(分)···········································1 分
故乙比甲晚 10 分钟到达李庄.
(2)设直线 BC 解析式为
y
k x b
2
,
由题意可得
60
k
80
k
2
2
b
b
,
··············································································· 1 分
10
15
1
k
,
解得 2
y
4
5
b
,
x
1
4
················································································· 1 分
5
5 5
x , 40
由图像可知甲 20 分钟行驶的路程为 5 千米,
1
4
故甲因事耽误了 20 分钟.
(3)分两种情况:
x , 40 20
20
(分)··························································1 分
①
②
1
6
1
6
x , 36
x ,·················································································· 1 分
5 1
x
1
4
x
5
1
, 48
x ,······································································· 1 分
当 x 为 36 或 48 时,乙行驶的路程比甲行驶的路程多 1 千米.·································· 1 分
.······························································ 2 分
26.(本题 8 分)
解:图 2 结论: PD PE PF AB
证明:过点 P 作 MN
BC∥ 分别交 AB AC, 于 M N, 两点,
由题意得 PE PF AM
四边形 BDPM 是平行四边形, MB PD
即 PD PE PF AB
图 3 结论: PE PF PD AB
27.(本题 10 分)
PD PE PF MB AM AB
.·············································································· 2 分
.···················································1 分
,
.················································································ 1 分
.···································································· 2 分
解:(1)设该店购进 A 种香油 x 瓶, B 种香油 (140
)x 瓶,
由题意得 6.5
x
8(140
x
) 1000
,···································································2 分
60
x
x ,140
解得 80
················································································1 分
该店购进 A种香油 80 瓶,B种香油 60 瓶.··························································· 1 分
(2)80 (8 6.5) 60 (10 8)
(元)····················································· 1 分
240
将购进的 140 瓶香油全部销售完可获利 240 元.···················································· 1 分
(3)设购进 A种香油 a 瓶,B种香油(200
)a 瓶,
由题意得
6.5
8(200
a
1.5
2(200
a
122
a≤ ≤ .
)
a
)
a
≤ ,
1420
339
≥ .
··································································2 分
200 a
相应取 80,79,78.·········································································· 1 分
解得120
a 为非负整数, a 取 120,121,122.
有三种购货方案:A种香油 120 瓶,B种香油 80 瓶;A种香油 121 瓶,B种香油 79 瓶;
A种香油 122 瓶,B种香油 78 瓶.······································································ 1 分
28.(本题 10 分)
6
3
,
ABC
由已知可得
解: (0 3)
.
S
△
OC
AB .
C , ,
m
OA OB
3 0
x
,解得 1
OB .······················································································ 1 分
1m .····················································1 分
, 4
4m ,
1
3
x
,
3
2 4
x
1
OA ,
.
3
4
4
x
△
OBC
为等腰直角三角形,
ABC
45
.···················································1 分
90
,
∽△
(2)由
DCO
AOC
ACD
, CAO
AOC
90
COD
,·······················································································1 分
OD ,··················································································· 1 分
△
AO OC
CO OD
点 D 的坐标为(9,0).················································································ 1 分
(3)存在.
,
,
9
直线 PD 的解析式为
y
x
2
3
或
6
y
2
3
x
.·················································4 分
6