巴中市二○○八高中阶段教育学校招生考试数学试题试题及答案

巴中市二○○八年高中阶段教育学校招生考试
数 学 试 卷
（全卷满分150分，120分钟完卷）
第Ⅰ卷 选择题（共30分）
注意事项：
1．考生姓名、考号、考试科目，应在答题卡上“先填后涂”． 2．每小题选出的答案，必须用2B 铅笔在答题卡上“对应涂黑”． 3．答题卡上答案若需改动，应用橡皮擦擦干净后再涂．
一、选择题：在每小题给出的四个选项中，只有一项是符合题目要求的，请将正确选项的番号涂卡．（本题共10个小题，每小题3分，共30分） 1．下列各式正确的是（ ） A ．
33-
-= B ．3
2
6-=- C ．(3)3--=
D ．0
(π2)0-=
2．在学校开展的“为灾区儿童过六一”的活动中，晶晶把自己最喜爱的铅笔盒送给了一位灾区儿童．这个铅笔盒（图1）的左视图是（ ）
A ．
B ．
C ．
D ．
3．如图2．在ABCD Y 中，对角线AC 和BD 相交于点O ， 则下面条件能判定ABCD Y 是矩形的是（ ） A ．AC BD =
B ．AC
BD ⊥
C ．AC B
D =且AC
BD ⊥
D ．AB AD =
4．在常温下向一定量的水中加入食盐Nacl ，则能表示盐水溶液的浓度与加入的Nacl 的量之间的变化关系的图象大致是（ ）
A ．
B ．
C ．
D ． 5．下列命题是真命题的是（ ）
A ．对于给定的一组数据，它的平均数一定只有一个
B ．对于给定的一组数据，它的中位数可以不只一个
C ．对于给定的一组数据，它的众数一定只有一个
D ．对于给定的一组数据，它的极差就等于方差
6．点(213)
P m -，在第二象限，则m 的取值范围是（ ） A ．12
m >
B ．12
m ≥
C ．12
m <
D ．12
m ≤
7．如图3，“寸”是电视机常用尺寸，1寸约为大拇指第一节的长，
图1
则7寸长相当于（ ） A ．一支粉笔的长度 B ．课桌的长度 C ．黑板的宽度 D ．数学课本的宽度
8．用计算器计算数据，，，，，，，，，，，，，，的平均数约为（ ） A ．
B ．14.16
C ．
D ．
9．二次函数2
(0)y ax bx c a =++≠的图象如图4所示， 则下列说法不正确的是（ ） A ．2
40b ac -> B ．0a >
C ．0c >
D ．02b
a
-
< 10．巴中日报讯：今年我市小春粮油再获丰收，全市产量预计由前年的45万吨提升到50万吨，设从前年到今年我市的粮油产量年平均增长率为x ，则可列方程为（ ） A ．45250x += B ．2
45(1)50x += C ．2
50(1)45x -=
D ．45(12)50x +=
巴中市二○○八年高中阶段教育学校招生考试
数 学 试 卷
说明：
1．全卷满分为150分，120分钟完卷．
2．本试卷分为第Ⅰ卷和第Ⅱ卷，第Ⅰ卷为选择题，答案涂卡；第Ⅱ卷为非选择题，考生用蓝、黑墨水钢笔或圆珠笔在试卷上作答．
3．考试结束后监考老师将答题卡装入专用袋，不装订第Ⅰ卷，只装订第Ⅱ卷．
第Ⅱ卷 非选择题（共120分）
二、填空题：（每小题3分，共30分，把正确答案直接填写在题中横线上）． 11．当x = 时，分式
3
3
x x --无意义． 12．唐家山堰塞湖是“5g 12汶川地震”形成的最大最险的堰塞湖，垮塌山体约达2037万立方米，把2037万立方米这个数用科学记数法表示为 立方米． 13．把多项式3
2
2
44x x y xy -+分解因式，结果为 ．
14．下面图形：四边形，三角形，正方形，梯形，平行四边形，圆，从中任取一个图形既是轴对称图形又是中心对称图形的概率为 ． 15．如图5，PA 为O e 的切线，A 为切点，PO 交O e 于点B ，4PA =，3OA =，则OP =
．
16．在长为a m ，宽为b m 的一块草坪上修了一条1m 宽的笔直小路，则余下草坪的面积可表示为
2m ；现为了增加美感，把这条小路改为宽恒为1m 的弯曲小路（如图6），
则此时余下草坪的面积为
2m ．
17．如图7，将一平行四边形纸片ABCD 沿AE EF ，折叠，使点E B C ，，在同一直线上，则
AEF ∠= ．
18．如图8，若点A 在反比例函数(0)k
y k x
=≠的图象上，AM x ⊥轴于点M ，AMO △的面积为3，
则k
= ． 19．若0234x y z ==≠，则23x y z
+= ．
20．大家一定熟知杨辉三角（Ⅰ），观察下列等式（Ⅱ）
根据前面各式规律，则5
()a b += ． 三、解答题（每题6分，共18分）
21．解方程：26160x x --= 22
．计算：2008
(1)
2tan 20cot 20-+o o
23．在解题目：“当1949x =时，求代数式222
4421
142x x x x x x x
-+-÷-+-+的值”时，聪聪认为x 只要任取一个使原式有意义的值代入都有相同结果．你认为他说的有理吗？请说明理由． 四、推理论证（24题10分，25题10分，共20分）
24．已知：如图9，梯形ABCD 中，AD BC ∥，点E 是CD 的中点，BE 的延长线与AD 的延长线相交于点F ．
（1）求证：BCD FDE △≌△．
（2）连结BD CF ，，判断四边形BCFD 的形状，并证明你的结论． 25．已知：如图10，在ABC △中，点D 是BAC ∠的角平分线上一点，BD AD ⊥
于点D ，过点D 作
DE AC ∥交AB 于点E ．求证：点E 是过A B D ，，三点的圆的圆心．
五、社会实践（10分）
26．国家主管部门规定：从2008年6月1日起，各商家禁止向消费者免费提供一次性塑料购物袋．为了了解巴中市市民对此规定的看法，对本市年龄在16—65岁之间的居民，进行了400个随机访问抽样调查，并根据每个年龄段的抽查人数和该年龄段对此规定的支持人数绘制了下面的统计图． 根据上图提供的信息回答下列问题：
（1）被调查的居民中，人数最多的年龄段是 岁．
（2）已知被调查的400人中有83%的人对此规定表示支持，请你求出31—40岁年龄段的满意人数，并补全图b ．
（3）比较21—30岁和41—50岁这两个年龄段对此规定的支持率的高低（四舍五入到1%，注：某年龄段的支持率100=
⨯该年龄段支持人数该年龄段被调查人数
%）．
六、实践应用（27题10分，28题10分，共20分）
1
1 1 1
2 1 1
3 3 1 1
4 6 4 1
．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．． Ⅰ
Ⅱ
27．为预防“手足口病”，某校对教室进行“药熏消毒”．已知药物燃烧阶段，室内每立方米空气中的含药量y （mg ）与燃烧时间x （分钟）成正比例；燃烧后，y 与x 成反比例（如图所示）．现测得药物10分钟燃完，此时教室内每立方米空气含药量为8mg ．据以上信息解答下列问题： （1）求药物燃烧时y 与x 的函数关系式． （2）求药物燃烧后y 与x 的函数关系式．
（3）当每立方米空气中含药量低于时，对人体方能无毒害作用，那么从消毒开始，经多长时间学生才可
以回教室？
28．又到了一年中的春游季节，某班学生利用周末到白塔山去参观“晏阳初博物馆”． 下面是两位同学的一段对话： 甲：我站在此处看塔顶仰角为60o
乙：我站在此处看塔顶仰角为30o 甲：我们的身高都是 乙：我们相距20m
请你根据两位同学的对话，计算白塔的高度（精确到1米）．
七、实践探索（10分）
29．王强在一次高尔夫球的练习中，在某处击球，其飞行路线满足抛物线
218
55
y x x =-+，其中y （m ）
是球的飞行高度，x （m ）是球飞出的水平距离，结果球离球洞的水平距离还有2m ． （1）请写出抛物线的开口方向、顶点坐标、对称轴． （2）请求出球飞行的最大水平距离．
（3）若王强再一次从此处击球，要想让球飞行的最大高度不变且球刚好进洞，则球飞行路线应满足怎样的抛物线，求出其解析式． 八、拓展探索（12分）
30．已知：如图14，抛物线
23
34y x =-+与x
轴交于点A ，点B ，与直线3
4y x b =-+相交
于点B ，点C ，直线3
4
y x b =-+与y 轴交于
点E ．
（1）写出直线BC 的解析式． （2）求ABC △的面积．
（3）若点M 在线段AB 上以每秒1个单位长度的速度从A 向B 运动（不与A B ，重合），同时，点N 在射线BC 上以每秒2个单位长度的速度从B 向C 运动．设运动时间为t 秒，请写出MNB △的面积S 与t 的函数关系式，并求出点M 运动多少时间时，MNB △的面积最大，最大面积是多少？
巴中市二○○八年高中阶段教育学校招生考试
数学试题参考答案及评分意见
请评卷老师注意：解答题中各题的解法并不唯一，此答案只是给出一种参考答案，在评卷中，请视具体情况给分．
一、选择题（每小题3分，共30分） 1．C
2．B
3．A
4．D
5．A
6．C
7．D
8．B
9．D
10．B
二、填空题（每小题3分，共30分） 11．3
12．7
2.03710⨯
13．2
(2)x x y -
14．
13
15．5 16．(1)a b -（或ab a -） (1)a b -（或ab a -）
17．90o
18．6-
19．
134
20．5
4322345510105a
a b a b a b ab b +++++
三、解答题（每小题6分，共18分） 21．2
6160x x --=
解：(8)(2)0x x -+= ························································································ 3分 80x -=或20x += ·
························································································· 5分 18x ∴=，22x =- ···························································································· 6分
22
．解：原式12)1=-+ ············································································· 2分
121=-+ ······························································································· 5分
4=-······································································································ 6分
23．解：聪聪说的有理． ······················································································ 1分
2(2)211(2)(2)(2)x x x x x x x
-+=⨯-++-- ······································································· 3分 11
1x x =
-+·
······································································································ 4分 1=·
················································································································· 5分 ∴只要使原式有意义，无论x 取何值，原式的值都相同，为常数1． ······························ 6分 四、推理论证（24题10分，25题10分，共20分） 24．（1）证明：Q 点E 是DC 中点
DE CE ∴= ···························································1分 又AD BC Q ∥，F 在AD 延长线上，
DFE EBC ∴∠=∠，FDE ECB ∠=∠ ·
······························································· 3分 在BCE △与FDE △中EBC DFE
ECB FDE CE DE ∠=∠⎧⎪
∠=∠⎨⎪=⎩ ····························································· 5分
(AAS)BCE FDE ∴△≌△ ················································································· 6分
（2）四边形BCFD 是平行四边形．理由如下： ························································ 7分 DE CE ∴=，FE BE = ·
··················································································· 9分 ∴四边形BCFD 是平行四边形． ·········································································· 10分
A
B C
D
E F
25．证明：Q 点D 在BAC ∠的平分线上
12∴∠=∠ ·
······································································································ 1分 又DE AC Q ∥
23∴∠=∠，13∴∠=∠ ·
·················································································· 2分 AE DE ∴= ·
···································································································· 3分 又BD AD ⊥Q 于点D ，90ADB ∴∠=o
······························································· 4分 1390EBD EDB ∴∠+∠=∠+∠=o ·
································ 5分 EBD EDB ∴∠=∠························································ 6分 BE DE ∴= ·
································································ 7分 AE BE DE ∴== ·
························································ 8分 Q 过A B D ，，三点确定一圆，又90ADB ∠=o
AB ∴是A B D ，，所在的圆的直径． ·
···································································· 9分 ∴点E 是A B D ，，所在的圆的圆心． ·································································· 10分 五、社会实践（26题10分，共10分）
26．（1）21－30 ···································3分
解：（2）40083332⨯=％（人） ·
··········4分 332(6015032135)72-++++=（人） 5分 （3）21－30岁的支持率：
150
1009640039⨯⨯％≈％％
·
···················7分 41－50岁的支持率：
32
1005340015⨯⨯％≈％％
·
····················9分 ∴20－30岁年龄段的市民比41－50岁年龄段的
市民对此规定的支持率高，约高43个百分点． ························································· 10分 六、实践应用（27题10分，28题10分，共20分）
27．解：（1）设药物燃烧阶段函数解析式为11(0)y k x k =≠，由题意得：
1810k = ··········································································································· 2分
145k =
．∴此阶段函数解析式为4
5
y x = ································································ 3分 （2）设药物燃烧结束后的函数解析式为22(0)k
y k x
=≠，由题意得：
2810
k = ············································································································ 5分
280k =．∴此阶段函数解析式为80
y x
=································································ 6分
（3）当 1.6y <时，得80
1.6x
< ············································································ 7分 0x >Q ············································································································ 8分 A B
C
D
E
1 2
3 支持人数
年龄段
各年龄段抽调支持人数条形图（b ）
20 －
30 －
40 －
50
－
60 －
65
－
A
M N P C
B 乙 甲
50x > ·
··········································································································· 9分 ∴从消毒开始经过50分钟后学生才可回教室． ························································ 10分 28．由题目可得：
30CAB ∠=o ，60CBD ∠=o ，20m AB =
1.5m AM BN DP === ·
··················································································· 1分 在ABC △中，CBD ACB CAB ∠=∠+∠ ····························································· 2分
ACB CAB ∴∠=∠ ···························································································· 3分 20m BC AB ∴==·
··························································································· 4分 在Rt CBD △中，20m BC =，60CBD ∠=o
sin CD CBD BC ∠=
·································6分 sin 6020CD
∴=o ···································7分 ·········································································· 8分 103 1.519m CP CD DP ∴=+=≈． 答：白塔的高度约为19米． ················································································ 10分 七、实践探索（29题10分，共10分） 29．解：（1）
218
55
y x x =-+
2116
(4)55
x =--+ ·
··························································································· 1分 ∴抛物线21855y x x =-+开口向下，顶点为1645⎛⎫
⎪⎝⎭
，，对称轴为4x = ························· 3分
（2）令0y =，得：
218
055
x x -+= ·
······························································································· 4分 解得：10x =，28x = ························································································ 5分
∴球飞行的最大水平距离是8m ． ··········································································· 6分
（3）要让球刚好进洞而飞行最大高度不变，则球飞行的最大水平距离为10m
∴抛物线的对称轴为5x =，顶点为1655⎛⎫ ⎪⎝⎭
， ···························································· 7分
设此时对应的抛物线解析式为
216
(5)5
y a x =-+
······················································ 8分 又Q 点(00)，
在此抛物线上，16
2505
a ∴+= 16
125
a =-
········································································································ 9分。