高中阶段学校招生考试数学试题(课改区)
本试卷分为第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.第Ⅰ卷1—2页,第Ⅱ卷3—10页,共120分.考试时间120分钟.
第Ⅰ卷(选择题 共30分)
注意事项:
1.数学考试允许使用科学计算器(凡符合大纲或课程标准要求的计算器都可带入考场).
2.数学考试允许考生进行剪、拼、折叠实验.
3.答第Ⅰ卷前,考试务必将自己的姓名、准考证号、考试科目填涂在答题卡上.
4.每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,再选涂其他答案,答案写在试卷上无效.
5.考试结束,监考人将本试卷和答题卡一并收回.
一、选择题:本大题共10小题,每小题3分,共30分,在每小题给出的四个选项中,只有一项是符合题目要求的.
1.
如图,数轴上
两点所表示的两数的( )
A.和为正数 B.和为负数
C.积为正数 D.积为负数
2.下列计算错误的是( )
A.
B.
C.
D.
3.如图,是一个正在绘制的扇形统计图,整个圆表示某班参加
体育活动的总人数,那么表示参加立定跳远训练的人数占总人数
的35%的扇形是( )
A. B.
C. D.
4.如图,直线与直线
互相平行,则
的值是( )
A.20 B.80
C.120 D.180
5.亮亮准备用自己节省的零花钱买一台英语复读机,他现在已存有45元,计划从现在起以后每个月节省30元,直到他至少有300元.设个月后他至少有300元,则可以用于计算所需要的月数
的不等式是( )
A. B.
C. D.
6.如图,雷达可用于飞机导航,也可用来监测飞
机的飞行.假设某时刻雷达向飞机发射电磁波,电
磁波遇到飞机后反射,又被雷达接收,两个
过程共用了秒.已知电磁波的传播速度为
米/秒,则该时刻飞机与雷达站的距离是( )
A.米 B.
米 C.
米 D.
米
|
A. B.
C.
D.
8.如图,用8个积木搭成了的立方休,其中
的长方体有3个,的长方体有2个,
的长方
体有1个,的立方体有2个.某人站在该立方体的左侧
观察,请你判断他看到的图形是( )
9.如图,直线是函数
的图象.若点
满足,且
,则
点的坐标可能是( )
A. B.
C. D.
![]() |
10.如图,是半径为6的
的
圆周,
点是
上
的任意一点,是等边三角形,则四边形
的周
长的取值范围是( )
A. B.
C. D.
第Ⅱ卷(非选择题 共90分)
注意事项:
1.第Ⅱ卷共8页,用钢笔或圆珠笔直接答在试卷上.
2.答卷前将密封线内的项目填写清楚.
二、填空题:本大题共6小题,每小题3分,共18分,把答案填写在题中的横线上.
11.若分式的值为零,则
的值为
.
12.根据如图的程序,计算当输入
时,输出的结果
.
13.如图,一根电线杆的接线柱部分在阳光下的投影
的长为1.2m,太阳光线与地面的夹角,则
的
长为 m.
14.如图,是反比例函数
在第一象限内的图象,且过点
与
关于
轴对称,那么图象
的函数解析式为
(
).
15.如图,矩形中,
,将矩形
在直线
上按顺时针方向不滑动的每秒转动
,转动3秒后停止,则顶点
经过的路线长为
.
16.现有若干张边长不相等但都大于4cm的正方形纸片,从中任选一张,如图从距离正方形的四个顶点2cm处,沿
角画线,将正方形纸片分成5部分,则中间阴影部分的面积是
cm
;若在上述正方形纸片中再任选一张重复上述过程,并计算阴影部分的面积,你能发现什么规律?
.
![]() |
三、解答题:本大题共11小题,共72分,解答应写出文字说明或演算步骤.
17.(本题5分)请你从下列各式,任选两式作差,并将得到的式子进行因式分解.
.
18.(本题5分)解方程.
19.(本题6分)小明和小丽用形状大小相同、面值不同的5张邮票设计了一个游戏,将面值1元、2元、3元的邮票各一张装入一个信封,面值4元、5元的邮票各一张装入另一个信封.游戏规定:分别从两个信封中各抽取1张邮票,若它们的面值和是偶数,则小明赢;若它们的面值和是奇数,则小丽赢,请你判断这个游戏对双方是否公平,并说明理由.
20.(本题7分)某高校共有5个餐厅和2个小餐厅,经过测试:同时开放1个大餐厅、2个小餐厅,可供1680名学生就餐;同时开放2个大餐厅、1个小餐厅,可供2280名学生就餐.
(1)求1个大餐厅、1个小餐厅分别可供多少名学生就餐;
(2)若7个大餐厅同时开放,能否供全校的5300名学生就餐?请说明理由.
21.(本题6分)元旦联欢会前某班布置教室,同学们利用彩纸条粘成一环套一环的彩纸链,小颖测量了部分彩纸链的长度,她得到的数据如下表:
纸环数 | 1 | 2 | 3 | 4 | …… |
彩纸链长度 | 19 | 36 | 53 | 70 | …… |
(1)把上表中的各组对应值作为点的坐标,在如图的平面直角坐标系中描出相应的点,猜想
与
的函数关系,并求出函数关系式;
(2)教室天花板对角线长10m,现需沿天花板对角线各拉一根彩纸链,则每根彩纸链至少用多少个纸环?
22.(本题6分)如图1,分别表示边长为
的等边三角形和正方形,
表示直径为
的圆.图2是选择基本图形
用尺规画出的图案,
.
(1)请你从图1中任意选择两种基本图形,按给定图形的大小设计一个新图案,还要选择恰当的图形部分涂上阴影,并计算阴影的面积;(尺规作图,不写作法,保留痕迹,作直角时可以使用三角板)
(2)请你写一句在完成本题的过程中感受较深且与数学有关的话.
23.(本题6分)某数学老师为了了解学生在数学学习中常见错误的纠正情况,收集了学生在作业和考试中的常见错误,编制了10道选择题,每题3分,对她所任教的初三(1)班和(2)班进行了检测.如图表示从两班各椭机抽取的10名学生的得分情况:
(1)利用图中提供的信息,补全下表:
班级 | 平均数(分) | 中位数(分) | 众数(分) |
(1)班 | 24 | 24 | |
(2)班 | 24 |
(2)若把24分以上(含24分)记为“优秀”,两班各有60名学生,请估计两班各有多少名学生成绩优秀;
(3)观察图中的数据分布情况,你认为哪个班的学生纠错的整体情况更好一些?
![]() |
24.(本题7分)如图,在与
中,
,
相交于点
,过点
作
交
的延长线于点
,过点
作
交
的延长线于点
相交于点
.
(1)图中有若干对三角形是全等的,请你任选一对进行证明;(不添加任何辅助线)
(2)证明四边形是菱形;
(3)若使四边形
是正方形,还需在
的边长之间添加一个什么条件?请你写出这个条件.(不必证明)
25.(本题7分)某校数学研究性学习小组准备设计一种高为60cm的简易废纸箱.如图1,废纸箱的一面利用墙,放置在地面上,利用地面作底,其它的面用一张边长为60cm的正方形硬纸板围成.经研究发现:由于废纸箱的高是确定的,所以废纸箱的横截面图形面积越大,则它的容积越大.
(1)该小组通过多次尝试,最终选定下表中的简便且易操作的三种横截面图形,如图2,是根据这三种横截面图形的面积
与
(见表中横面图形所示)的函数关系式而绘制出的图象.请你根据有关信息在表中空白处填上适当的数、式,并完成
取最大值时的设计示意图;
横截面图形 | | | |
| | | |
| 30 | 20 | |
|
| | |
| | |
(2)在研究性学习小组展示研究成果时,小华同学指出:图2中“底角为
的等腰梯形”的图象与其他两个图象比较,还缺少一部分,应该补画.你认为他的说法确吗?请简要说明理由.
26.(本题8分)如图1,以矩形的两边
和
所在的直线为
轴、
轴建立平面直角坐标系,
点的坐标为
点的坐标为
.将矩形
绕
点逆时针旋转,使
点落在
轴的正半轴上,旋转后的矩形为
相交于点
.
(1)求点的坐标与线段
的长;
(2)将图1中的矩形沿
轴向上平移,如图2,矩形
是平移过程中的某一位置,
相交于点
,点
运动到
点停止.设点
运动的距离为
,矩形
与原矩形
重叠部分的面积为
,求
关于
的函数关系式,并写出
的取值范围;
(3)如图3,当点
运动到点
时,平移后的矩形为
.请你思考如何通过图形变换使矩形
与原矩形
重合,请简述你的做法.
27.如图1,已知中,
,
,过点
作
,且
,连接
交
于点
.
(1)求的长;
(2)以点为圆心,
为半径作
,试判断
与
是否相切,并说明理由;
(3)如图2,过点作
,垂足为
.以点
为圆心,
为半径作
;以点
为圆心,
为半径作
.若
和
的大小是可变化的,并且在变化过程中保持
和
相切,且使
点在
的内部,
点在
的外部,求
和
的变化范围.
![]() |
高中阶段学校招生考试
数学试题参考答案及评分标准(课改区)
一、 选择题
1.D 2.C 3.C 4.A 5.B 6.A 7.A 8.D 9.B 10.C
二、填空题
11.1 12.2 13.2.1 14. 15.
16.8;························································································································· 2分
得到的阴影部分的面积是,即阴影部分的面积不变.···································· 3分
三、解答题
17.本题存在12种不同的作差结果,不同选择的评分标准分述如下:
;
;
;
;
;
这6种选择的评分范例如下:
例1:············································································································ 2分
.······················································································ 5分
;
;
;
;
;
这6种选择的评分范例如下:
例2:·········································································································· 2分
·················································································· 4分
.···················································································· 5分
提示:因式分解结果正确但没有中间步骤的不扣分.
18.方程两边同乘以,得
.························································· 2分
解这个方程,得.································································································ 4分
检验:将代入原方程,得左边
右边.
所以,是原方程的根.························································································· 5分
19.游戏对双方是公平的.··························································································· 1分
通过列表或树状图等方法,求得.································································ 3分
.································································ 5分
因为,所以游戏对双方是公平的.························································· 6分
20.(1)设1个大餐厅可供名学生就餐,1个小餐厅可供
名学生就餐,根据题意,得
································································································································· 得1分
··········································································································· 3分
解这个方程组,得
答:1个大餐厅可供960名学生就餐,1个小餐厅可供360名学生就餐.························ 5分
(2)因为,
所以如果同时开放7个餐厅,能够供全校的5300名学生就餐.····································· 7分
21.(1)在所给的坐标系中准确描点.·········································································· 1分
由图象猜想到与
之间满足一次函数关系.······························································· 2分
设经过,
两点的直线为
,则可得
解得
,
.即
.
当时,
;当
时,
.
即点都在一次函数
的图象上.
所以彩纸链的长度(cm)与纸环数
(个)之间满足一次函数关系
····································································································································· 4分
(2),根据题意,得
.·············································· 5分
解得.
答:每根彩纸链至少要用59个纸环.············································································ 6分
22.(1)正确运用两种基本图形进行组合设计.···························································· 3分
尺规作图运用恰当.························································································ 4分
阴影面积计算正确.························································································ 5分
参考举例:
(2)写出在解题过程中感受较深且与数学有关的一句话.······································· 6分
参考举例:
① 运用圆的半径,可以作正方形的边上的中点,这对于作图很有利.
② 这三个图形关系很密切,能组合设计许多美丽的图案,来装饰我们的生活.
③ 数学作图中要一丝不苟,否则产生的作图误差会影响图形的美观.
提示:本问题应积极评价学生富有个性和创造性的解答,只要回答合理,即可得分.
23.(1)
班级 | 平均数(分) | 中位数(分) | 众数(分) |
(1)班 | 24 | ||
(2)班 | 24 | 21 |
····································································································································· 3分
(2)(名),
(名).
答:(1)班有42名学生成绩优秀,(2)班有36名学生成绩优秀.·················· 5分
(3)(1)班的学生纠错的整体情况更好一些.······················································ 6分
24.(1).······················································································ 1分
,
.········································································· 3分
(2),
四边形
是平行四边形.······················· 4分
,
.··························· 5分
平行四边形
是菱形.···································································· 6分
(3)需要添加的条件是.······································································ 7分
25.(1)表中空白处填写项目依次为;15;450.································· 3分
表中取最大值时的设计示意图分别为
![]() |
····································································································································· 5分
(2)小华的说法不正确.······················································································ 6分
因为腰长大于30cm时,符合题意的等腰梯形不存在,所以
的取值范围不能超过30cm,因此研究性学习小组画出的图象是正确的.·································································································· 7分
26.(1)如图1,因为,所以点
的坐标为
.·············· 2分
.······················································································ 3分
(2)在矩形沿
轴向上平移到
点与
点重合的过程中,点
运动到矩形
的边
上时,求得
点移动的距离
.
当自变量的取值范围为
时,如图2,由
,
得,此时,
.
即(或
).······················································ 5分
当自变量的取值范围为
时,
求得(或
).··········································· 7分
(3)部分参考答案:···································································································· 8分
①把矩形沿
的角平分线所在直线对折.
②把矩形绕
点顺时针旋转,使点
与点
重合,再沿
轴向下平移4个单位长度.
③把矩形绕
点顺时针旋转,使点
与点
重合,再沿
所在的直线对折.
④把矩形沿
轴向下平移4个单位长度,再绕
点顺时针旋转,使点
与点
重合.
提示:本问只要求整体图形的重合,不必要求图形原对应点的重合.
27.(1)在
中,
,
.······················································································ 1分
,
.
.
,
.·························································· 3分
(2)与
相切.·························································································· 4分
在
中,
,
,
,
.····································· 5分
又,
,
与
相切.························································································ 6分
(3)因为,所以
的变化范围为
.····················· 7分
当与
外切时,
,所以
的变化范围为
;
······················································································································· 8分
当与
内切时,
,所以
的变化范围为
.
······················································································································· 9分