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高中阶段学校招生考试数学试题(课改区)

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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.

卷(选择题   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.

 
7.某快餐店用米饭加不同炒菜配制了一批盒饭,配土豆丝炒肉的有25盒,配芹菜炒肉丝的有30盒,配辣椒炒鸡蛋的有10盒,配芸豆炒肉片的有15盒.每盒盒饭的大小、外形都相同,从中任选一盒,不含辣椒的概率是(  )

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的圆周,点是

的任意一点,是等边三角形,则四边形的周

的取值范围是(  )

A.               B.

C.         D.

卷(非选择题 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

长为      m.

14.如图,是反比例函数在第一象限内的图象,且过点关于轴对称,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(个)

1

2

3

4

……

彩纸链长度(cm)

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,在中,

相交于点,过点的延长线于点,过点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(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

的函数关系式

取最大值时(cm)的值

30

20

取得的最大值

450

取最大值时的设计示意图

(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,且,连接于点

(1)求的长;

(2)以点为圆心,为半径作,试判断是否相切,并说明理由;

(3)如图2,过点,垂足为.以点为圆心,为半径作;以点为圆心,为半径作.若的大小是可变化的,并且在变化过程中保持相切,且使点在的内部,点在的外部,求的变化范围.


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

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

,此时,

(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)中,

    .······················································································ 1分

    

    

    .·························································· 3分

  (2)相切.·························································································· 4分

    中,

    .····································· 5分

    又

    相切.························································································ 6分

  (3)因为,所以的变化范围为.····················· 7分

    当外切时,,所以的变化范围为

······················································································································· 8分

    当内切时,,所以的变化范围为

······················································································································· 9分