初中毕业学业与升学考试数学试卷3新课程）
本试卷满分为150分，考试时间为120分钟．
一、选择题（每小题只有一个正确选项，将正确选项的序号填入题后的括号内．10小题，每小题3分，共30分）
1．
的值是（ ）
Ａ．1 Ｂ．2 Ｃ．
Ｄ．
2．下列运算中，不正确的是（ ）
Ａ．
Ｂ．
Ｃ．
Ｄ．
3．方程
的解是（ ）
Ａ．1 Ｂ．2 Ｃ．3 Ｄ．
4．下列物体中，是同一物体的为（ ）
Ａ．（1）与（2） Ｂ．（1）与（3）
Ｃ．（1）与（4） Ｄ（2）与（3）
5．“数轴上的点并不都表示有理数，如图中数轴上的点
所表示的数是
”，这种利用图形直观说明问题的方式体现的数学思想方法叫（ ）
Ａ．代入法
Ｂ．换元法
Ｃ．数形结合的思想方法
Ｄ．分类讨论的思想方法
6．在一幅长60cm，宽40cm的矩形风景画的四周镶一条金色纸边，制成一幅矩形挂图，如图所示．如果要使整个挂图的面积是
，设金色纸边的宽为
，那么
满足的方程是（ ）
Ａ．
Ｂ．
Ｃ．
Ｄ．
7．夏天，一杯开水放在桌子上，杯中水的温度
随时间
变化的关系的大致图象是（ ）
8．如图，
的边长都大于2，分别以它的顶点为圆心，1为半径画弧（弧的端点分别在三角形的相邻两边上），则这三条弧的长的和是（ ）
Ａ．
Ｂ．
Ｃ．
Ｄ．
9．一人乘雪橇沿如图所示的斜坡笔直滑下，滑下的距离
（米）与时间（秒）间的关系式为
，若滑到坡底的时间为2秒，则此人下滑的高度为（ ）
Ａ．24米 Ｂ．12米
Ｃ．
米 Ｄ．6米
10．某超市（商场）失窃，大量的商品在夜间被罪犯用汽车运走．三个嫌疑犯被警察局传讯，警察局已经掌握了以下事实：（1）罪犯不在甲、乙、丙三人之外；（2）丙作案时总得有甲作从犯；（3）乙不会开车．在此案中，能肯定的作案对象是（ ）
Ａ．嫌疑犯乙 Ｂ．嫌疑犯丙 Ｃ．嫌疑犯甲 Ｄ．嫌疑犯甲和丙
二、填空题（把答案填在题中的横线上．8小题，每小题4分，共32分）
11．某种感冒病毒的直径是0.米，用科学记数法表示为 米．
12．如图中两圆的位置关系是 （相交，外切，外离）．
13．反比例函数
的图象位于 象限．
14．下列是三种化合物的结构式及分子式，则按其规律第4个化合物的分子式为 ．
15．小明将一把钥匙放进自己家中的抽屉中，他记不清到底放进三个抽屉中的哪一个了，那么他一次选对抽屉的概率是 ．
16．如图，
，
平分
，若
，则
．
17．为等边三角形，
分别在边
上，且
，则
为 三角形．
18．二次函数
图象上部分点的对应值如下表：
|
|
|
|
| 0 | 1 | 2 | 3 | 4 |
|
| 6 | 0 |
|
|
|
| 0 | 6 |
则使
的的取值范围为 ．
三、解答题（解答时，必须写出必要的解题步骤．5小题，共38分）
19．（6分）先化简，再求值：
，其中
．
20．（8分）有7名同学测得某楼房的高度如下（单位：米）：
29，28.5，30，30，32，31，33．
（1）求这组数据的中位数、众数、平均数；
（2）你认为此楼房大概有多高？
21．（8分）一次函数图象如图所示，求其解析式．
22．（8分）如图，已知
．求证：
．
23．（8分）为节约用电，某学校在本学期初制定了详细的用电计划．如果实际每天比计划多用2度电，那么本学期的用电量将会超过2990度；如果实际每天比计划节约2度电，那么本学期的用电量将不超过2600度．若本学期的在校时间按130天计算，那么学校原计划每天用电量应控制在什么范围内？
四、解答题（解答时，必须写出必要的解题步骤．5小题，共50分）
24．（8分）如图是两个半圆，点
为大半圆的圆心，
是大半圆的弦关与小半圆相切，且
．问：能求出阴影部分的面积吗？若能，求出此面积；若不能，试说明理由．
25．（10分）一架长5米的梯子，斜立在一竖直的墙上，这时梯子底端距墙底3米．如果梯子的顶端沿墙下滑1米，梯子的底端在水平方向沿一条直线也将滑动1米吗？用所学知识，论证你的结论．
26．（10分）如图，
为正方形
边的中点，
是延长线上的一点，
，且交
的平分线于
．
（1）求证：
；
（2）若将上述条件中的“为边的中点”改为“为边上任意一点”，其余条件不变，则结论“”成立吗？如果成立，请证明；如果不成立，说明理由．
27．（10分）某公司现有甲、乙两种品牌的计算器，甲品牌计算器有
三种不同的型号，乙品牌计算器有
两种不同的型号，新华中学要从甲、乙两种品牌的计算器中各选购一种型号的计算器．
（1）写出所有的选购方案（利用树状图或列表方法表示）；
（2）如果（1）中各种选购方案被选中的可能性相同，那么
型号计算器被选中的概率是多少？
（3）现知新华中学购买甲、乙两种品牌计算器共40个（价格如图所示），恰好用了1000元人民币，其中甲品牌计算器为型号计算器，求购买的型号计算器有多少个？
28．（12分）如图，在
中，
所对的圆心角为
，已知圆的半径为2cm，并建立如图所示的直角坐标系．
（1）求圆心的坐标；
（2）求经过三点的抛物线的解析式；
（3）点
是弦所对的优弧上一动点，求四边形
的最大面积；
（4）在（2）中的抛物线上是否存在一点，使
和相似？若存在，求出点的坐标；若不存在，请说明理由．
初中毕业学业与升学考试（新课程）
数学试题参考答案与评分标准
一、选择题：每小题3分，共30分
ＡＣＤＢＣ ＡＢＤＢＣ
二、填空题：每小题4分，共32分
11．
12．外离 13．一，三 14．
15．
16．
17．正 18．
三、解答题：共38分
19．本小题满分6分
解：原式
，······································································································· 4分
当时，原式
．······················································································ 6分
说明：会运用分配律且对者给2分，会用公式
且对者给1分，若先通分且对者给2分．
20．本小题满分8分
解：（1）在这组数据中，中位数是30，········································································· 2分
众数是30，··················································································································· 4分
平均数是30.5；············································································································· 6分
（2）该楼房大概高30米（未写单位不扣分）······························································· 8分
21．本小题满分8分
解：设一次函数解析式为
，············································································ 1分
则
··········································································································· 5分
解得
所以，一次函数解析式为
．············································································ 8分
说明：只要求出无最后一步不扣分．
22．本小题满分8分
说明：
······································································································· 2分
．····································································································· 4分
．················································································································ 6分
又
，
．·················································································································· 8分
23．本小题满分8分
解：设学校原计划每天用电量为度，·········································································· 1分
依题意得
························································································· 5分
解得
．
即学校每天的用电量，应控制在21～22度（不包括21度）范围内．····························· 8分
说明：只要求出，无最后一步不扣分．
四、解答题：共50分
24．本小题满分8分
解法1：
能（或能求出阴影部分的面积）．·················································································· 1分
设大圆与小圆的半径分别为
，················································································ 2分
作辅助线如图所示（作对），························································································· 4分
可得
，······································································································· 6分
．··················································································· 8分
解法2：
能（或能求出阴影部分的面积）．·················································································· 1分
设大圆与小圆的半径分别为··················································································· 2分
平移小半圆使它的圆心与大半圆的圆心重合（如图）．··············································· 3分
作
于
，则
，
．··························································· 5分
，············································································································ 6分
．······································································· 8分
25．本小题满分10分
是．······························································································································ 2分
证明1：
在
中，
米．·········································· 4分
米．········································································································· 6分
在
中，
米．········································ 8分
．即梯子底端也滑动了1米．···························································· 10分
证明2：
在中，米．·········································· 4分
米．········································································································· 6分
可证
．························································································· 8分
米．
．即梯子底端也滑动了1米．···························································· 10分
26．本小题满分10分
证明：（1）取
的中点
，连结
．······································································ 1分
易证
，
······································································································· 3分
．·············································································································· 4分
（2）结论“
”仍成立．················································································ 5分
证明如下：
在上截取
，连结
．··········································································· 6分
，
．··············································································································· 7分
，
．···································································································· 8分
又
，
．··································································································· 9分
．············································································································· 10分
27．本小题满分10分
解：（1）树状图表示如下：
····································································································································· 2分
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
甲
列表表示如下：
····································································································································· 2分
有6种可能结果：
．··································· 3分
说明：用其它方式表达选购方案且正确者，只给1分．
（2）因为选中型号计算器有2种方案，即
，所以型号计算器被选中的概率是
． 5分
（3）由（2）可知，当选用方案
时，设购买型号，型号计算器分别为
个，
根据题意，得
解得
经检验不符合题意，舍去；··························································································· 7分
当选用方案
时，设购买型号、型号计算器分别为个，
根据题意，得
解得
································································ 9分
所以新华中学购买了5个型号计算器．···································································· 10分
说明：设购买型号计算器台，（或）型号计算器为
个，用一元一次方程解答，同样给分．
28．本小题满分12分
解：（1）如图（1），连结
．
则
，
，
．···················································································· 2分
，
．······················································································ 3分
（2）由三点的特殊性与对称性，
知经过三点的抛物线的解析式为
．·················································· 4分
，
，
．···································································································· 5分
．
．·············································································································· 6分
说明：只要求出
，无最后一步不扣分．
（3）
，又
与均为定值，··································· 7分
当
边上的高最大时，
最大，此时点为与
轴的交点，如图（1）．
····································································································································· 8分
．········· 9分
（4）方法1：
如图（2），
为等腰三角形，
，
等价于
．······················ 10分
设
且
，则
，
．······································································································ 11分
又
的坐标满足
，
在抛物线上，存在点
，
使
．
由抛物线的对称性，知点
也符合题意．
存在点，它的坐标为
或．···························································· 12分
说明：只要求出，，无最后一步不扣分．下面的方法相同．
方法2：
如图（3），当时，
，又由（1）知
，
点在直线
上．
设直线的解析式为，
将
代入，解得
直线的解析式为
．··········································································· 10分
解方程组
得．············································································ 11分
又
，
．
，．
在抛物线上，存在点，使．
由抛物线的对称性，知点也符合题意．
存在点，它的坐标为或．···························································· 12分
方法3：
如图（3），为等腰三角形，且
，设，则
等价于
，
．·········································· 10分
当时，得
解得．··········································································································· 11分
又
的坐标满足，
在抛物线上，存在点，使．
由抛物线的对称性，知点也符合题意．
存在点，它的坐标为或．