初中毕业、升学考试数学试题
（满分：150分 考试时间：120分钟）
一、填空题（每小题3分，计36分）
1．
的相反数是 ．
2．因式分解：
．
3．从《闽西日报》获悉：今年我市对农村九年制义务教育免除学杂费的中、小学生数约330000人．330000用科学记数法表示为 ．
4．不等式组
的解集是 ．
5．函数
中自变量
的取值范围是 ．
6．如图，已知
，则
．
7．化简：
．
8．如图，小明在操场上距离旗杆18米的
处，用测角仪测得旗杆
的顶端
的仰角为
，已知测角仪
的高为1.4米，那么旗杆的高为 米．（保留三个有效数字）
9．圆锥的侧面展开图的面积为
，母线长为3，则该圆锥的底面半径为 ．
10．如图，矩形
的对角线
和
相交于点
，过点的直线分别交
和
于点
，
，则图中阴影部分的面积为 ．
 | |||
 | |||
11．如图，已知
的半径为5，弦
是弦上一点，且
，则
．
12．已知实数
满足：
，那么
的值为 ．
二、选择题（每小题4分，计32分）在每小题给出的四个选项中，只有一个是符合题目要求的，请把正确选项的代号填入题后的括号内
13．下列计算正确的是（ ）
Ａ．
Ｂ．
Ｃ．
Ｄ．
14．下面四张扑克牌中，属于中心对称图形的是（ ）
 |
15．用换元法解分式方程
时，设
，则原方程可化为整式方程（ ）
Ａ．
Ｂ．
Ｃ．
Ｄ．
16．已知两圆的半径分别为3cm和7cm，圆心距为10cm，则这两圆的公切线条数是（ ）
Ａ．1条 Ｂ．2条 Ｃ．3条 Ｄ．4条
17．根据欧姆定律
，当电压
一定时，电阻
与电流
的函数图象大致是（ ）
18．下列说法错误的是（ ）
Ａ．矩形的四个角都相等 Ｂ．四条边都相等的四边形是菱形
Ｃ．等腰梯形的对角线相等 Ｄ．对角线互相垂直平分的四边形是正方形
19．如图，小亮用六块形状、大小完全相同的等腰梯形拼成一个四边形，则图中
的度数是（ ）
Ａ．
Ｂ．
Ｃ．
Ｄ．
20．已知抛物线
与轴交于
两点，则线段的长度为（ ）
Ａ．
Ｂ．
Ｃ．
Ｄ．
三、解答题（共8小题，计82分）
21．（8分）计算：
22．（8分）如图，在
中，点分别在
上，在不添加辅助线的情况下，请你添加一个适当的条件，使
和
全等，你添加的条件是 ，并给出你的证明．
证明：
23．（9分）我国宋朝数学家杨辉在他的著作《详解九章算法》中提出“杨辉三角”（如下图），此图揭示了
（
为非负整数）展开式的项数及各项系数的有关规律．
例如：
，它只有一项，系数为1；
，它有两项，系数分别为1，1，系数和为2；
，它有三项，系数分别为1，2，1，系数和为4；
，它有四项，系数分别为1，3，3，1，系数和为8；
……
根据以上规律，解答下列问题：
（1）
展开式共有 项，系数分别为 ；
（2）展开式共有 项，系数和为 ．
24．（10分）某县为了解初三6000名学生初中毕业考试数学成绩（分数为整数）从中抽取了200名学生的数学成绩进行分析，下面是200名学生数学成绩的频率分布表：
| 分组 | 频数 | 频率 |
| 89.5～99.5 | 12 |
|
| 99.5～109.5 | 24 | 0.12 |
| 109.5～119.5 | 36 | 0.18 |
| 119.5～129.5 | 68 | 0.34 |
| 129.5～139.5 |
| 0.2 |
| 139.5～149.5 | 20 | 0.1 |
| 合计 | 200 | 1 |
根据所给信息回答下列问题：
（1）频率分布表中的数据
；
（2）中位数落在 分数段内；
（3）若成绩不低于120分的为优秀，试估计该县初三学生初中毕业考试数学成绩优秀的学生有 人．
25．（10分）已知：关于的一元二次方程
．
（1）求证：不论
取何值，方程总有两个不相等的实数根；
（2）若方程的两个实数根
满足
，求的值．
26．（11分）某水果经销商上月份销售一种新上市的水果，平均售价为10元／千克，月销售量为1000千克．经市场调查，若将该种水果价格调低至元／千克，则本月份销售量
（千克）与（元／千克）之间满足一次函数关系
．且当
时，
；
时，
．
（1）求与之间的函数关系式；
（2）已知该种水果上月份的成本价为5元／千克，本月份的成本价为4元／千克，要使本月份销售该种水果所获利润比上月份增加
，同时又要让顾客得到实惠，那么该种水果价格每千克应调低至多少元？［利润＝售价－成本价］
27．（12分）如图，已知点为
斜边上一点，以为圆心，
为半径的圆与相切于点
，与相交于点
．
（1）试判断是否平分
？并说明理由；
（2）若
，求的半径．
28．（14分）如图，已知抛物线
与坐标轴交于
三点，点的横坐标为
，过点
的直线
与轴交于点
，点
是线段上的一个动点，
于点
．若
，且
．
（1）确定
的值：
；
（2）写出点
的坐标（其中
用含
的式子表示）：
；
（3）依点的变化，是否存在的值，使
为等腰三角形？若存在，求出所有的值；若不存在，说明理由．
初中毕业、升学考试
参考答案及评分标准
数 学
一、填空题（每小题3分，共36分）
1．
2．
3．
4．
5．
6．
7． 8．
9． 10． 11．
12．
二、选择题（每小题4分，共32分）
13．Ｄ 14．Ｂ 15．Ｄ 16．Ｃ 17．Ｂ 18．Ｄ 19．Ａ 20．Ｄ
三、解答题（8小题，共82分）
21．（8分）
解：原式
·································································· （5分）
······························································································ （7分）
··································································································· （8分）
（备注：第一步每个计算对均给1分，第二步
对1分，
对1分）
22．（8分）
解：条件
（答案不唯一） ································································ （3分）
证明：
四边形是平行四形
································································· （6分）
又
····················································································· （7分）
············································································· （8分）
（备注：给出的条件只要能证明结论就给3分）
23．（9分）
解：（1）
······································································································ （2分）
1，4，6，4，1 ·················································································· （4分）
（2）
································································································· （6分）
···································································································· （9分）
24．（10分）
解：（1）
，
························································································ （4分）
（2）
～
·················································································· （7分）
（3）
······························································································ （10分）
（备注：第1小题每空2分）
25．（10分）
解：（1）
····················································· （2分）
························································································· （3分）
不论取何值，方程总有两个不相等实数根 ············································· （4分）
（2）解法一：由原方程可得
或
················································· （6分）
··························································· （7分）
又
······················································· （8分）
····················································································· （9分）
经检验：
符合题意．
的值为． ········································································ （10分）
解法二：根据根与系数关系有
········· （5分）
又
················································· （6分）
·························································· （7分）
整理得
············································································ （8分）
解得
································································ （9分）
经检验
是增根舍去
的值为． ········································································ （10分）
26．（11分）
解：（1）由已知得
··························································· （3分）
解得
······································································· （5分）
（2）由题意可得
······················ （8分）
整理得
···························································· （9分）
得
（舍去） ························································ （10分）
答：该种水果价格每千克应调低至
元． ········································· （11分）
27．（12分）
（1）判断：平分． ······································································ （2分）
证法一：连结
切于
·············································································· （3分）
又
为
，且
················································································ （4分）
又
··············································································· （5分）
················································································ （6分）
证法二：连结
是直径
······························· （3分）
又
································································· （4分）
又
······························································· （5分）
················································································ （6分）
证法三：连结
是直径
·········································································· （3分）
又
··············································································· （4分）
又
·························································· （5分）
················································································ （6分）
（2）解法一：设
，则
········································ （7分）
据切割线定理得
······································· （8分）
得
··········································· （9分）
又
································································· （10分）
········································································· （11分）
的半径为 ··························································· （12分）
解法二：如图2，过作
，又
，
则四边形
为矩形．
又
·································· （10分）
（舍去） ······················································ （11分）
的半径为 ·································································· （12分）
备注：本解法是在解法一得的基础上进行的．
解法三：如图（1）
·········································································· （7分）
又
············································································· （8分）
又
··································································· （9分）
又
，则
························································· （10分）
在
中，
得
在
中，
·························· （11分）
的半径的长
··········································· （12分）
（备注：本大题第2小题解法较多，若出现其它解法，请根据评分标准量分）
28．（14分）
解：（1）
······························································································· （2分）
································································································ （4分）
（2）
····························································································· （5分）
···························································································· （6分）
······················································································ （8分）
（3）存在的值，有以下三种情况
①当
时
，则
························································································ （10分）
②当
时
得
························································································ （12分）
③当
时，如图
解法一：过作
，又
则
又
········································································ （14分）
解法二：作
斜边中线
则
，
此时
···································································· （14分）
解法三：在
中有
（舍去） ··········································· （14分）
又
当
或
或
时，为等腰三角形．