内蒙古鄂尔多斯市2006年中考数学试题课标卷
本试题满分120分,考试用时120分钟;
一、选择题(本大题10个小题,每小题3分,共30分.在每小题给出的四个选项中只有一个是正确的,请把正确选项的标号填在下面的选项栏内.) 1.计算2
3-的结果是( )
A.6- B.6 C.9- D.9
2.如图1,在一个长方体上放着一个小正方体,若这个组合体的俯视图是图2,则这个组合体的左视图是( )
3.若家用电冰箱冷藏室的温度是4℃,冷冻室比冷藏室的温度低22℃,则冷冻室的温度为( ) A.18-℃ B.18℃ C.26-℃ D.26℃
4.在等边三角形、等腰梯形、平行四边形、正五边形中,是轴对称图形的有( ) A.1个 B.2个 C.3个 D.4个
5.已知圆柱的底面半径为4,高为6,则这个圆柱的侧面积为( ) A.24 B.24π C.48 D.48π 6.将点(53)P ,向下平移1个单位后,落在函数k
y x
=
的图象上,则k 的值为( ) A.10k = B.12k =
C.18k = D.20k =
7.为了解决老百姓看病难的问题,卫生部门决定大幅度降低药品价格,某种常用药品降价40%后的价格为a 元,则降价前此药品价格为( ) A.40%a 元
B.60%a 元
C.
5
2
a 元
D.3
a 5元
8.图3是测量一物体体积的过程:
步骤一,将180ml 的水装进一个容量为300ml 的杯子中. 步骤二,将三个相同的玻璃球放入水中,结果水没有满. 步骤三,同样的玻璃球再加一个放入水中,结果水满溢出.
正面
图1 图
2 A. B. C. D.
根据以上过程,推测一颗玻璃球的体积在下列哪一范围内?3
(1ml 1cm )= A.3
10cm 以上,3
20cm 以下 B.320cm 以上,3
30cm 以下 C.3
30cm 以上,3
40cm 以下
D.3
40cm 以上,3
50cm 以下
9.国家为九年义务教育期间的学生实行“两免一补”政策,下表是我市某中学国家免费提供教科书补助的部分情况.
七
八
九
合计
每人免费补助金额(元)
110 90 50 —— 人数(人)
80 300 免费补助总金额(元)
4000 26200 如果要知道空白处的数据,可设七年级的人数为x ,八年级的人数为y ,根据题意列出方程
组为( )
A.3001109026200x y x y +=??+=?
B.300
11090400026200x y x y +=??++=?
C.80300
400026200
x y x y ++=??
++=?
D.80300
11090400026200
x y x y ++=??
++=?
10.如图4,在直角梯形ABCD 中,90ABC ∠=
,DC AB ∥,
3BC =,4DC =,5AD =.动点P 从B 点出发,由
B C D A →→→沿边运动,
则ABP △的最大面积为( ) A.10 B.12 C.14 D.16
二、填空题(本大题8个小题,每小题3分,共24分)
鄂尔多斯初中数学补习班手机187********
11.一元二次方程2
32x x =的根是 .
12.不等式组21
10
x x ->??
-
.
13.图5是根据我市2001年至2005年财政收入绘制的折线统计图,观察统计图可得:同上年相比我市财政收入增长速度最快的年份是 年,比它的前一年增加 亿年 级
项
目
图4
14.如图6,将一张等腰直角三角形纸片沿中位线剪开可以拼出不同形状的四边形,请写出其中两个不同的四边形的名称: . 15.如图7,在数轴上,A B ,两点之间表示整数的点有 个.
16.如图8,A B C ,,是O 上的三点,2AB =,30ACB ∠=
,那么O 的半径等于 .
17.如图9所示,某校宣传栏后面2米处种了一排树,每隔2米一棵,共种了6棵,小勇站在距宣传栏中间位置的垂直距离3米处,正好看到两端的树干,其余的4棵均被挡住,那么宣传栏的长为 米.(不计宣传栏的厚度)
18.如图10是小明用火柴搭的1条、2条、3条“金鱼” ,则搭n 条“金鱼”需要火柴 根.
三、解答题(本大题8个小题,共66分.解答时要写出必要的文字说明、演算步骤或推证过程)鄂尔多斯初中数学补习班手机187********
19.(本小题满分5分)
图8
2米 3米
图9 1条 2条 3条 图10
图7
化简:222
a a
b b a b a b
-+-+
20.(本小题满分7分)
高为12米的教学楼ED 前有一棵大树AB ,如图11(a ). (1)某一时刻测得大树AB 、教学楼ED 在阳光下的投影长分别是 2.5BC =米,7.5DF =米,求大树AB 的高度;
(2)现有皮尺和高为h 米的测角仪,请你设计另一种测量大树AB 高度的方案,要求: ①在图11(b )中,画出你设计的测量方案示意图,并将应测量的数据标记在图上(长度用字母m n ,表示,角度用希腊字母αβ ,表示)
; ②根据你所画出的示意图和标注的数据,求出大树的高度(用字母表示).
21.(本小题满分8分) 我市某中学为了解九年级300名学生的理化实验操作水平,从中随机抽取30名学生进行测试.下表是这30名学生的测试成绩(分): 4 5 7 6 3 7 9 4 5 7 7 3 7 5 6 8 7 5 6 8 6 7 4 10 5 6 7 3 9 4 (1)请你设计一张统计表,能够清楚反映出各成绩的人数分布情况; (2)求出这30名学生成绩的平均数、众数;
(3)如果测试成绩6分以上(包括6分)为合格,请估计300名学生中成绩合格的约有多少人?
A
B
C D F E A B 图11(a ) 图11(b )
如图12,在ABC △中,90ACB ∠=
,BC 的垂直平分线EF 交BC 于D ,交AB 于E ,且CF BE =.
(1)求证:四边形BECF 是菱形.
(2)当A ∠的大小满足什么条件时,菱形BECF 是正方形?请回答并证明你的结论. 23.(本小题满分7分) 如图13,有两个可以自由转动的均匀转盘A B ,.转盘A 被平均分成3等份,分别标上123,,三个数字;转盘B 被平均分成4等份,分别标上3456,,,四个数字.有人为甲、乙两人设计了一个游戏规则;自由转动转盘A 与B ,转盘停止后,指针各指向一个数字,将指针所指的两个数字相加,如果和是6,那么甲获胜,否则为乙获胜.
你认为这样的游戏规则是否公平?如果公平,请说明理由;如果不公平,怎样修改规则才能使游戏对双方公平?
24.(本小题满分9分)
某产品每件成本10元,在试销阶段每件产品的日销售价x (元)与产品的日销售量y (件)之间的关系如下表:
x (元)
20 25 30 35 y (件) 30 25 20 15
(1)在草稿纸上描点,观察点的分布,确定y 与x 的函数关系式.
(2)要使每日的销售利润最大,每件产品的销售价应定为多少元?此时每日销售利润是多少元? 25.(本小题满分10分)
如图14(a ),两个不全等的等腰直角三角形OAB 和OCD 叠放在一起,并且有公共的直角顶点O .
(1)将图14(a )中的OAB △绕点O 顺时针旋转90
角,在图14(b )中作出旋转后的OAB △(保留作图痕迹,不写作法,不证明).
(2)在图14(a )中,你发现线段AC ,BD 的数量关系是 ,直线AC ,BD 相交成 度角.
(3)将图14(a )中的OAB △绕点O 顺时针旋转一个锐角,得到图14(c ),这时(2)中的两个结论是否成立?作出判断并说明理由.若OAB △绕点O 继续旋转更大的角
A B
图13
时,结论仍然成立吗?作出判断,不必说明理由.
26.(本小题满分12分)
如图15,点P 在y 轴上,P 交x 轴于A B ,两点,连结BP 并延长交P 于C ,过点C 的直线2y x b =+交x 轴于D ,且P
4AB =.
(1)求点B P C ,,的坐标;
(2)求证:CD 是P 的切线;
(3)若二次函数2
(1)6y x a x =-+++
并写出使二次函数值小于一次函数2y x =+
2006年鄂尔多斯市初中毕业升学考试 数学试题参考答案及评分说明(新课标)
(一)阅卷评分说明
1、正式阅卷前先进行试评,在试评中认真阅读参考答案,明确评分标准,不得随意拔高或降低评分标准.试评的试卷必须在阅卷后期全部予以复查,防止阅卷前后期评分标准宽严不一致.
图14(a ) 图14(b )
图14(c )
图15
2、评分方式为分步累计评分,解答过程的某一步骤发生笔误,只要不降低后继部分的难度,而后继部分再无新的错误,后继部分可评应得分数的50%;若是几个相对独立的得分点,其中一处错误不影响其它得分点的评分.
3、最小记分单位为1分,不得将评分标准细化至1分以下(即不得记小数分)
4、解答题题头一律记该题的实际得分,不得用记负分的方式记分.对解题中的错误须用红笔标出,并继续评分,直至将解题过程评阅完毕,并在最后得分点处标上该题实际得分.
5、本参考答案只给出一至两种解法,凡有其它正确解法都应参照本评分说明分步确定得分点,并同样实行分步累计评分.
6、合理精简解题步骤者,其简化的解题过程不影响评分. (二)参考答案及评分标准
一、选择题(本大题10个小题,每小题3分,共30分.)
题号
1 2 3 4 5 6 7 8 9 10 选项 C B A C D A D C D B 二、填空题(本大题8个小题,每小题3分,共24分) 11.10x =,223
x =
12.3x >
13.2005 50(约50)(填对一空给2分)
14.等腰梯形、矩形(长方形)、平行四边形中任选两个即可(填对一个给2分)
15.4 16.2 17.6 18.62n + 三、解答题(本大题8个小题,共66分) 19.(本小题满分5分)
解:222
a a
b b a b a b -+-+ ()()()a a b b
a b a b a b
-=
+
+-+ ········································································································ 2分 a b
a b a b =
+
++ ······················································································································ 3分 a b
a b +=
+ ··································································································································· 4分 1= ··········································································································································· 5分 20.(本小题满分7分) 解:(1)连结AC ,EF ,则ABC EDF △∽△
2.5
127.5AB =
∴
··························································································································· 2分 4AB =∴
即大树AB 高是4米 ··············································································································· 3分 (2)解法一: ①如图1(b )(标注m ,α,画草图也可给相同的分) ··················································· 5分 ②在Rt CMA △中,tan tan AM CM m αα== ································································ 6分
tan AB m h α=+∴ ·
············································································································· 7分
E A
A
8
解法二:
①如图1(c )(标注m αβ,,,画草图也可给相同的分)··············································· 5分 ②cot cot AM AM m αβ-=
cot cot m
AM αβ
=
-∴ ········································································································· 6分
cot cot m
AB h αβ
=
+-∴ ·
···································································································· 7分 21(本小题满分8分) (1)统计表如下:
成绩
3 4 5 6 7 8 9 10 人数
3 4 5 5 8 2 2 1 ··········································································································································· 3分 (表格格式正确给1分;填数、计数正确给2分,错误一处扣1分,扣完2分为止.画正字统计以及其它符合题意的统计表均可按此标准给分.如果统计出现错误,下面(2),(3)的评分按“评分说明”中的第2条规定酌情评分.) (2)平均数1
(33445565788292101)30
x =
?+?+?+?+?+?+?+? ·
··················· 4分 1
180630
=
?= ·
··············································································································· 5分 众数为7 ·························································································································· 6分
(3)18
30018030
?= ·
············································································································ 7分 答:估计有180人合格. ······························································································· 8分 22.(本小题满分8分)
(1)证法一:如图2
EF ∵垂直平分BC ,BE EC =∴,BF CF = ························
C F B E =∵ B E E C C F ===∴ ····················· 3分 ∴四边形BECF 是菱形 ··················································· 4分 证法二:如图2
EF ∵垂直平分BC ,BD DC =∴,EF BC ⊥ ········· 1分 B E C F =∵ B E D C F
∴△≌△ ······················· 2分
D E D F
=∴ ····································································· 3分 ∴四边形BECF 是菱形 ··················································· 4分 (2)解法一:
当45A ∠=
时,菱形BECF 是正方形. ····································································· 6分 45A ∠=
∵,90ACB ∠=
45EBC ∠=
∴ ························································ 7分 2245
9
E B
F E B C ∠=∠=
?=
∴ ∴菱形BECF 是正方形. ······························································································ 8分
解法二:
当45A ∠=
时,菱形BECF 是正方形. ····································································· 6分 45A ∠=
∵,90ACB ∠=
45EBC =
∴ ··························································· 7分
BE EC = 45ECB EBC ∠=∠= ∴
90BEC ∠=
∴ 菱形BECF 是正方形. ··································································· 8分 23.(本小题满分7分) 解:不公平. ·························································································································· 2分 P ∵(和为6)31
124
=
=,甲、乙获胜的概率不相等
················································· 5分 ∴不公平.(无列表或树状图不扣分)
规则改为:和是6或7,甲胜;否则乙胜. ·································································· 7分
(和为奇数,甲胜;和为偶数,乙胜;或和小于7,甲胜;和大于等于7,乙胜.答案不唯一.) 列 表
3 4 5 6 1 4 5 6 7 2 5 6 7 8 3
6
7
8
9
24.(本小题满分9分)
解:(1)设函数关系式为y kx b =+,根据题意得(方程组较多):230
3020
x b x b +=??
+=? ········· 2分
解之得:1
50x b =-??=?
············································································································ 3分
50y x =-+∴ ················································································································ 4分
A
(2)设每日的销售利润为m 元,则:
(10)m y x =- ········································································································· 6分 2
(50)(10)60500x x x x =-+-=-+- ····························································· 7分 2
(30)400x =--+ ····························································································· 8分
∴当30x =时,400m =最大(当302b
x a
=-
=时,400m =最大,同样给分) ·
············· 9分 答:每件产品的销售价定为30元时,每日销售利润最大是400元.
25.(本小题满分10分) 解:(1)如图3(a )(A B ,字母位置互换扣1分,无弧扣1分,不连结AB 扣1分,扣完
为止) ······························································································································ 2分 (2)AC BD =;90(90)
(每空1分) ····································································· 4分
(3)成立.如图3(b )
90COD AOB ∠=∠= ∵
COA AOD AOD DOB ∠+∠=∠+∠∴
即:COA DOB ∠=∠(或由旋转得COA DOB ∠=∠) ············································ 5分
CO OD =∵ O A O B = C O A D O
∴△≌△ ··············································· 6分 AC BD =∴
···················································································································· 7分 延长CA 交OD 于E ,交BD 于F (下面的证法较多)
COA DOB ∵△≌△,ACO ODB ∠=∠∴ ·
······························································ 8分 CEO DEF ∠=∠∵ 90COE EFD ∠=∠= ∴ A C B D ∴⊥ ······················ 9分 旋转更大角时,结论仍然成立. ·················································································· 10分
26.(本小题满分12分) 解:(1)如图4,连结CA O P A B ∵⊥ 2OB OA ==∴ ············································································ 1分 2
2
2
OP BO BP +=∵ 2
541OP =-=∴,1OP = ··········································· 2分 BC ∵是P 的直径90CAB ∠=
∴(也可用勾股定理求得下面的结论)
图3(a )
图3(b )
C P B P =
∵,OB OA = 22AC OP ==∴ ·
····················································· 3分 (20)B ,∴,(01)P ,,(22)C -,(写错一个不扣分) ············································ 4分 (2)2y x b =+∵过C 点6b =∴ 26y x =+∴ ·············································· 5分 ∵当0y =时,3x =- (30)D -,∴ ∴1AD = ··············································· 6分 21OB AC AD OP ====,∵,90CAD POB ∠=∠= D A C
P O ∴△≌△ D C A A B
∠=∠∴ 90ACB CBA ∠+∠=
∵
90DCA ACB ∠+∠=
∴(也可用勾股定理逆定理证明) ·································· 7分 DC ∴是P 的切线 ······························································································· 8分 (3)2
(1)6y x a x =-+++∵过(20)B ,点
202(1)26a =-++?+∴ 2a =-∴ ································································· 9分 26y x x =--+∴ ································································································· 10分
因为函数2
6y x x =--+与26y x =+的图象交点是(06),和点(30)D -,(画图可得此结论) ············································································································· 11分
所以满足条件的x 的取值范围是3x <-或0x > ················································ 12分
图4
2007年鄂尔多斯市初中毕业升学考试
数 学(课标)
注意事项:
1.本试题满分120分,考试用时120分钟; 2.答题前将密封线内的项目填写清楚;
3.考试结束后将试卷按页码顺序排好,全部上交.
一、选择题(本大题10个小题,每小题3分,共30分.在每小题给出的四个选项中只有一个是正确的,请把正确选项的标号填在下面的选项栏内.)
题号
1 2 3 4 5 6 7 8 9 10 选项
1.3-的相反数是( ) A .3-
B .3
C .13
-
D .
13
2.图1是由6个相同的小正方体搭成的几何体,那么这个几何体的俯视图是( )
3.我市2006年财政收入近150亿元,居自治区首位.150亿用科学记数法可表示为( ) A .8
1.510?
B .9
1.510?
C .10
1.510?
D .11
1.510?
4.能够刻画一组数据离散程度的统计量是( ) A .平均数 B .众数 C .中位数 D .方差 5.将圆柱形纸筒沿母线AB 剪开铺平,得到一个
矩形(如图2).如果将这个纸筒沿线路
B M A →→剪开铺平,得到的图形是( ) A .平行四边形 B .矩形
C .三角形
D .半圆
6.鄂尔多斯市成陵旅游区到响沙湾旅游区之间的
距离为105公里,在一张比例尺为1:2000000的交通旅游图上,它们之间的距离大约相当于( )
A .一根火柴的长度
B .一支钢笔的长度
C .一支铅笔的长度
D .一根筷子的长度 7.下列说法正确的有( ) (1)如图3(a ),可以利用刻度尺和三角板测量圆形工件的直径;
图1
A .
B .
C .
D .
()A
)B 图2
(2)如图3(b ),可以利用直角曲尺检查工件是否为半圆形; (3)如图3(c ),两次使用丁字尺(CD 所在直线垂直平分线段AB )可以找到圆形工件的圆心;
(4)如图3(d ),测倾器零刻度线和铅垂线的夹角,就是从P 点看A 点时仰角的度数.
A .1个
B .2个
C .3个
D .4个
8.一种蔬菜加工后出售,单价可提高20%,但重量减少10%.现有未加工的这种蔬菜30千克,加工后可以比不加工多卖12元,则这种蔬菜加工前和加工后每千克各卖多少元?设这种蔬菜加工前每千克卖x 元,加工后每千克卖y 元,根据题意,所列方程组正确的是( ) A .(120)30(110)3012y x
y x =+??
+-=?
%%
B .(120)30(110)3012y x
y x =+??
--=?
%%
C .(120)30(110)3012
y x
y x =-??
--=?%%
D .(120)30(110)3012
y x
y x =-??
+-=?%%
9.如图4,一只蚂蚁以均匀的速度沿台阶12345A A A A A →→→→爬行,那么蚂蚁爬行的高度..h 随时间t 变化的图象大致是( )
10.观察表1,寻找规律.表2是从表1中截取的一部分,其中a b c ,,的值分别为( ) 表1 表2 1 2 3 4 …… 2 4 6 8 …… 3 6 9 12 …… 4 8 12 16 …… ……
……
……
……
……
A .20,25,24
B .25,20,24
C .18,25,24
D .20,30,25
二、填空题(本大题8个小题,每小题3分,共24分)
11.如图5,AB CD ∥,58B =
∠,20E =
∠,则D ∠的度数为 .
16 a 20
b
c
30
图3(a )
图3(b )
图3(c )
图3(d )
A
A
B
C
D
P
图4 1
A
2A 3A 4A 5A A . B . C . D .
12.若
43x y =,则y x y
=+ . 13.如图6,双曲线1
k y x
=
与直线2y k x =相交于A B ,两点,如果A 点的坐标是(12),,那么B 点的坐标为 .
14.不等式组30
240x x -??+>?
≤的解集是 .
15.如图7,以O 为圆心的两个同心圆中,大圆的弦AB 切小圆于P ,如果4cm AB =,则图中阴影部分的面积为 2
cm
(结果用π表示).
16.如图8
,点P 在AOB ∠的平分线上,若使AOP BOP △≌△,则需添加的一个条件是
(只写一个即可,不添加辅助线). 17.在边长为a 的正方形纸片中剪去一个边长为
b 的小正方形()a b >(如图9(1)),把余下的部分沿虚线剪开,拼成一个矩形(如图9(2)),分别计算这两个图形阴影部分的面积,可以验证的乘法公式是 (用字母表示).
18.如图10,房间里有一只老鼠,门外蹲着一只小猫,如果每块正方形地砖的边长为1米,那么老鼠在地面上能避开小猫视线的活动范围为 平方米(不计墙的厚度).
三、解答题(本大题8个小题,共66分.解答时要写出必要的文字说明、演算步骤或推证过程) 19.(本小题满分8分)
图7 图8
A
B
P O
图9(1) 图9(2) 图10 1
(1)计算:
1
1
(14
2
-
??
++--
?
??
.
(2)化简:
2
121
1
1
a a
a
a a
-+
??
+-
?-
??
.
20.(本小题满分6分)
某市教育行政部门为了解初中学生参加综合实践活动的情况,随机抽取了本市初一、初二、初三年级各500名学生进行了调查.调查结果如图11所示,请你根据图中的信息回答问题.
(1)在被调查的学生中,参加综合实践活动的有多少人?参加科技活动的有多少人?(2)如果本市有3万名初中学生,请你估计参加科技活动的学生约有多少名?
21.(本小题满分6分)
有四张背面相同的纸牌A B C D
,,,,其正面分别画有四个不同的几何图形(如图12).小明将这4张纸牌背面朝上洗匀后摸出一张,将剩余3张洗匀后再摸出一张.
(1)用树状图(或列表法)表示两次摸牌所有可能出现的结果(纸牌用A B C D
,,,表示);
(2)求摸出的两张牌面图形既是轴对称图形又是中心对称图形纸牌的概率.
22.(本小题满分6分)
如图13,A B
,两镇相距60km,小山C在A镇的北偏东60 方向,在B镇的北偏西30 方向.经探测,发现小山C周围20km的圆形区域内储有大量煤炭,有关部门规定,该区域内
年级
参加综合实践活动人数统计图
文体活动
社会调查
社区服务
科技活动
参加综合实践活动人数分布统计图
图11
图12
禁止建房修路.现计划修筑连接A B ,两镇的一条笔直的公路,试分析这条公路是否会经过该区域? 23.(本小题满分9分) 如图14,在ABC △中,90ACB = ∠,D 是AB 的中点,以DC 为直径的O 交ABC △的边于G F E ,,点. 求证:(1)F 是BC 的中点;(2)A GEF =∠∠.
24.(本小题满分10分)
有甲、乙两家通迅公司,甲公司每月通话的收费标准如图15所示;乙公司每月通话收费标准如表3所示.
表3
(1)观察图15,甲公司用户月通话时间不超过100分钟时应付话费金额是 元;甲公司用户通话100分钟以后,每分钟的通话费为 元;
(2)李女士买了一部手机,如果她的月通话时间不超过100分钟,她选择哪家通迅公司更合算?如果她的月通话时间超过100分钟,又将如何选择? 25.(本小题满分9分) 我们给出如下定义:若一个四边形中存在相邻两边的平方和等于一条对角线的平方,则称这个四边形为勾股四边形,这两条相邻的边称为这个四边形的勾股边.
(1)写出你所学过的特殊四边形中是勾股四边形的两种图形的名称 , ; (2)如图16(1),已知格点(小正方形的顶点)(00)O ,,(30)A ,,(04)B ,,请你画出以格点为顶点,OA OB ,为勾股边且对角线相等的勾股四边形OAMB ; 月租费 通话费 2.5元 0.15元/分钟
A B C D E F G
O
图14
图15
(3)如图16(2),将ABC △绕顶点B 按顺时针方向旋转60
,得到DBE △,连结
AD DC ,,30DCB = ∠.
求证:222
DC BC AC +=,即四边形ABCD 是勾股四边形. 26.(本小题满分12分)
如图17,抛物线2
2
29y x nx n =-++-(n 为常数)经过坐标原点和x 轴上另一点C ,顶点在第一象限.
(1)确定抛物线所对应的函数关系式,并写出顶点坐标;
(2)在四边形OABC 内有一矩形MNPQ ,点M N ,分别在OA BC ,上,点Q P ,在x 轴上.当MN 为多少时,矩形MNPQ 的面积最大?最大面积是多少?
2007年鄂尔多斯市初中毕业升学考试
A
图16(2)
图17
数学试题参考答案及评分说明(课标)
(一)阅卷评分说明
1.正式阅卷前先进行试评,在试评中认真阅读参考答案,明确评分标准,不得随意拔高或降低评分标准.试评的试卷必须在阅卷后期全部予以复查,防止阅卷前后期评分标准宽严不一致.
2.评分方式为分步累计评分,解答过程的某一步骤发生笔误,只要不降低后继部分的难度,而后继部分再无新的错误,后继部分可评应得分数的50%;若是几个相对独立的得分点,其中一处错误不影响其它得分点的评分.
3.最小记分单位为1分,不得将评分标准细化至1分以下(即不得记小数分).
4.解答题题头一律记该题的实际得分,不得用记负分的方式记分.对解题中的错误须用红笔标出,并继续评分,直至将解题过程评阅完毕,并在最后得分点处标上该题实际得分.
5.本参考答案只给出一至两种解法,凡有其它正确解法都应参照本评分说明分步确定得分点,并同样实行分步累计评分.
6.合理精简解题步骤者,其简化的解题过程不影响评分. (二)参考答案及评分标准
一、选择题(本大题10个小题,每小题3分,共30分.)
题号
1 2 3 4 5 6 7 8 9 10 选项
B C C D A A D B B A 二、填空题(本大题8个小题,每小题3分,共24分.) 11.38
(或38)
12.
3
7
13.(12)--, 14.23x -<≤ 15.4π
16.OA OB =(或OAP OBP =∠∠或APO BPO =∠∠)
17.2
2
()()a b a b a b -=+-(或2
2
()()a b a b a b +-=-)
18.17(填空正确给3分,图形不正确不扣分;图形正确,计算不正确可给1分.) 三、解答题(本大题8个小题,共66分.) 19.(本小题满分8分)
(1)计算:1
1(142-??
++-- ???
解:原式124=+- ······································································ 3分(一处计算正确给1分) 1=- ························································································································· 4分
(2)化简:2
121
11a a a a a -+??+- ?-??
解:原式2
(1)(1)1
a a a -=+-- ······················································· 2分(一处计算正确给1分)
(1)(1)a a =+-- ···································································································· 3分 2= ···································································································································· 4分
20.(本小题满分6分) 解:(1)450350150950++=(人) ················································ 1分(无单位不扣分) 950(1601614)95?---=%%%(人) ·
············································ 3分(无单位不扣分) 答:参加综合实践活动的有950人,参加科技活动的有95人. ········································ 4分
(2)950
30000105003
?
??% ·
································································································ 5分 95201900=?=(人) ·
······························································· 6分(无单位不扣分) 答:参加科技活动的学生估计有1900人. 21.(本小题满分6分)
树状图: 列表:
······························································································· 4分 注:出现3处(共12处)错误扣1分,扣完为止.
(2)21
126P =
= ·
·················································································································· 6分 答:概率是1
6
.
22.(本小题满分6分)
解:作CD AB ⊥于D ,由题意知:30CAB =
∠ 60CBA =
∠ 90ACB =
∠ ·
········································ 1分 30DCB ∴= ∠ ·
······························································· 2分 ∴在Rt ABC △中,1
302
BC AB =
= ·
················································································ 3分 在Rt DBC △中,cos30CD BC =
················································································· 4分
30= ····················································································· 5分
20=> ················································································· 6分 答:这条公路不经过该区域.
A B C D
A A B
A C A D
B A B B
C B
D C A C B C D C D A D B D D C A
B C D D B C A D C A B D A C 1
23.(本小题满分9分) 证法一:
(1)连结DF ,
90ACB = ∠,D 是AB 的中点
1
2
BD DC AB ∴==
························································ 2分 DC 是O 的直径
DF BC ∴⊥ ····································································· 4分 BF FC ∴=,即F 是BC 的中点. ······························· 5分 (2)D F ,分别是AB BC ,的中点
DF AC ∴∥ ·························································································································· 6分 A BDF ∴=∠∠ ·
···················································································································· 7分 BDF GEF ∴=∠∠ ·············································································································· 8分 A GEF ∴=∠∠ ·
···················································································································· 9分 证法二:
(1)连结DF DE , DC 是O 直径
90DEC DFC ∴== ∠∠ ·
···································································································· 1分 90ECF = ∠ ∴四边形DECF 是矩形
EF CD ∴=,DF EC = ·
··············································· 2分 D 是AB 的中点,90ACB =
∠
1
2
EF CD BD AB ∴=== ·············································· 3分 DBF EFC ∴△≌△ ·
······················································· 4分 BF FC ∴=,即F 是BC 的中点. ·
······························ 5分 (2)DBF EFC △≌△
BDF FEC ∴=∠∠,B EFC =∠∠ ·
················································································· 6分 90ACB = ∠(也可证AB EF ∥,得A FEC =∠∠)
A FEC ∴=∠∠ ·
···················································································································· 7分 FEG BDF = ∠∠ ·············································································································· 8分 A GEF ∴=∠∠ ·
···················································································································· 9分 (此题证法较多,大纲卷参考答案中,又给出了两种不同的证法,可供参考.)
24.(本小题满分10分) (1)20;0.2 ································································································ 4分(每空2分) (2)通话时间不超过100分钟选甲公司合算 ······································································· 5分 解:设通话时间为t 分钟(100t >),甲公司用户通话费为1y 元,乙公司用户通话费为2y 元. 则:1200.2(100)0.2y t t =+-= ·································· 6分(条件100t >没有写出不扣分)
A B
C D E
F
G
O
A B
C
D E F G
O