2009年南充市高中阶段学校招生统一考试数学试卷

南充市二○○九年高中阶段学校招生统一考试
数 学 试 卷
(满分100分,考试时间90分钟)
一、细心选一选(本大题共8个小题,每小题3分,共24分)每小题下面都有代号为A 、B 、C 、D 的四个选项,其中只有一个选项是正确的,请把正确选项的代号填在该题后面的括号内．填写正确记3分,不填、填错或填出的代号超过一个均记0分． 1．计算2009
(1)-的结果是( )
A ．1-
B ．1
C ．2009-
D ．2009
2．在平面直角坐标系中,点(25)A ，与点B 关于y 轴对称,则点B 的坐标是( ) A ．(52)--，
B ．(25)--，
C ．(25)-，
D ．(25)-，
3．某物体的展开图如图1,它的左视图为( )
4．方程(3)(1)3x x x -+=-的解是( )
A ．0x =
B ．3x =
C ．3x =或1x =-
D ．3x =或0x =
5．已知一组数据2,1,x ,7,3,5,3,2的众数是2,则这组数据的中位数是( ) A ．2 B ．2.5 C ．3 D ．5 6．化简12
3
()x x -的结果是( ) A ．5
x
B ．4
x
C ．x
D ．
1
x
7．抛物线(1)(3)(0)y a x x a =+-≠的对称轴是直线( )
A ．1x =
B ．1x =-
C ．3x =-
D ．3x = 8．如图2,AB 是O ⊙的直径,点C 、D 在O ⊙上,110BOC ∠=°,
AD OC ∥,则AOD ∠=( )
A ．70°
B ．60°
C ．50°
D ．40°
二、认真填一填(本大题共4个小题,每小题3分,共12分)请将答案直接填写在题中横线上． 9．不等式5(1)31x x -<+的解集是 ．
10．某校为了举办“庆祝建国60周年”的活动,调查了本校所有学生,调查的结果如图3所示,根据图中给出的信息,这所学校赞成举办演讲比赛的学生有 人．
A ．
B ．
C ．
D ．
图1 （图2）
O
B
D
A
C
11．如图4,等腰梯形ABCD 中,AD BC ∥, 6047B AD BC ∠===°，，,则梯形ABCD 的周长是 ．
12．ABC △中,10cm 8cm 6cm AB AC BC ===，，, 以点B 为圆心、6cm 为半径作B ⊙,则边AC 所在的直线 与B ⊙的位置关系是 ．
三、(本大题共2个小题,每小题6分,共12分)
13．计算
:0
(π2009)|2|-+．
14．化简:22
1211
241
x x x x x x --+÷++--．
四、(本大题共2个小题,每小题6分,共12分)
15．如图5,ABCD 是正方形,点G 是BC 上的任意一点,DE AG ⊥于E ,BF DE ∥,交AG 于F ． 求证:AF BF EF =+．
16．甲口袋中装有两个相同的小球,它们分别写有1和2；乙口袋中装有三个相同的小球,它们分别写有3、4和5；丙口袋中装有两个相同的小球,它们分别写有6和7．从这3个口袋中各随机地取出1个小球．
(1)取出的3个小球上恰好有两个偶数的概率是多少？ (2)取出的3个小球上全是奇数的概率是多少？
D
C
A
B
（图4）
（图5）
D C B A
E F
G
五、(本大题共2个小题,每小题8分,共16分)
17．在达成铁路复线工程中,某路段需要铺轨．先由甲工程队独做2天后,再由乙工程队独做3天刚好完成这项任务．已知乙工程队单独完成这项任务比甲工程队单独完成这项任务多用2天,求甲、乙工程队单独完成这项任务各需要多少天？
18．如图6,在平面直角坐标系中,已知点(42)B ，,BA x ⊥轴于A ． (1)求tan BOA ∠的值；
(2)将点B 绕原点逆时针方向旋转90°后记作点C ,求点C 的坐标；
(3)将OAB △平移得到O A B '''△,点A 的对应点是A ',点B 的对应点B '的坐标为(22)-，,在坐标系中作出O A B '''△,并写出点O '、A '的坐标．
六、(本大题8分)
19．某电信公司给顾客提供了两种手机上网计费方式:
方式A 以每分钟0.1元的价格按上网时间计费；方式B 除收月基费20元外,再以每分钟0.06元的价格按上网时间计费．假设顾客甲一个月手机上网的时间共有x 分钟,上网费用为y 元． (1)分别写出顾客甲按A 、B 两种方式计费的上网费y 元与上网时间x 分钟之间的函数关系式,并在图7的坐标系中作出这两个函数的图象； (2)如何选择计费方式能使甲上网费更合算？
七、(本大题8分
)
（图7）
20．如图8,半圆的直径10AB =,点C 在半圆上,6BC =． (1)求弦AC 的长；
(2)若P 为AB 的中点,PE AB ⊥交AC 于点E ,求PE 的长．
八、(本大题8分)
21．如图9,已知正比例函数和反比例函数的图象都经过点(33)A ，． (1)求正比例函数和反比例函数的解析式；
(2)把直线O A 向下平移后与反比例函数的图象交于点(6)B m ，,求m 的值和这个一次函数的解析式；
(3)第(2)问中的一次函数的图象与x 轴、y 轴分别交于C 、D ,求过A 、B 、D 三点的二次函数的解析式；
(4)在第(3)问的条件下,二次函数的图象上是否存在点E ,使四边形O ECD 的面积1S 与四边形O ABD 的面积S 满足:12
3
S S =？若存在,求点E 的坐标；若不存在,请说明理由．
南充市二○○九年高中阶段学校招生统一考试
数学试卷参考答案及评分意见
说明:
1． 全卷满分100分,参考答案和评分意见所给分数表示考生正确完成当前步骤时应得的累
加分数．
2． 参考答案和评分意见仅是解答的一种,如果考生的解答与参考答案不同,只要正确就应该
参照评分意见给分．合理精简解答步骤,其简化部分不影响评分．
3． 要坚持每题评阅到底．如果考生解答过程发生错误,只要不降低后继部分的难度且后继
部分再无新的错误,可得不超过后继部分应得分数的一半,如果发生第二次错误,后面部
P B
C E
A
（图8）
分不予得分；若是相对独立的得分点,其中一处错误不影响其它得分点的评分．
一、细心选一选(本大题共8个小题,每小题3分,共24分)
1．A 2．C 3．B 4．D 5．B 6．C 7．A 8．D 二、认真填一填(本大题共4个小题,每小题3分,共12分) 9．3x < 10．100 11．17 12．相切 三、(本大题共2个小题,每小题6分,共12分)
13．解:
原式12=+ ················································ [共4分,分项给分:
0(π2009)1-=(1分
=分
),2|2=-分)]
(12)=++ (结果正确,没有此步不扣分)
3= ··········································································· (6分)
14．解:原式22141
2211
x x x x x x --=++-+- ····························································· (1分)
21(2)(2)1
2(1)1
x x x x x x -+-=
++-- ······················································· (3分)
21
11x x x -=
+
-- ·········································································· (4分) 21
1x x -+=-
1
1
x x -=- ···················································································· (5分)
1= ·························································································· (6分)
四、(本大题共2个小题,每小题6分,共12分) 15．证明:ABCD 是正方形,
90AD AB BAD ∴=∠=，°． ·
······································································· (1分) DE AG ⊥,
90DEG AED ∴∠=∠=°． 90ADE DAE ∴∠+∠=°．
又90BAF DAE BAD ∠+∠=∠=°,
ADE BAF ∴∠=∠． ·
·················································································· (2分) BF DE ∥,
AFB DEG AED ∴∠=∠=∠． ·
······································································ (3分) 在ABF △与DAE △中,AFB AED
ADE BAF AD AB ∠=∠⎧⎪
∠=∠⎨⎪=⎩
,
(AAS)ABF DAE ∴△≌△． ·
········································································ (4分)
BF AE ∴=． ·
···························································································· (5分) AF AE EF =+,
AF BF EF ∴=+． ·
···················································································· (6分) 16．解:根据题意,画出如下的“树形图”:
从树形图看出,所有可能出现的结果共有12个． ·················································· (2分) (1)取出的3个小球上恰好有两个偶数的结果有4个,即1,4,6；2,3,6；2,4,7；2,5,6．所以
P (两个偶数)41
123
=
=． ·
·············································································· (4分) (2)取出的3个小球上全是奇数的结果有2个,即1,3,7；1,5,7．所以
P (三个奇数)21
126
=
=． ·
·············································································· (6分) 五、(本大题共2个小题,每小题8分,共16分)
17．解:设甲工程队单独完成任务需x 天,则乙工程队单独完成任务需(2)x +天,
························································· (1分)
依题意得
23
12
x x +=+． ·
·············································································· (4分) 化为整式方程得
2340x x --= ···························································································· (5分)
解得1x =-或4x =． ··················································································· (6分) 检验:当4x =和1x =-时,(2)0x x +≠,
4x ∴=和1x =-都是原分式方程的解． 但1x =-不符合实际意义,故1x =-舍去； ························································ (7分) ∴乙单独完成任务需要26x +=(天)．
答:甲、乙工程队单独完成任务分别需要4天、6天． ··········································· (8分) 18．解:(1)
点(42)B ，,BA x ⊥轴于A ,
42OA BA ∴==，,
21
tan 42
AB BOA OA ∴∠===． ·····························(3分)
(2)如图,由旋转可知:24CD BA OD OA ====，,
∴点C 的坐标是(24)-，． ·
··································(5分) (3)O A B '''△如图所示,
(24)O '--，,(24)A '-，． ·
···································(8分)
6 7 6 7 6 7 3 4 5 1 6 7 6 7 6 7
3 4
5 2 甲
乙 丙
六、(本大题8分)
19．(1)方式A :0.1(0)y x x =≥, ······································································ (1分) 方式B :0.0620(0)y x x =+≥, ······································································· (2分) 两个函数的图象如图所示． ············································································ (4分)
(2)解方程组0.10.0620y x y x =⎧⎨
=+⎩ 得500
50
x y =⎧⎨=⎩
所以两图象交于点P (500,50)． ········································································· (5分)
由图象可知:当一个月内上网时间少于500分时,选择方式A 省钱；当一个月内上网时间等于
500分时,选择方式A 、方式B 一样；当一个月内上网时间多于500分时,选择方式B 省钱．(8分) 七、(本大题8分)
20．解:AB 是半圆的直径,点C 在半圆上, 90ACB ∴∠=°． 在Rt ABC △中
,8AC ===
········································· (3分) (2)
PE AB ⊥,
90APE ∴∠=°．90ACB ∠=°, APE ACB ∴∠=∠． 又PAE CAB ∠=∠,
AEP ABC ∴△∽△, ·
···················································································· (6分) PE AP
BC AC
∴=
······························································································· (7分) 110268
PE ⨯∴= 3015
84
PE ∴==． ······················································································· (8分)
八、(本大题8分)
21．解:(1)设正比例函数的解析式为11(0)y k x k =≠, 因为1y k x =的图象过点(33)A ，,所以
133k =,解得11k =．
这个正比例函数的解析式为y x =． ································································· (1分) 设反比例函数的解析式为2
2(0)k y k x
=≠． 因为2
k y x
=
的图象过点(33)A ，,所以 2
33
k =
,解得29k =． 这个反比例函数的解析式为9y x
=．································································· (2分) (2)因为点(6)B m ，在9
y x
=
的图象上,所以 9362m =
=,则点362B ⎛⎫
⎪⎝⎭
，． ············································································ (3分) 设一次函数解析式为33(0)y k x b k =+≠． 因为3y k x b =+的图象是由y x =平移得到的, 所以31k =,即y x b =+．
又因为y x b =+的图象过点362B ⎛
⎫ ⎪⎝⎭
，,所以
362b =+,解得92
b =-, ∴一次函数的解析式为92
y x =-． ·································································· (4分)
(3)因为92y x =-
的图象交y 轴于点D ,所以D 的坐标为902⎛
⎫- ⎪⎝
⎭，． 设二次函数的解析式为2
(0)y ax bx c a =++≠．
因为2
y ax bx c =++的图象过点(33)A ，、362B ⎛
⎫ ⎪⎝⎭，、和D 902⎛⎫- ⎪⎝⎭
，
, 所以933336629.2
a b c a b c c ⎧⎪++=⎪
⎪
++=⎨⎪⎪=-⎪⎩，， ···················· (5分) 解得1249
.2a b c ⎧=-⎪⎪=⎨⎪⎪=-⎩，，
这个二次函数的解析式为219
422
y x x =-
+-． ·
················································· (6分)
(4)
92y x =-
交x 轴于点C ,∴点C 的坐标是902⎛⎫
⎪⎝⎭
，
, 如图所示,151131
66633322222S =
⨯-⨯⨯-⨯⨯-⨯⨯ 99
451842
=---
814
=． 假设存在点00()E x y ，,使1281227
3432
S S ==⨯=
． 四边形CDOE 的顶点E 只能在x 轴上方,∴00y >,
1OCD OCE S S S ∴=+△△
019919
22222y =⨯⨯+⨯ 0819
84y =+． 081927842y ∴+=
,03
2
y ∴=． ········································································ (7分) 00()E x y ，在二次函数的图象上,
2001934222
x x ∴-+-=．
解得02x =或06x =．
当06x =时,点362E ⎛
⎫ ⎪⎝⎭
，与点B 重合,这时CDOE 不是四边形,故06x =舍去,
∴点E 的坐标为322⎛⎫
⎪⎝⎭
，． ·············································································· (8分)。