2006年广西贵港市数学试卷及答案(大纲)

2006年中等学校招生贵港市统一考试
数 学
（考试时间120分钟，赋分120分）
一、细心填一填：本大题共10小题，每小题2分，共20分．请将答案填写在题中的横线
上．
1．3-的相反数是 ．
2．分解因式：3
2
2x x x -+= ．
3．人体中约有13
2.510⨯个红细胞，则13
2.510⨯中有 个有效数字． 4．计算：
342x y
y x
⨯= ． 5．如图，直线y x =是线段AB 的垂直平分线，若A 点的坐标是(02)，，则B 点的坐标 是 ． 6
．在函数y =
中，自变量x 的取值范围是 ． 7．如图，在Rt ABC △中，90B =o
∠，D E ，分别是边AB AC ，的中点，
410DE AC ==，，则AB = ．
8．如图，在O e 中，弦AD 平行于弦BC ，若80AOC =o
∠，则DAB ∠= 度． 9．如图，正六边形ABCDEF 的边长为1cm ，则图中阴影部分的面积为 2
cm . 10．观察下列各等式：
111111111121223233434
=-=-=-⨯⨯⨯L ，，， 根据你发现的规律，计算：
2222122334(1)
n n ++++=⨯⨯⨯⨯+L （n 为正整数）． 二、精心选一选：本大题共8小题，每小题3分，共24分．在每小题给出的四个选项中，
只有一个是正确的，请将正确答案前的字母填入题后的括号内．每小题选对得3分，选错、不选或多选均得零分．
第5题
B
C
D
第7
题
第8
题
第9题
11．下列计算中，正确的是（ ） A ．2
3
6
x x x =g
B ．235a b ab +=
C ．321a a -=
D ．236
()a a =
12．已知12x x ，是方程2
250x x +-=的两根，则12
11
x x +的值为（ ） A ．25
-
B ．
25 C ．52 D ．52
- 13．如图，PAB 为O e 的割线，且3PA AB ==， PO 交O e 于点C ，若2PC =，则O e 的半径的
长为（ ） A ．
72
B ．
92
C ．
94
D ．7
14．下列二次根式中，最简二次根式是（ ） A
B
C
D
15．在同一直角坐标系中，函数y kx k =-+与(0)k
y k
=
≠的图象大致是（ ）
16．下列命题：①平行四边形对角线一定相等；②等腰梯形在同一底上的两个角相等；③四
边形的内角和等于360o
；④关于中心对称的两个图形是全等形．其中正确命题的个数有（ ） A ．1个 B ．2个 C ．3个 D ．4个
17．如图，CA
CB ，分别与O e 相切于点D B ，，圆心O 在 AB 上，AB 与O e 的另一交点为E ，2AE =，O e 的半径 为1，则BC 的长为（ ） A
B ．
C ．
2
D
18．已知菱形的周长为，面积为16，则这个菱形较短的对角线长为（ ） A ．4
B ．8
C ．
D ．10
三、解答题：本大题共8小题，满分76分．
19．计算下列各题（本题满分11分，第（1）题5分，第（2）题6分）
（1）01
|1|sin 30(5tan 60)2-+--+o o
第13题
第17题
Ａ． Ｂ． Ｃ． Ｄ．
（2）
2
1224
x
x x --- 20．（本题满分8分）
我市某初中举行“八荣八耻”知识抢答赛，总共50道抢答题．抢答规定：抢答对1题得3分，抢答错1题扣1分，不抢答得0分．小军参加了抢答比赛，只抢答了其中的20道题，要使最后得分不少于50分，问小军至少要答对几道题？ 21．（本题满分8分）
如图，AB CD ∥，AB CD =，点B E F D ，，，在同一直线上，BAE DCF =∠∠． （1）求证：AE CF =；
（2）连结AF EC ，，试猜想四边形AECF 是什么四边形，并证明你的结论．
22．（本题满分8分）
在一次投篮比赛中，甲、乙两人共进行五轮比赛，每轮各投10个球，他们每轮投中的球数如下表：
（1）甲在五轮比赛中投中球数的平均数是 ，方差是 ； （2）乙在五轮比赛中投中球数的平均数是 ，方差是 ； （3）通过以上计算，你认为在比赛中甲、乙两人谁的发挥更稳定些？
A B C
D E F
23．（本题满分9分）
小文家与学校相距1000米．某天小文上学时忘了带一本书，走了一段时间才想起，于是返回家拿书，然后加快速度赶到学校．下图是小文与家的距离y （米）关于时间x （分钟）的函数图象．请你根据图象中给出的信息，解答下列问题： （1）小文走了多远才返回家拿书？
（2）求线段AB 所在直线的函数解析式； （3）当8x =分钟时，求小文与家的距离．
24．（本题满分10分）
如图所示，AB 是O e 的直径，AD 是弦，DBC A =∠∠． （1）求证：BC 与O e 相切；
（2）若OC AD ∥，OC 交BD 于点E ，6BD =，4CE =，求AD 的长．
25．（本题满分10分）
如图所示，在一笔直的公路MN 的同一旁有两个新开发区A
B ，，已知10AB =千米，直线AB 与公路MN 的夹角30AON =o ∠，新开发区B 到公路MN 的距离3B
C =千米．
（1）求新开发区A 到公路MN 的距离；
（2）现要在MN 上某点P 处向新开发区A
B ，修两条公路PA PB ，，使点P 到新开发区A B ，的距离之和最短．请你用尺规作图在图中找出点P 的位置（不用证明，不写作法，保留作图痕迹），并求出此时PA PB +的值．
x （分钟）
26．（本题满分12分）
如图，已知抛物线2
y x bx c =-++与x 轴的两个交点分别为12(0)(0)A x B x ，，
，，且11221
43
x x x x +==，．
（1）分别求出A B ，两点的坐标； （2）求此抛物线的函数解析式；
（3）设此抛物线与y 轴的交点为C ，过C 作直线l 与抛物线交于另一点D （点D 在x 轴上方），连结AC CB BD DA ，，，，当四边形ACBD 的面积为4时，求点D 的坐标和直线l 的函数解析式．
2006年中等学校招生贵港市统一考试
数学参考答案及评分标准
一、1．3 2．()2
1x x - 3．2 4．
22
x
5．()20， 6．2x > 7．6 8．40 9
．
4 10．21
n
n + 二、11．Ｄ 12．Ｂ 13．Ａ 14．Ｂ 15．Ｃ 16．Ｃ 17．Ａ 18．Ａ 三、19．解：（1
）原式11
1122
=+
-+ ·
······································ 4分
1= ····················································· 5分
（2）原式()()
12222x
x x x =
--+- ··········································· 1分 ()()()
2222x x x x +-=
+-
······················································· 3分
()()
222x
x x -=
+- ······················································· 5分
1
2
x =-
+ ································································ 6分 20．解：设小军答对x 道题，依题意得： ··········································· 1分
()32050x x --≥ ································································ 4分 解得：1
17
2
x ≥ ···································································· 6分 x Q 为正整数
x ∴的最小正整数为18 ··························································· 7分 答：小军至少要答对18道题． ················································· 8分 21．证明：（1）AB CD Q ∥ B D ∴=∠∠ ······································································· 1分 又AB CD BAE DCF ==Q ，∠∠ ABE CDF ∴△≌△ ······························································ 2分 AE CF ∴= ········································································· 3分 （2）四边形AECF 是平行四边形 ············································· 4分 证明：由（1）ABE CDF △≌△得AE CF AEB CFD ==，∠∠
····································· 5分
180180AEB CFD ∴-=-o o
∠∠
即AEF CFE =∠∠ ······························································· 6分
AE CF ∴∥ ·
······································································· 7分 AE CF =Q
∴四边形AECF 是平行四边形 ················································· 8分 22．解：（1）7（1分），2（2分） ················································ 3分 （2）7（1分），0.4（2分） ·················································· 6分
（3）22S S <Q 乙甲
∴在比赛中乙的发挥更稳定些． ··············································· 8分 23．解：（1）200米 ···································································· 1分
（2）设直线AB 的解析式为：y kx b =+ ··································· 2分 由图可知：()()50101000A B ，，，
50
101000k b k b +=⎧∴⎨
+=⎩
································································ 4分 解得200
1000
k b =⎧⎨
=-⎩ ··································································· 6分
∴直线AB 的解析式为：2001000y x =- ·································· 7分
（3）当8x =时，600y =（米）
即8t =分钟时，小文离家600米． ··········································· 9分 24．（1）证明：AB Q 是直径 90D AD BD ∴=⊥o
∠， ························································ 1分 90A ABD ∴+=o
∠∠ ···························································· 2分 又DBC A =Q ∠∠
90DBC ABD ∴+=o
∠∠，即90ABC =o
∠，OB BC ∴⊥ ··········· 3分 OB Q 是半径
BC ∴与O e 相切． ······························································· 4分 （2）解：90OC AD D =o
Q ∥，∠
90OEB D OC BD ∴==∴⊥o
∠∠， ········································· 5分
1
32
BE DE BD ∴==
= ·
························································ 6分 90BE OC OBC ⊥=o
Q ，∠
OBE BCE ∴△∽△ ·
····························································· 7分 OE BE BE EC ∴
=即3
34
OE =
9
4
OE ∴= ············································································ 9分
OA OB DE EB ==Q ，
9
22
AD EO ∴== ································································· 10分
25．解：（1）330BC AOC ==o
Q ，∠ 6OB ∴= ·
··········································································· 2分 过点A 作AE MN ⊥于点E ，16AO AB OB =+=
8AE ∴= ·
··········································································· 3分 即新开发区A 到公路的距离为8千米． ······································ 4分
（2）画图正确． ··································································· 6分
过D 作DF AE ⊥的延长线（点D 是点B 关于MN 的对称点），垂足为F ，过B 作BG AE ⊥于G
3BG DF EF CD BC ∴====，．
cos30BG AB ==o
Q g ····················································· 7分
又8311AF AE EF =+=+=． ·············································· 8分
14AD ∴=
=== ·
················· 9分
连结PB ，则PB PD =
14PA PB PA PD AD ∴+=+==（千米） ································ 10分 26．解：（1）由1122
143
x
x x x +==
，得 1213x x ==， ······································································ 1分 ()()1030A B ∴，，，． ····························································· 3分 （2）把()()1030A B ，，，的坐标代入2
y x bx c =-++得
10
930b c b c -++=⎧⎨
-++=⎩
··································································· 4分
解得43b c ==-， ································································ 5分 ∴所求抛物线的函数解析式为2
43y x x =-+- ··························· 6分 （3）由题意，设点D 的坐标为()f h ， 2
43y x x =-+-Q ，∴点C 的坐标为()03-，
4ADB ABC S S +=△△
即
11
223422h ⨯+⨯⨯= ·
························································· 7分 1h ∴= ················································································ 8分
2
431f f ∴-+-=
解得122f f == ···································································· 9分 ()21D ∴， ············································································ 10分 设l 的解析式为y kx m =+
则321m k m =-⎧⎨
+=⎩ 2
3
k m =⎧∴⎨=-⎩ ················································· 11分
l ∴的函数解析式为23y x =- ················································· 12分。