2008年新疆乌鲁木齐市高中招生统一考试答案

2008年新疆乌鲁木齐市高中招生统一考试
数学试卷参考答案及评分建议
一、选择题（本大题共7小题，每小题4分，共28分）
二、填空题（本大题共6小题，每小题4分，共24分） 8．(00)，
9．90A ∠=或AD BC =或AB CD ∥
10．2
5786(1)8058.9x +=
11．4.8
12．
2
3
13．
15π4
三、解答题（本大题Ⅰ—Ⅴ题，共10小题，共98分） Ⅰ．（本题满分12分，第14题6分，第15题6分）
14．解：由239x x ++≥，得6x ≥ ····································································· 2分
由2593x x +>-，得4
5
x >
··································································· 4分 所以，不等式组的解集是6x ≥ ································································· 6分
15．解：原式1
)1()1)(1(1112
+-⋅-+-+=x x x x x ······················································· 2分 2211(1)(1)1(1)(1)x x x x x x -+--=
-=+++ ······································································ 4分 2
2
(1)x =
+ ······················
···········
······································································· 5分
当1x =
时，原式2
3=
= ······························································ 6分 Ⅱ．（本题满分28分，第16题7分，第17题10分，第18题11分）
16．已知：①③（或①④，或②③，或②④） ··························································· 2分 证明：在ABE △和DCE △中，
B C
AEB DEC AB DC ∠=∠⎧⎪
∠=∠⎨⎪=⎩
，ABE DCE ∴
△≌△ ·························································· 6分 AE DE ∴=，即AED △是等腰三角形 ·
································································ 7分
17．解：设该厂原来每天生产x 顶帐篷 ···································································· 1分 据题意得：
1500300120041.5x x
x ⎛⎫
-+= ⎪⎝⎭ ··································································· 5分 解这个方程得100x = ························································································· 8分 经检验100x =是原分式方程的解 ·········································································· 9分 答：该厂原来每天生产100顶帐篷． ······································································ 10分 18．解：（1）
······················································································································· 3分 4009100x =+ ································································································· 6分 （2）
30x -≥且150x -≥即315x ≤≤，又y 随x 增大而增大 ·
··························· 9分 ∴当3x =时，能使运这批挖掘机的总费用最省，运送方案是A 地的挖掘机运往甲地3台，运往
乙地13台；B 地的挖掘机运往甲地12台，运往乙地0台 ··········································· 11分 Ⅲ．（本题满分36分，第19题12分，第20题12分，第21题12分） 19．解：树形图如下：
或列表如下：
共20种情况 ······································································································ 6分
贝贝 甲 乙 丙 宝宝 甲 乙 丙 宝宝 贝贝 乙 丙 甲 丙 甲 宝宝 贝贝 乙
宝宝 贝贝 宝宝
贝贝
甲
丙
乙
（1）宝宝和贝贝同时入选的概率为
21
2010
= ·
··························································· 9分 （2）宝宝和贝贝至少有一人入选的概率为147
2010
= ·
················································· 12分 20．解：过点C 作CE AD ∥，交AB 于E
CD AE ∥，CE AD ∥ ·
·················································································· 2分 ∴四边形AECD 是平行四边形 ·············································································· 4分
50AE CD ∴==m ，50EB AB AE =-=m ，30CEB DAB ∠=∠= ·
······················ 6分 又60CBF ∠=，故30ECB ∠=，50CB EB ∴==m ··········································· 8分
∴在Rt CFB △中，sin 50sin 6043CF CB CBF =∠=≈m ·································· 11分
答：河流的宽度CF 的值为43m ． ········································································· 12分 21．证明：（1）在BEC △中，
G F ，分别是BE BC ，的中点
GF EC ∴∥且1
2
GF EC =
·
··············································································· 3分 又H 是EC 的中点，1
2
EH EC =，
GF EH ∴∥且GF EH = ·
················································································· 4分 ∴四边形EGFH 是平行四边形 ············································································· 6分
（2）证明：
G H ，分别是BE EC ，的中点
GH BC ∴∥且1
2GH BC =
·
·············································································· 8分 又EF BC ⊥，且1
2
EF BC =，EF GH ∴⊥，且EF GH = ································ 10分
∴平行四边形EGFH 是正方形．
Ⅳ．（本题满分8分）
22．他的推断是正确的． ······················································································ 1分 因为“两点确定一条直线”，设经过A B ，两点的直线解析式为y kx b =+ ·
···················· 2分 由(1
2)(34)A B ，，，，得234k b k b +=⎧⎨+=⎩解得1
1k b =⎧⎨=⎩
························································ 4分
∴经过A B ，两点的直线解析式为1y x =+ ····························································· 5分
把1x =-代入1y x =+中，由116-+≠，可知点(1
6)C -，不在直线AB 上， 即A B C ，，三点不在同一直线上 ·
········································································· 7分
所以A B C ，，三点可以确定一个圆． ···································································· 8分 Ⅴ．（本题满分14分） 23．解：
（1）作CH
x ⊥轴，H 为垂足，
1CH =，半径2CB = ·
················································ 1分 60BCH ∠=，120ACB ∴∠= ··································· 3分
（2）
1CH =，半径2CB =
HB ∴=
(1A ， ········································ 5分
(1B ·································································· 6分
（3）由圆与抛物线的对称性可知抛物线的顶点P 的坐标为(13)， ··································· 7分
设抛物线解析式2
(1)3y a x =-+ ·········································································· 8分
把点(1B 代入上式，解得1a =- ································································· 9分
222y x x ∴=-++ ··························································································· 10分
（4）假设存在点D 使线段OP 与CD 互相平分，则四边形OCPD 是平行四边形 ············ 11分
PC OD ∴∥且PC OD =．
PC y ∥轴，∴点D 在y 轴上． ······································································· 12分
又
2PC =，2OD ∴=，即(02)D ，
． 又(02)D ，
满足2
22y x x =-++， ∴点D 在抛物线上 ····························································································· 13分
所以存在(02)D ，
使线段OP 与CD 互相平分． 14分
生于忧患，死于安乐《孟子•告子》
舜发于畎亩之中，傅说举于版筑之间，胶鬲举于鱼盐之中，管夷吾举于士，孙叔敖举于海，百里奚举于市。

故天将降大任于是人也，必先苦其心志，劳其筋骨，饿其体肤，空乏其身，行拂乱其所为，所以动心忍性，曾益其所不
能。

人恒过，然后能改；困于心，衡于虑，而后作；征于色，发于声，而后喻。

入则无法家拂士，出则无敌国外患者，国恒亡。

然后知生于忧患，而死于安乐也。