山西省2007年高中阶段教育学校招生统一考试答案

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山西省2007年高中阶段教育学校招生统一考试答案

数 学

一、填空题

5) 1．8 2．圆柱 3．6 4．71 5．0 6．5 7．(?3，8．1 9．?2 10．10 11．15 12．4

二、选择题 13．Ｃ 14．Ｄ 15．Ｃ 16．Ａ 18．Ｂ 19．Ｂ 20．Ａ 三、解答题 21．（1）解：原式?17．Ａ

(a?b)[(a?b)?(a?b)](a?b)[a?b?a?b](a?b)?2b??

2b2b2b?a?b． ························································································································ 6分

当a?3，b?2时，

原式?3?2． ·············································································································· 8分 （2）答：?CPD?60?(或?AOC)时，直线PD与直线AB垂直． ························ 2分

?于点D． 证明：过点P作直线PD?AB于E，交劣弧BC?PC是?O的切线，

??OCP?90?． ··········································································································· 4分

················································································ 5分 ?四边形PCOE内角和为360?， ·

又??CPE??CPD?60，?EOC??BOC?120，

????PEO?360??120??90??60??90?． ································································ 7分

···················································· 8分 ?当?CPD?60?时，直线PD与直线AB垂直． ·

（说明：若将直线PD与直线AB垂直当成题设进行证明，则最多给3分．但若用分析法或

逆证法进行证明并正确的应该给满分．）

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22．解：（1）60?30%?200（人）． ········································································· 3分 （2）200?60?30?110（人）． ················································································ 5分 统计图如右图． ·············································································································· 7分

人数（人）

120

90

60

30

A B

（第22题）

C

类型

30?300（人）（3）2000?． ················································································· 10分 20023．（本题10分） 解：（1）所求概率是

21?． ························································································ 4分 422 1

3

4

1

3 2

4

1

4 2

3

（2）解法一（树状图）： 第一次抽取 1

第二次抽取 2 3 4

共有12种可能的结果：

3)，(2，4)， (1，2)，(1，3)，(1，4)，(2，1)，(2，2)，(3，4)，(4，2)，(4，3)． ·(31)，，(3，1)，(4，························································· 8分

，2)和(2，1)是符合条件的． 其中有两种结果(1所以贴法正确的概率是解法二（列表法）：

第一次取出 第二次 一张 再取出一张 1 2 3 4 1 （1，2） （1，3） （1，4） 2 （2，1） （2，3） （2，4） 3 （3，1） （3，2） （3，4） 4 （4，1） （4，2） （4，3） 21?． ················································································ 10分 126 ········································································································································· 8分

2）和（2，1）是符合条件的． 共有12种结果，其中有两种结果（1，学数学 用专页 第 2 页 共 6 页 搜资源 上网站

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?所求的概率是

21?． ··························································································· 10分 12624．解：（1）据题意得：y?20x?15(700?x)，

即y?5x?10500．······································································································· 3分 （2）据题意得：50x?35(700?x)≥30000， ························································· 5分 解得x≥11002?366． ······························································································ 6分 33?x是整数，

?取x?367代入y?5x?10500得：

y?12335．即每天至少获得利12335元． ································································· 8分 20215323?，?，??（或用百分数近似表示）． 50535757······························ 10分 ?要使每天的利润最大，应生产A种酒0瓶，B种酒700瓶． ·

（3）?（说明：如果有学生用函数模型：利润率r?20x?15(700?x)1?1400???1??对题

50x?35(700?x)3?3x?4900?（3）进行分析解决，则给4分附加分，并且计入总分，但总分不超过120分．） 25．解：（1）△ADC≌△ABC，△ADF≌△ABF，△CDF≌△CBF． ····· 3分 （2）AE?DF． ········································································································· 4分 证法一：设AE与DF相交于点H． Ｅ Ｃ Ｄ ?四边形ABCD是正方形， 5 7 6 1

Ｈ Ｆ Ｍ

3 4 2 Ｂ Ａ (第25题)

?AD?AB，?DAF??BAF．又?AF?AF，

?△ADF≌△ABF．??1??2． ··········································································· 6分

?又?AD?BC，?ADE??BCE?90，DE?CE，

?△ADE≌△BCE．??3??4． ··········································································· 8分

??2??4?90?，??1??3?90?．??AHD?90?．

?AE?DF． ············································································································· 10分 证法二：设AE与DF相交于点H． ?四边形ABCD是正方形，

?DC?BC，?DCF??BCF．又?CF?CF，

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?△DCF≌△BCF．??4??5． ·········································································· 6分

?又?AD?BC，?ADE??BCE?90，DE?CE，

?△ADE≌△BCE． ??6??7． ················································································································· 8分

??4??6?90?，??5??7?90?．??EHD?90?．

?AE?DF． ············································································································· 10分 证法三：同“证法一”得△ADE≌△BCE． ?EA?EB．

??EAB??2． ··········································································································· 8分 ??EAB??1．

??EAB??3?90?，??1??3?90?．??AHD?90?．

?AE?DF． ············································································································· 10分 （3）BM?MC． ······································································································ 12分

26．解：（1）据题意得：k?4?0，

2?k??2．

当k?2时，2k?2?2?0． 当k??2时，2k?2??6?0．

又抛物线与y轴的交点在x轴上方，?k?2．

········································································· 4分 ?抛物线的解析式为：y??x2?2． ·

函数的草图如图所示．（只要与坐标轴的三个交点的位置及图象大致形状正确即可）

········································································································································· 6分 （2）解：令?x?2?0，得x??2． 不0?x?22时，A1D1?2x，A1B1??x2?2，

2····································································· 8分 ?l?2(A1B1?A1D1)??2x?4x?4． ·4 当x?y 2时，A2D2?2x，

223 A2B2??(?x?2)?x?2．

?l?2(A2D2?A2B2)?2x?4x?4．

?l关于x的函数关系是：

当0?x?当x?2D1 C2 C1 ?4 ?3 ?2 ?1 2 1 A1 B2 1 2 3 4 x ?1 B1 2时，l??2x2?4x?4；

?2 ?3 2时，l?2x2?4x?4． ································ 10分

D2 ?4 A2 ?5 学数学 用专页 第 4 页 共 6 页 搜资源 上网站 ?6 ?7 （第26题）