发布时间 : 星期六 文章江苏省南通中学高一数学第一学期期末考试试卷更新完毕开始阅读
1····························8分 ?(1??)AC??AM?(1??)AC??AB·
2?1???1??,??3····························11分 由平面向量基本定理可知：?，
1??????224 ∴??,??·····················································13分
5521221 ∴AP?a?(1?)b=a?b·····································15分
53555ANANMPBMPCCB
π3π19．已知平面直角坐标系内有三点A(sinx,1),B(cosx,2a),C(a,1)，x?[?,]，若函数
44f(x)?ACBC的最大值为g(a)，求函数g(a)的最小值．
解：f(x)?ACBC?(a?sinx,0)(a?cosx,1?2a)·····························2分 ?(a?sinx)(a?cosx)
?sinxcosx?a(sinx?cosx)?a2····································3分
t2?1 令sinx?cosx?t，则sinxcosx?································4分
2t2?2at?2a2?1(t?a)2?a2?1 ∴y?f(x)?·························5分 ?22π3ππ ∵x?[?,]∴t?sinx?cosx?2sin(x?)?[0,2]················6分
44 422?2a2?2a2?122 当a?时，ymax?···················8分 ?(a?)·
22222a2?121 当a?时，ymax?································10分 ?a?·
2222?22),?(a??2 ∴g(a)???21a?,??2 当a? 当a?a?22······································11分 2a?222时，[g(a)]min?g()?0，·······························13分 222时，g(a)?0···········································14分 2 ∴[g(a)]min?0····················································16分
20．设?ABC的外心为O（三角形外接圆的圆心），其外接圆半径为1，以线段OA、OB为邻边作平行四边形，第四个顶点为D，再以OC，OD为邻边作平行四边形，它的第四个顶点为
CH．若OA?a,OB?b,OC?c． （1）用a,b,c表示OH； （2）求证：点H为?ABC的垂心；
（3）设?ABC中，?A?60,?B?45，求|OH|． A 解：（1） OH?OC?OD?OC?(OA?OB)?a?b?c MBOGHD ··································3分
（2）方法一：AH?AO?OH??a?(a?b?c)?b?c· ····················4分 BC??b?c·····················································5分 则AH?BC?(b?c)?(c?b)?|c|2?|b|2·····························6分 因为O为外心，且外接圆半径为1，所以|c|?|b|?1 ∴AH?BC?|c|2?|b|2?0
则AH?BC，即AH?BC·········································7分 同理可得：CH?AB···············································8分 所以，点H为?ABC的垂心；·········································9分 方法二：OH?a?b?c?OA?OB?OC
设G为?ABC的重心，则GA?GB?GC?GA?2GM?0
OH?OA?OB?OC?GA?GO?GB?GO?GC?GO?GA?GB?GC?3GO ?3OG·························································6分 则O，H，G三点共线，且OH?3OG··································7分
由欧拉线可知，点H为?ABC的垂心；··································9分 （3）|OH|2?··········10分 （a?b?c)2?a2?b2?c2?2a?b?2b?c?2c?a· ?1?1?1?2?1?1?cos?AOB?2?1?1?cos?BOC?2?1?1?cos?COA ?3?2(cos?AOB?cos?BOC?cos?COA)·····················11分 ∵?A?60 ∴?BOC?120 （圆心角是圆周角的两倍）··········12分 1 ∴cos?BOC??
2 同理可得，cos?AOC?0,cos?AOB??3·······················14分 2 ∴|OH|2?2?3···············································15分 ∴|OH|?2?3?6?2·····································16分 2 （说明：若没有化为
6?2也可以给分） 2