n(n?1)·········· 14分 lna?ln(a?1)?ln(2a?1)?ln(3a?1)??ln(na?1)(n?N*)成立． ·
2n(n?1)解法四：······· 4分 lna?ln(a?1)?ln(2a?1)?ln(3a?1)??ln(na?1)(n?N*)．
2理由如下：因为a?e，
n(n?1)欲证lna?ln(a?1)?ln(2a?1)?ln(3a?1)??ln(na?1)(n?N*)成立，
2只要证an(n?1)2?(a?1)(2a?1)(3a?1)(na?1)(n?N*)，
只需证a?a2?a3??an?(a?1)(2a?1)(3a?1)·································· 6分 (na?1)， ·
即证an?na?1(n?N*)． ·········································································· 8分
用数学归纳法证明如下： ①当n?1时，a?a?1成立，
②当n?k时，假设ak?ka?1成立， ····························································· 9分 那么当n?k?1时，ak?1?ak?a ?(ka?1)?a， 下面只需证明(ka?1)?a?(k?1)a?1， ·························································· 10分 注意到a?e?2且k?2，则
(ka?1)?a?2(ka?1)?(k?2)a?2?(k?1)a?(a?1)?1?(k?1)a?1， ·················· 12分 所以当n?k?1时，ak?1?(k?1)a?1成立，
由①和②可知，对一切n?N*，an?na?1成立． ·········································· 13分 所以当a?e时， n(n?1)·········· 14分 lna?ln(a?1)?ln(2a?1)?ln(3a?1)??ln(na?1)(n?N*)成立． ·
221（1）本小题主要考查矩阵及其逆矩阵、求曲线在矩阵所对应的线性变换作用下的曲线的方程等基础知识，考查运算求解能力，考查化归与转化思想．满分7分． 解：（Ⅰ）因为矩阵A是矩阵A的逆矩阵， 且A?1??1222?2?······················································ 2分 ??????1?0， ·??222?2????所以A?????2??2?． ·
·········································································· 3分
2??2?（Ⅱ）解法一：设xy?1上任意一点?x,y?在矩阵A所对应的线性变换作用下的像为点?x?,y??，则
2222??22??x???2?x??2?1?x???······························································ 4分 ???A???????， ·???yyy22??????????22??2x??x??y??,??2由此得? ·············································································· 5分
2?y??y??x??,??2代入方程xy?1，得y??x??2. ································································ 6分
22
所以xy?1在矩阵A所对应的线性变换作用下的曲线方程为y2?x2?2． ············ 7分
解法二：设xy?1上任意一点?x,y?在矩阵A所对应的线性变换作用下的像为点?x?,y??，则
??x?????????y????2222?2??2??x?， ············································································ 4分 ???y2???2??22?2x?y,x??x???x??y??,???222其坐标变换公式为?由此得? ·························· 5分
2??y??2x?2y,y??y??x??,???2?22代入方程xy?1，得y?2?x?2?2. ································································ 6分 所以xy?1在矩阵A所对应的线性变换作用下的曲线方程为y2?x2?2． ············· 7分
（2）本小题主要考查参数方程、极坐标方程、直角坐标方程、极点坐标等基础知识，考查运算求解能力，考查数形结合思想、化归与转化思想．满分7分．
?x?2?2cos?,2解：（Ⅰ）将?消去参数?，得?x?2??y2?4，
?y?2sin?所以C1的普通方程为：x2?y2?4x?0． ····················································· 1分 ?x??cos?,222将?代入x?4x?y?0得??4?cos??0， ······························· 2分
y??sin??所以C1的极坐标方程为??4cos?． ···························································· 3分
（Ⅱ）将曲线C2的极坐标方程化为直角坐标方程得：x?y?4?0． ·················· 4分
?x2?y2?4x?0,由? ·················································································· 5分 ?x?y?4?0,?x?4,?x?2,解得?或? ·············································································· 6分
y?0y??2.??7???所以C1与C2交点的极坐标分别为?4,0?或?22,·································· 7分 ?． ·
4??（3）本小题主要考查平均值不等式、解含有绝对值号的不等式等基础知识，考查推理论证能力,
考查化归与转化思想．满分7分．
解：（Ⅰ）因为a?0,x?0，根据三个正数的算术—几何平均不等式，得
2aaaaaaaa22，当且仅当x2?，即x?3时等号成f?x??x??x???33x???332x2x2x2x2x2x42立， ········································································································ 1分 a2?3(a?0)， ·又因为函数f?x?的最小值为3，所以3································· 2分 4解得a?2． ····························································································· 3分 （Ⅱ）解法一：由（Ⅰ）得：x?2?x?1?4．
3?x…2,??1?x?2,?x??1,原不等式等价于?或?或? ·············· 5分
2?x?x?1?4.x?2?x?1?42?x?x?1?4???
53或?1?x?2或??x??1， ·················································· 6分 2235?35?解得??x?．所以原不等式解集为?x??x??． ·································· 7分
2222??所以2剟x解法二：由（Ⅰ）得：x?2?x?1?4．
由绝对值的几何意义，可知该不等式即求数轴上到点2和点?1的距离之和不大于4的点的集合．
·············································································································· 5分 故原不等式解集为?x???35?·························································· 7分 ?x??． ·
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