?12?k． ·········································································· 10分 23?4k12?1???k6k1?所以y1?y2?，解得， ············································· 12分 ??3?4k23?4k221所以PF?AQ?． ·················································································· 13分
220．本小题主要考查导数的几何意义、导数的应用（单调性、最值）、用点斜式求直线方程、比较不等式、证明不等式、数学归纳法等基础知识，考查推理论证能力、运算求解能力等，考查函数与方程思想、化归与转化思想、数形结合思想、有限与无限思想等．满分14分． 解：（Ⅰ）当a?e时，函数f(x)?ex?e?x，则f?(x)?ex?e， ··························· 1分 所以f?(1)?0，且f(1)?0， ········································································ 2分 于是f(x)在点(1,f(1))处的切线方程为y?f(1)?f?(1)?(x?1)， ··························· 3分 故所求的切线方程为y?0． ········································································ 4分
n(n?1)（Ⅱ）解法一：lna?ln(a?1)?ln(2a?1)?ln(3a?1)??ln(na?1)(n?N*)．
2 ·············································································································· 5分
e， 理由如下：因为a…所以y1???y2?构造函数u(x)?ax?ax?1(a?e,x?[1,??))， ·················································· 6分 所以u?(x)?axlna?a，
e，所以lna…lne?1，所以u?(x)?axlna?a?ax?a…0． 因为a…························ 7分
所以函数u(x)?ax?ax?1在[1,??)上单调递增，且u(1)?a?a?1?1?0，
u(1)?0， ·所以u(x)…·················································································· 9分 e时，ax?ax?1?0恒成立， 即当x?[1,??)，且a…所以ax?ax?1． ····················································································· 10分 取x?n(n?N*)，得an?na?1成立． ······················································· 11分 所以a?a2?a3?n(n?1)2?an?(a?1)(2a?1)(3a?1)···································· 12分 (na?1)， ·
?(a?1)(2a?1)(3a?1)(na?1)， ·所以a················································· 13分
n(n?1)所以··· 14分 lna?ln(a?1)?ln(2a?1)?ln(3a?1)??ln(na?1)(n?N*)成立． ·
2n(n?1)解法二：lna?ln(a?1)?ln(2a?1)?ln(3a?1)??ln(na?1)(n?N*)．
2 ·············································································································· 5分
e， 理由如下：因为a…n(n?1)欲证lna?ln(a?1)?ln(2a?1)?ln(3a?1)??ln(na?1)(n?N*)成立，
2只需证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
ax?xaxlna1?xlnax构造函数g(x)?x，则g?(x)?． ··································· 10分 ?aa2xaxe，所以lna?1． 因为a…
1； lna1令g?(x)?0，得x?．
lna11所以函数g(x)在(??,)单调递增；在(,??)上单调递减．
lnalnax111所以函数g(x)的最大值为g(．所以x?， ···························· 11分 )?alnaelnaelnax?11所以x?1?，即ax?1…e(x?1)lna，则 aelnaxa?ax?1?a[ax?1?(x?1)]?1?a?a[e(x?1)lna?(x?1)]?1?a ?a[2(x?1)?(x?1)]?1?a?a(x?2)?1?0， ············································ 12分 令g?(x)?0，得x?所以ax?ax?1．
取x?n(n?N*)，得an?na?1成立． ······················································· 13分
e时， 所以当a…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)解法三：lna?ln(a?1)?ln(2a?1)?ln(3a?1)??ln(na?1)(n?N*)．
2 ·············································································································· 5分
e， 理由如下：因为a…n(n?1)欲证lna?ln(a?1)?ln(2a?1)?ln(3a?1)??ln(na?1)(n?N*)成立，
2只需证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 只需证明k(a2?a)?2a?1，
e，所以a2?a?0，所以只需证明k?因为a…2a?1， a2?a2a?1，只需证明a(a?3)??1， a2?ae恒成立即可． ·只需证明a2?3a?1?0对a…················································· 11分 所以只需证明1?构造函数h(a)?a2?3a?1(a…e)，
3535因为h(a)?(a?)2?在[e,??)单调递增，所以h(a)?h(e)?(e?)2??0． ······ 12分
2424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． ·
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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