(ⅱ)当a?3时，x,g'(x),g(x)的变化如下表：
(??,2a?3) 3 2a?3 30 (2a?3,1) 3 (1,??) g'(x) g(x)
0 ∴增区间为：(1,??),(??,2a?32a?3); 减区间为：(,1). ?????12分 3320．(本小题满分12分)
解：(Ⅰ)依题意得：A(?2,0),B(2,0),
∴曲线的方程为y2?8x. ??????4分
??y?k(x?1)(Ⅱ)由?得：kx2?(2k?8)x?k?0, 2??y?8x???(2k?8)2?4k2?0?k?0 ????7分 由??k?02k?8,x1x2?1, 设M(x1,y1),N(x2,y2),则：x1?x2?k∴AM?AN?(x1?2,y1)(x2?2,y2)?(x1?2)(x2?2)?y1y2 ????9分
16?(k?1)x1x2?(2?k)(x1?x2)?4?k??1?17
k?∴0?k?1, ∴??(0,]. ??????12分
421．(本小题满分12分)
Snan?1解：(Ⅰ)?, ∴2Sn?an2?an(n?1)……①, 2Sn?1?an?12?an?1(n?2)……②
an2①-②得：2an?an2?an?12?an?an?1?(an?an?1)(an?an?1?1)?0
an?0, ∴an?an?1?1, 故{an}为等差数列，又在①中令n?1得a1?1, ∴an?1?(n?1)?1?n ??????4分
n(n?1), (Ⅱ)an?n,∴Sn?2111222∴Tn? ???????S1S2Sn1?22?3n(n?1)111112n?2[(1?)?(?)??(?)]?. ??????8分
223nn?1n?12(Ⅲ)m?n?2p, ∴mn?p, ??????9分
112∴SmSn?mn[(a1?am)(a1?an)]?mn[a1?a1(am?an)?aman]
441mn(a1?ap)p2222?mn(a1?2a1ap?ap)?2[]?Sp, ??????11分 4p2∴
1121112??. ????????12分 即??2?,22222222SmSnSpSmSnSmSnSp