即c?2c?24?0， 解得c??6（舍去），c?4. （2）由题设可得?CAD?2?2，
所以?BAD??BAC??CAD??6.
故?ABD面积与?ACD面积的比值为
1?AB?AD?sin26?1. 1AC?AD21又?ABC的面积为?4?2sin?BAC?23，
2所以?ABD的面积为83. 19．解：（1）当n?1时，S1?1?a1，得a1?当n?2时，由Sn?1?an，① 得Sn?1?1?an?1，② ①—②，得2an?an?1，又a1?1. 21a1?0，∴an?0，∴n??n?2?， 2an?12n?1?∴?an?是等比数列，∴an???.
?2??1??1?（2）由an???，则cn?4an?1?4????1，
?2??2?则Tn?c1?c2?c3?L?cn?4?a1?a2?a3?L?an??n
nn1?1?1???42?2n??4??n?4?n?n.
121?2urur20．解：（1）若m∥p，得3sinx?23cosx?0，∴sinx?2cosx；
因为cosx?0，所以tanx?2.
urr3sinxcosx?cos2x3tanx?123?12所以m?n?3sinxcosx?cosx?. ??222sinx?cosxtanx?15（2）在?ABC中，由正弦定理得
cosBbsinB???? cosC2a?c2sinA?sinC?2sinAcosB?cosBsinC??sinBcosC
?2sinAcosB???cosBsinC?sinBcosC???sin?B?C???sinA.
又A??0,??，故sinA?0，得cosB??因为0?B??，所以B?又
1. 22??，则0?A?. 33urrf?x??m?n?3sinxcosx?cosxcosx?3sin2x1?cos2x??1???sin?2x???. 226?2?所以f?A??sin?2A?????1????0?A????. 6?2?3????5???,6?66??. ?因为A??0,????3??，所以2A?所以sin?2A??????1????,1?. 6??2??3???所以f?A???1,?，即函数f?A?的值域为?1,?.
2221．（1）证明：由bn?1??3???an?cna?bn，cn?1?n两式相减，得 22cn?1?bn?1??1?cn?bn?， 2又c1?b1?2?0，∴cn?bn?0， ∴
cn?1?bn?11??为常数.
cn?bn2∴?cn?bn?是等比数列. （2）解：由bn?1?an?cn，得an?2bn?1?cn， 2∴a1?a2?L?an?2?b2?b3?L?bn?1???c1?c2?L?cn?，
∴Tn?2Sn?1???a1?a2?L?an??2b1??∴不等式Tn?2Sn?1?3n?2，可化为?a1?*a1?1?2n?1?2?2??a1?2n?1??2，
3n. n2?13?n?1?2n?3?3n??33n?n?n?1?0， ∵n?N时，n?1n2?12?1?2?1??2?1?∴数列??3n??是递减数列， n?2?1?3n取最大值3. 2n?1n?1时
∴?a1?3，a1??3. ∴整数a1的最大值是-4.
12x2?3x?122．解：（1）由f?x??lnx?x?3x，?x?0?，则f??x??，
2x当x?3x?1?0时，得x?23?53?52或0?x?；当x?3x?1?0时，得223?53?5?x?. 22即函数f?x?在?0,????3?53?5?3?5?,上单调递增，在上单调递减，在??????2?2??2?3?5?,?????2?上单调递增，
???3?5?3?511?35?ln?∴f?x?的极大值为f?， ??2?24???3?5?3?511?35?ln?. f?x?的极小值为f??2??24??（2）f?x2??f?x1??lnx212??x2?x12??a?x2?x1?， x121x2?ax?1又f??x???x?a??x?0?，所以x1,x2是方程x2?ax?1?0的两个实根，
xx由韦达定理得：x1?x2?a，x1x2?1， ∴f?x2??f?x1??lnx212??x2?x12??a?x2?x1? x12?lnx212x121x1?xx???x2?x12??ln2??x2?x12??ln2??2?1?. x12x12x1x2x12?x1x2?2x1?1??t?1??0. 11?1?设t?2?t?e?，令g?t??lnt??t??,t?e，g??t????1?2???x12?t?t2?t?2t2∴g?t?在?e,???上是减函数，g?t??g?e??1?故f?x2??f?x1?的最大值为1?
e1?， 22ee1?. 22e