xxxfxfx?????????.因为(2)()fxfxm??对于[0,)x???恒成立,且()0fx?,2()(2)()[()]2mfxfxfxfx?????.又0x?,∴由(2)知()fx最小值为2,∴()2fx?时,m最小为2-4+2=0.…………………………………………………………………………12分22.解:(1)依题意,'()2lnfxxaxa???,故'(1)21fa???,解得1a?.(2)依题意,0[1,]xe??,使得0001ln0axaxx????成立,即函数1()lnahxxaxx????在[1,]e上的最小值min[()]0hx?.22221(1)(1)[(1)]'()1aaxaxaxxahxxxxx????????????,当10a??,即1a??时,令'()0hx?,∵0x?,∴1xa??,令'()0hx?,∵0x?,-10-∴01xa???,∴()hx的单调增区间为[1,)a???,单调减区间为(0,1]a?.当10a??,即1a??时,'()0hx?恒成立,∴()hx的单调增区间为(0,)??.……………………6分①当1ae??,即1ae??时,()hx在[1,]e上单调递减,∴min1[()]()0ahxheeae??????,∴211eae???,∵2111eee????,∴211eae???;②当11a??,即0a?时,()hx在[1,]e上单调递增,∴min[()](1)110hxha?????,∴2a??;③当11ae???,即01ae???时,∴min[()](1)2ln(1)0hxhaaaa???????,∵0ln(1)1a???,∴0ln(1)aaa???,∴(1)2ha??,此时不存在0x,使0()0hx?成立.综上可得所求a的范围为21(,2][,)1ee????????.…………………………………………………………12分
全文预览