∴当x?(1,??)时h??x??0,∴h?x?在x?(1,??)上为减函数, ∴h?x?max?h?1??2e
x?Qa??4?2xe????max
∴a?2e, 即a??2e,??? .
?Ⅱ?方法一:因为g(x)?ex(x2?4x?5)?a,
所以g'(x)?e(x?1)?0, 所以g(x) 在???,???上为增函数,
因为g(x1)?g(x2)?2g(m),即g(x1)?g(m)?g(m)?g(x2),
x2g(x1)?g(m)和g(m)?g(x2)同号,
所以不妨设x1?m?x2,设h(x)?g(2m?x)?g(x)?2g(m)(x?m?1),…8分 所以h'(x)??e2m?x(2m?x?1)2?ex(x?1)2,
22因为e2m?x?ex,(2m?x?1)?(x?1)?(2m?2)(2m?2x)?0, 所以h'(x)?0,所以h(x)在(m,??)上为增函数,
所以h(x)?h(m)?0,所以h(x2)?g(2m?x2)?g(x2)?2g(m)?0, 所以g(2m?x2)?2g(m)?g(x2)?g(x1), 所以2m?x2?x1,即x1?x2?2m. 方法二:
Qg?x??exf?x??x2?4x?5ex?a
??g?x1??g?x2??2g?m? m??1,???,
x2?4x2?5ex2?a?2m2?4m?5em?2a ∴x12?4x1?5e1?a?x2x2?4x2?5ex2?2m2?4m?5em ∴x12?4x1?5e1?x2????????????∴设??x??x2?4x?5ex x?R,则??x1????x2??2??m?,
x∴???x???x?1?e?0 ???x?在x?R上递增且???1??0
2??令x1????,m?,x2??m,???
设F?x????m?x????m?x?, x??0,???,
m?xm?x∴F??x???m?x?1?e??m?x?1?e
22Qx?0
∴em?x?em?x?0,?m?x?1???m?x?1???2m?2?2x?0
22∴F??x??0, F?x?在x??0,???上递增, ∴F?x??F?0??2??m?,
∴??m?x????m?x??2??m?,x??0,???
令x?m?x1
∴??m?m?x1????m?m?x1??2??m?
即:??2m?x1????x1??2??m? 又Q??x1????x2??2??m?,
∴??2m?x1??2??m????x2??2??m?即:??2m?x1????x2? Q??x?在x?R上递增
∴2m?x1?x2,即:x1?x2?2m得证.
22.?Ⅰ?解:联立???cos??33,cos???,
2???4cos??0????2,???6,
??23,
交点坐标?23,?????. 6??0,?Ⅱ?设P??,??,Q??0,?0?且?0?4cos?0,?0???2?,
???2uuur2uuur?
????
由已知OQ?QP,得?05,
3?
??0??2???∴?=4cos?,点P的极坐标方程为??10cos?,???0,?. 5?2???4x?1??23.解:?Ⅰ?当m=-2时,f?x??2x?2x?3-2=?1????4x?5?当??x?0??3??<x<0??, ?2?3??x????2???4x?1?313解得0?x?;当?<x<0,1?3恒成立
22x?0???4x?5?33?当?解得-2?x?? 32x????2此不等式的解集为?-2,?.
2??1????3?m?Ⅱ?当x????,0?时f?x??2x?2x?3?m=????4x?3?m??当?<x<0时,不等式化为3+m?x??3??<x<0???2?, 3??x????2??322. x由x?222??[(?x)?(?)]??2(?x)(?)??22 xxx2即x??2时等号成立. x当且仅当?x??∴m?3??22,∴m??3?22.
32时,不等式化为?4x?3?m?x?.2x223,?]. ∴m?5x??3,令y?5x??3,x?(?? 2xx23Qy??5?2?0,x?(??,?],
x223∴y?5x??3在(??,?]上是增函数.
2x3235∴当x??时,y?5x??3取到最大值为?.
6 2x35?∴m??.
6当x??综上m??3?22.