設(shè)函數(shù)f(x)=(2-a)lnx+
+2ax.
(Ⅰ)當(dāng)a=0時,求f(x)的極值;
(Ⅱ)當(dāng)a≠0時,求f(x)的單調(diào)區(qū)間;
(Ⅲ)當(dāng)a=2時,對任意的正整數(shù)n,在區(qū)間[
,6+n+
]上總有m+4個數(shù)使得f(a
1)+f(a
2)+f(a
3)+…+f(a
m)<f(a
m+1)+f(a
m+2)+f(a
m+3)+f(a
m+4)成立,試問:正整數(shù)m是否存在最大值?若存在,求出這個最大值;若不存在,說明理由.