时,v′(x )≤0 ,则v (x )在[0 ,+∞)上单调递减,∴v (x )=u′(x )≤v (0 )=0 , ∴u (x )在[0 ,+∞)上单调递减,∴u (x )≤u (0 )=0 ,此时f (x )≤g (x )在[0 ,+∞)上恒成立; ⅱ. 若2a-1>0 ,即a> 12 时,若0<x< 2a-1 a 时,v′(x )>0 ,则v (x )在0 , 2a-1 ?è ? ?? ÷ a 上单调递增, ∴v (x )=u′(x )>v (0 )=0 ,∴u (x )在0 , 2a-1 ?è ? ?? ÷ a 上也单调递增, ∴u (x )>u (0 )=0 ,即f (x )>g (x ),不满足条件. 综上,不等式f (x )≤g (x )在[0 ,+∞)上恒成立时,实数a 的取值范围是0 , é? êê ù? úú 12 . (10 分) ………(3 )由(2 )知,当a= 12 时,则1-e x≤ x 12 x+1 ?e x≥ 2-x 2+x , 当x∈[0 ,2 )时,e x≥ 2-x 2+x ?x≤ln 2+x 2-x ,令 2+x 2-x =n ,则x= 2n+2 n+1 =2- 4 n+1 , ∴lnn>2- 4 n+1 (n∈N ),∴∑ n k lnk≥2n-∑ n k 4 k+1 ,∴ln (n !)≥2n-∑ n k 4 k+1 , 又由(1 )得h (x )≤h (1 ),即xe x≤ 1 e ,当x>0 时,ln (xe x)≤ln 1 e =-1 ,∴lnx≤x-1 , ln (n !)=ln2+ln3+ …+lnn≤1+2+ …+ (n-1 )= n (n-1 ) 2 , 综上得2n-∑ n k 4 k+1 ≤ln (n !)≤ n-n 2 ,即e n-∑ n k k≤n !≤e n (n-1 ) 2 . (14 分) …………………
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