1. (本小题满分12分) 已知常数a > 0, n为正整数,f n ( x ) = x n – ( x + a)n ( x > 0 )是关于x的函数. (1) 判定函数f n ( x )的单调性,并证明你的结论. (2) 对任意n ³ a , 证明f `n + 1 ( n + 1 ) < ( n + 1 )fn`(n) 解: (1) fn `( x ) = nx n – 1 – n ( x + a)n – 1 = n [x n – 1 – ( x + a)n – 1 ] , ∵a > 0 , x > 0, ∴ fn `( x ) < 0 , ∴ f n ( x )在(0,+∞)单调递减. 4分(2)由上知:当x > a>0时, fn ( x ) = xn – ( x + a)n是关于x的减函数, ∴ 当n ³ a时, 有:(n + 1 )n– ( n + 1 + a)n n n – ( n + a)n. 2分又 ∴f `n + 1 (x ) = ( n + 1 ) [xn –( x+ a )n ] , ∴f `n + 1 ( n + 1 ) = ( n + 1 ) [(n + 1 )n –( n + 1 + a )n ] < ( n + 1 )[ nn – ( n + a)n] = ( n + 1 )[ nn – ( n + a )( n + a)n – 1 ] 2分 ( n + 1 )fn`(n) = ( n + 1 )n[n n – 1 – ( n + a)n – 1 ] = ( n + 1 )[n n – n( n + a)n – 1 ], 2分 ∵( n + a ) > n , ∴f `n + 1 ( n + 1 ) < ( n + 1 )fn`(n) . 2分