Question

(本小題滿分12分)已知常數(shù)a > 0, n為正整數(shù)，f _n ( x ) = x ⁿ – ( x + a)ⁿ ( x > 0 )是關(guān)于x的函數(shù).(1) 判定函數(shù)f _n ( x )的單調(diào)性，并證明你的結(jié)論.(2) 對(duì)任意n ?? a , 證明f `<sub>n + 1</sub> ( n + 1 ) < ( n + 1 )f<sub>n</sub>`(n)

Answer 1

(Ⅰ) f _n ( x )在（0，+∞）單調(diào)遞減  (Ⅱ) 略

解析:

： (1)  f_n `( x ) = nx ^{n – 1} – n ( x + a)^{n – 1} = n [x ^{n – 1} – ( x + a)^{n – 1} ] ,

  ∵a > 0 , x > 0,  ∴ f_n `( x ) < 0 , ∴ f _n ( x )在（0，+∞）單調(diào)遞減.      4分

（2）由上知：當(dāng)x > a>0時(shí), f_n ( x ) = xⁿ – ( x + a)ⁿ是關(guān)于x的減函數(shù),

∴ 當(dāng)n ?? a時(shí), 有：(n + 1 )ⁿ– ( n + 1 + a)ⁿ ?? n ⁿ – ( n + a)ⁿ.     2分

又 ∴f `<sub>n + 1</sub> (x ) = ( n + 1 ) [x<sup>n</sup>  –( x+ a )<sup>n</sup> ] ,∴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 )f_n`(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 `<sub>n + 1</sub> ( n + 1 ) < ( n + 1 )f<sub>n</sub>`(n) . 2分