備戰(zhàn)2010高考數(shù)學(xué)：壓軸題跟蹤演練（二）

備戰(zhàn)2010高考數(shù)學(xué)――壓軸題跟蹤演練系列二
    1. (本小題滿(mǎn)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 `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，+∞）單調(diào)遞減. 4分
    （2）由上知：當(dāng)x > a>0時(shí), fn ( 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 `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) ......
    點(diǎn)擊下載試題