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备战2010高考数学:压轴题追踪二

备战2010高考数学――压轴题追踪系列二

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 ≥ a , 证明f `n + 1 ( n + 1 ) < ( n + 1 )f n `(n) 解: (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,+∞)单调递减. 4分 (2)由上知:当x > a>0时, f n ( x ) = x n – ( x + a)n 是关于x 的减函数,

∴ 当n ≥ a 时, 有:(n + 1 )n

– ( n + 1 + a)n

≤ n n

– ( n + a)n

. 2分

又 ∴f `n + 1 (x ) = ( n + 1 ) [x

n –( x+ a )n

] ,

∴f `n + 1 ( n + 1 ) = ( n + 1 ) [(n + 1 )n

–( n + 1 + a )n ] < ( n + 1 )[ n n – ( n + a)n ] = ( n + 1 )[ n n

– ( 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 `n + 1 ( n + 1 ) < ( n + 1 )f n `(n) . 2分

2. (本小题满分12分)

已知:y = f (x) 定义域为[–1,1],且满足:f (–1) = f (1) = 0 ,对任意u ,v ∈[–1,1],都有|f (u) – f (v) | ≤ | u –v | .

(1) 判断函数p ( x ) = x 2 – 1 是否满足题设条件? (2) 判断函数g(x)=1,[1,0]1,[0,1]

x x x x +∈-??

-∈?,是否满足题设条件?

解: (1) 若u ,v ∈ [–1,1], |p(u) – p (v)| = | u 2

– v 2

|=| (u + v )(u – v) |,

取u =

4

3∈[–1,1],v =

2

1∈[–1,1],

则 |p (u) – p (v)| = | (u + v )(u – v) | = 4

5| u – v | > | u – v |,

所以p( x)不满足题设条件. (2)分三种情况讨论:

10

. 若u ,v ∈ [–1,0],则|g(u) – g (v)| = |(1+u) – (1 + v)|=|u – v |,满足题设条件; 20

. 若u ,v ∈ [0,1], 则|g(u) – g(v)| = |(1 – u) – (1 – v)|= |v –u|,满足题设条件; 30. 若u ∈[–1,0],v ∈[0,1],则:

|g (u) –g(v)|=|(1 – u) – (1 + v)| = | –u – v| = |v + u | ≤| v – u| = | u –v|,满足题设条件; 40 若u ∈[0,1],v ∈[–1,0], 同理可证满足题设条件.

综合上述得g(x)满足条件.