文档介绍：2010高考数学――压轴题
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
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 ,vÎ[–1,1],都有|f (u) – f (v) | ≤| u –v | .
(1) 判断函数p ( x ) = x2 – 1 是否满足题设条件?
(2) 判断函数g(x)=,是否满足题设条件?
解: (1) 若u ,v Î [–1,1], |p(u) – p (v)| = | u2 – v2 |=| (u + v )(u – v) |,
取u = Î[–1,1],v = Î[–1,1],
则|p (u) – p (v)| = | (u + v )(u – v) | = | 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 Î [–1,0],则|g(u) – g (v)| = |(1+u) –(1 + v)|=|u – v |,满足题设条件;
20. 若u ,v Î [0,1], 则|g(u) – g(v)| = |(1 – u) –(1 – v)|= |v –u|,满足题设条件;
30. 若uÎ[–1,0],vÎ[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;
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