文档介绍：2014高考数学压轴题二
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 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;
40 若uÎ[0,1],vÎ[–1,0], 同理可证满足题设条件.
综合上述得g(x)满足条件.
3. (本小题满分14分)
已知点P ( t , y )在函数f ( x ) = (x ¹ –1)的图象上,且有t2 – c2at + 4c2 = 0 ( c ¹ 0 ).
(1) 求证:| ac | ³ 4;
(2) 求证:在(–1,+∞)上f ( x )单调递增.
(3) (仅理科做)求证:f ( | a | ) + f ( | c | ) > 1.
证:(1) ∵ tÎR, t ¹ –1,
∴⊿= (–c2a)2 – 16c2 = c4a2 – 16c2 ³ 0 ,
∵ c ¹ 0, ∴c2a2 ³ 16 , ∴| ac | ³ 4.
(2) 由 f ( x ) = 1 –,
法1. 设–1 < x1 < x2, 则f (x2) – f ( x1) = 1––1 + = .
∵–1 < x1 < x2, ∴ x1 – x2 < 0, x1 + 1 > 0, x2 + 1 > 0 ,
∴f (x2) – f ( x1) < 0 , 即f (x2) < f ( x1) , ∴x ³ 0时,f ( x )单调递增.
法2. 由f ` ( x ) = > 0 得x ¹ –1,
∴x > –1时,f ( x )单调递增.
(3)(仅理科做)∵f ( x