题目

若f ( x ) = a x 2 + b x + c，( a，b，c∈R )在区间[ 0，1 ]上恒有| f ( x ) | ≤ 1。（1）对所有这样的f ( x )，求 | a | + | b | + | c | 的最大值；（2）试给出一个这样的f ( x )，使 | a | + | b | + | c | 确实取到上述最大值。 答案：解析：（1）依题设有| f ( 0 ) | = | c | ≤ 1，| f ( 1 ) | = | a + b + c | ≤ 1，| f () | = |++ c | ≤ 1，于是| a + b | = | a + b + c c | ≤ | a + b + c | + | c | ≤ 2，| a b | = | 3 ( a + b + c ) + 5 c 8 (++ c ) | ≤ 3 | a + b + c | + 5 | c | + 8 |++ c | ≤ 3+5+8 = 16，从而，当a b ≥ 0时，| a | + | b | = | a + b |，∴ | a | + | b | + | c | = | a + b | + | c | ≤ 2 + 1 = 3；当a b < 0时，| a | + | b | = | a b |，∴ | a | + | b | + | c | = | a b | + | c | ≤ 16 + 1 = 17。∴ max { | a | + | b | + | c | } = 17。（2）当a = 8，b = 8，c = 1时，f ( x ) = 8 x 2 8 x + 1 = 8 ( x ) 2 1，∴ 当x∈[ 0，1 ]时，有| 8 x 2 8 x + 1 | ≤ 1，此时| a | + | b | + | c | = 8 + 8 + 1 = 17。

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