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一九八三年省、市自治区联合数学竞赛第一试有题;已知函数f(x)=ax~2-c满足-4≤f(1)≤-1,-1≤f(2)≤5,那么,f(3)应满足;(A)7≤f(3)≤26;(B)-4≤f(3)≤15;(C)-1≤f(3)≤20;(D)28/3≤f(3)≤35/3答案是(C) (Ⅰ)联想到这样一道题:设f(x)=x~2+ax+b,证明|f(1)|f(2)|,|f(3)|中至少有一个不小于1/2。 (Ⅱ) 我们发现(Ⅰ)、(Ⅱ)是同一类型题,解这类题的关键是如何从f(1)、f(2)、f(3)中消去参系数, 题(Ⅰ)中,由条件可得-4≤c-c≤-1 ①-1≤4a-c<5 ②问题是要求出9a-c。即由
The first trial of the joint mathematics competition of the provinces, municipalities and autonomous regions in 1983 has a problem; the known function f(x)=ax~2-c satisfies -4≤f(1)≤-1, -1≤f(2) ≤ 5, then, f (3) should be satisfied; (A) 7 ≤ f (3) ≤ 26; (B) - 4 ≤ f (3) ≤ 15; (C) - 1 ≤ f (3) ≤ 20; (D) 28/3 ≤ f (3) ≤ 35/3 The answer is (C) (I) Think of such a question: Let f(x) = x~2+ax+b, prove |f(1)| At least one of f(2)|,|f(3)| is not less than 1/2. (II) We have found that (I) and (II) are the same type of questions. The key to solving such questions is how to eliminate the coefficients from f(1), f(2), and f(3). (I) , from the conditions available -4 ≤ cc ≤ -1 ≤ ≤ ≤ 4a-c <5 2 The problem is 9a-c. By