论文部分内容阅读
1.对于函数f(x)=x2+bx+c(b、c∈R),不论α、β为任何实数恒有f(sinx)≥O,f(2+cosβ)≤0.(1)求证:b+c=-1;(2)求证:c≥3;(3)若f(sinx)的最大值为8,求b、c的值.(2000年山东省济南卷) 简答:(1)只有f(1)=1+b+c=0;(2)根据(1),可得f(x)=(x-1)(x-c),-1≤x≤c,(3)c=3,b=-4. 略解:(1)∵sinx∈[-1,1],(2+cosβ)∈[1,3],只有f(1)=1+b+c=0成立;(2)∵1≤x≤3,∴f(x)=x2-(1+c)x+1=(x-1)(x-c)≤0,∴1≤x≤c,则3≤c;(3)当sinx=-1代入f(-1)=1-b+c=2+2c=8,∴c=3,b=-4. 评:难易适中的函数题,数形结合处理,题型较新.
1. For the function f(x)=x2+bx+c(b,c∈R), whether α,β is any real constant f(sinx)≥O,f(2+cosβ)≤0.(1) Proof: b+c=-1; (2) Proof: c≥3; (3) If the maximum value of f(sinx) is 8, find the values of b and c. (2000 Jinan, Shandong Province) (1) Only f(1)=1+b+c=0; (2) According to (1), f(x)=(x-1)(xc), -1≤x≤c, (3) )c=3, b=-4. Slightly solved: (1) ∵sinx∈[-1,1],(2+cosβ)∈[1,3], only f(1)=1+b+c=0 Established; (2) ∵ 1 ≤ x ≤ 3, ∴ f (x) = x2 - (1 + c) x + 1 = (x - 1) (xc) ≤ 0, ∴ 1 ≤ x ≤ c, then 3 ≤ c; (3) When sinx = -1 is substituted into f(-1) = 1 - b + c = 2 + 2c = 8, ∴c = 3, b = -4. Comments: moderate difficulty function, number shape Combined treatment, the question type is relatively new.