题目

设函数f (x)＝x2－x＋b拜托各位了 3Q
设函数f (x)＝x2－x＋b,已知log 2f (a)＝2,且f (log 2a)＝b (a>0且a≠1),(1) 求a,b的值；(2) 试在f (log 2x)>f (1)且log 2f (x)

提问时间：2021-02-17

答案

(1) logf(a)=log(a^2-a+b)=2 ==> a^2-a+b=4 f[log(a)]=[log(a)]^2-log(a)+b=log(a)[log(a)-1]+b=b ∴a=1 或 2,∵a≠1,∴a=2,b=2 (2)f[log(x)]＞f(1) [log(x)]^2-log(x)＞0 解得log(x)＞1 或 log(x)＜0,即x＞2 或 0＜x＜1 logf(x)＜f(1) log(x^2-x+2)＜2 即0＜x^2-x+2＜4,解得-1＜x＜2 取交集得 x∈(0,1)