若f(x) = x^2 - x + b,且f(log_{2}a) = b,log_{2} f(a) = 2,(a≠1).
①求f(log_{2}x)的最小值及对应的x值；
②x取何值时,f(log_{2}x) > f(1)且 log_{2} f(x) < f(1).
由,
b = f(log_{2}a) = [log_{2}a]^2 - log_{2}a + b,
0 = [log_{2}a]^2 - log_{2}a = log_{2}a[log_{2} - 1],
0 = log_{2}a - 1,[因a≠1,所以 log_{2} a ≠ 0]
1 = log_{2}a,
a = 2.
f(a) = a^2 - a + b = 2 + b.
由,
log_{2}f(a) = 2,
log_{2}(2 + b) = 2,
2 + b = 2^2 = 4,
b = 2.
因此,
f(x) = x^2 - x + 2
= x^2 - x + 1/4 + 7/4
= (x - 1/2)^2 + 7/4.
所以,f(x) 在 x = 1/2处取得最小值 7/4.
由于,log_{2}x = 1/2,x = 2^(1/2).
因此,
① f(log2x)的最小值为 7/4,对应的x值为 2^(1/2).
f(1) = 1^2 - 1 + 2 = 2,
f(log_{2}x) = [log_{2}x]^2 - log_{2}x + 2 > f(1) = 2,x > 0.
[log_{2}x]^2 - log_{2}x > 0,
log_{2}x[log_{2}x - 1] > 0,
则,
log_{2}x > 0 且 log_{2}x > 1.
或者,log_{2}x < 0 且log_{2}x < 1.
也就是,
x > 1,且,x > 2,
或者,0 < x < 1,且,0 < x < 2.
所以,
x > 2,或者,0 < x < 1..（1）
再由,
log_{2} f(x) < f(1) = 2,
0 < f(x) < 2^2 = 4,
0 < x^2 - x + 2 < 4,
因,x^2 - x + 2 = (x - 1/2)^2 + 7/4 >= 7/4 > 0.
所以,只要考虑,
x^2 - x + 2 < 4,
x^2 - x - 2 < 0,
(x - 2)(x + 1) < 0.
-1 < x < 2.
但 由于（1）,x > 2,或者,0 < x < 1.
因此,
② 0 < x < 1时,f(log_{2}x) > f(1)且log_{2}f(x) < f(1).
