1.f(x)=x^2-x+b,且f[log2(a)]=b
所以log2 a)^2-log2 a +b=b
log2a(log2a-1)=0,a≠1
所以a=2
log2 [f(a)]=2,即log2 f（2）=2
log2 (2+b)=2,所以b=2
f(log2 x)=(log2 x)^2-log2 x+2
=(log2 x-0.5)^2+7/4
當(dāng)log2 x=0.5,x=√2時(shí),f（log2 x）取最小值7/4
2.f(log2 x)>f(1)
即(log2 x)^2-log2 x+2>2
log2 x>2 或log2 x<0
x>2或0log2 [f(x)]即log2 (x^2-x+2)<2
0-1綜上可得0