已知函数f(x)=x2-x+k,且log2 f(a)=2,就是log以2为底f(a)的对数为2..f(log2 a)=k

f(log2 a)=k
(log2 a)^2-log2 a+k=k
log2 a(log2 a-1)=0
log2 a=0,或,log2 a=1
a=1,或,a=2
因为,a≠1
所以,a=2
log2 f(a)=2
f(a)=4
a^2-a+k=4
4-2+k=4
k=2
1)
f(log2 x)=(log2 x)^2-log2 x+2=(log2 x-1/2)^2+7/4
所以,log2 x=1/2,即x=√2时,f(log2 x)有最小值7/4
2)
f(1)=1-1+k=k=2
f(log2 x)>f(1),
(log2 x)^2-log2 x+2>2
log2 x(log2 x-1)>0
log2 x>1,或,log2 x2,或,0