Question

设f(x)=x^2-x+k若log2f(a)=2,f(log2a)=K(a>0且a不等于1),求使f(log2x)>f(

Answer 1

log2f(a)=2,

f(a)=2^2=4=a^2-a+k

f(log2a)=k

(log2a)^2-log2a+k=k

(log2a)^2=log2a

所以log2a=0或log2a=1

即a=1或a=2

因为a大于0且a不=1

所以a=2

代入4=a^2-a+k

k=2

所以

f(x)=x2-x+2

f(log2x)=(log2x)^2-log2x+2>f(1)=2

(log2x)^2-log2x>0

log2x>1或log2x2或0