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若f(x)=x^2-x+b,且f[log2(a)]=b log2[f(a)]=2 (a不等于1)
1.求f[log2(x)]的最小值及对应的x值
2.x取何值时,f[loga(x)]大于f(1)且log2[f(x)]小于f(1)
人气:453 ℃ 时间:2019-09-19 02:52:12
解答
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
当log2 x=0.5,x=√2时,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综上可得0
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