题目

解关于x的不等式2(log1/2 X）平方+7log1/2X+3≤0,并求函数f（x）=log2 x/2 *log2 x/4最小值及相应x值

提问时间：2020-09-18

答案

2(log1/2x)^2+7log1/2x+3≤0,
令t=log1/2(x),有
2t^2+7t+3≤0,
(2t+1)(t+3)≤0.
-3≤t≤-1/2.
当t=-3时,log1/2(x)=-3,x=(1/2)^-3=8.
当t=-1/2时,log1/2(x)=-1/2,x=(1/2)^(-1/2)=√2.
∵log1/2(x),是减函数,有
√2≤X≤8.
又∵log2(x)为增函数,
∴1/2≤log2(x)≤3.
f(x)=[log2(x/4)]*[log2(x/2)]
=[log2(x)-log2(4)]*[log2(x)-log2(2)]
=[log2(x)-2]*[log2(x)-1]
=[log2(x)]^2-3[log2(x)]+2
=[log2(x)-3/2]^2-1/4
由于log2(x)属于[1/2,3]
则：当log2(x)=3/2时,f(x)取最小值为-1/4.此时x=2√2.
当log2(x)=3时,f(x)有最大值2.此时x=8.