题目内容

2.已知函数$f(x)=\left\{\begin{array}{l}{-2,0<x<1}\\{1,x≥1}\end{array}\right.$则不等式$lo{g}_{2}x-(lo{g}_{\frac{1}{4}}4x-1)f(lo{g}_{3}x+1)≤5$的解集为(  )
A.($\frac{1}{3}$,1)B.[1,4]C.($\frac{1}{3}$,4]D.[1,+∞)

分析 不等式$lo{g}_{2}x-(lo{g}_{\frac{1}{4}}4x-1)f(lo{g}_{3}x+1)≤5$?$\left\{\begin{array}{l}{lo{g}_{3}x+1≥1}\\{lo{g}_{2}x-(lo{g}_{\frac{1}{4}}4x-1)≤5}\end{array}\right.$,或$\left\{\begin{array}{l}{0<lo{g}_{3}x+1<1}\\{lo{g}_{2}x+2(lo{g}_{\frac{1}{4}}4x-1)≤5}\end{array}\right.$,解出即可得出.

解答 解:不等式$lo{g}_{2}x-(lo{g}_{\frac{1}{4}}4x-1)f(lo{g}_{3}x+1)≤5$?$\left\{\begin{array}{l}{lo{g}_{3}x+1≥1}\\{lo{g}_{2}x-(lo{g}_{\frac{1}{4}}4x-1)≤5}\end{array}\right.$,或$\left\{\begin{array}{l}{0<lo{g}_{3}x+1<1}\\{lo{g}_{2}x+2(lo{g}_{\frac{1}{4}}4x-1)≤5}\end{array}\right.$,
解得1≤x≤4,或$\frac{1}{3}<x<1$,
∴原不等式的解集为$(\frac{1}{3},4]$.
故选:C.

点评 本题考查了分段函数的性质、对数函数的单调性、不等式的解法,考查了推理能力与计算能力,属于中档题.

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