A) \[x\in (1,2)\]
B) \[x\in \left( \frac{1}{3},1 \right)\]
C) \[x\in \left( -\infty \frac{1}{3}, \right]\cup \left[ 1,\infty \right)\]
D) \[x\in \left[ \frac{1}{3},1 \right]\]
Correct Answer: C
Solution :
(c): \[\frac{x-1}{3x-1}>0\] \[\Rightarrow (x-1)>0\] and \[3x-1>0\Rightarrow x>1\] Or, \[(x-1)<0\] and \[(3x-1)<0\Rightarrow x<\frac{1}{3}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,x\in \left( -\infty ,\frac{1}{3} \right]\cup \left[ 1,\infty \right)\]
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