A) \[x\in (2,\infty )\]
B) \[x\in (-\infty ,2)\]
C) \[x\in \left( -\infty ,\left. \frac{1}{2} \right]\cup [1,\infty ) \right.\]
D) \[x\in \left[ \frac{1}{2},1 \right)\]
Correct Answer: C
Solution :
(c): Upon simplifying, we get,\[\frac{1}{(x-1)(2x-1)}\ge 0\]; which actually becomes \[(x-1)(2x-1)\ge 0\] So, this is again greater than inequality \[\Rightarrow (x-1)\left( x-\frac{1}{2} \right)\ge 0\Rightarrow x\in \left( -\infty ,\frac{1}{2} \right)\cup \left[ 1,\infty \right)\]
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