question_answer

Solution set of the inequality \[\frac{1}{{{2}^{x}}-1}>\frac{1}{1-{{2}^{x-1}}}\] is

A) \[(1,\infty )\]

B) \[(0,lo{{g}_{2}}(4/3))\]

C) \[(-1,\infty )\]

D) \[(0,lo{{g}_{2}}(4/3))\cup (1,\infty )\]

Correct Answer: D

Solution :

[d] Put \[{{2}^{x}}=t.\] Then \[t>0.\] Now, given inequality becomes \[\frac{1}{t-1}>\frac{2}{2-t}\Rightarrow \frac{1}{t-1}-\frac{2}{2-t}>0\Rightarrow \frac{2-t-2t+2}{(t-1)(2-t)}>0\] From sign scheme we get \[1<t<4/3\,\,or\,\,t>2.\] \[\Rightarrow 1<{{2}^{x}}<4/3\]or \[{{2}^{x}}>2\] \[\Rightarrow x\in (0,\,\,lo{{g}_{2}}(4/3))\cup (1,\infty )\]