A) \[{{\left( \frac{1}{2} \right)}^{x(x-1)}}\]
B) \[\frac{1}{2}(1+\sqrt{1+4{{\log }_{2}}x})\]
C) \[\frac{1}{2}(1-\sqrt{1+4{{\log }_{2}}x})\]
D) Not defined
Correct Answer: B
Solution :
Given \[f(x)={{2}^{x(x-1)}}\,\Rightarrow \,\,x\,(x-1)={{\log }_{2}}f(x)\] \[\Rightarrow \,\,{{x}^{2}}-x-{{\log }_{2}}f(x)=0\,\,\Rightarrow \,\,x=\frac{1\pm \sqrt{1+4{{\log }_{2}}f(x)}}{2}\] Only \[x=\frac{1+\sqrt{1+4{{\log }_{2}}f(x)}}{2}\] lies in the domain \[\therefore \,\,{{f}^{-1}}(x)=\frac{1}{2}\,[1+\sqrt{1+4\,{{\log }_{2}}x}]\].
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