A) \[{{\left( \frac{1}{2} \right)}^{x(x-1)}}\]
B) \[\frac{1}{2}(1+\sqrt{1+4{{\log }_{2}}y})\]
C) \[\frac{1}{2}(1-\sqrt{1+4{{\log }_{2}}y})\]
D) \[\infty \]
Correct Answer: B
Solution :
Given that, \[f:[1,-\infty )\xrightarrow{{}}[1,\,\infty )\] and \[f(x)={{2}^{x(x-1)}}=y\] \[\Rightarrow \] \[{{\log }_{2}}y={{x}^{2}}-x\] \[\Rightarrow \] \[{{x}^{2}}-x-{{\log }_{2}}y=0\] \[\Rightarrow \] \[x=\frac{1\pm \sqrt{1+4\,{{\log }_{2}}y}}{2}\] \[\Rightarrow \] \[x=\frac{1+\sqrt{1+4\,{{\log }_{2}}y}}{2}\] (Neglect \[-ve\] sign due to the domain)
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