A) \[{{\left( \frac{1}{2} \right)}^{x(x-1)}}\]
B) \[\frac{1}{2}\left( 1+\sqrt{1+4{{\log }_{2}}x} \right)\]
C) \[\frac{1}{2}\left( 1-\sqrt{1+4{{\log }_{2}}x} \right)\]
D) Not defined
Correct Answer: B
Solution :
| Let \[y={{2}^{x\,(x-1)}}\] where \[y\ge 1\] as \[x\ge 1\] |
| Taking \[{{\log }_{2}}\] of both side, \[{{\log }_{2}}y={{\log }_{2}}\,{{2}^{x\,(x-1)}}\] |
| \[\Rightarrow \,\,\,{{\log }_{2}}y=x\,\left( x-1 \right)\] \[\left( \because \,\,{{\log }_{e}}{{a}^{x}}=x \right)\] |
| \[\Rightarrow \,\,{{x}^{2}}-x-{{\log }_{2}}y=0\Rightarrow x=\frac{1\pm \sqrt{1+4{{\log }_{2}}y}}{2}\] |
| For \[y\ge 1,\,{{\log }_{2}}y\ge 0\Rightarrow \,4{{\log }_{2}}y\ge 0\Rightarrow 1+4{{\log }_{2}}y\ge 1\] |
| \[\Rightarrow \,\,\sqrt{1+4{{\log }_{2}}y}\ge 1\Rightarrow -\sqrt{1+4{{\log }_{2}}y}\le -1\] |
| But \[x\ge 1\] |
| \[\therefore \,\,\,x=1-\sqrt{1+4{{\log }_{2}}y}\] is not possible |
| Therefore \[x=\frac{1}{2}\,(1+\sqrt{1+4{{\log }_{2}}y)}\] |
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