A) \[{{\left( \frac{1}{2} \right)}^{x(x-1)}}\]
B) \[\frac{1}{2}(1-\sqrt{1+4{{\log }_{2}}x})\]
C) \[\frac{1}{2}\sqrt{1+4{{\log }_{2}}x}\]
D) \[\frac{1}{2}[1+\sqrt{1+4{{\log }_{2}}x}]\]
E) not defined
Correct Answer: D
Solution :
\[f:[1,\infty ]\to [1,\infty )\]and \[f(x)={{2}^{x(x-1)}}\] Let \[y={{2}^{x(x-1)}}\] \[\Rightarrow \] \[{{\log }_{2}}y=x(x-1)\] \[\Rightarrow \] \[{{x}^{2}}-x-{{\log }_{2}}y=0\] \[\Rightarrow \] \[x=\frac{1\pm \sqrt{1+4{{\log }_{2}}y}}{2}={{f}^{-1}}(y)\] \[[\because x={{f}^{-1}}(y)]\] \[\Rightarrow \] \[{{f}^{-1}}(x)=\frac{1\pm \sqrt{1+4{{\log }_{2}}x}}{2}\] Hence, \[{{f}^{-1}}(x)=\frac{1+\sqrt{1+4{{\log }_{2}}x}}{2}\] \[[\because x\ge 1]\]
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