Question
If the function $$f:\left[ {1,\, + \infty } \right) \to \left[ {1,\, + \infty } \right)$$ is defined by $$f\left( x \right) = {2^{x\left( {x - 1} \right)}}$$ then $${f^{ - 1}}\left( x \right)$$ is :
A.
$${\left( {\frac{1}{2}} \right)^{x\left( {x - 1} \right)}}$$
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
Answer :
$$\frac{1}{2}\left( {1 + \sqrt {1 + 4\,{{\log }_2}x} } \right)$$
Solution :
$$\eqalign{
& y = {2^{x\left( {x - 1} \right)}} \cr
& \Rightarrow x\left( {x - 1} \right) = {\log _2}y \cr
& \Rightarrow {x^2} - x - {\log _2}y = 0 \cr
& \therefore \,\,x = \frac{{1 \pm \sqrt {1 + 4\,{{\log }_2}y} }}{2}\, \cr
& {\text{But}}\,\,x \geqslant 1.\,\,\,\,\,\,\,\,\therefore \,\,x = \frac{{1 + \sqrt {1 + 4\,{{\log }_2}y} }}{2}\, \cr
& \therefore \,\,{f^{ - 1}}\left( x \right) = \frac{1}{2}\left\{ {1 + \sqrt {1 + 4\,{{\log }_2}y} } \right\} \cr} $$