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 :
$$\eqalign{
& {\text{Let}}\,y = {2^{x\left( {x - 1} \right)}} \Rightarrow {x^2} - x - {\log _2}y = 0; \cr
& x = \frac{1}{2}\left( {1 \pm \sqrt {1 + 4{{\log }_2}y} } \right) \cr} $$
Since $$x$$ is + ive, we choose only + out of $$ \pm \,{\text{for}}\,\left( {y \geqslant 1,{{\log }_2},y \geqslant 0} \right)$$
$$\eqalign{
& \therefore x = \frac{1}{2}\left( {1 + \sqrt {1 + 4{{\log }_2}y} } \right) \cr
& {\text{or}}\,{f^{ - 1}}\left( x \right) = \frac{1}{2}\left( {1 + \sqrt {1 + 4{{\log }_2}x} } \right) \cr} $$