If the function `f:[1,oo)->[1,oo)` is defined by `f(x)=2^(x(x-1)),` then `f^-1(x)` is

Options

A. `((1)/(2))^(x(x-1))`
B. `(1)/(2)(1+sqrt(1+4log_(2)x))`
C. `(1)/(2)(1-sqrt(1+4log_(2)x))`
D. not defined

Answer

B

Explanation

`y=2^(x(x-1)) or x^(2)-x-log_(2) y=0`
`or x=(1)/(2)(1+-sqrt(1+4 log_(2)y))`
Since `x in [1, oo), " we choose " x=(1)/(2)(1+sqrt(1+4 log_(2)y))`
` or f^(-1) (x) =(1)/(2)(1+sqrt(1+4 log_(2)x)).`