If the function,`f:[1,oo]to [1,oo]` is defined by `f(x)=3^(x(x-1))`, then `f^(-1)(x)` is

Options

A. `((1)/(3))^(x^((x-1)))`
B. `(1)/(2){1-sqrt(1+4log_(3)x)}`
C. `(1)/(2){1+sqrt(1+4log_(3)x)}`
D. not defined

Answer

C

Explanation

It can be checked that f is a bijection and hence invertible.
Now, `fof^(-1)=x`
`Rightarrow f(f^(-1)(x))=x`
`Rightarrow 3^(f^(-1)(x)(f^(-1)(x)-1))=x`
`Rightarrow f^(-1)(x)(f^(-1)(x)-1)=log_(3)x`
`Rightarrow (f^(-1)(x))^(2)-f^(-1)(x)-log_(3)x=0`
`Rightarrow f^(-1)(x)=(1)/(2){1+sqrt(1+4 log _(3)x)}" "[therefore f^(-1)(x) ge 1]`