If the function `f:[1,oo)->[1,oo)` is defined by `f(x)=2^(x(x-1)),` then `f^-1(x)` is (A) `(1/2)^(x(x-1))` (B) `1/2 sqrt(1+4log_2x)` (C) `1/2(1-sqrt(1+4log_2x))` (D) not defined

Explanation

`f(x)=2^{s(x+1)} text {, i.e., } y=2^{x(x+1)} `


`Rightarrow log _{2} y=x(x-1)=x^{2}-x`


`Rightarrow x^{2}-x-log _{2} y=0 `


`Rightarrow x=frac{1 pm sqrt{1+4 log _{2} y}}{2}`



Since `x in[1, infty) therefore`-ve sign is ruled out.



`therefore Rightarrow x=frac{1}{2}(1+sqrt{1+4 log _{2} y})`


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