Browse Questions

If $A=\{x:x\in R\:and\:x\geq 1\}$ and $f:A\rightarrow A$ is defined as $f(x)=2^{x(x-1)},$ then find $f^{-1}(x).$

(A) $\large\frac{1+\sqrt{1+4log_2x}}{2}$

(B) $\large\big(\frac{1}{2}\big)^{x(x-1)}$

(C) $\large\frac{1-\sqrt{1+4log_2x}}{2}$

(D) none of these

Toolbox:
• $loga^n=nloga$
• $\large\frac{loga}{logb}$ $=log_b a$
Let $y=f(x)=2^{x(x-1)}$
$\Rightarrow x=f^{-1}(y)$
Taking log on both the sides
$logy=x(x-1)log2$
$\Rightarrow\:x(x-1)=\large\frac{logy}{log2}$ $=log_2 y$
$\Rightarrow\:x^2-x-log_2y=0$
$x=\large\frac{1\pm\sqrt {1+4log_2y}}{2}$ $=f^{-1}(y)$
But for inverse to exist $f$ should be 1-1. and $(1-\sqrt{1+4log_2x})$ is not in the domain.
$\therefore \:f^{-1}(x)=\large\frac{1+\sqrt{1+4log_2x}}{2}$

+1 vote