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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
Can you answer this question?
 
 

1 Answer

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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.
$\therefore \:f^{-1}(x)=\large\frac{1+\sqrt{1+4log_2x}}{2}$
answered May 25, 2013 by rvidyagovindarajan_1
 

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