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If the function $f:[1, \infty) \rightarrow[1, \infty)$ is defined by $f(x)=2^{x(x-1)}$, then $f^{-1}(x)$ is
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$\frac{1}{2}\left(1+\sqrt{1+4 \log _2 x}\right)$
$\begin{aligned} & \text { Given } f(x)=2^{x(x-1)} \Rightarrow x(x-1)=\log _2 f(x) \\ & \Rightarrow x^2-x-\log _2 f(x)=0 \Rightarrow x=\frac{1 \pm \sqrt{1+4 \log _2 f(x)}}{2} \\ & \text { Only } x=\frac{1+\sqrt{1+4 \log _2 f(x)}}{2} \text { lies in the domain } \\ & \therefore f^{-1}(x)=\frac{1}{2}\left[1+\sqrt{1+4 \log _2 x}\right]\end{aligned}$
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