If the function f given by $\mathrm{f}(\mathrm{x})=\mathrm{x}^3-3(\mathrm{a}-2) \mathrm{x}^2+3 \mathrm{ax}+7$, for some $\mathrm{a} \in \mathrm{R}$ is increasing in $(0,1]$ and decreasing in $[1,5)$, then a root of the equation, $\frac{f(x)-14}{(x-1)^2}=0(x \neq 1)$ is :
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