If the function $f$ is given by $f(x)=x^3-3(a-2)x^2+3ax+7$,for some $a \in R$,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

  • A
    $-7$
  • B
    $6$
  • C
    $7$
  • D
    $5$

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