If the function $f$ 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
    $5$
  • C
    $7$
  • D
    $6$

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