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f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7 increases on (0, 1] and decreases on [1, 5): a root of (f(x) - 14)/(x - 1)^2 = 0

The same question, asked 2 times: May 2022 S1, 11 May 2024 S2. Every wording below, oldest first.

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  1. Question 1

    May 2022 S1

    104
    If the function f(x)=x3−3(a−2)x2+3ax+7f(x) = x^3 - 3(a-2)x^2 + 3ax + 7, for some a∈Ra \in \mathbb{R} is increasing in (0,1](0,1] and decreasing in [1,5)[1,5), then a root of the equation f(x)−14(x−1)2=0\frac{f(x)-14}{(x-1)^2} = 0 (x≠1)(x \neq 1) is

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  2. Question 2

    11 May 2024 S2

    120
    If the function f is given by f(x)=x3−3(a−2)x2+3ax+7f(x)=x^3-3(a-2)x^2+3ax+7, for some a∈Ra\in\mathbb{R}, is increasing in (0,1](0,1] and decreasing in [1,5)[1,5), then a root of the equation f(x)−14(x−1)2=0  (x≠1)\frac{f(x)-14}{(x-1)^2}=0\;(x\neq1) is

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