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(C) Given $f(t) = \frac{|t+1|}{t^2}$ for $t < 0$.For $t \in (-1, 0)$, $f(t) = \frac{t+1}{t^2} = \frac{1}{t} + \frac{1}{t^2}$. Then $f'(t) = -\frac{1}{

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$4$

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(C) Given $f(t) = \frac{|t+1|}{t^2}$ for $t For $t \in (-1, 0)$, $f(t) = \frac{t+1}{t^2} = \frac{1}{t} + \frac{1}{t^2}$. Then $f'(t) = -\frac{1}{t^2} - \frac{2}{t^3} = -\frac{t+2}{t^3}$. Since $t  0$ for $t \in (-1, 0)$.For $t \in (-\infty, -1)$, $f(t) = \frac{-(t+1)}{t^2} = -\frac{1}{t} - \frac{1}{t^2}$. Then $f'(t) = \frac{1}{t^2} + \frac{2}{t^3} = \frac{t+2}{t^3}$.For $f(t)$ to be strictly decreasing, $f'(t)  0$, i.e.,$t > -2$. Thus, $f(t)$ is strictly decreasing on $(-2, -1)$.Comparing $(-2, -1)$ with $(2\alpha, \alpha)$, we get $\alpha = -1$.Now, $g(x) = 2\log_e(x-2) - x^2 + 4x + 1$ for $x > 2$.$g'(x) = \frac{2}{x-2} - 2x + 4 = \frac{2 - 2x(x-2) + 4(x-2)}{x-2} = \frac{2 - 2x^2 + 4x + 4x - 8}{x-2} = \frac{-2x^2 + 8x - 6}{x-2} = \frac{-2(x^2 - 4x + 3)}{x-2} = \frac{-2(x-3)(x-1)}{x-2}$.For $x > 2$, $g'(x) = 0$ at $x = 3$.For $x \in (2, 3)$, $g'(x) > 0$ and for $x > 3$, $g'(x) The local maximum value is $g(3) = 2\log_e(3-2) - 3^2 + 4(3) + 1 = 2\log_e(1) - 9 + 12 + 1 = 0 - 9 + 13 = 4$.

Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2\log_e(x-2) + \alpha x^2 + 4x - \alpha, x > 2$, is

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