Let $(2, 3)$ be the largest open interval in which the function $f (x) = 2 \log_{e} (x - 2) - x^{2} + a x + 1$ is strictly increasing and (b, c) be the largest open interval, in which the function $g (x) = {(x - 1)}^{3} {(x + 2 - a)}^{2}$ is strictly decreasing. Then $100 (a + b - c)$ is equal to:

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160

280

360

420

answer is B.

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Detailed Solution

$\begin{array}{l} f^{'} (x) = \frac{2}{x - 2} - 2 x + a \geq 0 \\ f^{''} (x) = \frac{- 2}{(x - 2)^{2}} - 2 < 0 \end{array}$
$\begin{array}{l} f^{'} (x) ↓ \\ f^{'} (3) \geq 0 \\ 2 - 6 + a \geq 0 \\ a \geq 4 \\ a_{min} = 4 \\ g (x) = (x - 1)^{3} (x + 2 - a)^{2} \\ g (x) = (x - 1)^{3} (x - 2)^{2} \\ g^{'} (x) = (x - 1)^{3} 2 (x - 2) + (x - 2)^{2} 3 (x - 1)^{2} \\ = (x - 1)^{2} (x - 2) (2 x - 2 + 3 x - 6) \\ = (x - 1)^{2} (x - 2) (5 x - 8) < 0 \\ x \in (\frac{8}{5}, 2) \end{array}$
$100 (a + b - c) = 100 (4 + \frac{8}{5} - 2) = 360$