Let (2, 3) be the largest open interval in wh | vedclass

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(B) For $f(x)$ to be strictly increasing on $(2, 3)$,we require $f'(x) \geq 0$ for $x \in (2, 3)$.$f'(x) = \frac{2}{x-2} - 2x + a \geq 0$.Since $f''(x

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$360$

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(B) For $f(x)$ to be strictly increasing on $(2, 3)$,we require $f'(x) \geq 0$ for $x \in (2, 3)$.$f'(x) = \frac{2}{x-2} - 2x + a \geq 0$.Since $f''(x) = -\frac{2}{(x-2)^2} - 2 Thus,$f'(x) \geq 0$ on $(2, 3)$ implies $f'(3) \geq 0$.$f'(3) = \frac{2}{3-2} - 2(3) + a = 2 - 6 + a = a - 4 \geq 0$,so $a \geq 4$. The smallest value is $a = 4$.Now,$g(x) = (x-1)^3(x+2-4)^2 = (x-1)^3(x-2)^2$.$g'(x) = 3(x-1)^2(x-2)^2 + (x-1)^3 \cdot 2(x-2) = (x-1)^2(x-2)[3(x-2) + 2(x-1)]$.$g'(x) = (x-1)^2(x-2)(3x - 6 + 2x - 2) = (x-1)^2(x-2)(5x - 8)$.For $g(x)$ to be strictly decreasing,$g'(x) Since $(x-1)^2 > 0$ for $x 
eq 1$,we need $(x-2)(5x-8) The roots are $x = 8/5$ and $x = 2$. The interval is $(8/5, 2)$.Thus,$b = 8/5$ and $c = 2$.$100(a + b - c) = 100(4 + 8/5 - 2) = 100(2 + 1.6) = 100(3.6) = 360$.

Let $(2, 3)$ be the largest open interval in which the function $f(x) = 2 \log_e(x-2) - x^2 + ax + 1$ 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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