Key Concepts and Formulas

• Monotonicity and Derivatives: A function $f(x)$ is strictly increasing where $f'(x) > 0$ and strictly decreasing where $f'(x) < 0$.
• Product Rule: The derivative of a product of two functions $u(x)$ and $v(x)$ is given by $(uv)' = u'v + uv'$.
• Chain Rule: The derivative of a composite function $f(g(x))$ is given by $f'(g(x)) \cdot g'(x)$.

Step 1: Analyzing Function $f(x)$ for Strict Increase

We are given $f(x) = 2 \log_{\mathrm{e}}(x-2) - x^2 + ax + 1$ and that it is strictly increasing on the largest open interval $(2,3)$. Our goal is to find the value of $a$.

1.1 Determine the Domain of $f(x)$:

The logarithm is defined only for positive arguments, so $x-2 > 0$, which means $x > 2$. The domain of $f(x)$ is $(2, \infty)$.

1.2 Calculate the First Derivative $f'(x)$:

To find where $f(x)$ is strictly increasing, we need to find its first derivative: $f'(x) = \frac{d}{dx} \left( 2 \log_{\mathrm{e}}(x-2) - x^2 + ax + 1 \right)$ Using the derivative rules: $f'(x) = 2 \cdot \frac{1}{x-2} - 2x + a$ $f'(x) = \frac{2}{x-2} - 2x + a$

1.3 Apply the Condition for Strict Increase:

Since $(2,3)$ is the largest open interval where $f(x)$ is strictly increasing, $f'(x) > 0$ for $x \in (2,3)$ and $f'(3) = 0$. This is because if $f'(3) > 0$, then $f'(x)$ would be positive on some interval $(2, 3 + \epsilon)$ for some $\epsilon > 0$, contradicting that $(2,3)$ is the largest open interval.

1.4 Solve for 'a':

Substitute $x=3$ into the expression for $f'(x)$ and set it equal to 0: $f'(3) = \frac{2}{3-2} - 2(3) + a = 0$ $2 - 6 + a = 0$ $a = 4$

Step 2: Analyzing Function $g(x)$ for Strict Decrease

We are given $g(x) = (x-1)^3(x+2-a)^2$. We found $a=4$, so $g(x) = (x-1)^3(x-2)^2$. We want to find the largest open interval $(b, c)$ where $g(x)$ is strictly decreasing.

2.1 Calculate the First Derivative $g'(x)$:

Using the product rule and chain rule: $g'(x) = \frac{d}{dx} \left[ (x-1)^3(x-2)^2 \right]$ $g'(x) = 3(x-1)^2(x-2)^2 + (x-1)^3 \cdot 2(x-2)$ $g'(x) = (x-1)^2(x-2) \left[ 3(x-2) + 2(x-1) \right]$ $g'(x) = (x-1)^2(x-2)(3x - 6 + 2x - 2)$ $g'(x) = (x-1)^2(x-2)(5x - 8)$

2.2 Find the Critical Points:

To find the intervals where $g(x)$ is decreasing, we need to find where $g'(x) < 0$. First, we find the critical points by setting $g'(x) = 0$: $(x-1)^2(x-2)(5x-8) = 0$ The critical points are $x = 1, x = 2, x = \frac{8}{5} = 1.6$.

2.3 Analyze the Sign of $g'(x)$:

We analyze the sign of $g'(x)$ in the intervals determined by the critical points. Since $(x-1)^2$ is always non-negative, it only affects where $g'(x) = 0$. We only need to consider the sign of $(x-2)(5x-8)$.

• $x < \frac{8}{5}$: $(x-2) < 0$ and $(5x-8) < 0$, so $g'(x) > 0$.
• $\frac{8}{5} < x < 2$: $(x-2) < 0$ and $(5x-8) > 0$, so $g'(x) < 0$.
• $2 < x$: $(x-2) > 0$ and $(5x-8) > 0$, so $g'(x) > 0$.

Since $g'(x) < 0$ on the interval $(\frac{8}{5}, 2)$, the largest open interval where $g(x)$ is strictly decreasing is $(\frac{8}{5}, 2)$. Thus, $b = \frac{8}{5}$ and $c = 2$.

Step 3: Calculate the Final Result

We have $a = 4$, $b = \frac{8}{5}$, and $c = 2$. We need to find $100(a + b - c)$: $100 \left( 4 + \frac{8}{5} - 2 \right) = 100 \left( 2 + \frac{8}{5} \right) = 100 \left( \frac{10 + 8}{5} \right) = 100 \left( \frac{18}{5} \right) = 20 \cdot 18 = 360$

Common Mistakes & Tips

• Remember to consider the domain of logarithmic functions.
• Don't forget the chain rule when differentiating composite functions.
• When analyzing the sign of the derivative, pay attention to squared terms, as they are always non-negative.

Summary

We found the value of $a$ by analyzing the first derivative of $f(x)$ and using the given interval where $f(x)$ is strictly increasing. Then, we used this value of $a$ to analyze the first derivative of $g(x)$ and determine the largest open interval where $g(x)$ is strictly decreasing. Finally, we calculated the value of $100(a+b-c)$.

Final Answer

The final answer is \boxed{360}, which corresponds to option (A).