Animated Solution for Mathematics - Differentiation: 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:

The Calculus of Prediction: Unlocking the Mystery of Intervals. Welcome, future IITian. Today, we are not just solving a problem; we are learning to read the 'DNA' of functions. Calculus is the language of change, and in this problem, we are tasked with deciphering the behavior of two distinct functions, $f(x)$ and $g(x)$.

We begin with $f(x)=2\ln (x-2)-x^2+ax+1$. The first thing any seasoned mathematician does is check the domain. The natural logarithm $\ln (x-2)$ demands that $x-2>0$, or $x>2$. This is our playground.

The problem states that $f(x)$ is strictly increasing on the interval $(2,3)$. To understand 'increasing' behavior, we must look at the rate of change—the derivative. Differentiating $f(x)$, we get:

Now, here is the crux of the matter: if $(2,3)$ is the largest interval of increase, the function must stop increasing exactly at $x=3$. This implies that at the boundary $x=3$, the derivative must be zero.

Setting $f^{\prime }(3)=0$, we get:

This simplifies beautifully to $2-6+a=0$. Thus, we find our first key: $a=4$. We have successfully unlocked the first gate.

With $a=4$ in our pocket, we turn to $g(x)=(x-1{)}^3(x+2-a{)}^2$. Substituting our value, it becomes $g(x)=(x-1{)}^3(x-2{)}^2$. We need to find where this function is strictly decreasing, which means we need $g^{\prime }(x)<0$.

Many students would immediately expand this polynomial, but that is a trap! It leads to a long, tedious expression. Instead, let us use the Product Rule: $\frac{d}{dx}[u\cdot v]=u^{\prime }v+u{v}^{\prime }$.

Applying this, we get:

Now, look at the elegance of the algebra. We can factor out $(x-1{)}^2(x-2)$. This leaves us with:

Simplifying the bracket, we get $3x-6+2x-2=5x-8$. Our derivative is now perfectly factored:

Now, we find the critical points by setting $g^{\prime }(x)=0$. The roots are $x=1$, $x=2$, and $x=1.6$ (or $\frac{8}{5}$). We place these on the number line.

This is where precision matters. We start from the rightmost interval ($x>2$), where the derivative is positive. As we cross $x=2$, the sign changes to negative because the power of $(x-2)$ is odd.

Crossing $x=1.6$, the sign changes back to positive. But watch out at $x=1$! The factor $(x-1{)}^2$ has an even power. This means the sign does not change; it remains positive.

We are looking for the interval where $g(x)$ is strictly decreasing, meaning $g^{\prime }(x)<0$. Looking at our sign scheme, the only region where the curve dips below the axis is $(1.6,2)$. Therefore, $b=1.6$ and $c=2$.

We have all our pieces: $a=4$, $b=1.6$, and $c=2$. The problem asks for $100(a+b-c)$.

Substituting our values:

There it is. The complexity melts away when you approach it with logic and patience. You have not just solved a problem; you have mastered the behavior of functions. Keep this clarity, and you will conquer any challenge the JEE throws your way. The final answer is 360.