Animated Solution for Mathematics - Functions: Let $f(x)=\frac{x^2-6x+5}{x^2-5x+6}$. Match of expressions/statements in Column I with expressions/statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4\times 4$ matrix given in the ORS.

Welcome, future engineer! Today, we are not just solving a problem; we are mapping a landscape. When you look at a rational function like

most students see a scary fraction. I want you to see a story of growth, decay, and boundaries. This is a classic JEE Advanced challenge, and to conquer it, we need to move beyond rote memorization and into the realm of true mathematical intuition.

The Art of Factorization

Before we do anything, we must simplify. The expression

is like a locked chest. The key is factorization.

Look at the numerator: $x^2-6x+5$. It cries out to be written as $(x-1)(x-5)$.

Now look at the denominator: $x^2-5x+6$. That is clearly $(x-2)(x-3)$.

Suddenly, our function transforms into

This is our master key. We now see the 'DNA' of the function—its roots at $x=1$ and $x=5$, and its vertical asymptotes at $x=2$ and $x=3$.

The Comparison Tool

Now, the problem asks us to compare $f(x)$ with $1$. How do we do that? We could try to guess, but in JEE Advanced, guessing is the path to the dark side.

Instead, we build a tool. Let's look at the difference $f(x)-1$. By subtracting $1$ from our function, we get

When we find a common denominator, the $x^2$ terms vanish, leaving us with the elegant expression

This is the secret weapon. The sign of this expression tells us everything. If it is positive, $f(x)>1$. If it is negative, $f(x)<1$.

The Wavy Curve Odyssey

Now, we traverse the number line. Let's take the interval $-1<x<1$. In this region, the factors $(x-1)$, $(x-5)$, $(x-2)$, and $(x-3)$ are all negative.

A ratio of four negatives is positive, so $f(x)>0$. Now check our comparison tool:

With $x$ between $-1$ and $1$, the numerator is negative, and the denominator is positive. The negative sign in front makes the whole thing negative. Thus, $f(x)<1$. We have found our first treasure: $0<f(x)<1$.

As we move to $1<x<2$, the function crosses the root at $x=1$, changing its sign. Now, $f(x)<0$. If the function is negative, it is automatically less than $1$.

We continue this logic across all intervals, treating the vertical asymptotes at $x=2$ and $x=3$ with the respect they deserve—they are the points where the function's behavior flips violently. By the time we reach $x>5$, we find that the function has returned to positive territory, but remains below $1$.

Look at what we have achieved. We didn't just calculate; we visualized. We broke down the function, built a comparison tool, and navigated the number line like explorers.

This is the essence of JEE Advanced mathematics—not just finding the answer, but understanding the soul of the equation. Keep this mindset, and no problem will ever be too complex for you.