Animated Solution for Mathematics - Functions: Let $f:(1,3)\to R$ be a function defined by $f(x)=\frac{x[x]}{x^2+1}$, where $[x]$ denotes the greatest integer $\le x$. Then the range of $f$ is:

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dissect a function that, at first glance, might seem like a chaotic mess of brackets and variables. We are looking at $f(x)=\frac{x[x]}{x^2+1}$ on the domain $x\in (1,3)$.

Many students see the Greatest Integer Function, denoted by $[x]$, and immediately feel a sense of dread. But I want you to shift your perspective. Do not see it as a barrier; see it as a gatekeeper. It is simply telling us that the function behaves differently depending on which 'room' of the number line we are in.

Imagine you are walking along the $x$-axis, starting from $1$ and heading toward $3$. As you walk, you are constantly checking the value of $[x]$. For any number between $1$ and $2$ (like $1.5$), the greatest integer is $1$.

The moment you step onto $2$, the gatekeeper changes the rule. Now, for any number from $2$ up to (but not including) $3$, the greatest integer is $2$. This is the core of our problem; we cannot treat this as one single, monolithic function because it is not.

It is a piecewise function in disguise. To find the range, we must split our journey into two distinct paths:

1. Path A: $x\in (1,2)$, where $[x]=1$. Our function simplifies to:

2. Path B: $x\in [2,3)$, where $[x]=2$. Our function simplifies to:

By splitting the domain, we have transformed a scary, discontinuous expression into two clean, manageable rational functions. This is the power of mathematical decomposition.

Now that we have our two paths, we need to know how they behave. Are they climbing up a hill, or are they sliding down into a valley? To answer this, we turn to the most powerful tool in our arsenal: the derivative.

Let us consider the general form $g(x)=\frac{kx}{x^2+1}$, where $k$ is a constant. Using the quotient rule, which states that the derivative of $\frac{u}{v}$ is $\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$, we get:

Look closely at this result. The denominator $(x^2+1{)}^2$ is always positive, and the term $k$ is just a positive constant ($1$ or $2$). The fate of our derivative rests entirely on the term $(1-x^2)$.

Since our entire domain is $x>1$, it follows that $x^2>1$, which means $1-x^2$ is always negative. A negative derivative means our function is strictly decreasing. It is always sliding downwards.

For a strictly decreasing function, we do not need to worry about local maxima or minima in the middle of the interval. The maximum value will always be at the start of the interval, and the minimum value will always be at the end.

We are now ready to find the range. Let us tackle our two paths one by one.

For Path A ($x\in (1,2)$): Since the function $f(x)=\frac{x}{x^2+1}$ is decreasing, the range is determined by the limits at the endpoints: - As $x\to 1^{+}$, $f(x)\to \frac{1}{1^2+1}=\frac{1}{2}$. - As $x\to 2^{-}$, $f(x)\to \frac{2}{2^2+1}=\frac{2}{5}$.

Because the endpoints $1$ and $2$ are excluded, the range for this piece is the open interval $(\frac{2}{5},\frac{1}{2})$.

For Path B ($x\in [2,3)$): Here, our function is $f(x)=\frac{2x}{x^2+1}$. Again, it is decreasing: - At $x=2$, $f(2)=\frac{2(2)}{2^2+1}=\frac{4}{5}$. This is a solid, included point. - As $x\to 3^{-}$, $f(x)\to \frac{2(3)}{3^2+1}=\frac{6}{10}=\frac{3}{5}$. This is an excluded limit.

So, the range for this piece is $(\frac{3}{5},\frac{4}{5}]$.

The Union

To find the total range of the original function, we simply unite the ranges of our two paths. Putting them together, we get the final range:

Look at what we have achieved. We took a function that seemed intimidating, broke it down into logical pieces, applied the rigors of calculus to understand its behavior, and reconstructed the solution with precision. This is the essence of JEE Advanced mathematics. You have mastered the gatekeeper, and you have found the path.