Animated Solution for Mathematics - Functions: If the domain of the function $f(x)=\frac{[x]}{1+x^2}$, where $[x]$ is greatest integer $\le x$, is $(2,6)$, then its range is

Welcome, JEE Warrior. Today, we are going to dismantle a problem that often trips up even the brightest students. We are looking at the function $f(x)=\frac{[x]}{1+x^2}$ over the domain $(2,6)$.

At first glance, it looks like a standard rational function, but that Greatest Integer Function, $[x]$, is a wolf in sheep's clothing. It introduces discontinuities, and if you treat it like a continuous function, you will fall into the trap. Let us embark on this journey of logical decomposition.

The Art of Partitioning

Imagine you are walking along the $x$-axis from $2$ to $6$. The Greatest Integer Function $[x]$ is like a staircase; it stays flat, then jumps, stays flat, then jumps. It is constant between integers.

Therefore, our strategy is simple: we must break the domain $(2,6)$ into the intervals where $[x]$ is constant. We have four distinct regions: $[2,3)$, $[3,4)$, $[4,5)$, and $[5,6)$.

In each of these regions, the numerator $[x]$ is no longer a variable; it is a constant integer. This is the moment the problem becomes solvable.

Analyzing the Staircase

Let us take the first interval, $x\in [2,3)$. Here, $[x]=2$. Our function simplifies beautifully to:

As $x$ increases from $2$ to $3$, $x^2$ increases, which means $1+x^2$ increases, and the entire fraction decreases. This is a strictly decreasing function.

To find the range, we evaluate the endpoints: As $x\to 2^{+}$, $f(x)\to \frac{2}{1+2^2}=\frac{2}{5}$. As $x\to 3^{-}$, $f(x)\to \frac{2}{1+3^2}=\frac{2}{10}=\frac{1}{5}$. So, for this first interval, the range is $(\frac{1}{5},\frac{2}{5})$.

We repeat this logic for the next interval, $x\in [3,4)$. Here, $[x]=3$, so $f(x)=\frac{3}{1+x^2}$. At $x=3$, $f(x)=\frac{3}{1+3^2}=\frac{3}{10}$. As $x\to 4^{-}$, $f(x)\to \frac{3}{1+4^2}=\frac{3}{17}$. Thus, the range for this segment is $(\frac{3}{17},\frac{3}{10}]$.

We continue this for $x\in [4,5)$, where $[x]=4$. The function is $f(x)=\frac{4}{1+x^2}$. At $x=4$, $f(x)=\frac{4}{1+4^2}=\frac{4}{17}$. As $x\to 5^{-}$, $f(x)\to \frac{4}{1+5^2}=\frac{4}{26}=\frac{2}{13}$. The range is $(\frac{2}{13},\frac{4}{17}]$.

Finally, for $x\in [5,6)$, $[x]=5$, so $f(x)=\frac{5}{1+x^2}$. At $x=5$, $f(x)=\frac{5}{1+5^2}=\frac{5}{26}$. As $x\to 6^{-}$, $f(x)\to \frac{5}{1+6^2}=\frac{5}{37}$. The range is $(\frac{5}{37},\frac{5}{26}]$.

The Synthesis

Now, we have four ranges: $R_1=(0.2,0.4)$, $R_2\approx (0.176,0.3]$, $R_3\approx (0.153,0.235]$, and $R_4\approx (0.135,0.192]$.

Look at these numbers. $R_1$ ends at $0.2$, but $R_2$ starts at $0.176$. Since $0.176<0.2$, the intervals overlap. This pattern continues across all segments.

Because every subsequent interval starts at a value lower than the previous one ended, there are no gaps. The entire collection of ranges merges into one continuous interval.

The absolute minimum value approached is the lower bound of the final interval, $\frac{5}{37}$, and the absolute maximum value approached is the upper bound of the first interval, $\frac{2}{5}$.

Since the domain $(2,6)$ is open, the final range is $(\frac{5}{37},\frac{2}{5})$.