JEE Mathematics 2023 Question with Solution

Given: $f(x) = \frac{\lfloor x \rfloor}{1 + x^2}$ with domain $[2,6)$.

Find: The range of the function.](streamdown:incomplete-link)

From the solution, the function behaves piecewise as

On each interval, the numerator is constant and the denominator increases as $x$ increases, so each branch is decreasing.

Therefore, the largest value occurs at $x=2$:

This value is included because $2 \in [2,6)$.](streamdown:incomplete-link)

The smallest value is approached on the interval $[5,6)$ as $x \to 6^-$:

Since $x=6$ is not included, $\frac{5}{37}$ is not attained.](streamdown:incomplete-link)

Hence the range is

So, the correct option is A.

The listed raw options contain a mismatch with the source correct-answer field, but the solution explicitly concludes option A with range $\left(\frac{5}{37}, \frac{2}{5}\right]$.

Given: The greatest integer function $\lfloor x \rfloor$ changes value at integers.

Find: How the range is formed over the domain $[2,6)$.](streamdown:incomplete-link)

Break the domain into intervals where $\lfloor x \rfloor$ is constant:

• For $x \in [2,3)$, $\lfloor x \rfloor = 2$
• For $x \in [3,4)$, $\lfloor x \rfloor = 3$
• For $x \in [4,5)$, $\lfloor x \rfloor = 4$
• For $x \in [5,6)$, $\lfloor x \rfloor = 5$](streamdown:incomplete-link)

This gives four decreasing branches. Their endpoint values are:

Combining these intervals, the overall highest value is $\frac{2}{5}$ and the overall lowest approached value is $\frac{5}{37}$, not included.

Therefore, the range is $\left(\frac{5}{37}, \frac{2}{5}\right]$, which matches Option A.