JEE MainMathematics

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Range of a Piecewise Rational Function with Greatest Integer Function

If the domain of the function f(x) = ([x])/(1+x^2), where [x] is greatest integer ≤ x, is [2,6), then its range is

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A(5/26, 2/5]B(5/37, 2/5] - {9/29, 27/109, 18/89, 9/53}C(5/37, 2/5]D(5/26, 2/5] - {9/29, 27/109, 18/89, 9/53}

Step-by-step solutionView

Correct answer

By analyzing the function piecewise on each interval $[k, k + 1)$ for $k \in {2, 3, 4, 5}$ , the union of the resulting continuous image intervals is $(\frac{5}{37}, \frac{2}{5}]$ .

Option analysis

Why each option works or fails

A · (5/26, 2/5]

Assumed the minimum value is reached at $x \to 5$ from the left instead of evaluating the final piecewise interval $[5, 6)$ where $x \to 6$ . Check the entire domain $[2, 6)$ : the last piece is $x \in [5, 6)$ with $[x] = 5$ , giving values down to $\lim_{x \to 6^{-}} \frac{5}{1 + x^{2}} = \frac{5}{37}$ .

B · (5/37, 2/5] - {9/29, 27/109, 18/89, 9/53}

Believed the function has holes or excluded points within the continuous intervals of each piece due to the step nature of $[x]$ . Within each interval $[k, k + 1)$ , $f (x) = \frac{k}{1 + x^{2}}$ is continuous and strictly decreasing, covering every real value in $(\frac{k}{1 + (k + 1)^{2}}, \frac{k}{1 + k^{2}}]$ without any isolated missing points.

C · (5/37, 2/5]

This is the correct option. Each sub-interval $[k, k + 1)$ maps to $(\frac{k}{1 + (k + 1)^{2}}, \frac{k}{1 + k^{2}}]$ . Since these intervals overlap (e.g., $\frac{2}{10} < \frac{3}{10}$ ), their union is the contiguous interval $(\frac{5}{37}, \frac{2}{5}]$ .

D · (5/26, 2/5] - {9/29, 27/109, 18/89, 9/53}

Combined missing the last interval $[5, 6)$ with incorrectly assuming points must be punctured from the range. Evaluate each piecewise segment over the entire domain up to $6$ , and note that rational functions on intervals map to connected intervals with no isolated points removed.

Reviewed route

Solution

StepWorking

01Given

$f (x) = \frac{[x]}{1 + x^{2}}$ with domain $x \in [2, 6)$ , where $[x]$ denotes the greatest integer function.

02Goal

Find the complete range of $f (x)$ on $[2, 6)$ .

03Approach

Break the domain $[2, 6)$ into unit intervals $[n, n + 1)$ where $[x] = n$ is constant. On each sub-interval, $f (x) = \frac{n}{1 + x^{2}}$ is strictly decreasing. Find the range on each piece and take the union.

04Execute

Partition $[2, 6)$ into four intervals: 1. For $x \in [2, 3)$ , $[x] = 2$ , so $f (x) = \frac{2}{1 + x^{2}}$ . As $x$ increases from $2$ to $3$ , $f (x)$ decreases strictly from $\frac{2}{1 + 2^{2}} = \frac{2}{5}$ to $\lim_{x \to 3^{-}} \frac{2}{1 + x^{2}} = \frac{2}{10} = \frac{1}{5}$ . Thus, $R_{1} = (\frac{1}{5}, \frac{2}{5}] = (0.2, 0.4]$ .

05Execute

2. For $x \in [3, 4)$ , $[x] = 3$ , so $f (x) = \frac{3}{1 + x^{2}}$ . As $x$ increases from $3$ to $4$ , $f (x)$ decreases strictly from $\frac{3}{10}$ to $\frac{3}{17}$ . Thus, $R_{2} = (\frac{3}{17}, \frac{3}{10}] \approx (0.1765, 0.3]$ . Note that $\frac{3}{10} < \frac{2}{5}$ and $\frac{3}{17} < \frac{1}{5} = \frac{3}{15}$ , so $R_{1} \cup R_{2} = (\frac{3}{17}, \frac{2}{5}]$ . There is an overlap since $\frac{3}{10} > \frac{1}{5}$ .

06Execute

3. For $x \in [4, 5)$ , $[x] = 4$ , so $f (x) = \frac{4}{1 + x^{2}}$ . As $x$ increases from $4$ to $5$ , $f (x)$ decreases strictly from $\frac{4}{17}$ to $\frac{4}{26} = \frac{2}{13}$ . Since $\frac{4}{17} > \frac{3}{17}$ , the intervals overlap continuously without gaps. The union is $(\frac{2}{13}, \frac{2}{5}]$ .

07Execute

4. For $x \in [5, 6)$ , $[x] = 5$ , so $f (x) = \frac{5}{1 + x^{2}}$ . As $x$ increases from $5$ to $6$ , $f (x)$ decreases strictly from $\frac{5}{26}$ to $\frac{5}{1 + 6^{2}} = \frac{5}{37}$ . Since $\frac{5}{26} > \frac{2}{13} = \frac{4}{26}$ , this sub-interval also overlaps with the previous one. The overall union of all four range segments is $⋃_{k = 1}^{4} R_{k} = (\frac{5}{37}, \frac{2}{5}]$ .

✓Verify

At each integer transition $k \in {3, 4, 5}$ , $f (k) = \frac{k}{1 + k^{2}} > \lim_{x \to k^{-}} f (x) = \frac{k - 1}{1 + k^{2}}$ . Because each interval $[k, k + 1)$ starts at a value higher than where the previous interval ended, the successive ranges overlap and form a single continuous interval with no holes.

Hints that build this answer step by step

How should the function $f (x) = \frac{[x]}{1 + x^{2}}$ be decomposed over the domain $[2, 6)$ ?
Partition $[2, 6)$ into unit intervals $[2, 3), [3, 4), [4, 5), [5, 6)$ where $[x]$ is constant.
For $x \in [k, k + 1)$ , what is the range of $f_{k} (x) = \frac{k}{1 + x^{2}}$ ?
$(\frac{k}{1 + (k + 1)^{2}}, \frac{k}{1 + k^{2}}]$
What is the union of the intervals $(\frac{2}{10}, \frac{2}{5}]$ , $(\frac{3}{17}, \frac{3}{10}]$ , $(\frac{4}{26}, \frac{4}{17}]$ , and $(\frac{5}{37}, \frac{5}{26}]$ ?
$(\frac{5}{37}, \frac{2}{5}]$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why are there no points subtracted from the range even though f(x) has jump discontinuities?

A jump discontinuity in f(x) creates a gap in the range under a specific condition. This happens only when the lower limit of one piece exceeds the upper limit of the next piece. Here, the pieces overlap. For example, [3/17, 3/10] and (1/5, 2/5] overlap on (1/5, 3/10]. Their union is connected with no missing values.

Answer

By analyzing the function piecewise on each interval $[k, k + 1)$ for $k \in {2, 3, 4, 5}$ , the union of the resulting continuous image intervals is $(\frac{5}{37}, \frac{2}{5}]$ .

Why each option works or fails

A: (5/26, 2/5] - Assumed the minimum value is reached at $x \to 5$ from the left instead of evaluating the final piecewise interval $[5, 6)$ where $x \to 6$ . Check the entire domain $[2, 6)$ : the last piece is $x \in [5, 6)$ with $[x] = 5$ , giving values down to $\lim_{x \to 6^{-}} \frac{5}{1 + x^{2}} = \frac{5}{37}$ .
B: (5/37, 2/5] - {9/29, 27/109, 18/89, 9/53} - Believed the function has holes or excluded points within the continuous intervals of each piece due to the step nature of $[x]$ . Within each interval $[k, k + 1)$ , $f (x) = \frac{k}{1 + x^{2}}$ is continuous and strictly decreasing, covering every real value in $(\frac{k}{1 + (k + 1)^{2}}, \frac{k}{1 + k^{2}}]$ without any isolated missing points.
C · correct: (5/37, 2/5] - This is the correct option. Each sub-interval $[k, k + 1)$ maps to $(\frac{k}{1 + (k + 1)^{2}}, \frac{k}{1 + k^{2}}]$ . Since these intervals overlap (e.g., $\frac{2}{10} < \frac{3}{10}$ ), their union is the contiguous interval $(\frac{5}{37}, \frac{2}{5}]$ .
D: (5/26, 2/5] - {9/29, 27/109, 18/89, 9/53} - Combined missing the last interval $[5, 6)$ with incorrectly assuming points must be punctured from the range. Evaluate each piecewise segment over the entire domain up to $6$ , and note that rational functions on intervals map to connected intervals with no isolated points removed.

Step-by-step solution

Given: $f (x) = \frac{[x]}{1 + x^{2}}$ with domain $x \in [2, 6)$ , where $[x]$ denotes the greatest integer function.
Goal: Find the complete range of $f (x)$ on $[2, 6)$ .
Approach: Break the domain $[2, 6)$ into unit intervals $[n, n + 1)$ where $[x] = n$ is constant. On each sub-interval, $f (x) = \frac{n}{1 + x^{2}}$ is strictly decreasing. Find the range on each piece and take the union.
Execute: Partition $[2, 6)$ into four intervals: 1. For $x \in [2, 3)$ , $[x] = 2$ , so $f (x) = \frac{2}{1 + x^{2}}$ . As $x$ increases from $2$ to $3$ , $f (x)$ decreases strictly from $\frac{2}{1 + 2^{2}} = \frac{2}{5}$ to $\lim_{x \to 3^{-}} \frac{2}{1 + x^{2}} = \frac{2}{10} = \frac{1}{5}$ . Thus, $R_{1} = (\frac{1}{5}, \frac{2}{5}] = (0.2, 0.4]$ .
Execute: 2. For $x \in [3, 4)$ , $[x] = 3$ , so $x \in [2, 3)$ . As $x$ increases from $3$ to $4$ , $f (x)$ decreases strictly from $\frac{3}{10}$ to $\frac{3}{17}$ . Thus, $R_{2} = (\frac{3}{17}, \frac{3}{10}] \approx (0.1765, 0.3]$ . Note that $\frac{3}{10} < \frac{2}{5}$ and $\frac{3}{17} < \frac{1}{5} = \frac{3}{15}$ , so $R_{1} \cup R_{2} = (\frac{3}{17}, \frac{2}{5}]$ . There is an overlap since $\frac{3}{10} > \frac{1}{5}$ .
Execute: 3. For $x \in [4, 5)$ , $[x] = 4$ , so $f (x) = \frac{4}{1 + x^{2}}$ . As $x$ increases from $4$ to $5$ , $f (x)$ decreases strictly from $\frac{4}{17}$ to $\frac{4}{26} = \frac{2}{13}$ . Since $\frac{4}{17} > \frac{3}{17}$ , the intervals overlap continuously without gaps. The union is $(\frac{2}{13}, \frac{2}{5}]$ .
Execute: 4. For $x \in [5, 6)$ , $[x] = 5$ , so $f (x) = \frac{5}{1 + x^{2}}$ . As $x$ increases from $5$ to $6$ , $f (x)$ decreases strictly from $\frac{5}{26}$ to $\frac{5}{1 + 6^{2}} = \frac{5}{37}$ . Since $\frac{5}{26} > \frac{2}{13} = \frac{4}{26}$ , this sub-interval also overlaps with the previous one. The overall union of all four range segments is $⋃_{k = 1}^{4} R_{k} = (\frac{5}{37}, \frac{2}{5}]$ .
Verify: At each integer transition $k \in {3, 4, 5}$ , $f (k) = \frac{k}{1 + k^{2}} > \lim_{x \to k^{-}} f (x) = \frac{k - 1}{1 + k^{2}}$ . Because each interval $[k, k + 1)$ starts at a value higher than where the previous interval ended, the successive ranges overlap and form a single continuous interval with no holes.

Range of a Piecewise Rational Function with Greatest Integer Function

What feels right?

By analyzing the function piecewise on each interval $[k, k + 1)$ for $k \in {2, 3, 4, 5}$ , the union of the resulting continuous image intervals is $(\frac{5}{37}, \frac{2}{5}]$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

07Execute

✓Verify

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