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Sets, Relations and Functions: Mathematics | JEE Main

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

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\left(\frac{5}{26}, \frac{2}{5}\right]

\left(\frac{5}{37}, \frac{2}{5}\right] - \left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}

\left(\frac{5}{37}, \frac{2}{5}\right]

\left(\frac{5}{26}, \frac{2}{5}\right] - \left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}

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 $\left(\frac{5}{37}, \frac{2}{5}\right]$ .

Option analysis

Why each option works or fails

A ·

\left(\frac{5}{26}, \frac{2}{5}\right]

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 ·

\left(\frac{5}{37}, \frac{2}{5}\right] - \left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}

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 $\left(\frac{k}{1+(k+1)^2}, \frac{k}{1+k^2}\right]$ without any isolated missing points.

C ·

\left(\frac{5}{37}, \frac{2}{5}\right]

None. This is the correct option. Each sub-interval $[k, k+1)$ maps to $\left(\frac{k}{1+(k+1)^2}, \frac{k}{1+k^2}\right]$ . Since these intervals overlap (e.g., $\frac{2}{10} < \frac{3}{10}$ ), their union is the contiguous interval $\left(\frac{5}{37}, \frac{2}{5}\right]$ .

D ·

\left(\frac{5}{26}, \frac{2}{5}\right] - \left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}

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, $R_1 = \left(\frac{1}{5}, \frac{2}{5}\right] = (0.2, 0.4]$ .

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, $R_2 = \left(\frac{3}{17}, \frac{3}{10}\right] \approx (0.1765, 0.3]$ . Note that $\frac{3}{10} < \frac{2}{5}$ and $\frac{3}{17} < \frac{1}{5} = \frac{3}{15}$ , so $R_1 \cup R_2 = \left(\frac{3}{17}, \frac{2}{5}\right]$ . 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 $\left(\frac{2}{13}, \frac{2}{5}\right]$ .

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 $\bigcup_{k=1}^4 R_k = \left(\frac{5}{37}, \frac{2}{5}\right]$ .

✓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 $f_k(x) = \frac{k}{1+x^2}$ ?
$\left(\frac{k}{1+(k+1)^2}, \frac{k}{1+k^2}\right]$
What is the union of the intervals $\left(\frac{2}{10}, \frac{2}{5}\right]$ , $\left(\frac{3}{17}, \frac{3}{10}\right]$ , $\left(\frac{4}{26}, \frac{4}{17}\right]$ , and $\left(\frac{5}{37}, \frac{5}{26}\right]$ ?
$\left(\frac{5}{37}, \frac{2}{5}\right]$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $\left(\frac{5}{37}, \frac{2}{5}\right]$ .

Why each option works or fails

A: $\left(\frac{5}{26}, \frac{2}{5}\right]$ - 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: $\left(\frac{5}{37}, \frac{2}{5}\right] - \left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$ - 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 $\left(\frac{k}{1+(k+1)^2}, \frac{k}{1+k^2}\right]$ without any isolated missing points.
C · correct: $\left(\frac{5}{37}, \frac{2}{5}\right]$ - None. This is the correct option. Each sub-interval $[k, k+1)$ maps to $\left(\frac{k}{1+(k+1)^2}, \frac{k}{1+k^2}\right]$ . Since these intervals overlap (e.g., $\frac{2}{10} < \frac{3}{10}$ ), their union is the contiguous interval $\left(\frac{5}{37}, \frac{2}{5}\right]$ .
D: $\left(\frac{5}{26}, \frac{2}{5}\right] - \left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$ - 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, $R_1 = \left(\frac{1}{5}, \frac{2}{5}\right] = (0.2, 0.4]$ .
execute: 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, $R_2 = \left(\frac{3}{17}, \frac{3}{10}\right] \approx (0.1765, 0.3]$ . Note that $\frac{3}{10} < \frac{2}{5}$ and $\frac{3}{17} < \frac{1}{5} = \frac{3}{15}$ , so $R_1 \cup R_2 = \left(\frac{3}{17}, \frac{2}{5}\right]$ . 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 $\left(\frac{2}{13}, \frac{2}{5}\right]$ .
execute: 4. For $x \in [5, 6)$ , $[x] = 5$ , so $f(x) = \frac{5}{1+x^2}$ . As $x$ increases from $5$ to $6$ , $f(x) = \frac{3}{1+x^2}$ 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 $\bigcup_{k=1}^4 R_k = \left(\frac{5}{37}, \frac{2}{5}\right]$ .
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.

Sets, Relations and Functions: Mathematics | JEE Main

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 $\left(\frac{5}{37}, \frac{2}{5}\right]$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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