Special Functions Reading Time: 15 min JEE Focus

The Greatest Integer Function [x]: Tackling Its Tricky Domain & Range for JEE

Master the step-function that appears in 80% of JEE papers. Understand discontinuities, range limitations, and problem-solving strategies.

80%

JEE Papers

3-8

Marks/Paper

∞

Discontinuities

100%

Masterable

Why the Greatest Integer Function is Crucial for JEE

The Greatest Integer Function (also called Floor Function) appears in JEE Main and Advanced every year. Its unique step-like behavior and infinite discontinuities make it a favorite for testing conceptual understanding.

🎯 Definition

The Greatest Integer Function, denoted by $[x]$ or $\lfloor x \rfloor$, gives the largest integer less than or equal to x.

$$ [x] = n \quad \text{where} \quad n \leq x < n+1, \quad n \in \mathbb{Z} $$

Examples:

$[2.8] = 2$
$[5] = 5$
$[-1.3] = -2$
$[\pi] = 3$

📚 Quick Navigation

Definition Domain & Range Discontinuities JEE Problems

Domain & Range Analysis

📥 Domain

The domain of $[x]$ is all real numbers because we can find the greatest integer for any real number.

$$ \text{Domain} = \mathbb{R} = (-\infty, \infty) $$

No restrictions - works for positive, negative, integers, and decimals alike.

📤 Range

The range consists of all integer values because the output is always an integer.

$$ \text{Range} = \mathbb{Z} = \{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\} $$

Every integer appears infinitely many times in steps of length 1.

📈 Visualizing the Step Function

The function creates a "step" at every integer value:

For $x \in [0,1)$: $[x] = 0$ At $x=1$: jump to $[x] = 1$

For $x \in [1,2)$: $[x] = 1$ At $x=2$: jump to $[x] = 2$

For $x \in [-1,0)$: $[x] = -1$ At $x=0$: jump to $[x] = 0$

Understanding the Discontinuities

⚠️ Key Property: Discontinuous at All Integers

The Greatest Integer Function is discontinuous at every integer point. This creates infinite jump discontinuities.

Left-Hand Limit

As we approach an integer from the left:

$$ \lim_{x \to n^-} [x] = n-1 $$

Example: $\lim_{x \to 2^-} [x] = 1$

Right-Hand Limit

As we approach an integer from the right:

$$ \lim_{x \to n^+} [x] = n $$

Example: $\lim_{x \to 2^+} [x] = 2$

🎯 Jump Discontinuity Magnitude

At each integer $x = n$, the function jumps by exactly 1 unit:

$$ \lim_{x \to n^+} [x] - \lim_{x \to n^-} [x] = n - (n-1) = 1 $$

This consistent jump size is why it's called a "step" function.

Essential Properties for JEE

✓ Algebraic Properties

$[x] \leq x < [x] + 1$ (Definition)
$[x + n] = [x] + n$ for $n \in \mathbb{Z}$
$[x] + [-x] = \begin{cases} 0 & \text{if } x \in \mathbb{Z} \\ -1 & \text{if } x \notin \mathbb{Z} \end{cases}$
$[x] + [y] \leq [x + y] \leq [x] + [y] + 1$

✓ Inequality Properties

If $[x] = n$, then $n \leq x < n+1$
$[x] \geq n \Rightarrow x \geq n$
$[x] \leq n \Rightarrow x < n+1$
$[x] > n \Rightarrow x \geq n+1$
$[x] < n \Rightarrow x < n$

JEE-Level Solved Problems

JEE Main 2023 Medium

Problem 1: Finding Specific Values

Evaluate: $[3.7] + [-2.4] + [\pi] + [-\sqrt{2}]$

Solution:

$[3.7] = 3$ (largest integer ≤ 3.7)

$[-2.4] = -3$ (largest integer ≤ -2.4)

$[\pi] = 3$ (largest integer ≤ π)

$[-\sqrt{2}] = [-1.414...] = -2$

Sum = $3 + (-3) + 3 + (-2) = 1$

JEE Advanced 2022 Hard

Problem 2: Solving Equations

Solve for $x$: $[x]^2 - 3[x] + 2 = 0$

Solution:

Let $[x] = t$, where $t \in \mathbb{Z}$

Equation becomes: $t^2 - 3t + 2 = 0$

Factor: $(t-1)(t-2) = 0 \Rightarrow t = 1$ or $t = 2$

Case 1: $[x] = 1 \Rightarrow 1 \leq x < 2$

Case 2: $[x] = 2 \Rightarrow 2 \leq x < 3$

Solution: $x \in [1, 2) \cup [2, 3) = [1, 3)$

JEE Main 2021 Advanced

Problem 3: Domain with Composition

Find domain of $f(x) = \sqrt{[x]^2 - 4[x] + 3}$

Solution:

For square root: $[x]^2 - 4[x] + 3 \geq 0$

Factor: $([x]-1)([x]-3) \geq 0$

Solution: $[x] \leq 1$ or $[x] \geq 3$

Case 1: $[x] \leq 1 \Rightarrow x < 2$

Case 2: $[x] \geq 3 \Rightarrow x \geq 3$

Domain: $(-\infty, 2) \cup [3, \infty)$

Common JEE Mistakes to Avoid

❌ Mistake 1: Confusing with Rounding

Wrong: Thinking $[x]$ rounds to nearest integer

Correct: $[x]$ always rounds down (toward -∞)

Example: $[-1.5] = -2$ (not -1)

❌ Mistake 2: Domain Misunderstanding

Wrong: Thinking domain is limited due to discontinuities

Correct: Domain is all real numbers $\mathbb{R}$

Discontinuities don't affect domain, only continuity

❌ Mistake 3: Inequality Errors

Wrong: $[x] = n \Rightarrow n \leq x \leq n+1$

Correct: $[x] = n \Rightarrow n \leq x < n+1$

The upper bound is strictly less than n+1

📋 Quick Revision Checklist

Must Remember:

Domain: $\mathbb{R}$ (All real numbers)
Range: $\mathbb{Z}$ (All integers)
Discontinuous at all integers
Jump size = 1 at each discontinuity
$[x+n] = [x] + n$ for integer n

Problem Solving:

Use substitution $t = [x]$
Remember $t \in \mathbb{Z}$
Convert back: $t \leq x < t+1$
Check boundary conditions
Watch for negative values

🎯 Test Your Understanding

Try these JEE-style problems:

1. Find range of $f(x) = [x] + \sqrt{x - [x]}$

Hint: Consider fractional part function

2. Solve: $[x]^2 = 4[x] - 3$

Hint: Substitute and solve quadratic

3. Find domain of $f(x) = \frac{1}{\sqrt{[x]^2 - 5[x] + 6}}$

Hint: Denominator constraints

🚀 Pro Tip for JEE Success

The Greatest Integer Function always appears in JEE. Master it thoroughly - it's often combined with limits, continuity, and other functions.

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