Differentiability of the Greatest Integer Function [x] & Fractional Part Function {x}
Complete analysis of step functions' differentiability with graphical interpretations and JEE-focused problem solving.
Why These Functions Matter in JEE
The Greatest Integer Function (GIF) and Fractional Part Function (FPF) are among the most frequently tested non-differentiable functions in JEE Main and Advanced. Understanding their differentiability is crucial because:
- They appear in limits, continuity, and differentiability questions
- They help test conceptual understanding of piecewise functions
- They're used in composite functions with other elementary functions
- They frequently appear in integer-type questions in JEE Advanced
🎯 Quick Navigation
1. Greatest Integer Function [x]
Mathematical Definition
The Greatest Integer Function (also called floor function) is defined as:
Alternative notation: $\lfloor x \rfloor$
Key Properties
- $[x] \leq x < [x] + 1$ for all real $x$
- $[n] = n$ for all integers $n$
- $[x + n] = [x] + n$ for any integer $n$
- $[x] + [-x] = \begin{cases} 0 & \text{if } x \in \mathbb{Z} \\ -1 & \text{if } x \notin \mathbb{Z} \end{cases}$
Graphical Representation
The graph consists of horizontal line segments with jump discontinuities at all integer points.
2. Fractional Part Function {x}
Mathematical Definition
The Fractional Part Function is defined as:
This represents the decimal or fractional part of $x$.
Key Properties
- $0 \leq \{x\} < 1$ for all real $x$
- $\{n\} = 0$ for all integers $n$
- $\{x + n\} = \{x\}$ for any integer $n$ (Periodic with period 1)
- $\{x\} + \{-x\} = \begin{cases} 0 & \text{if } x \in \mathbb{Z} \\ 1 & \text{if } x \notin \mathbb{Z} \end{cases}$
Graphical Representation
The graph shows a sawtooth pattern with discontinuities at integer points.
3. Differentiability Analysis
🎯 Fundamental Differentiability Rule
A function is differentiable at a point if:
- It is continuous at that point
- Left-hand derivative = Right-hand derivative at that point
Differentiability of [x]
Analysis:
At non-integer points:
- Function is constant in small neighborhoods
- Derivative exists and equals 0
- $f'(x) = 0$ for $x \notin \mathbb{Z}$
At integer points:
- Function has jump discontinuities
- Left-hand derivative = 0
- Right-hand derivative = 0
- But function is not continuous
Conclusion:
$[x]$ is differentiable for all $x \notin \mathbb{Z}$ and non-differentiable at all integer points.
Differentiability of {x}
Analysis:
At non-integer points:
- Function behaves like $f(x) = x - c$ where $c$ is constant
- Derivative exists and equals 1
- $f'(x) = 1$ for $x \notin \mathbb{Z}$
At integer points:
- Left-hand limit: $\lim_{h \to 0^-} \frac{\{n+h\} - \{n\}}{h} = \lim_{h \to 0^-} \frac{(n+h - (n-1)) - 0}{h} = 1$
- Right-hand limit: $\lim_{h \to 0^+} \frac{\{n+h\} - \{n\}}{h} = \lim_{h \to 0^+} \frac{h - 0}{h} = 1$
- But function is not continuous at integers
Conclusion:
$\{x\}$ is differentiable for all $x \notin \mathbb{Z}$ with derivative 1, and non-differentiable at all integer points due to discontinuity.
💡 JEE Shortcut
Remember this pattern:
- [x]: Differentiable except at integers → Derivative = 0 where it exists
- {x}: Differentiable except at integers → Derivative = 1 where it exists
- Both are never differentiable at integers due to discontinuity
4. JEE Problems & Solutions
Problem 1: Composite Function
Discuss the differentiability of $f(x) = [x] + \sqrt{\{x\}}$ at $x = 1$
Solution:
Step 1: Analyze continuity at $x = 1$
LHL: $\lim_{x \to 1^-} f(x) = [1^-] + \sqrt{\{1^-\}} = 0 + \sqrt{1} = 1$
RHL: $\lim_{x \to 1^+} f(x) = [1^+] + \sqrt{\{1^+\}} = 1 + \sqrt{0} = 1$
$f(1) = [1] + \sqrt{\{1\}} = 1 + 0 = 1$
✓ Function is continuous at $x = 1$
Step 2: Check differentiability
LHD: $\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{(0 + \sqrt{1+h}) - 1}{h}$
Using binomial expansion: $\frac{(1 + \frac{h}{2} + \cdots) - 1}{h} = \frac{1}{2}$
RHD: $\lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{(1 + \sqrt{h}) - 1}{h} = \lim_{h \to 0^+} \frac{\sqrt{h}}{h} = \infty$
Step 3: Conclusion
LHD = $\frac{1}{2}$, RHD = $\infty$ → Not equal
$f(x)$ is continuous but not differentiable at $x = 1$
Problem 2: Absolute Value with GIF
Find all points where $f(x) = |[x] - x|$ is not differentiable
Solution:
Step 1: Simplify the function
$|[x] - x| = | -\{x\} | = \{x\}$ (since $\{x\} \geq 0$)
So $f(x) = \{x\}$
Step 2: Differentiability of $\{x\}$
As established earlier, $\{x\}$ is differentiable for all $x \notin \mathbb{Z}$
At integer points, function is discontinuous
Step 3: Conclusion
$f(x)$ is not differentiable at all integer points $x \in \mathbb{Z}$
Problem 3: Trigonometric with FPF
Check differentiability of $f(x) = \sin(\pi\{x\})$ at $x = \frac{3}{2}$
Solution:
Step 1: Analyze the function around $x = \frac{3}{2}$
For $x$ near $\frac{3}{2}$, $\{x\} = x - 1$
So $f(x) = \sin(\pi(x - 1)) = \sin(\pi x - \pi) = -\sin(\pi x)$
Step 2: Check continuity
LHL: $\lim_{x \to \frac{3}{2}^-} -\sin(\pi x) = -\sin(\frac{3\pi}{2}) = 1$
RHL: $\lim_{x \to \frac{3}{2}^+} -\sin(\pi x) = -\sin(\frac{3\pi}{2}) = 1$
$f(\frac{3}{2}) = -\sin(\frac{3\pi}{2}) = 1$
✓ Continuous at $x = \frac{3}{2}$
Step 3: Check differentiability
$f(x) = -\sin(\pi x)$ in neighborhood of $\frac{3}{2}$
$f'(x) = -\pi\cos(\pi x)$
$f'(\frac{3}{2}) = -\pi\cos(\frac{3\pi}{2}) = 0$
$f(x)$ is differentiable at $x = \frac{3}{2}$ with derivative 0
📊 Quick Comparison Summary
| Property | [x] | {x} |
|---|---|---|
| Domain | $\mathbb{R}$ | $\mathbb{R}$ |
| Range | $\mathbb{Z}$ | $[0, 1)$ |
| Continuity | Discontinuous at $\mathbb{Z}$ | Discontinuous at $\mathbb{Z}$ |
| Differentiability Points | $\mathbb{R} - \mathbb{Z}$ | $\mathbb{R} - \mathbb{Z}$ |
| Derivative where exists | 0 | 1 |
🎯 Test Your Understanding
Try these problems to reinforce your learning:
1. Is $f(x) = [x^2]$ differentiable at $x = 1$? Explain.
2. Find all points where $f(x) = \{x\} + \{ -x \}$ is not differentiable.
3. Discuss differentiability of $f(x) = [\sin x]$ at $x = \frac{\pi}{2}$.
💡 Key Takeaways for JEE
For [x]:
- Always check integer points first
- Discontinuous ⇒ Non-differentiable at integers
- Derivative = 0 where it exists
- Watch for composite functions like $[g(x)]$
For {x}:
- Periodic with period 1
- Discontinuous at all integers
- Derivative = 1 where it exists
- Often appears with trigonometric functions
🎯 Exam Strategy:
When you see [x] or {x} in a differentiability question, immediately check:
- Continuity at the point
- Left-hand and right-hand derivatives
- Special behavior at integer points
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