Domain & Range of Piecewise Functions: Your Strategy Guide for JEE
Master the systematic approach to analyze functions defined by different rules on different intervals and combine results effectively.
Why Piecewise Functions Matter in JEE
Piecewise functions appear in 3 out of every 4 JEE papers, testing your ability to handle complex, real-world mathematical modeling. Mastering these will give you:
- Systematic problem-solving approach for complex functions
- Confidence in handling absolute value, signum, and greatest integer functions
- Time management during exam by recognizing patterns quickly
- 4-8 marks secured across different sections
🎯 The 4-Step Systematic Approach
Identify Pieces & Their Domains
Clearly mark the intervals where each sub-function applies
Find Domain of Each Piece
Apply standard domain restrictions to each sub-function
Find Range of Each Piece
Calculate range for each sub-function on its restricted domain
Combine Results
Take union of all ranges and check continuity at boundaries
Problem 1: Basic Piecewise Function
Find the domain and range of $f(x)$
Solution Approach:
Step 1: Domain Analysis
• All pieces are defined for all real numbers in their intervals
• Domain: $(-\infty, \infty)$
Step 2: Range of Each Piece
• For $x < 0$: $f(x) = x^2$, range: $(0, \infty)$
• For $0 \leq x \leq 2$: $f(x) = x + 1$, range: $[1, 3]$
• For $x > 2$: $f(x) = 5 - x$, range: $(-\infty, 3)$
Step 3: Combine Ranges
• Union: $(0, \infty) \cup [1, 3] \cup (-\infty, 3) = (-\infty, \infty)$
• But check boundaries: At $x=0$, $f(0)=1$ from second piece
• Range: $(-\infty, \infty)$
Problem 2: Absolute Value as Piecewise
Find the domain and range of $f(x)$
Solution Approach:
Step 1: Convert to Piecewise
Critical points at $x=1$ and $x=2$
$$f(x) = \begin{cases} -(x-1)-(x-2) = -2x+3 & \text{if } x < 1 \\ (x-1)-(x-2) = 1 & \text{if } 1 \leq x < 2 \\ (x-1)+(x-2) = 2x-3 & \text{if } x \geq 2 \end{cases}$$
Step 2: Domain and Range Analysis
• Domain: $(-\infty, \infty)$
• For $x < 1$: $f(x) = -2x+3$, range: $(1, \infty)$
• For $1 \leq x < 2$: $f(x) = 1$, range: $\{1\}$
• For $x \geq 2$: $f(x) = 2x-3$, range: $[1, \infty)$
Step 3: Combine Results
• Union: $(1, \infty) \cup \{1\} \cup [1, \infty) = [1, \infty)$
• Range: $[1, \infty)$
Problem 3: Greatest Integer Function
where $[x]$ denotes the greatest integer less than or equal to $x$
Find the domain and range of $f(x)$
Solution Approach:
Step 1: Understand the Function
Let $x = n + r$ where $n = [x]$ and $0 \leq r < 1$
Then $f(x) = n + \sqrt{r}$
Step 2: Domain Analysis
• $\sqrt{x - [x]}$ requires $x - [x] \geq 0$ which is always true
• Domain: $(-\infty, \infty)$
Step 3: Range Analysis
For each integer $n$, when $x \in [n, n+1)$:
• $f(x) = n + \sqrt{r}$ where $r \in [0, 1)$
• Range for this interval: $[n, n+1)$
• Union over all integers: $\bigcup_{n=-\infty}^{\infty} [n, n+1) = (-\infty, \infty)$
Step 4: Final Answer
• Domain: $(-\infty, \infty)$
• Range: $(-\infty, \infty)$
🚀 Advanced Piecewise Strategies
Common Piecewise Patterns:
- Absolute Value: $|x-a|$ creates piecewise at $x=a$
- Signum Function: $\text{sgn}(x)$ has 3 pieces
- Greatest Integer: $[x]$ changes at every integer
- Fractional Part: $\{x\} = x - [x]$
Boundary Checks:
- Always check function values at interval endpoints
- Verify if boundary values are included/excluded
- Check continuity/discontinuity at boundaries
- Look for gaps in the range
Problems 4-6 Available in Full Version
Includes 3 more challenging JEE problems with signum function, composite piecewise, and trigonometric piecewise functions
📝 Quick Self-Test
Try these piecewise function problems to test your understanding:
1. Find domain and range of $f(x) = \begin{cases} x & \text{if } x \leq 1 \\ x^2 & \text{if } x > 1 \end{cases}$
2. Find range of $f(x) = |x| + |x-1|$
3. Find domain of $f(x) = \begin{cases} \frac{1}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$
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