How to Solve: Sets, Relations, and Functions (IIT JEE Guide)

How to Solve: Sets, Relations, and Functions (IIT JEE Guide)

Hook: Mastering sets, relations, and functions unlocks 10-15% of IIT JEE Maths—that’s 30-45 marks in JEE Main and 50+ marks in JEE Advanced. Whether it’s counting functions, finding domains, or composing functions, this topic appears in every major exam, often as a high-scoring, time-saving question.

What You Need To Know First

1. Basic Set Theory – Union, intersection, subsets, Venn diagrams.
2. Cartesian Product – Ordered pairs, A × B notation.
3. Basic Function Concepts – Input-output mapping, one-one/many-one.

(If you’re shaky on these, pause and review them first.)

Key Vocabulary

Formulas To Know

1. Number of Functions

• Total functions from A to B: |B|^|A| (MEMORISE THIS)
• |A| = number of elements in set A (domain)

• |B| = number of elements in set B (codomain)

• One-one (injective) functions: P(|B|, |A|) = |B|! / (|B| - |A|)! (MEMORISE THIS)

• Only if |B| ≥ |A| (otherwise, 0)

• Onto (surjective) functions: |B|^|A| - C(|B|,1)(|B|-1)^|A| + C(|B|,2)(|B|-2)^|A| - ... (Given on exam sheet)

• Simpler for small sets: Count manually (see Example 3).

2. Domain & Range Shortcuts

3. Composition Rules

• (f ∘ g)(x) = f(g(x)) (MEMORISE THIS)
• Domain of (f ∘ g): All x in g’s domain where g(x) is in f’s domain.
• Range of (f ∘ g): f’s range applied to g’s range.

Step-by-Step Method

How to Solve Any Function Problem (IIT JEE Style)

Step 1: Identify the type of problem. - Is it about counting functions? → Use formulas. - Is it about domain/range? → Check restrictions. - Is it about composition? → Work inside-out.

Step 2: Write down given sets/functions clearly. - For sets: A = {1, 2}, B = {a, b, c} - For functions: f(x) = x², g(x) = 2x + 1

Step 3: Apply the correct formula or logic. - Counting functions? → Use |B|^|A| or permutations. - Domain? → Find where function is undefined (denominator = 0, √negative, log(≤0)). - Range? → Find possible y-values (inverse method or graph analysis). - Composition? → Compute f(g(x)) and simplify.

Step 4: Check for edge cases. - Empty sets? → 0 functions. - Repeated elements? → Not a function. - One-one/onto? → Verify conditions.

Step 5: Write the final answer in the required format. - For counting: Just the number. - For domain/range: Interval notation (e.g., [2, ∞)). - For composition: Simplified function.

Worked Example (Using Steps Above)

Problem: Let A = {1, 2}, B = {a, b, c}. Find: 1. Total number of functions from A to B. 2. Number of one-one functions. 3. Domain and range of f(x) = √(x-1) + 1/(x-2).

Solution (Step-by-Step):

Part 1: Total functions - Step 1: Counting functions problem. - Step 2: A = {1, 2} → |A| = 2, B = {a, b, c} → |B| = 3. - Step 3: Total functions = |B|^|A| = 3² = 9.

Part 2: One-one functions - Step 1: Counting injective functions. - Step 2: |A| = 2, |B| = 3 → |B| ≥ |A|, so possible. - Step 3: P(3,2) = 3! / (3-2)! = 6 / 1 = 6.

Part 3: Domain & Range of f(x) = √(x-1) + 1/(x-2) - Step 1: Domain/range problem. - Step 2: Break into two parts: - √(x-1) → x - 1 ≥ 0 → x ≥ 1 - 1/(x-2) → x - 2 ≠ 0 → x ≠ 2 - Step 3: Domain = x ≥ 1 and x ≠ 2 → [1, 2) ∪ (2, ∞) - Step 4: Range: - √(x-1) → y ≥ 0 - 1/(x-2) → y ≠ 0 (since x ≠ 2) - Combined: y ≥ 0, but y ≠ 1/(x-2) when x > 2. - For x ∈ [1,2), 1/(x-2) ∈ (-∞, -1] → f(x) ∈ [0, ∞) + (-∞, -1] = (-∞, ∞) - For x ∈ (2, ∞), 1/(x-2) ∈ (0, ∞) → f(x) ∈ [0, ∞) + (0, ∞) = (0, ∞) - Range = ℝ (since all real numbers are covered).

Final Answers: 1. 9 functions. 2. 6 one-one functions. 3. Domain = [1, 2) ∪ (2, ∞), Range = ℝ.

Worked Examples

Example 1 – Basic (Counting Functions)

Problem: Let A = {1, 2, 3}, B = {x, y}. Find the number of: 1. Total functions from A to B. 2. One-one functions. 3. Onto functions.

Solution: 1. Total functions = |B|^|A| = 2³ = 8. 2. One-one functions = P(2,3) = 0 (since |B| < |A|). 3. Onto functions = Total - functions missing at least one element.
- Functions missing x: 1 (only y for all inputs).
- Functions missing y: 1 (only x for all inputs).
- Onto = 8 - 2 = 6.

What we did and why: - Used |B|^|A| for total functions. - P(|B|,|A|) for one-one (0 if |B| < |A|). - Inclusion-exclusion for onto (easier than formula for small sets).

Example 2 – Medium (Domain & Range)

Problem: Find the domain and range of: f(x) = log₂(x² - 4x + 3)

Solution: Domain: - Inside log > 0 → x² - 4x + 3 > 0 - Factor: (x-1)(x-3) > 0 - Critical points: x = 1, 3 - Test intervals: - x < 1 → (x-1)(x-3) > 0 (e.g., x=0 → 3 > 0) - 1 < x < 3 → (x-1)(x-3) < 0 (e.g., x=2 → -1 < 0) - x > 3 → (x-1)(x-3) > 0 (e.g., x=4 → 3 > 0) - Domain = (-∞, 1) ∪ (3, ∞)

Range: - Let y = log₂(x² - 4x + 3) - x² - 4x + 3 = 2^y - x² - 4x + (3 - 2^y) = 0 - For real x, discriminant ≥ 0: - 16 - 4(1)(3 - 2^y) ≥ 0 → 16 - 12 + 4·2^y ≥ 0 → 4 + 4·2^y ≥ 0 → Always true. - But x² - 4x + 3 > 0 → Minimum value of x² - 4x + 3 is at x = 2 (vertex). - f(2) = 4 - 8 + 3 = -1 → But log(-1) is undefined. - Wait: We already restricted domain to where x² - 4x + 3 > 0. - Minimum value in domain:
- For x < 1, as x → -∞, x² - 4x + 3 → ∞.
- For x > 3, as x → ∞, x² - 4x + 3 → ∞.
- At x=1, f(1) = 0 → log₂(0) is undefined (but x=1 is not in domain).
- At x=3, f(3) = 0 → log₂(0) is undefined (but x=3 is not in domain).
- Minimum value in domain is just above 0 (e.g., x=0.9 → f(0.9) ≈ 0.01 → log₂(0.01) ≈ -6.64).
- Range = ℝ (since log can take any real value as input approaches 0⁺ or ∞).

What we did and why: - Domain: Solved inequality for log’s argument. - Range: Used inverse function and discriminant (but had to check domain restrictions).

Example 3 – Exam-Style (Composition & Tricky Counting)

Problem (JEE Advanced 2018): Let A = {1, 2, 3, 4, 5}, B = {1, 2, 3, 4, 5, 6}. A function f: A → B is called good if f(i) ≠ i for any i ∈ A. Find the number of good functions.

Solution: - Total functions = 6⁵ = 7776. - Good functions = Total - functions where at least one f(i) = i. - Use inclusion-exclusion: - Functions where f(1)=1: 6⁴ = 1296 (fix f(1)=1, others free). - Similarly, f(2)=2: 6⁴, ..., f(5)=5: 6⁴. - Total for single fixed point: C(5,1)·6⁴ = 5·1296 = 6480. - Functions where f(1)=1 and f(2)=2: 6³ = 216. - Total for two fixed points: C(5,2)·6³ = 10·216 = 2160. - Continue:
- Three fixed points: C(5,3)·6² = 10·36 = 360.
- Four fixed points: C(5,4)·6¹ = 5·6 = 30.
- Five fixed points: C(5,5)·6⁰ = 1·1 = 1. - Good functions = Total - (single) + (double) - (triple) + (four) - (five) = 7776 - 6480 + 2160 - 360 + 30 - 1 = 3125.

What we did and why: - Recognized this as a derangement-like problem (counting functions with no fixed points). - Used inclusion-exclusion to subtract bad cases. - Key trap: Forgetting to add/subtract higher-order overlaps.

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up—this is your 60-second crash course for sets, relations, and functions in IIT JEE:

1. Counting functions?
2. Total = |B|^|A|.
3. One-one = P(|B|,|A|) (0 if |B| < |A|).

4. Onto = Total - bad cases (use inclusion-exclusion for small sets).

5. Domain?

6. Denominator ≠ 0, √(≥0), log(>0).

7. For f(g(x)), g(x) must be in f’s domain.

8. Range?

9. For √f(x), find f(x)’s minimum.
10. For log f(x), find f(x)’s range (must be >0).

11. For rational functions, find y where x exists.

12. Composition?

13. (f ∘ g)(x) = f(g(x))—work inside-out.

14. Domain: x in g’s domain and g(x) in f’s domain.

15. Common traps?

16. |B| < |A| → 0 one-one/onto functions.
17. f(g(x)) ≠ g(f(x))—order matters!
18. Always check edge cases (e.g., x=2 in 1/(x-2)).

You’ve got this. Now go ace that exam!