Why this topic matters · 8 min read
Sets, Relations, and Functions form the foundation of discrete maths tested heavily in Agniveer Navy SSR/MR written exam (typically 2-4 questions). Expect direct definition-based MCQs, Venn diagram interpretation, function type identification (one-one, onto, bijective), and domain-range problems. This topic is moderate difficulty but high-accuracy potential — most aspirants score well here if basics are clear.
Sets — Basics and Operations
A set is a well-defined collection of distinct objects. In Agniveer exams, you need to recognize set notation, perform union/intersection/complement operations, and count elements using Venn diagrams. Sets are either finite (countable elements) or infinite (like natural numbers). Roster form lists elements explicitly; set-builder form uses a condition. Key operations: Union (A ∪ B) combines all elements; Intersection (A ∩ B) keeps only common elements; Complement (A') includes elements NOT in A.
- Roster form: {1, 2, 3}; Set-builder form: {x | x is a natural number < 4}
- Union A ∪ B: all elements from A or B or both
- Intersection A ∩ B: only elements in both A and B
- Complement A': all elements in universal set U except A
- De Morgan's Laws: (A ∪ B)' = A' ∩ B' and (A ∩ B)' = A' ∪ B'
- Cardinality |A| = number of elements in set A
Key formulas
Inclusion-Exclusion Principle
|A ∪ B| = |A| + |B| - |A ∩ B|
When: Finding total elements in union of two sets; very common in Agniveer MCQs
Three Sets Inclusion-Exclusion
|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |A ∩ C| + |A ∩ B ∩ C|
When: Counting problems with three overlapping sets; expect 1 question per paper
Worked examples
If |A| = 30, |B| = 25, |A ∩ B| = 10, find |A ∪ B|. Solution: 30 + 25 - 10 = 45.
In a class of 100 students, 60 play cricket, 50 play football, 30 play both. How many play at least one? Answer: 60 + 50 - 30 = 80.
Relations — Definition and Types
A relation from set A to set B is any subset of the Cartesian product A × B. If A = {1, 2} and B = {a, b}, then A × B = {(1,a), (1,b), (2,a), (2,b)}. A relation R ⊆ A × B can have special properties: Reflexive (every element relates to itself), Symmetric (if x relates to y, then y relates to x), Transitive (if x relates to y and y relates to z, then x relates to z). An Equivalence Relation satisfies all three properties. Agniveer exams test your ability to identify these properties from given relations.
- Reflexive: (a, a) ∈ R for all a ∈ A (example: 'equal to' relation)
- Symmetric: if (a, b) ∈ R then (b, a) ∈ R (example: 'is sibling of')
- Transitive: if (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R (example: 'less than')
- Equivalence Relation: reflexive + symmetric + transitive (example: 'congruent modulo n')
- Number of relations from A to B: 2^(|A| × |B|) because each ordered pair can be included or excluded
Key formulas
Total Relations from A to B
2^(m × n) where |A| = m, |B| = n
When: Counting total possible relations; rarely asked directly but useful for elimination
Functions — Definition, Types, and Properties
A function f: A → B is a special relation where each element of A maps to exactly one element of B. A is the domain, B is the codomain. The range is the set of all outputs actually achieved. Three critical types: (1) One-One (Injective): different inputs give different outputs — no two elements of A map to same element of B. (2) Onto (Surjective): every element of B is mapped to by at least one element of A — range = codomain. (3) Bijective: both one-one and onto — creates perfect pairing, inverse function exists. Agniveer tests all three types heavily.
- One-One (Injective): f(a1) = f(a2) implies a1 = a2; graphically, horizontal line cuts graph at most once
- Onto (Surjective): for every b in B, there exists a in A such that f(a) = b; range = codomain
- Bijective: one-one AND onto; inverse function f^(-1) exists; used in permutations
- Composition f ∘ g means apply g first, then f: (f ∘ g)(x) = f(g(x))
- Identity function f(x) = x; constant function f(x) = c for all x
Key formulas
Composition of Functions
(f ∘ g)(x) = f(g(x))
When: Finding composite function; verify domain of g is subset of domain of f
Total Functions from A to B
n^m where |A| = m, |B| = n
When: Counting total functions; each of m elements can map to any of n elements independently
Worked examples
f(x) = 2x from {1, 2, 3} to {2, 4, 6, 8}. Is it one-one? Yes (different inputs give different outputs). Is it onto? No (8 is not mapped to). Is it bijective? No (not onto).
f(x) = x^2 from R to R. One-one? No (f(-2) = f(2) = 4). Onto? No (negative numbers not in range). If domain is [0, ∞) and codomain is [0, ∞), then it's bijective.
Venn Diagrams and Practical Problem Solving
Venn diagrams visually represent sets and their relationships. Circles represent sets; overlapping regions show intersections; area outside all circles shows complement. In Agniveer exams, you'll see 2-3 set Venn diagrams with shaded regions and must identify which operation (union, intersection, complement, symmetric difference) is shown. Symmetric difference A Δ B = (A ∪ B) - (A ∩ B) = elements in A or B but not both. Always label regions clearly and count systematically.
- Symmetric Difference A Δ B: elements in exactly one of A or B, not in both
- Shaded region interpretation: identify which sets are included/excluded in each region
- Three-set Venn: 8 regions total (2^3); label center, pairwise intersections, single sets, outside
- Practical: use Venn to solve word problems about surveys, preferences, classifications
⚠ Common mistakes to avoid
- Confusing codomain with range — codomain is given, range is what's actually achieved. A function can be onto only if range = codomain.
- Mixing up one-one and onto — one-one is about injectivity (no two inputs map to same output); onto is about surjectivity (every output is hit). A function can be one-one but not onto, or vice versa.
- Forgetting De Morgan's Laws — (A ∪ B)' = A' ∩ B' NOT A' ∪ B'. This is tested in complement questions.
- Counting relations incorrectly — total relations from A to B is 2^(|A| × |B|), not 2^|A| or 2^|B|. Each ordered pair is independent.
- Misinterpreting Venn diagram regions — always identify which sets each region belongs to before counting or shading.
🧠 Memory aids
- RIOT for Relation Properties: Reflexive (self-loop), Irreflexive (no self-loop), Symmetric (bidirectional), Transitive (chain rule). Equivalence = R + S + T.
- IOB for Function Types: Injective (one-to-one, I), Onto (surjective, O), Bijective (both, B). Remember: Bijective = I + O = Invertible.
- UIC for Set Operations: Union (Include all), Intersection (Include common), Complement (Include outside). De Morgan flips: (A ∪ B)' = A' ∩ B'.
- Venn Diagram Zones: For two sets A and B, four zones exist: only A, only B, both A and B, neither. For three sets, expand to 8 zones.
🎯 AGNIVEER NAVY exam tips
- Agniveer Navy SSR/MR typically has 1-2 MCQs on set operations (inclusion-exclusion), 1 on relation properties (identify reflexive/symmetric/transitive), 1-2 on function types (one-one/onto/bijective). Total ~4 questions, 4-5 marks.
- Venn diagram questions appear in 30-40% of papers — practice shading and region counting. These are quick if you label regions first.
- Function composition questions are less common but high-accuracy if you remember (f ∘ g)(x) = f(g(x)) and check domain restrictions.
- Word problems on surveys/classifications use inclusion-exclusion principle — always draw Venn diagram first, then apply formula. This saves time and reduces errors.
- Time management: Sets/Relations/Functions should take 4-5 minutes total per question. If stuck on identifying relation properties, eliminate options by testing reflexivity first (easiest to verify).
Q1 · medium · AI-verified
If A and B are two sets such that A ⊂ B, then A ∩ B equals:
- ∅
- A
- A ∪ B
- B
Q2 · medium · AI-verified
If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, then n(A ∪ B) is:
- 7
- 8
- 6
- 4
Q3 · medium · AI-verified
The number of subsets of a set with 4 elements is:
- 32
- 16
- 12
- 8
Q4 · medium · AI-verified
The Cartesian product A × B where A = {1, 2} and B = {a, b} has how many elements?
- 4
- 8
- 6
- 2
Q5 · hard · AI-verified
If n(A) = 3 and n(B) = 4, then the total number of relations from A to B is:
- 2048
- 144
- 4096
- 512