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Sets Relations Functions Solutions | KalviGavani

TN Class 11 Maths Ch 1 Sets, Relations and Functions exercise solutions with worked examples, set operations, Cartesian product, types of functions.

Updated 15 June 202618 min read

Sets, Relations and Functions — Exercise Solutions

1. How to Use This Page

This page covers every exercise question from TN Class 11 Maths Chapter 1 — Sets, Relations and Functions — with step-by-step worked solutions.

Study strategy:

StepActionWhy it works
1Read Key Concepts to refresh definitionsActivates prior knowledge
2Try each question on your own firstBuilds problem-solving muscle
3Check your answer against the worked solutionImmediate feedback loop
4Note errors in Common Mistakes sectionPrevents repeat errors
5Use Exam Checklist before testsConfidence boost

Every numerical solution is worked out fully — no skipped steps.


2. Key Concepts (Quick Recap)

2.1 Set Notation

ConceptNotationMeaning
Belongs toa ∈ Aa is an element of A
Does not belong toa ∉ Aa is not an element of A
SubsetA ⊂ BEvery element of A is in B
Proper subsetA ⊊ BA ⊂ B and A ≠ B
SupersetA ⊃ BB is a subset of A
Empty set∅ or {}Set with no elements
Universal setUContains all elements under discussion
Power setP(A)Set of all subsets of A

2.2 Set Operations and Formulas

OperationNotationFormula / Definition
UnionA ∪ B{x : x ∈ A or x ∈ B}
IntersectionA ∩ B{x : x ∈ A and x ∈ B}
DifferenceA − B{x : x ∈ A and x ∉ B}
ComplementA′ or AᶜU − A
Symmetric differenceA Δ B(A − B) ∪ (B − A)

2.3 Key Counting Formulas

n(A ∪ B) = n(A) + n(B) − n(A ∩ B) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C) n(A − B) = n(A) − n(A ∩ B) n(A Δ B) = n(A) + n(B) − 2·n(A ∩ B) |P(A)| = 2ⁿ where n = |A| n(A × B) = n(A) × n(B)

2.4 De Morgan's Laws

(A ∪ B)′ = A′ ∩ B′ (A ∩ B)′ = A′ ∪ B′

2.5 Cartesian Product

For sets A and B:

A × B = {(a, b) | a ∈ A, b ∈ B} n(A × B) = n(A) · n(B) A × B ≠ B × A (in general)

2.6 Types of Relations

TypeCondition
Reflexive(a, a) ∈ R for all a ∈ A
Symmetric(a, b) ∈ R ⟹ (b, a) ∈ R
Transitive(a, b) ∈ R and (b, c) ∈ R ⟹ (a, c) ∈ R
EquivalenceReflexive + Symmetric + Transitive

2.7 Types of Functions

TypeCondition
One-one (Injective)f(a) = f(b) ⟹ a = b
Onto (Surjective)For every b ∈ B, ∃ a ∈ A such that f(a) = b
BijectionBoth one-one and onto

3. I. Choose the Best Answer (MCQs)

Q1. If A = {1, 2, 3}, then the number of subsets of A is

(a) 4   (b) 6   (c) 8   (d) 9

Answer(c) 8
Explanation
n(A) = 3 Number of subsets = 2ⁿ = 2³ = 8

The 8 subsets are: ∅, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}


Q2. If A = {1, 2} and B = {3, 4}, then A × B has how many elements?

(a) 2   (b) 4   (c) 8   (d) 16

Answer(b) 4
Explanation
n(A × B) = n(A) × n(B) = 2 × 2 = 4 A × B = {(1,3), (1,4), (2,3), (2,4)}

Q3. If n(A) = 10, n(B) = 15, n(A ∪ B) = 20, then n(A ∩ B) is

(a) 5   (b) 10   (c) 15   (d) 25

Answer(a) 5
Explanation
n(A ∪ B) = n(A) + n(B) − n(A ∩ B) 20 = 10 + 15 − n(A ∩ B) n(A ∩ B) = 25 − 20 = 5

Q4. Which of the following relations on A = {1, 2, 3} is transitive?

(a) R = {(1,1), (2,2)}   (b) R = {(1,2), (2,1)}   (c) R = {(1,2), (2,3), (1,3)}   (d) R = {(1,2), (2,3)}

Answer(c) R = {(1,2), (2,3), (1,3)}
ExplanationFor transitivity, whenever (a,b) ∈ R and (b,c) ∈ R, we must have (a,c) ∈ R.
  • (1,2) and (2,3) are in R, and (1,3) is also in R. The condition is satisfied.
  • Option (d) fails because (1,2) and (2,3) are present but (1,3) is missing.

Q5. If f : A → B is one-one and onto, then f is called

(a) Injective   (b) Surjective   (c) Bijective   (d) Constant

Answer(c) Bijective
ExplanationA function that is both one-one (injective) and onto (surjective) is called a bijection or bijective function.

4. II. Fill in the Blanks

1. If A = ∅, then n(P(A)) = 1 (since P(∅) = {∅}, andP(∅)= 2⁰ = 1)

2. A ∪ A′ = U (the universal set)

3. A ∩ A′ = ∅ (the empty set)

4. If n(A) = 5, then the number of proper subsets of A is 31 (2⁵ − 1 = 32 − 1 = 31, excluding A itself)

5. A × (B ∪ C) = (A × B) ∪ (A × C)

6. If f(x) = 2x + 3 and g(x) = x − 1, then (f ∘ g)(x) = 2x + 1

(f ∘ g)(x) = f(g(x)) = f(x − 1) = 2(x − 1) + 3 = 2x − 2 + 3 = 2x + 1

7. The identity relation on A = {a, b, c} is {(a,a), (b,b), (c,c)}

8. If A has 4 elements, then n(A × A) = 16 (4 × 4 = 16)

9. (A′)′ = A (complement of complement is the original set)

10. A − B = A ∩ B′ (set difference expressed using complement)

11. A ∪ ∅ = A (identity for union)

12. The number of relations from set A (with m elements) to set B (with n elements) is 2^(mn) (since A × B has mn ordered pairs, each may or may not be in the relation)


5. III. True or False (with corrections for false statements)

1. True or False: Every set is a subset of itself.

True. By definition, A ⊂ A for any set A (every element of A is in A).

2. True or False: A × B = B × A for all sets A and B.

False. Correction: A × B = B × A only if A = B or one of the sets is empty. In general, A × B ≠ B × A because ordered pairs have direction: (1, 3) ≠ (3, 1).

3. True or False: Every function is a relation.

True. A function f : A → B is a special type of relation from A to B where each a ∈ A has exactly one image in B.

4. True or False: An equivalence relation is always reflexive, symmetric, and transitive.

True. This is the definition of an equivalence relation.

5. True or False: If f is one-one, then f⁻¹ always exists as a function.

False. Correction: f⁻¹ exists as a function only if f is both one-one and onto (bijective). A one-one function that is not onto does not have an inverse function from B to A.

6. True or False: The empty set is a subset of every set.

True. ∅ ⊂ A for any set A because there is no element in ∅ that is not in A (vacuously true).

7. True or False: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

True. This is the distributive law of union over intersection.


6. IV. Match the Following

Column AColumn B
1. n(P(A)) where n(A) = 4(a) A ∩ B′
2. A − B(b) 16
3. A Δ B(c) (A − B) ∪ (B − A)
4. f : R → R, f(x) = x²(d) Neither one-one nor onto
5. f : R → R, f(x) = 3x + 1(e) Bijective
6. Reflexive relation(f) (a, a) ∈ R for all a ∈ A
Answer1 → b, 2 → a, 3 → c, 4 → d, 5 → e, 6 → f

Explanations:

  • 1 → b: n(P(A)) = 2⁴ = 16
  • 2 → a: A − B = {x : x ∈ A and x ∉ B} = A ∩ B′
  • 3 → c: By definition, symmetric difference A Δ B = (A − B) ∪ (B − A)
  • 4 → d: f(x) = x² is not one-one (f(1) = f(−1) = 1) and not onto (negative numbers have no pre-image in R)
  • 5 → e: f(x) = 3x + 1 is linear with non-zero slope — both one-one and onto, hence bijective
  • 6 → f: The defining property of a reflexive relation

7. V. Answer Briefly (2-mark questions)

Q1. Write the set A = {x : x is a positive integer and x² < 20} in roster form.

Answer
x² < 20 and x is a positive integer x = 1 → 1² = 1 < 20 ✓ x = 2 → 2² = 4 < 20 ✓ x = 3 → 3² = 9 < 20 ✓ x = 4 → 4² = 16 < 20 ✓ x = 5 → 5² = 25 < 20 ✗ A = {1, 2, 3, 4}

Q2. If A = {1, 2, 3, 4, 5} and B = {2, 4, 6}, find A − B and A ∩ B.

Answer
A − B = {x : x ∈ A and x ∉ B} = {1, 3, 5} A ∩ B = {x : x ∈ A and x ∈ B} = {2, 4}

Q3. If A = {a, b} and B = {1, 2, 3}, find A × B and n(A × B).

Answer
A × B = {(a,1), (a,2), (a,3), (b,1), (b,2), (b,3)} n(A × B) = n(A) × n(B) = 2 × 3 = 6

Q4. Check whether the relation R = {(1,1), (1,2), (2,1)} on A = {1, 2} is symmetric.

Answer
For symmetry: if (a, b) ∈ R, then (b, a) must also ∈ R. (1,1) ∈ R → (1,1) ∈ R ✓ (1,2) ∈ R → (2,1) ∈ R ✓ (2,1) ∈ R → (1,2) ∈ R ✓ R is symmetric.

Check transitivity: (1,2) and (2,1) ∈ R, so (1,1) must ∈ R. Yes, (1,1) ∈ R. Also (2,1) and (1,2) ∈ R, so (2,2) must ∈ R. But (2,2) ∉ R. So R is not transitive.


8. VI. Answer in Detail (5-mark questions)

Q1. In a group of 400 students, 250 can speak Tamil, 200 can speak English, and 100 can speak both. Find how many can speak either Tamil or English and how many can speak only Tamil.

Worked Solution:

Given: n(T) = 250, n(E) = 200, n(T ∩ E) = 100, n(U) = 400 Step 1: Number who can speak either Tamil or English n(T ∪ E) = n(T) + n(E) − n(T ∩ E) = 250 + 200 − 100 = 350 Step 2: Number who can speak only Tamil n(Only Tamil) = n(T) − n(T ∩ E) = 250 − 100 = 150 Step 3: Number who can speak only English n(Only English) = n(E) − n(T ∩ E) = 200 − 100 = 100 Step 4: Number who can speak neither n((T ∪ E)′) = n(U) − n(T ∪ E) = 400 − 350 = 50 Answer: 350 students speak either language; 150 speak only Tamil.

Q2. Let A = {1, 2, 3, 4, 5, 6} and R be a relation on A defined by R = {(a, b) : a + b ≤ 5}. Find R, its domain, range, and check if R is reflexive, symmetric, or transitive.

Worked Solution:

Step 1: List all ordered pairs (a, b) where a + b ≤ 5 a = 1: b can be 1,2,3,4 → (1,1),(1,2),(1,3),(1,4) a = 2: b can be 1,2,3 → (2,1),(2,2),(2,3) a = 3: b can be 1,2 → (3,1),(3,2) a = 4: b can be 1 → (4,1) a = 5: no b satisfies → ∅ a = 6: no b satisfies → ∅ R = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)} Step 2: Domain Domain = {a : (a,b) ∈ R for some b} = {1, 2, 3, 4} Step 3: Range Range = {b : (a,b) ∈ R for some a} = {1, 2, 3, 4} Step 4: Check Reflexive (1,1) ∈ R ✓, (2,2) ∈ R ✓ (3,3): 3+3=6 > 5, so (3,3) ∉ R ✗ R is NOT reflexive. Step 5: Check Symmetric (1,2) ∈ R. Is (2,1) ∈ R? Yes ✓ (1,3) ∈ R. Is (3,1) ∈ R? Yes ✓ (1,4) ∈ R. Is (4,1) ∈ R? Yes ✓ (2,3) ∈ R. Is (3,2) ∈ R? Yes ✓ All pairs check out. R IS symmetric. Step 6: Check Transitive (4,1) ∈ R and (1,4) ∈ R. Is (4,4) ∈ R? 4+4=8 > 5 → No ✗ R is NOT transitive. Answer: R is symmetric only — not reflexive, not transitive.

Q3. Let f(x) = x² and g(x) = 2x + 3 be two functions from R → R. Find (f ∘ g)(x), (g ∘ f)(x), and determine whether each composition is one-one or onto.

Worked Solution:

Step 1: Find (f ∘ g)(x) (f ∘ g)(x) = f(g(x)) = f(2x + 3) = (2x + 3)² = 4x² + 12x + 9 Step 2: Find (g ∘ f)(x) (g ∘ f)(x) = g(f(x)) = g(x²) = 2x² + 3 Step 3: Check if f ∘ g is one-one Suppose (f ∘ g)(a) = (f ∘ g)(b) Then (2a + 3)² = (2b + 3)² So 2a + 3 = ±(2b + 3) Case 1: 2a + 3 = 2b + 3 → a = b ✓ Case 2: 2a + 3 = −(2b + 3) → 2a + 2b + 6 = 0 Since Case 2 gives a ≠ b (e.g., a=0, b=−3), the composition is NOT one-one. Step 4: Check if f ∘ g is onto (f ∘ g)(x) = (2x + 3)² ≥ 0 for all x ∈ R Negative numbers have no pre-image. NOT onto (from R to R). Step 5: Check if g ∘ f is one-one (g ∘ f)(x) = 2x² + 3 g ∘ f(1) = 2(1) + 3 = 5 g ∘ f(−1) = 2(1) + 3 = 5 Since g ∘ f(1) = g ∘ f(−1) but 1 ≠ −1, NOT one-one. Step 6: Check if g ∘ f is onto 2x² + 3 ≥ 3 for all x ∈ R Values less than 3 have no pre-image. NOT onto. Answer: (f ∘ g)(x) = 4x² + 12x + 9 — neither one-one nor onto (g ∘ f)(x) = 2x² + 3 — neither one-one nor onto

Q4. Verify De Morgan's Laws using Venn diagram reasoning for A = {1, 2, 3, 4}, B = {2, 4, 6, 8}, U = {1, 2, 3, 4, 5, 6, 7, 8}.

Worked Solution:

Given: A = {1,2,3,4}, B = {2,4,6,8}, U = {1,2,3,4,5,6,7,8} ═══ Law 1: (A ∪ B)′ = A′ ∩ B′ ═══ Step 1: A ∪ B = {1, 2, 3, 4, 6, 8} Step 2: (A ∪ B)′ = U − (A ∪ B) = {5, 7} Step 3: A′ = U − A = {5, 6, 7, 8} Step 4: B′ = U − B = {1, 3, 5, 7} Step 5: A′ ∩ B′ = {5, 7} (A ∪ B)′ = {5, 7} = A′ ∩ B′ ✓ Verified! ═══ Law 2: (A ∩ B)′ = A′ ∪ B′ ═══ Step 6: A ∩ B = {2, 4} Step 7: (A ∩ B)′ = U − (A ∩ B) = {1, 3, 5, 6, 7, 8} Step 8: A′ ∪ B′ = {5, 6, 7, 8} ∪ {1, 3, 5, 7} = {1, 3, 5, 6, 7, 8} (A ∩ B)′ = {1, 3, 5, 6, 7, 8} = A′ ∪ B′ ✓ Verified!

9. Common Mistakes to Avoid

#MistakeCorrect Approach
1Writing A × B = B × AA × B ≠ B × A in general; ordered pairs are directional
2Saying n(P(A)) = n(A)n(P(A)) = 2ⁿ where n = \A\; e.g., n(A)=3 gives 8 subsets, not 3
3Confusing A ⊂ B with A ∈ B⊂ means "is a subset of" (every element); ∈ means "is an element of"
4Forgetting the empty set when listing subsets∅ is always a subset; a set with n elements has 2ⁿ subsets total
5Saying f(x₁) = f(x₂) ⟹ x₁ = x₂ proves one-oneThe correct test is: f(x₁) = f(x₂) ⟹ x₁ = x₂ — this IS correct, but many students reverse the direction
6Assuming every onto function is one-oneOnto and one-one are independent properties; verify each separately
7Writing A − B = B − AThey are different: A − B = {x ∈ A : x ∉ B}, B − A = {x ∈ B : x ∉ A}
8Omitting (a, a) when checking reflexivityEvery element must map to itself; check all a ∈ A
9Using n(A ∪ B) = n(A) + n(B) without intersection correctionAlways use n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
10Confusing "number of subsets" with "number of proper subsets"Proper subsets exclude the set itself: 2ⁿ − 1

10. Exam Checklist

Before you walk into the exam, confirm you can do each of these:

  • [ ] Convert between roster form and set-builder form quickly and accurately
  • [ ] Apply the four set operation formulas: union, intersection, difference, complement
  • [ ] Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B) for 2-set problems
  • [ ] Use the 3-set formula for n(A ∪ B ∪ C) — know all 7 terms
  • [ ] Apply De Morgan's Laws: (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′
  • [ ] Compute Cartesian products and state n(A × B) = n(A) · n(B)
  • [ ] Identify relation types: check reflexive, symmetric, transitive systematically
  • [ ] Check one-one: assume f(a) = f(b) and show a = b (or find a counterexample)
  • [ ] Check onto: for every b ∈ codomain, solve for a ∈ domain such that f(a) = b
  • [ ] Compute compositions: (f ∘ g)(x) = f(g(x)) — always apply the inner function first
  • [ ] Find domain and range of a relation from its ordered pairs
  • [ ] Count subsets and power set elements: |P(A)| = 2ⁿ, proper subsets = 2ⁿ − 1
  • [ ] Draw Venn diagrams to verify set identity answers
  • [ ] Distinguish between relations and functions: every function is a relation, not vice versa

Frequently Asked Questions

A relation is any set of ordered pairs; a function is a relation in which every input has exactly one output.
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