Class 12 Maths Notes
~8 min readRelations and Functions Class 12 Notes
Complete, exam-ready notes on relations and functions: reflexive, symmetric, transitive and equivalence relations, one-one and onto functions, composition, invertibility and binary operations — written for CBSE boards and JEE revision.
What is this chapter about in one line?
Relations and Functions studies how sets are connected — equivalence relations classify elements, and well-defined functions that are one-one and onto can be inverted.
Types of Relations
Relation
A relation R from set A to set B is a subset of A × B, the Cartesian product. A relation on A means R ⊆ A × A. The number of such relations is .
- Reflexive: a R a for every
- .
- Symmetric: a R b implies b R a.
- Transitive: a R b and b R c imply a R c.
- An equivalence relation is reflexive, symmetric and transitive — e.g. being in the same class among students.
Equivalence = all three
Types of Functions
Function
A function f: A → B assigns exactly one output in B to each input in A. Its domain is A; its range is the set of actual outputs, and B is the codomain.
- One-one (injective):
- .
- Onto (surjective): every element of B is the image of some element of A, i.e. range = codomain.
- Bijective: both one-one and onto; only bijective functions are invertible.
- Number of functions from A to B:
- .
Composition and Invertible Functions
Composition
For f: A → B and g: B → C, the composition maps A to C. Composition is associative but generally not commutative.
- A function is invertible if and only if it is bijective.
- The inverse satisfies
- and
- .
- To find the inverse: solve y = f(x) for x and swap the roles of x and y.
Binary Operations
- A binary operation * on a set A maps each ordered pair (a, b) to an element of A.
- Commutative:
- .
- Associative:
- .
- Identity element e satisfies
- ; the inverse of a is the element that gives the identity.
Solved Examples
Example: Show that the function on the reals is invertible, and find .
Solution: For one-one, . It is onto since every real y is hit. So f is bijective and invertible. Let , giving , hence .
NCERT Exercise Questions
From the textbook
Exercises (NCERT Q1–Q54)
- 1) Determine whether each of the following relations are reflexive, symmetric and transitive: (i) R on with ; (ii) R on N with ; (iii) R on with ; (iv) R on Z with ; (v) R on the set of people in a town: (a) same workplace, (b) same locality, (c) x exactly 7 cm taller than y, (d) x is wife of y, (e) x is father of y.
- 2) Determine whether the relation in the set defined as is reflexive, symmetric and transitive.
- 3) Check whether the relation defined in the set as is reflexive, symmetric or transitive.
- 4) Show that the relation in the set of real numbers defined as is reflexive and transitive but not symmetric.
- 5) Check whether the relation in the real numbers defined by is reflexive, symmetric or transitive.
- 6) Show that the relation in the set given by is symmetric but neither reflexive nor transitive.
- 7) Show that the relation in the set A of all books in a library, given by is an equivalence relation.
- 8) Show that the relation in given by is an equivalence relation. Show that all elements of {1,3,5} are related to each other and all elements of {2,4} are related to each other, but no element of {1,3,5} is related to any element of {2,4}.
- 9) Show that each of the relations on : (i) and (ii) is an equivalence relation. Find the set of all elements related to 1 in each case.
- 10) Give an example of a relation which is (i) symmetric but neither reflexive nor transitive, (ii) transitive but neither reflexive nor symmetric, (iii) reflexive and symmetric but not transitive, (iv) reflexive and transitive but not symmetric, (v) symmetric and transitive but not reflexive.
- 11) Show that the relation in the set A of points in a plane, given by is an equivalence relation. Further, the set of all points related to a point is the circle through P with origin as centre.
- 12) Show that the relation in the set A of all triangles as is an equivalence relation. Three right triangles (3,4,5), (5,12,13), (6,8,10) — which are related?
- 13) Show that the union of two equivalence relations on a set is not necessarily an equivalence relation.
- 14) Let L be the set of all lines in the XY-plane and R = {(L₁,L₂) : L₁ is parallel to L₂}. Show R is an equivalence relation. Find the set of all lines related to y = 2x + 4.
- 15) Let R be the relation on {1,2,3,4} given by R = {(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)}. Choose the correct answer: (A) reflexive and symmetric but not transitive; (B) reflexive and transitive but not symmetric; (C) symmetric and transitive but not reflexive; (D) an equivalence relation.
- 16) Let R be the relation in N given by R = {(a,b) : a = b − 2, b > 6}. Choose the correct answer: (A) (2,4) ∈ R; (B) (3,8) ∈ R; (C) (6,8) ∈ R; (D) (8,7) ∈ R.
- 17) Show that defined by is one-one and onto ( = non-zero reals). Is the result true if the domain is N?
- 18) Check injectivity and surjectivity of: (i) f: N → N, f(x) = x²; (ii) f: Z → Z, f(x) = x²; (iii) f: R → R, f(x) = x²; (iv) f: N → N, f(x) = x³; (v) f: Z → Z, f(x) = x³.
- 19) Prove that the Greatest Integer Function with is neither one-one nor onto.
- 20) Show that the Modulus Function with is neither one-one nor onto.
- 21) Show that the Signum Function with f(x) = 1 for x > 0, 0 for x = 0, −1 for x < 0, is neither one-one nor onto.
- 22) Let A = {1,2,3}, B = {4,5,6,7} and f = {(1,4),(2,5),(3,6)} be a function from A to B. Show that f is one-one.
- 23) State whether f is one-one, onto or bijective: (i) , ; (ii) , .
- 24) Let A and B be sets. Show that with f(a,b) = (b,a) is bijective.
- 25) Let be defined by f(n) = (n+1)/2 if n is odd and n/2 if n is even. State whether f is bijective. Justify.
- 26) Let A = ℝ − {3} and B = ℝ − {1}. Consider f: A → B defined by Is f one-one and onto? Justify.
- 27) Let be defined as Choose the correct answer. (A) one-one onto (B) many-one onto (C) one-one but not onto (D) neither one-one nor onto.
- 28) Let be defined as Choose: (A) one-one onto (B) many-one onto (C) one-one not onto (D) neither.
- 29) Let f: {1,3,4} → {1,2,5} and g: {1,2,5} → {1,3} be given by f = {(1,2),(3,5),(4,1)} and g = {(1,3),(2,3),(5,1)}. Write down gof.
- 30) Let f, g and h be functions from R to R. Show that (f + g)oh = (foh) + (goh) and (f·g)oh = (foh)·(goh).
- 31) Find gof and fog when (i) f(x) = |x|, g(x) = |5x − 2| and (ii) f(x) = 8x³, g(x) = x^{1/3}.
- 32) If x ≠ 2/3, show that fof(x) = x for all x ≠ 2/3. What is the inverse of f?
- 33) State, with reason, whether each function has an inverse: (i) f: {1,2,3,4} → {10} with f = {(1,10),(2,10),(3,10),(4,10)}; (ii) g: {5,6,7,8} → {1,2,3,4,5} with g = {(5,4),(6,3),(7,4),(8,2)}; (iii) h: {2,3,4,5} → {7,9,11,13} with h = {(2,7),(3,9),(4,11),(5,13)}.
- 34) Show that with is one-one, and find its inverse as a map to Range f.
- 35) Consider given by f(x) = 4x + 3. Show f is invertible and find its inverse.
- 36) Consider with Show f is invertible with = non-negative reals.
- 37) Consider with Show f is invertible with .
- 38) Let be an invertible function. Show that f has a unique inverse.
- 39) Consider f: {1,2,3} → {a,b,c} given by f(1) = a, f(2) = b, f(3) = c. Find f⁻¹ and show (f⁻¹)⁻¹ = f.
- 40) Let be invertible. Show that the inverse of f⁻¹ is f, i.e. (f⁻¹)⁻¹ = f.
- 41) If is given by then fof(x) is (A) x^{1/3} (B) x³ (C) x (D) 3 − x³.
- 42) Let be The inverse is g: Range f → ℝ, given by (A) 3y/(3-4y) (B) 4y/(4-3y) (C) 4y/(3-4y) (D) 3y/(4-3y).
- 43) Determine whether each definition of ∗ gives a binary operation: (i) on ℤ⁺, a∗b = a − b; (ii) on ℤ⁺, a∗b = ab; (iii) on ℝ, a∗b = ab²; (iv) on ℤ⁺, a∗b = |a − b|; (v) on ℤ⁺, a∗b = a.
- 44) For each binary operation, determine commutativity/associativity: (i) on ℤ, a∗b = a−b; (ii) on ℚ, a∗b = ab+1; (iii) on ℚ, a∗b = ab/2; (iv) on ℤ⁺, a∗b = 2^(ab); (v) on ℤ⁺, a∗b = a^b; (vi) on ℝ−{−1}, a∗b = a/(b+1).
- 45) Consider the binary operation ∧ on {1,2,3,4,5} defined by a ∧ b = min{a,b}. Write the operation table.
- 46) Consider the binary operation ∗ on {1,2,3,4,5} given by the table (operation = max). Compute (2∗3)∗4 and 2∗(3∗4). Is ∗ commutative? Associative?
- 47) Let ∗′ on {1,2,3,4,5} be a∗′b = HCF of a and b. Is ∗′ the same operation as ∗ in Exercise 4 (max)? Justify.
- 48) Let ∗ be the binary operation on N given by a∗b = LCM(a,b). (i) Is ∗ commutative? (ii) Associative? (iii) Find the identity. (iv) Which elements of N are invertible?
- 49) Is ∗ defined on ℚ by a∗b = a − b a binary operation on ℚ? Justify.
- 50) Let ∗ be the binary operation on the set of rational numbers defined by a∗b = a + 4b². Find (2∗3)∗4 and 2∗(3∗4). Are these equal? Does this contradict associativity?
- 51) On ℚ define a∗b = a/4 + b/4. Show that ∗ is commutative and associative, and find the identity and invertible elements.
- 52) Show that the operation ∗ on ℤ⁺ given by a∗b = 3^(a+b) is a binary operation. Is it commutative? Associative?
- 53) Consider the binary operation ∗ on the set {1,2,3,4,5} defined by a∗b = a. Write the operation table and answer: is ∗ commutative? Associative?
- 54) Let A = N × N and ∗ be the binary operation on A defined by (a,b)∗(c,d) = (a+c, b+d). Show that ∗ is commutative and associative, find the identity element, and show that the invertible elements are exactly the pairs (a,a).
Answers at a glance
- 1Q1 → (i) none; (ii) none (R = {(1,6),(2,7),(3,8)}); (iii) reflexive & transitive, not symmetric; (iv) equivalence relation; (v) equivalence for (a),(b); none for (c),(d),(e).
- 2Q2 → None of the three properties.
- 3Q3 → R = {(1,2),(2,3),(3,4),(4,5),(5,6)} — not reflexive, not symmetric, not transitive.
- 4Q4 → Reflexive and transitive, not symmetric.
- 5Q5 → None. Not reflexive (a = ½ fails since ½ ≤ ⅛ is false); not symmetric ((1,2) ∈ R, (2,1) ∉ R); not transitive ((28,4) ∈ R and (4,3) ∈ R, but (28,3) ∉ R since 28 ≰ 27).
- 6Q6 → Symmetric: (1,2) ∈ R ⇒ (2,1) ∈ R and vice versa. Not reflexive ((1,1),(2,2),(3,3) ∉ R). Not transitive ((1,2) ∈ R and (2,1) ∈ R, but (1,1) ∉ R).
- 7Q7 → Equivalence relation — its equivalence classes are the books grouped by page count.
- 8Q8 → Equivalence relation with classes {1,3,5} and {2,4}; no cross-class relation.
- 9Q9 → (i) {1, 5, 9}; (ii) {1}.
- 10Q10 → On A = {1,2,3}: (i) R = {(1,2),(2,1)}; (ii) R = {(1,2)}; (iii) R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}; (iv) R = {(1,1),(2,2),(3,3),(1,2)}; (v) R = {(1,1),(2,2),(1,2),(2,1)}.
- 11Q11 → Equivalence: equality of distances is reflexive, symmetric, transitive. Class of P = {Q : OQ = OP} = the circle centred at O passing through P.
- 12Q12 → Similarity is an equivalence relation. T₁ and T₃ are related: (6,8,10) = 2 × (3,4,5). T₂ is related to neither.
- 13Q13 → Union need not be transitive, hence not an equivalence relation in general.
- 14Q14 → Parallel lines treating a line as parallel to itself gives reflexivity; symmetry and transitivity clear. Related lines: every line of the family y = 2x + c, c ∈ ℝ.
- 15Q15 → Option (B) — reflexive and transitive but not symmetric.
- 16Q16 → Option (C) — (6,8) ∈ R.
- 17Q17 → Bijective ℝ* → ℝ*; not true for domain N (onto fails).
- 18Q18 → (i) one-one, not onto (e.g. 2 has no preimage). (ii) not one-one (f(1)=f(−1)), not onto. (iii) not one-one, not onto. (iv) one-one, not onto (x³ = 2 has no natural solution). (v) bijective.
- 19Q19 → Neither one-one nor onto.
- 20Q20 → Not one-one: |1| = |−1|. Not onto: no x gives |x| = −1.
- 21Q21 → Not one-one: f(2) = f(3) = 1. Not onto: no x gives ½.
- 22Q22 → f is one-one; not onto (7 unreached).
- 23Q23 → (i) Bijective — linear with slope −4 ≠ 0, onto since x = (3 − y)/4 for any y. (ii) Neither: not one-one (f(1) = f(−1) = 2), not onto (values ≥ 1).
- 24Q24 → f is a bijection.
- 25Q25 → Not bijective — f is onto but not one-one.
- 26Q26 → f is bijective (one-one and onto).
- 27Q27 → Option (D).
- 28Q28 → Option (A) — f is a bijection: f(x₁)=f(x₂) ⇒ x₁=x₂ and x = y/3 covers ℝ.
- 29Q29 → gof = {(1,3),(3,1),(4,3)}.
- 30Q30 → Both distributivity identities; verified pointwise.
- 31Q31 → (i) gof = |5|x|−2|, fog = |5x−2|. (ii) gof(x) = 2x, fog(x) = 8x.
- 32Q32 → f = f⁻¹ (f is its own inverse).
- 33Q33 → (i) No — not one-one. (ii) No — not one-one (5 and 7 both map to 4). (iii) Yes — h is bijective.
- 34Q34 → f⁻¹(y) = 2y/(1 − y).
- 35Q35 → f⁻¹(x) = (x − 3)/4.
- 36Q36 → f⁻¹(y) = √(y − 4), verified.
- 37Q37 → f⁻¹(y) = (√(y+6) − 1)/3.
- 38Q38 → The inverse is unique.
- 39Q39 → f⁻¹(a) = 1, f⁻¹(b) = 2, f⁻¹(c) = 3; applying the same reversal twice returns f.
- 40Q40 → f⁻¹: Y → X is itself bijective, and (f⁻¹)⁻¹ exists; following f∘f⁻¹ = I_Y and f⁻¹∘f = I_X shows (f⁻¹)⁻¹ = f.
- 41Q41 → Option (C) — x.
- 42Q42 → Option (B) — g(y) = 4y/(4 − 3y).
- 43Q43 → Binary operations: (ii), (iii), (v). Not: (i), (iv).
- 44Q44 → (i) neither; (ii) commutative, not associative; (iii) both (associative since a(bc)/2·1/2 = abc/4 from either side); (iv) commutative, not associative; (v) neither; (vi) neither.
- 45Q45 → Table rows/columns 1..5 with entry min(row, col): column j of row i is min(i,j); e.g. row 1 → 1,1,1,1,1; row 2 → 1,2,2,2,2; row 3 → 1,2,3,3,3; row 4 → 1,2,3,4,4; row 5 → 1,2,3,4,5.
- 46Q46 → (2∗3)∗4 = 4, 2∗(3∗4) = 4; ∗ is commutative and associative.
- 47Q47 → No — different operations, e.g. 2∗4 = 4 yet 2∗′4 = 2.
- 48Q48 → Commutative, associative; identity 1; only 1 is invertible.
- 49Q49 → Yes — for all a,b ∈ ℚ, a − b ∈ ℚ, so the operation is closed on ℚ.
- 50Q50 → (2∗3)∗4 = 102, 2∗(3∗4) = 17958; not equal — ∗ is not associative.
- 51Q51 → ∗ is commutative but not associative; identity and inverses exist only in the (a+b)/4 reading isn't an operation with identity here — identity would need a∗e = a ⇒ (a+e)/4 = a ⇒ e = 3a, which depends on a; so no identity in ℚ.
- 52Q52 → Binary operation, commutative, not associative.
- 53Q53 → ∗ is associative but not commutative.
- 54Q54 → Commutative and associative on N × N; identity (0,0) does not exist in N × N, so no invertible elements there.
CBSE Pattern Questions
How CBSE asks this chapter
MCQ Questions
1 Mark Questions
- 1Define an equivalence relation.
- 2State the condition (in terms of pre-images) for a function f: A → B to be onto.
- 3Give one example of a function that is one-one but not onto.
- 4What is the inverse of f(x) = x + 5?
2 Mark Questions
- 1Show that the relation R = {(a, b) : a − b is even} on the set of integers is an equivalence relation.
- 2If f(x) = 3x − 5, find f⁻¹(x) and verify f(f⁻¹(x)) = x.
- 3Check whether f(x) = x² defined on R → (0, ∞) is onto. (Ans: yes)
3 Mark Questions
- 1Prove that f: R − {2} → R − {4} given by f(x) = 4x/(x − 2) is bijective and find its inverse. (Ans: f⁻¹(x) = 2x/(x − 4))
- 2Show that the relation R = {(a, b) : 3 divides a − b} is an equivalence relation on Z and write the equivalence class of 1. (Ans: [1] = {3k + 1})
- 3If f(x) = 5x − 2 and g(x) = x² + 1, find (f∘g)(x), (g∘f)(x) and (f∘g)(2). (Ans: gives 5x² + 3; 25x² − 20x + 5; 23)
4 Mark Questions
- 1Let f: R → R be given by f(x) = 1 + x². Show that f is neither one-one nor onto. Then define g: [1, ∞) → [0, ∞) by g(x) = √(x − 1) and show that (g∘f)(x) = x for x ≥ 0, proving f has a right inverse.
- 2Define an equivalence relation. Let R = {(a, b) : a + b is even} on the set of even integers. Show R is an equivalence relation, list two elements of the equivalence class of 2, and explain why the partition splits the evens into congruence classes mod 2.
PDF: Assignments & Tests
Printable study material
Free PDF Downloads & Assignments
Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty with full solutions), then attempt the timed tests.
Covers all four topics: types of relations; types of functions; composition and invertible functions; and binary operations.
Higher-difficulty set with multi-step problems (equivalence relations, bijectivity proofs, inverse functions, composition, binary-operation properties) and full worked answers. Includes an answer key.
Full board-pattern test: Section A (MCQ/VSA 1M × 10), Section B (2M × 4), Section C (3M × 4), Section D (5M × 2) = 40 marks. Time: 90 minutes. Includes answer key.
Second full paper with different questions, same 40-mark / 90-minute blueprint. Includes answer key.
Quick Revision
Key formulas at a glance
Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.
Composition
Inverse condition
One-one test
Bijective
Number of functions
Exam Strategy
How this chapter is asked
High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.
- A relation is an equivalence relation only if it is reflexive AND symmetric AND transitive.
- Only bijective functions are invertible.
- f⁻¹∘f = I_A; composition is associative but not commutative.
- Range of f equals the domain of f⁻¹.
- Check one-one by setting f(a) = f(b) and proving a = b.
FAQ
Frequently asked questions
What is an equivalence relation?
A relation that is reflexive, symmetric and transitive. The classic example is congruence of line segments or students belonging to the same class.
When is a function invertible?
A function is invertible exactly when it is bijective — both one-one (injective) and onto (surjective). Only then does a well-defined inverse exist.
What is the difference between domain, codomain and range?
The domain is the set of inputs, the codomain is the set into which outputs are declared, and the range is the set of outputs actually produced — a subset of the codomain.
Is composition of functions commutative?
No. In general (g∘f)(x) ≠ (f∘g)(x). Composition is associative, but order matters.
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