Home › Free notes › Class 12 Mathematics › Relations and Functions
Relations and Functions – Class 12 Mathematics notes, important questions and MCQs
NCERT Class 12 Mathematics, chapter 1. Notes, key terms, important questions with answers and MCQs.
Chapter overview
Relations connect elements of sets; special types (reflexive, symmetric, transitive) make equivalence relations. Functions can be one-one, onto or both (bijective).
Relations
- Reflexive: (a, a) ∈ R for all a; symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R; transitive: (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R.
- Equivalence relation = reflexive + symmetric + transitive; it partitions a set into equivalence classes.
Functions
- One-one (injective): f(x₁) = f(x₂) ⇒ x₁ = x₂; onto (surjective): range = codomain.
- Bijective functions have inverses; composition (g∘f)(x) = g(f(x)).
- Number of functions from A (m elements) to B (n elements) = nᵐ; one-one functions (m ≤ n) = ⁿPₘ.
Key terms
- Equivalence relation
- relation that is reflexive, symmetric and transitive
- Bijection
- function that is both one-one and onto
- Invertible function
- a function f for which there is g with g∘f = identity and f∘g = identity; exists exactly when f is bijective
Formulas
- Functions A → B: nᵐ
- (g∘f)(x) = g(f(x))
Important questions with answers
Check whether f: ℕ → ℕ, f(x) = 2x is one-one and onto.2 marks
One-one: 2x₁ = 2x₂ gives x₁ = x₂. Not onto: odd numbers such as 1 have no pre-image in ℕ.
Show that the relation R on ℤ given by R = {(a, b): a − b is divisible by 5} is an equivalence relation.3 marks
Reflexive: a − a = 0 is divisible by 5. Symmetric: if 5 | (a − b) then 5 | (b − a) = −(a − b). Transitive: if 5 | (a − b) and 5 | (b − c), then 5 | (a − b) + (b − c) = a − c. Hence R is an equivalence relation.
Show that f: ℝ → ℝ, f(x) = 4x + 3 is invertible and find f⁻¹. Also find (f∘f)(x) and verify f∘f⁻¹ = I.5 marks
One-one: 4x₁ + 3 = 4x₂ + 3 ⇒ x₁ = x₂. Onto: for any y ∈ ℝ, x = (y − 3)/4 ∈ ℝ gives f(x) = y. So f is bijective and invertible, with f⁻¹(y) = (y − 3)/4. (f∘f)(x) = 4(4x + 3) + 3 = 16x + 15. f(f⁻¹(y)) = 4 × (y − 3)/4 + 3 = y, so f∘f⁻¹ = I.
Quick revision
- Equivalence = R + S + T.
- Only bijections are invertible.
- Onto: every element of codomain has a pre-image.
Exam tips
- f(x) = x² on ℝ is not one-one (f(2) = f(−2)); on ℕ it is.
- To show onto, take any y in the codomain and solve f(x) = y for x inside the domain.
Practice MCQs
A relation that is reflexive, symmetric and transitive is
- identity
- universal only
- empty
- equivalence
Show answer
Answer: (d) equivalence. Equivalence relation.
f: ℝ → ℝ, f(x) = 2x + 3 is
- one-one only
- onto only
- bijective
- neither
Show answer
Answer: (c) bijective. Linear with non-zero slope.
f(x) = x² from ℝ to ℝ is
- one-one
- onto
- neither one-one nor onto
- bijective
Show answer
Answer: (c) neither one-one nor onto. f(1) = f(−1) and negatives have no pre-image.
If f(x) = x + 1 and g(x) = 2x, then (g∘f)(3) is
- 7
- 6
- 8
- 10
Show answer
Answer: (c) 8. g(4) = 8.
Number of reflexive relations on a set with 3 elements is
- 2⁶
- 2⁹
- 2³
- 3²
Show answer
Answer: (a) 2⁶. Diagonal fixed; 6 other pairs free.
Number of functions from a set of 3 elements to a set of 2 elements is
- 6
- 9
- 8
- 5
Show answer
Answer: (c) 8. 2³.
Relation R on {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2)} is
- reflexive and transitive but not symmetric
- equivalence
- symmetric only
- not reflexive
Show answer
Answer: (a) reflexive and transitive but not symmetric. (2, 1) missing.
If f(x) = (3x + 2)/(x − 1), x ≠ 1, then f⁻¹(x) is
- (x + 3)/(x − 2)
- (x − 2)/(x + 3)
- (x + 2)/(x − 3)
- (3x − 2)/(x + 1)
Show answer
Answer: (c) (x + 2)/(x − 3). Solve y = (3x + 2)/(x − 1) for x.
All chapters of Class 12 Mathematics
Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.