Relations and Functions
Why This Chapter Matters
Relations and Functions is the foundation of Class 12 Maths and the first chapter in NCERT. Board exams test definitions, types of functions, composition, and inverse — 5-8 marks every year.
Core Concepts
1. Types of Relations
A relation R from set A to set B is a subset of A × B.
Empty relation: R = {} (no element of A is related to any element of B)
Universal relation: R = A × B (every element related to every other)
Identity relation: R = {(a,a) : a ∈ A}
Equivalence relation: Must satisfy all three:
2. Types of Functions
One-one (Injective): f(a) = f(b) → a = b. No two elements map to same image.
Test: different inputs give different outputs.
Onto (Surjective): Every element of codomain has a pre-image. Range = Codomain.
Bijective: Both one-one AND onto. Has inverse function.
Into: Some elements of codomain have NO pre-image (range ≠ codomain).
3. Composition of Functions
(gof)(x) = g(f(x)) — apply f first, then g.
Order matters: gof ≠ fog generally.
If f: A→B and g: B→C, then gof: A→C.
4. Inverse Functions
f⁻¹ exists only if f is bijective.
If f(x) = y then f⁻¹(y) = x.
Domain of f⁻¹ = Range of f. Range of f⁻¹ = Domain of f.
f⁻¹(f(x)) = x and f(f⁻¹(x)) = x.
5. Binary Operations
A binary operation on set A: : A × A → A.
Commutative: a b = b a.
Associative: (a b) c = a (b c).
Identity element e: a e = e a = a.
Inverse of a: a * a⁻¹ = e.
Board Questions
Q1: Show that f: R→R defined by f(x) = 2x+3 is bijective.
One-one: f(a)=f(b) → 2a+3=2b+3 → a=b ✓
Onto: For any y∈R, x=(y-3)/2 ∈ R gives f(x)=y ✓. Bijective.
Q2: If f(x)=x²+4 and g(x)=2x-1, find gof(x) and fog(x).
gof(x)=g(f(x))=g(x²+4)=2(x²+4)-1=2x²+7.
fog(x)=f(g(x))=f(2x-1)=(2x-1)²+4=4x²-4x+5.
Q3: Is f: N→N defined by f(n)=n² one-one? Onto?
One-one: f(1)=f(-1)=1 but 1≠-1... but domain is N. In N: f(a)=f(b)→a²=b²→a=b (since both positive) ✓ One-one.
Onto? f(N)={1,4,9,16...}≠N (2 has no pre-image). NOT onto.
PYQs (CBSE)
CBSE 2023: Let R be relation on Z defined by R={(a,b): a-b is divisible by 5}. Show R is equivalence.
Reflexive: a-a=0 divisible by 5 ✓.
Symmetric: a-b divisible by 5 → b-a=-(a-b) also divisible by 5 ✓.
Transitive: a-b div by 5, b-c div by 5 → a-c=(a-b)+(b-c) div by 5 ✓. Equivalence relation.
CBSE 2022: f: R→R, f(x)=3-4x. Find f⁻¹.
Let y=3-4x → 4x=3-y → x=(3-y)/4. So f⁻¹(y)=(3-y)/4 or f⁻¹(x)=(3-x)/4.