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CBSE • Class XII • Mathematics • Ch 1
Estimated Time: 45 Mins
Study Progress: In Progress

Relations and Functions

In Class 12 Mathematics, "Relations and Functions" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

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🔄 Have You Ever Wondered?

How do cryptographers prove that an RSA encryption function scrambles a secret message such that every encrypted output corresponds to exactly one uni...

How do cryptographers prove that an RSA encryption function scrambles a secret message such that every encrypted output corresponds to exactly one unique plain-text input? Equivalence relations and bijective mappings establish mathematical certitude.

Why This Chapter Matters

In Class 12 Mathematics, "Relations and Functions" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Before You Begin (Prerequisites)

  • Relations and functions from Class 11.
  • Domain and range.
  • Cartesian product.

What You Will Learn (Core Objectives)

  • Classify types of Relations: Reflexive, Symmetric, Transitive, and Equivalence relations.
  • Partition sets into disjoint Equivalence Classes $[a]$.
  • Classify types of Functions: One-One (Injective), Onto (Surjective), and Bijective.
  • Understand Inverse of a bijective function.
  • Evaluate Composition of Functions: $(g \circ f)(x)$ and $(f \circ g)(x)$.

Chapter Roadmap & Progression

1 1. Types of Relations & Equivalence
2 2. Injective, Surjective & Bijectiv...
3 3. Function Composition & Invertibi...

Complete Concept Guide (100% Curriculum Coverage)

1. Types of Relations & Equivalence

A relation $R$ on set $A$ is:
• Reflexive: $(a, a) \in R$ for every $a \in A$.
• Symmetric: $(a, b) \in R \implies (b, a) \in R$.
• Transitive: $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.
A relation satisfying all three conditions is an Equivalence Relation, partitioning $A$ into disjoint Equivalence Classes.

2. Injective, Surjective & Bijective Functions

  • One-One (Injective): $f(x_1) = f(x_2) \implies x_1 = x_2$. (Distinct elements have distinct images; Horizontal Line Test!).
  • Onto (Surjective): For every $y \in Y$, there exists $x \in X$ such that $f(x) = y$ (Range = Codomain!).
  • Bijective: Both One-One and Onto simultaneously (invertible!).

3. Function Composition & Invertibility

For $f: A \to B$ and $g: B \to C$: $$\mathbf{(g \circ f)(x) = g(f(x))}$$ If $f$ is bijective, there exists a unique inverse function $\mathbf{f^{-1}: B \to A}$ such that $f^{-1} \circ f = I_A$ and $f \circ f^{-1} = I_B$.

Visual Learning & Conceptual Map

Relations and Functions Master Matrix

Conceptual framework, core mechanisms, and analytical relationships
Academic Architecture

1. Types of Relations & Equivalence • 2. Injective, Surjective & Bijective Functions

Chapter Summary & 10 Key Takeaways

Takeaway 1
Equivalence Relation: Reflexive, Symmetric, and Transitive.
Takeaway 2
Equivalence Class: Subset of mutually equivalent elements partitioning the domain.
Takeaway 3
Injective: One-to-one mapping verified via $f(x_1) = f(x_2) \implies x_1 = x_2$.
Takeaway 4
Surjective: Onto mapping where range encompasses the entire codomain.
Takeaway 5
Bijective Invertibility: A function possesses an inverse if and only if it is bijective.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Show that the relation $R$ in the set $\mathbb{Z}$ of integers given by $R = \{(a, b) : 2 \text{ divides } (a - b)\}$ is an equivalence relation.
Reveal Answer & Explanation
Answer: (1) Reflexive: $a - a = 0$ is divisible by 2; $(a, a) \in R$. (2) Symmetric: If $2|(a-b)$, then $2|-(a-b) \implies 2|(b-a)$; $(b, a) \in R$. (3) Transitive: If $2|(a-b)$ and $2|(b-c)$, then $(a-b) + (b-c) = a - c$ is divisible by 2; $(a, c) \in R$. Hence, $R$ is an equivalence relation.
Proved reflexive, symmetric, and transitive.
2
Let $A = \mathbb{R} - \{3\}$ and $B = \mathbb{R} - \{1\}$. Consider $f: A \to B$ defined by $f(x) = \frac{x-2}{x-3}$. Is $f$ one-one and onto? Justify.
Reveal Answer & Explanation
Answer: One-One: $\frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3} \implies (x_1-2)(x_2-3) = (x_2-2)(x_1-3) \implies -3x_1-2x_2 = -3x_2-2x_1 \implies x_1 = x_2$. Onto: For any $y \in B$, set $y = \frac{x-2}{x-3} \implies xy - 3y = x - 2 \implies x(y-1) = 3y - 2 \implies x = \frac{3y-2}{y-1} \in A$. Since $x$ exists for every $y \ne 1$, $f$ is onto. Hence, $f$ is bijective.
Proved both injective and surjective; bijective.
3
State the Horizontal Line Test for function injectivity.
Reveal Answer & Explanation
Answer: If every horizontal line intersects the graph of a function in at most one point, the function is one-one (injective); if any horizontal line cuts the graph in two or more points, it is many-one.
Horizontal line cuts graph at most once.
4
What is the total number of bijective functions from a set containing $n$ elements to itself?
Reveal Answer & Explanation
Answer: The total number of bijections from a set of $n$ elements onto itself is $n!$ (factorial $n$).
n! bijections.
5
Find the equivalence class $[0]$ for the relation $R = \{(a, b) : a - b \text{ is a multiple of } 4\}$ on $\mathbb{Z}$.
Reveal Answer & Explanation
Answer: $[0] = \{x \in \mathbb{Z} : (x, 0) \in R\} = \{x : x - 0 \text{ is a multiple of } 4\} = \{\dots, -8, -4, 0, 4, 8, \dots\} = 4\mathbb{Z}$.
[0] = {..., -4, 0, 4, 8, ...}.
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