Home Class 12 Relations & Functions

Chapter 1: Relations and Functions

Overview

This page provides an Overview of Chapter Wise Notes for CBSE Class 12. Key topics covered: Exam Weightage & Blueprint, Last 24-Hour Checklist, Types of Relations, Properties of Relations, Types of Functions, Quick Reference: Function Types, Composition of Functions, Solved Examples (Board Marking Scheme). All content is aligned with the NCERT syllabus for Board Exam preparation.

Complete Board Exam Focused Notes with PYQs and Strategies

Exam Weightage & Blueprint

Total: ~10 Marks

Relations and Functions is a fundamental chapter that appears in various forms throughout the Board exam. Concepts from this chapter are also used in Calculus and other areas.

Question Type Marks Frequency Focus Topic
MCQ 1 Very High Types of Relations, One-One/Onto identification
Short Answer (2M) 2 High Proving relations (reflexive/symmetric/transitive)
Short Answer (3M) 3 Very High Equivalence Relations, One-One/Onto proofs
Long Answer 4 Medium Composition of Functions, Invertible Functions

Last 24-Hour Checklist

Types of Relations

Basic Definitions

Relation: A relation $R$ from set $A$ to set $B$ is a subset of $A \times B$. If $(a, b) \in R$, we write $a R b$.

Empty Relation

$R = \phi \subset A \times A$

No element is related to any element

Universal Relation

$R = A \times A$

Every element is related to every element

Equivalence Relation

Reflexive, Symmetric & Transitive

Most important for exams!

Properties of Relations

Definition: A relation $R$ in set $A$ is:
  • Reflexive if $(a, a) \in R$ for every $a \in A$
  • Symmetric if $(a_1, a_2) \in R \Rightarrow (a_2, a_1) \in R$
  • Transitive if $(a_1, a_2) \in R$ and $(a_2, a_3) \in R \Rightarrow (a_1, a_3) \in R$
Memory Aid - RST:
Reflexive: Every element relates to itself (like looking in a mirror)
Symmetric: If A relates to B, then B relates to A (friendship is mutual)
Transitive: If A→B and B→C, then A→C (chain of connections)

Equivalence Relation

A relation is an Equivalence Relation if and only if it is:
Reflexive AND Symmetric AND Transitive
Common Examples of Equivalence Relations:
  • "is equal to" in any set
  • "is congruent to" for triangles
  • "is similar to" for triangles
  • "has same number of pages as" for books
  • $R = \{(a, b) : |a - b|$ is even$\}$ in integers

Equivalence Class $[a]$

Definition: Given an equivalence relation $R$ in set $A$, the equivalence class of an element $a \in A$ is the set of all elements related to $a$.
Denoted by $[a] = \{x \in A : (x, a) \in R\}$.

Quick Check: If $R = \{(1,1), (2,2), (1,2), (2,1)\}$ on $\{1,2\}$, what is $[1]$?

Answer: $[1] = \{1, 2\}$ because 1 is related to 1 and 1 is related to 2.

Types of Functions

One-One Function (Injective)

A function $f: X \to Y$ is one-one (injective) if:
$$f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \text{ for all } x_1, x_2 \in X$$

In other words: Different inputs give different outputs

Testing One-One:
• Algebraic Method: Assume $f(x_1) = f(x_2)$ and prove $x_1 = x_2$
• Graphical Test: Horizontal line test (any horizontal line cuts graph at most once)

Onto Function (Surjective)

A function $f: X \to Y$ is onto (surjective) if:
For every $y \in Y$, there exists $x \in X$ such that $f(x) = y$

In other words: Range = Co-domain (every element in Y is mapped)

Testing Onto:
• Find the range of $f$ and check if it equals the co-domain
• For any arbitrary $y$ in co-domain, solve $f(x) = y$ for $x$
• If solution exists for all $y$, function is onto

Bijective Function (One-One and Onto)

A function is bijective if it is both:
One-One (Injective) AND Onto (Surjective)

Important: Only bijective functions have inverses!

One-One Examples

• $f: \mathbb{R} \to \mathbb{R}, f(x) = 2x$

• $f: \mathbb{N} \to \mathbb{N}, f(x) = x^2$

• $f: \mathbb{R} \to \mathbb{R}, f(x) = x^3$

Many-One Examples

• $f: \mathbb{R} \to \mathbb{R}, f(x) = x^2$

• $f: \mathbb{R} \to \mathbb{R}, f(x) = |x|$

• $f: \mathbb{R} \to \mathbb{R}, f(x) = \cos x$

Quick Reference: Function Types

Function Type Condition Inverse Exists?
One-One (Injective) $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$ No
Onto (Surjective) Range = Co-domain No
Bijective Both One-One and Onto Yes
Many-One Different inputs can give same output No

Composition of Functions

Definition: Let $f: A \to B$ and $g: B \to C$ be two functions. The composition of $f$ and $g$, denoted by $gof$, is defined as: $$(gof)(x) = g(f(x)) \text{ for all } x \in A$$
Key Points:
  • Read $gof$ as "g circle f" or "g composed with f"
  • First apply $f$, then apply $g$ to the result
  • $gof \neq fog$ (composition is NOT commutative)
  • $gof$ is defined only if Range of $f \subseteq$ Domain of $g$

Invertible Functions

Definition: A function $f: X \to Y$ is invertible if there exists a function $g: Y \to X$ such that: $$gof = I_X \text{ and } fog = I_Y$$ where $I_X$ and $I_Y$ are identity functions. The function $g$ is called the inverse of $f$ and is denoted by $f^{-1}$.
Theorem: A function is invertible if and only if it is bijective (one-one and onto)
Finding Inverse:
1. Replace $f(x)$ with $y$
2. Solve for $x$ in terms of $y$
3. Replace $y$ with $x$ to get $f^{-1}(x)$
4. Verify: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$

Solved Examples (Board Marking Scheme)

Q1. Show that the relation $R$ in $\mathbb{Z}$ given by $R = \{(a, b): 2$ divides $a - b\}$ is an equivalence relation. (3 Marks)

Step 1: Reflexive 1 Mark

For any $a \in \mathbb{Z}$, we have $a - a = 0 = 2 \times 0$.

Therefore, 2 divides $(a - a)$, so $(a, a) \in R$.

Hence, $R$ is reflexive.

Step 2: Symmetric 1 Mark

Let $(a, b) \in R$. Then 2 divides $(a - b)$.

So, $a - b = 2k$ for some integer $k$.

Therefore, $b - a = -2k = 2(-k)$, where $-k$ is an integer.

Hence, 2 divides $(b - a)$, so $(b, a) \in R$.

Therefore, $R$ is symmetric.

Step 3: Transitive 1 Mark

Let $(a, b) \in R$ and $(b, c) \in R$.

Then 2 divides $(a - b)$ and 2 divides $(b - c)$.

So, $a - b = 2m$ and $b - c = 2n$ for integers $m$ and $n$.

Adding: $(a - b) + (b - c) = 2m + 2n \Rightarrow a - c = 2(m + n)$.

Hence, 2 divides $(a - c)$, so $(a, c) \in R$.

Therefore, $R$ is transitive.

Conclusion: $R$ is reflexive, symmetric, and transitive. Hence, $R$ is an equivalence relation.

Q2. Show that $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = 2x$ is one-one and onto. (3 Marks)

Step 1: One-One 1.5 Marks

Let $f(x_1) = f(x_2)$ where $x_1, x_2 \in \mathbb{R}$.

Then $2x_1 = 2x_2$

Dividing both sides by 2: $x_1 = x_2$

Therefore, $f$ is one-one.

Step 2: Onto 1.5 Marks

Let $y \in \mathbb{R}$ (co-domain) be arbitrary.

We need to find $x \in \mathbb{R}$ such that $f(x) = y$.

From $f(x) = y$, we have $2x = y \Rightarrow x = \frac{y}{2}$.

Since $y \in \mathbb{R}$, we have $\frac{y}{2} \in \mathbb{R}$ (domain).

Therefore, for every $y$ in co-domain, there exists $x = \frac{y}{2}$ in domain such that $f(x) = y$.

Hence, $f$ is onto.

Q3. Let $f: \{2, 3, 4, 5\} \to \{3, 4, 5, 9\}$ and $g: \{3, 4, 5, 9\} \to \{7, 11, 15\}$ be given by $f(2) = 3, f(3) = 4, f(4) = f(5) = 5$ and $g(3) = g(4) = 7, g(5) = g(9) = 11$. Find $gof$. (2 Marks)

Solution: 2 Marks

$(gof)(x) = g(f(x))$

$(gof)(2) = g(f(2)) = g(3) = 7$

$(gof)(3) = g(f(3)) = g(4) = 7$

$(gof)(4) = g(f(4)) = g(5) = 11$

$(gof)(5) = g(f(5)) = g(5) = 11$

Therefore: $gof = \{(2, 7), (3, 7), (4, 11), (5, 11)\}$

Previous Year Questions (PYQs)

2023 (1 Mark MCQ): Let $R$ be a relation on the set $N$ of natural numbers defined by $R = \{(x, y): x + 2y = 8\}$. The range of $R$ is:
(A) $\{1, 2, 3, ...\}$   (B) $\{2, 3\}$   (C) $\{1, 2, 3\}$   (D) $\{1, 2, 3, 4\}$
Ans: (C) $\{1, 2, 3\}$. For $x + 2y = 8$: when $y = 1, x = 6$; when $y = 2, x = 4$; when $y = 3, x = 2$. So range = $\{1, 2, 3\}$.
2022 (2 Marks): Check whether the relation $R$ in the set $\mathbb{R}$ of real numbers defined by $R = \{(a, b): a \leq b^2\}$ is reflexive, symmetric, or transitive.
Ans:
• Not Reflexive: For $a = -1$, we have $-1 \not\leq (-1)^2 = 1$ is false. So $(-1, -1) \notin R$.
• Not Symmetric: $(1, 2) \in R$ since $1 \leq 4$, but $(2, 1) \notin R$ since $2 \not\leq 1$.
• Not Transitive: $(3, 2) \in R$ and $(2, 1.5) \in R$, but $(3, 1.5) \notin R$ since $3 \not\leq 2.25$.
2020 (3 Marks): Show that the function $f: \mathbb{N} \to \mathbb{N}$ given by $f(1) = f(2) = 1$ and $f(x) = x - 1$ for every $x > 2$, is onto but not one-one.
Ans:
• Not One-One: $f(1) = 1 = f(2)$, but $1 \neq 2$. Hence not one-one.
• Onto: For any $y \in \mathbb{N}$, if $y = 1$, then $f(1) = 1$. If $y > 1$, choose $x = y + 1$, then $f(y + 1) = (y + 1) - 1 = y$. Hence onto.
2019 (4 Marks): Let $A = \mathbb{R} - \{3\}$ and $B = \mathbb{R} - \{1\}$. Consider the function $f: A \to B$ defined by $f(x) = \frac{x - 2}{x - 3}$. Show that $f$ is one-one and onto and hence find $f^{-1}$.
Solution Outline:
1. Prove one-one by assuming $f(x_1) = f(x_2)$ and showing $x_1 = x_2$
2. Prove onto by taking arbitrary $y \in B$ and finding $x = \frac{3y - 2}{y - 1} \in A$
3. Since bijective, $f^{-1}$ exists: $f^{-1}(y) = \frac{3y - 2}{y - 1}$

Exam Strategy & Mistake Bank

Common Mistakes

Mistake 1: Confusing "one-one" with "onto". Remember: One-one is about inputs (no two inputs give same output), Onto is about outputs (all outputs are covered).
Mistake 2: Not checking all three properties for equivalence relation. You MUST prove reflexive, symmetric, AND transitive.
Mistake 3: Writing $(gof)(x) = f(g(x))$ instead of $g(f(x))$. First apply $f$, then $g$!
Mistake 4: Assuming $gof = fog$. Composition is NOT commutative!
Mistake 5: For finite sets, forgetting that one-one implies onto and vice versa.

Scoring Tips

Tip 1: For equivalence relation proofs, always use the standard three-step format: Check Reflexive → Check Symmetric → Check Transitive.
Tip 2: When proving one-one, always start with "Let $f(x_1) = f(x_2)$" and end with "$x_1 = x_2$".
Tip 3: For onto proofs, write "Let $y \in$ co-domain be arbitrary" and find corresponding $x$ in domain.
Tip 4: In composition problems, show your work step-by-step: $(gof)(a) = g(f(a)) = g(b) = c$.
Tip 5: For MCQs on relations, quickly check with specific examples rather than general proof.

Practice Problems (Self-Assessment)

Level 1: Basic (1-2 Marks Each)

Q1. Check whether the relation $R$ in $\{1, 2, 3\}$ given by $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}$ is reflexive, symmetric, or transitive.

Answer: Reflexive (all $(a,a)$ present). Not Symmetric ($(1,2) \in R$ but $(2,1) \notin R$). Transitive (Check all pairs, e.g., $(1,2), (2,3) \in R \Rightarrow (1,3) \notin R$, so Not Transitive).

Q2. Show that the function $f: \mathbb{N} \to \mathbb{N}$ given by $f(x) = x^2$ is one-one but not onto.

Answer: One-One: $x_1^2 = x_2^2 \Rightarrow x_1 = x_2$ (since $x \in \mathbb{N}$). Not Onto: $3 \in \mathbb{N}$ has no pre-image in $\mathbb{N}$ (since $\sqrt{3} \notin \mathbb{N}$).

Q3. If $f(x) = x + 1$ and $g(x) = 2x$, find $(gof)(x)$ and $(fog)(x)$. Are they equal?

Answer: $(gof)(x) = g(x+1) = 2(x+1) = 2x+2$.
$(fog)(x) = f(2x) = 2x+1$.
No, they are not equal.

Level 2: Intermediate (3 Marks Each)

Q4. Show that the relation $R$ in the set $A = \{1, 2, 3, 4, 5, 6\}$ given by $R = \{(a, b): |a - b|$ is even$\}$ is an equivalence relation.

Hint:
Reflexive: $|a-a|=0$ (even).
Symmetric: $|a-b| = |b-a|$.
Transitive: Sum of two even numbers is even. $|a-b|$ even, $|b-c|$ even $\Rightarrow |a-c|$ even.

Q5. Show that the function $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^3$ is bijective.

Hint:
One-One: $x_1^3 = x_2^3 \Rightarrow x_1 = x_2$.
Onto: For any $y \in \mathbb{R}$, $x = y^{1/3} \in \mathbb{R}$.

Level 3: Advanced (4-5 Marks Each)

Q6. Let $f: \mathbb{R} - \{-\frac{4}{3}\} \to \mathbb{R}$ be a function defined as $f(x) = \frac{4x}{3x + 4}$. Show that $f$ is one-one. Find the inverse of $f$.

Hint: For inverse, let $y = \frac{4x}{3x + 4}$ and solve for $x$.

Answer: $f^{-1}(y) = \frac{4y}{4-3y}$.

Q7. Consider $f: \{1, 2, 3\} \to \{a, b, c\}$ and $g: \{a, b, c\} \to \{apple, ball, cat\}$ defined as $f(1) = a, f(2) = b, f(3) = c$ and $g(a) = apple, g(b) = ball, g(c) = cat$. Show that $f, g$ and $gof$ are invertible. Find $(gof)^{-1}$.

Answer: $(gof)^{-1} = \{(apple, 1), (ball, 2), (cat, 3)\}$.

Formula Sheet (Must Remember!)

Relations

1. Empty Relation: $R = \phi \subset A \times A$
2. Universal Relation: $R = A \times A$
3. Reflexive: $(a, a) \in R$ for all $a \in A$
4. Symmetric: $(a, b) \in R \Rightarrow (b, a) \in R$
5. Transitive: $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$
6. Equivalence Relation: Reflexive + Symmetric + Transitive

Functions

7. One-One: $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$
8. Onto: Range = Co-domain
9. Bijective: One-One + Onto
10. Composition: $(gof)(x) = g(f(x))$
11. Invertible: $f$ is invertible $\Leftrightarrow$ $f$ is bijective
12. Inverse Property: $(gof)^{-1} = f^{-1}og^{-1}$
13. Identity: $fof^{-1} = I_Y$ and $f^{-1}of = I_X$

Key Theorems:
• Intersection of two equivalence relations is also an equivalence relation.
• For finite set $A$, $f: A \to A$ is one-one $\Leftrightarrow$ $f$ is onto.
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