Class 12 Relations & Functions Notes

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 - Simply Explained

Thinking Strategy: Imagine the set $A$ is a group of people.

1. Reflexive (Selfie)

Cue: "Everyone takes a selfie."

Every single person in the group must be related to themselves.

$(a, a) \in R$ for ALL $a$

2. Symmetric (Friends)

Cue: "If I choose you, you choose me."

If A is related to B, then B must be related back to A.

$(a, b) \in R \Rightarrow (b, a) \in R$

3. Transitive (Chain)

Cue: "Pass it on."

If A connects to B, and B connects to C, does A connect to C?

$(a, b), (b, c) \in R \Rightarrow (a, c) \in R$

Official Definition (For Exams):

Reflexive: $(a, a) \in R$ for every $a \in A$.
Symmetric: If $(a_1, a_2) \in R$, then $(a_2, a_1) \in R$.
Transitive: If $(a_1, a_2) \in R$ and $(a_2, a_3) \in R$, then $(a_1, a_3) \in R$.

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 $[a] = \{x \in A : (x, a) \in R\}$.

Important Properties of Equivalence Classes:

All elements of $A$ related to each other belong to the same equivalence class.
No element of one equivalence class is related to any element of another equivalence class.
Two equivalence classes are either identical or disjoint (no overlapping elements).
The union of all equivalence classes gives the original set $A$ (Partition of a set).

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 - Simply Explained

Thinking Strategy: Imagine Domain (Set X) are Passengers and Co-domain (Set Y) are Bus Seats.

1. One-One (Injective)

Cue: "Reserved Seating" • Every element has a unique image ($f(x_1) = f(x_2) \Rightarrow x_1 = x_2$).

Distinct elements → Distinct images

2. Onto (Surjective)

Cue: "Full House" • Every element in Co-domain is mapped ($\text{Range} = \text{Co-domain}$).

No unmapped elements in Co-domain

Bijective Function:

It's a "Perfect Match"! Everyone has a unique seat (One-One) AND all seats are full (Onto).

Only Bijective functions have an Inverse ($f^{-1}$).

One-One Examples

• $f(x) = 2x$ (Linear is usually 1-1)

• $f(x) = x^3$ (Odd powers usually 1-1)

Not One-One (Many-One)

• $f(x) = x^2$ (Parabola: Two inputs give same output)

• $f(x) = |x|$ (Modulus: Positive & negative give same result)

• $f(x) = \sin x$ (Periodic functions repeat values)

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 $f^{-1}(f(x)) = x$

Properties of Inverse and Composition:

Reversal Rule: $(gof)^{-1} = f^{-1}og^{-1}$
The inverse of a bijection is also a bijection.
$(f^{-1})^{-1} = f$

How to Solve Problems (Step-by-Step)

Proving Relations (The 3-Step Check)

Step 1 (Reflexive?):
Check if $(a,a)$ works for every $a$.
Pass: Yes, for all.
Fail: Find just ONE example where it fails.
Step 2 (Symmetric?):
Assume $(a,b) \in R$. Check if $(b,a) \in R$.
Pass: If equation stays same after swapping.
Step 3 (Transitive?):
Assume $(a,b)$ and $(b,c) \in R$. Check $(a,c)$.
Pass: If you can derive a relation between a & c.

Proving Functions (The 2-Step Proof)

Step 1 (One-One?):
Start with: Let $f(x_1) = f(x_2)$
Solve the equation.
Pass: If you get ONLY $x_1 = x_2$.
Fail: If you get $x_1 = \pm x_2$ etc.
Step 2 (Onto?):
Put $f(x) = y$.
Find $x$ in terms of $y$.
Pass: If this $x$ is valid in Domain for ALL $y$.

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.

Step 1: Reflexive? (Selfie Check)

We need $(1,1), (2,2), (3,3)$.

Are they present? Yes.

$\therefore$ Reflexive.

Step 2: Symmetric? (Friend Check)

We have $(1,2)$. Is $(2,1)$ present? No.

We have $(2,3)$. Is $(3,2)$ present? No.

$\therefore$ Not Symmetric.

Step 3: Transitive? (Chain Check)

We have $(1,2)$ and $(2,3)$.

Transitive requires $(1,3)$ to be present.

Is $(1,3) \in R$? No.

$\therefore$ 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.

Step 1: Check One-One

Let $f(x_1) = f(x_2)$

$\Rightarrow x_1^2 = x_2^2$

$\Rightarrow x_1 = x_2$ (Since $x \in \mathbb{N}$, no negative numbers allowed)

$\therefore$ One-One.

Step 2: Check Onto

Co-domain is $\mathbb{N} = \{1, 2, 3, 4, ...\}$.

Range is $\{1^2, 2^2, 3^2, ...\} = \{1, 4, 9, ...\}$.

Are numbers like 2 or 3 in the range? No.

$\therefore$ Range $\neq$ Co-domain.

$\therefore$ Not Onto.

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

Step 1: Find $(gof)(x)$

$(gof)(x) = g(f(x)) = g(x + 1)$

$= 2(x + 1) = 2x + 2$

Step 2: Find $(fog)(x)$

$(fog)(x) = f(g(x)) = f(2x)$

$= (2x) + 1 = 2x + 1$

Step 3: Compare

$2x + 2 \neq 2x + 1$

$\therefore$ 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.

Step 1: Reflexive

$|a - a| = |0| = 0$, which is even.

$\therefore (a, a) \in R$. Reflexive.

Step 2: Symmetric

Let $|a - b|$ be even.

$|b - a| = |-(a - b)| = |a - b|$, which is also even.

$\therefore (b, a) \in R$. Symmetric.

Step 3: Transitive

Let $|a - b|$ (even) and $|b - c|$ (even).

Sum of two even numbers is even: $(a - b) + (b - c) = a - c$.

So $a - c$ is even $\Rightarrow |a - c|$ is even.

$\therefore (a, c) \in R$. Transitive.

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

Step 1: Check One-One

Let $x_1^3 = x_2^3$. Taking cube root on both sides.

$x_1 = x_2$. (No $\pm$ issue for odd powers)

$\therefore$ One-One.

Step 2: Check Onto

Let $y \in \mathbb{R}$. We need $x^3 = y \Rightarrow x = y^{1/3}$.

Is cube root of any real number real? Yes (e.g., $\sqrt[3]{-8} = -2$).

$\therefore$ Range = $\mathbb{R}$. Onto.

Conclusion

Since One-One & Onto, it is Bijective.

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 $f$.

Step 1: Show One-One

Let $f(x_1) = f(x_2)$

$\frac{4x_1}{3x_1 + 4} = \frac{4x_2}{3x_2 + 4}$

Cross-multiply: $4x_1(3x_2 + 4) = 4x_2(3x_1 + 4)$

$12x_1x_2 + 16x_1 = 12x_1x_2 + 16x_2$

$16x_1 = 16x_2 \Rightarrow x_1 = x_2$

$\therefore$ One-One.

Step 2: Find Inverse ($f^{-1}$)

Let $y = \frac{4x}{3x + 4}$. We need to find $x$.

$y(3x + 4) = 4x$

$3xy + 4y = 4x$

$4y = 4x - 3xy = x(4 - 3y)$

$x = \frac{4y}{4 - 3y}$

$\therefore$ Inverse is $f^{-1}(y) = \frac{4y}{4 - 3y}$.

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 $(gof)^{-1}$.

Step 1: Check Invertibility

For $f$: 1-to-a, 2-to-b, 3-to-c (Bijective) $\Rightarrow$ Invertible.

For $g$: a-to-apple, b-to-ball, c-to-cat (Bijective) $\Rightarrow$ Invertible.

Step 2: Find $gof$

$(gof)(1) = g(f(1)) = g(a) = apple$

$(gof)(2) = g(f(2)) = g(b) = ball$

$(gof)(3) = g(f(3)) = g(c) = cat$

$\therefore gof$ is also Bijective $\Rightarrow$ Invertible.

Step 3: Find Inverse $(gof)^{-1}$

Just reverse the arrows:

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

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.

📝 10-Mark Self-Assessment Test

Time: 25 Mins

Solve these high-frequency CBSE Board questions on paper, then reveal step-by-step solutions to check your marks.

Q1 (1 Mark - MCQ): Let $R$ be the relation in the set $\{1, 2, 3, 4\}$ given by $R = \{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\}$. Then $R$ is:
(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

Q2 (2 Marks - VSA): Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = x^4$. Check whether $f$ is injective (one-one) or surjective (onto). Give reasons.

Q3 (3 Marks - SA): Show that the relation $R$ in the set $A = \{x \in \mathbb{Z} : 0 \le x \le 12\}$, given by $R = \{(a, b) : |a - b| \text{ is a multiple of } 4\}$ is an equivalence relation. Find the set of all elements related to 1.

Q4 (4 Marks - LA): Show that the function $f: \mathbb{R} - \{3\} \to \mathbb{R} - \{1\}$ defined by $f(x) = \frac{x - 2}{x - 3}$ is one-one and onto (bijective). Hence, find $f^{-1}(x)$.

Sol 1:
- $(1,1), (2,2), (3,3), (4,4) \in R \implies$ Reflexive.
- $(1,2) \in R$ but $(2,1) \notin R \implies$ Not symmetric.
- For any $(a,b), (b,c) \in R$, $(a,c) \in R$ holds $\implies$ Transitive.
Correct Option: (b) reflexive and transitive but not symmetric [1 Mark]

Sol 2: $f: \mathbb{R} \to \mathbb{R}$, $f(x) = x^4$.
- One-One: $f(1) = 1^4 = 1$ and $f(-1) = (-1)^4 = 1$. Since $f(1) = f(-1)$ but $1 \neq -1$, $f$ is not one-one [1 Mark].
- Onto: Consider $-2 \in \mathbb{R}$ (co-domain). Since $x^4 \ge 0$ for all real $x$, there is no $x \in \mathbb{R}$ such that $x^4 = -2$. Range $\neq$ Co-domain, so $f$ is not onto [1 Mark].
Conclusion: $f$ is neither one-one nor onto. [2 Marks]

Sol 3:
1. Reflexive: For any $a \in A$, $|a - a| = 0 = 0 \times 4$ (multiple of 4) $\implies (a,a) \in R$ [0.5 Mark].
2. Symmetric: Let $(a,b) \in R \implies |a - b| = 4k \implies |b - a| = |-(a - b)| = |a - b| = 4k \implies (b,a) \in R$ [0.5 Mark].
3. Transitive: Let $(a,b) \in R$ and $(b,c) \in R \implies a - b = \pm 4k_1$ and $b - c = \pm 4k_2$. Adding: $a - c = \pm 4(k_1 \pm k_2) \implies |a - c|$ is a multiple of 4 $\implies (a,c) \in R$ [1 Mark].
Hence $R$ is an equivalence relation.
4. Equivalence class of 1: $\{x \in A : |x - 1| \text{ is a multiple of } 4\} = \{1, 5, 9\}$ [1 Mark].
Answer: Set of elements related to 1 is $\{1, 5, 9\}$. [3 Marks]

Sol 4:
- One-One: Let $f(x_1) = f(x_2) \implies \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}$
$$(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$$ $$x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6 \implies -x_1 = -x_2 \implies x_1 = x_2 \text{ [1.5 Marks]}$$ - Onto: Let $y \in \mathbb{R} - \{1\}$ be arbitrary. Set $y = \frac{x - 2}{x - 3}$.
$$y(x - 3) = x - 2 \implies xy - 3y = x - 2 \implies x(y - 1) = 3y - 2 \implies x = \frac{3y - 2}{y - 1}$$ Since $y \neq 1$, $x$ exists in $\mathbb{R}$. Also if $x = 3 \implies 3y - 2 = 3y - 3 \implies -2 = -3$ (impossible), so $x \neq 3$. Hence $x \in \text{Domain}$ for all $y \in \text{Co-domain}$. Thus $f$ is onto [1.5 Marks].
- Inverse: $f^{-1}(x) = \frac{3x - 2}{x - 1}$ [1 Mark].
Answer: $f$ is bijective and $f^{-1}(x) = \frac{3x - 2}{x - 1}$. [4 Marks]

Chapter 1: Relations and Functions

Overview

Exam Weightage & Blueprint

Last 24-Hour Checklist

Types of Relations

Basic Definitions

Empty Relation

Universal Relation

Equivalence Relation

Properties of Relations - Simply Explained

1. Reflexive (Selfie)

2. Symmetric (Friends)

3. Transitive (Chain)

Equivalence Relation

Equivalence Class $[a]$

Types of Functions - Simply Explained

1. One-One (Injective)

2. Onto (Surjective)

One-One Examples

Not One-One (Many-One)

Quick Reference: Function Types

Composition of Functions

Invertible Functions

How to Solve Problems (Step-by-Step)

Proving Relations (The 3-Step Check)

Proving Functions (The 2-Step Proof)

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)

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

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)

Previous Year Questions (PYQs)

Exam Strategy & Mistake Bank

Common Mistakes

Scoring Tips

Practice Problems (Self-Assessment)

Level 1: Basic (1-2 Marks Each)

Level 2: Intermediate (3 Marks Each)

Level 3: Advanced (4-5 Marks Each)

Formula Sheet (Must Remember!)

Relations

Functions

📝 10-Mark Self-Assessment Test