Chapter 1: Relations and Functions - Board Exam 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

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

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

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

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

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

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.

Important Theorems & Results

Theorem 1: If $R_1$ and $R_2$ are equivalence relations in a set $A$, then $R_1 \cap R_2$ is also an equivalence relation.

Theorem 2: A function $f: X \to Y$ is invertible if and only if $f$ is bijective (one-one and onto).

Important Results:
1. For finite sets $X$: If $f: X \to X$, then $f$ is one-one ⟺ $f$ is onto
2. Empty relation and universal relation are called trivial relations
3. If $f: A \to B$ and $g: B \to C$, then $(gof)^{-1} = f^{-1}og^{-1}$
4. Identity function is always bijective

Quick Reference Table

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

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.

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

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

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

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

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 $(gof)^{-1}$.

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 ⟺ $f$ is bijective
12. Inverse Property: $(gof)^{-1} = f^{-1}og^{-1}$
13. Identity: $fof^{-1} = I_Y$ and $f^{-1}of = I_X$