Class 12 Maths Chapter Notes: Relations And Functions

Key Definitions

Relation (R)

A subset of the Cartesian product A × B, where A and B are two non-empty sets. It defines a relationship between elements of A and B.

Function (f)

A special type of relation from set A to set B (denoted f: A → B) such that every element of set A is associated with exactly one element of set B.

Domain

The set of all first elements (inputs) of the ordered pairs in a relation or function. For f: A → B, the domain is set A.

Codomain

The set B in a function f: A → B. It's the set of all possible output values.

Range

The set of all second elements (outputs) of the ordered pairs in a relation or function. It is a subset of the codomain.

Inverse of a Function (f⁻¹)

If f: A → B is a bijective function, then its inverse f⁻¹: B → A exists such that for every y ∈ B, f⁻¹(y) = x if and only if f(x) = y.

Binary Operation (*)

A function : A × A → A, which assigns to each ordered pair (a, b) of elements in A a unique element a b in A.

Types of Relations: Reflexivity, Symmetry, Transitivity, and Equivalence

Understanding the properties of relations is fundamental. Let R be a relation on a set A (i.e., R ⊆ A × A).

1. Reflexive Relation: A relation R on set A is said to be reflexive if every element of A is related to itself. Mathematically, for all a ∈ A, (a, a) ∈ R. For example, "is equal to" (=) is a reflexive relation on any set of numbers because a = a for all 'a'.

1. Symmetric Relation: A relation R on set A is said to be symmetric if whenever an element 'a' is related to an element 'b', then 'b' is also related to 'a'. Mathematically, for all a, b ∈ A, if (a, b) ∈ R, then (b, a) ∈ R. For instance, "is a sister of" is symmetric if we consider a set of females, as if A is a sister of B, then B is a sister of A. However, on a mixed-gender set, it's not symmetric.

1. Transitive Relation: A relation R on set A is said to be transitive if whenever an element 'a' is related to 'b', and 'b' is related to 'c', then 'a' is also related to 'c'. Mathematically, for all a, b, c ∈ A, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R. The relation "is less than" (<) on the set of real numbers is transitive: if a < b and b < c, then a < c.

1. Equivalence Relation: A relation R on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive. Equivalence relations partition the set A into disjoint subsets called equivalence classes. Each element belongs to exactly one equivalence class. A classic example is the relation "is congruent to" on the set of triangles.

Types of Functions

Composition of Functions and Inverse of a Function

Composition of Functions

If f: A → B and g: B → C are two functions, then the composition of f and g, denoted by g o f (read as 'g of f'), is a function from A to C defined as (g o f)(x) = g(f(x)) for all x ∈ A.

• Key Condition: The codomain of the first function (f) must be equal to the domain of the second function (g) for g o f to be defined.
• Order Matters: In general, f o g ≠ g o f. Composition of functions is not necessarily commutative.
• Associativity: Function composition is always associative: h o (g o f) = (h o g) o f.

Inverse of a Function

A function f: X → Y is defined to be invertible if there exists a function g: Y → X such that g o f = Iₓ (identity function on X) and f o g = Iᵧ (identity function on Y). The function g is called the inverse of f and is denoted by f⁻¹.

• Existence Condition: A function is invertible if and only if it is bijective (both one-to-one and onto).
• Uniqueness: The inverse of a bijective function, if it exists, is unique.
• Finding Inverse: To find f⁻¹(x), set y = f(x) and solve for x in terms of y. Then swap x and y. For example, if f(x) = 2x+3, let y = 2x+3. Then 2x = y-3, so x = (y-3)/2. Thus, f⁻¹(y) = (y-3)/2 or f⁻¹(x) = (x-3)/2.

Binary Operations

Worked Example: Checking Function Types

• {"title":"Example: Check if f(x) = x² is one-to-one and onto for f: R → R.","bodyMarkdown":"One-to-one (Injectivity):\nLet f(x₁) = f(x₂).\nx₁² = x₂²\n√(x₁²) = √(x₂²)\n|x₁| = |x₂|\nThis implies x₁ = ±x₂. For example, f(2) = 4 and f(-2) = 4, but 2 ≠ -2. So, f is not one-to-one.\n\nOnto (Surjectivity):\nThe range of f(x) = x² for x ∈ R is [0, ∞), which means f(x) can only produce non-negative real numbers. However, the codomain is R (all real numbers). Since the range [0, ∞) is not equal to the codomain R (e.g., -5 is in R but not in the range), f is not onto."}

Exam Tip for Relations and Functions

When proving a function is one-to-one, always start by assuming f(x₁) = f(x₂) and algebraically show that x₁ must equal x₂. For onto, take an arbitrary element 'y' from the codomain and show that there exists an 'x' in the domain such that f(x) = y. Often, this involves solving for x in terms of y. For relations, carefully check the definitions for all elements or for specific counterexamples.

Key Points to Remember

• A relation R on set A is an equivalence relation if and only if it is reflexive, symmetric, and transitive.
• To prove a function f: A → B is one-to-one, show f(x₁) = f(x₂) ⇒ x₁ = x₂.
• To prove a function f: A → B is onto, show that for every y ∈ B, there exists x ∈ A such that f(x) = y (i.e., Range = Codomain).
• An inverse function f⁻¹ exists if and only if the function f is bijective (both one-to-one and onto).
• The composition (g o f)(x) = g(f(x)) is defined only if the range of f is a subset of the domain of g.
• For binary operations, always check the properties (commutativity, associativity, identity, inverse) against the definition for ALL elements in the given set.
• The identity element for addition is 0, and for multiplication is 1.
• If a function is defined on specific domains (e.g., N, Z, R), its injectivity and surjectivity can change. Always pay attention to the given domain and codomain.

Practice Questions with Solutions

• Q1: If R is a relation on set A = {1, 2, 3} defined by R = {(1,1), (2,2), (3,3), (1,2), (2,1)}, is R an equivalence relation? A1: No. R is reflexive and symmetric, but it is not transitive. For instance, (1,2) ∈ R and (2,1) ∈ R, but if it were transitive, (1,1) would imply that the original (1,2) and (2,1) form a transitive chain that ends in (1,1) which is present. However, consider if it had (1,2) and (2,3) but not (1,3). Here, (1,2) and (2,1) does not break transitivity because (1,1) is present. Let's re-evaluate: (1,2) ∈ R, (2,1) ∈ R. For transitivity, (1,1) must be in R (which it is). Also (2,1) ∈ R, (1,2) ∈ R. For transitivity, (2,2) must be in R (which it is). So, R IS an equivalence relation. My initial thought process was incorrect for the specific example.
• Q2: Give an example of a function that is one-to-one but not onto from R to R. A2: f(x) = eˣ. It is one-to-one (since eˣ₁ = eˣ₂ ⇒ x₁ = x₂) but not onto because the range is (0, ∞), which is a proper subset of the codomain R.
• Q3: For a binary operation defined as a b = a + 2b on Z, is it commutative? A3: No. For example, 1 2 = 1 + 2(2) = 5. But 2 1 = 2 + 2(1) = 4. Since 5 ≠ 4, the operation is not commutative.
• Q4: When does the inverse of a function f: A → B exist? A4: The inverse of a function f: A → B exists if and only if the function f is bijective, meaning it is both one-to-one (injective) and onto (surjective).

Frequently Asked Questions