Class 12 — Mathematics (NCERT)

Relations and Functions

Class 12

✓By the end of this lesson students will be able to define and identify different types of relations (reflexive, symmetric, transitive, equivalence).
✓By the end of this lesson students will be able to define and identify different types of functions (one-one, onto, bijective).
✓By the end of this lesson students will be able to find the composition of two functions.
✓By the end of this lesson students will be able to determine if a function is invertible and find its inverse.
✓By the end of this lesson students will be able to solve problems involving types of relations, types of functions, composition of functions, and inverse of functions.

Key concepts

Relation

A relation R from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B. If (a, b) ∈ R, we say that 'a is related to b'. When A = B, we speak of a relation on A.

Types of Relations

Let R be a relation on a set A:\n1. Reflexive: R is reflexive if (a, a) ∈ R for every a ∈ A.\n2. Symmetric: R is symmetric if (a, b) ∈ R implies (b, a) ∈ R, for all a, b ∈ A.\n3. Transitive: R is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R, for all a, b, c ∈ A.\n4. Equivalence Relation: A relation R is an equivalence relation if it is reflexive, symmetric, and transitive.

Function

A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. We write f: A → B. A is the domain, B is the codomain, and the set of all images of elements of A is the range.

Types of Functions

Let f: A → B be a function:\n1. One-one (Injective): A function f: A → B is said to be one-one if distinct elements of A have distinct images in B. That is, for all x₁, x₂ ∈ A, f(x₁) = f(x₂) implies x₁ = x₂.\n2. Onto (Surjective): A function f: A → B is said to be onto if every element of B is the image of some element of A under f. That is, for every y ∈ B, there exists an x ∈ A such that f(x) = y. The range of f is equal to the codomain B.\n3. Bijective: A function f: A → B is said to be bijective if it is both one-one and onto.

Composition of Functions

Let f: A → B and g: B → C be two functions. Then the composition of f and g, denoted by gof, is defined as the function gof: A → C given by (gof)(x) = g(f(x)), for all x ∈ A.

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

Inverse of a Function

A function f: A → B is said to be invertible if there exists a function g: B → A such that gof = I_A and fog = I_B, where I_A and I_B are identity functions on sets A and B respectively. The function g is called the inverse of f and is denoted by f⁻¹. A function is invertible if and only if it is bijective (one-one and onto).

f⁻¹(y) = x if f(x) = y

Key facts to remember

1An empty relation (ϕ) on a non-empty set A is always symmetric and transitive, but never reflexive.
2The identity relation I_A = {(a, a) : a ∈ A} on a set A is always an equivalence relation.
3A function f: A → B is invertible if and only if it is bijective (both one-one and onto).
4If f: A → B and g: B → C are two bijective functions, then their composition gof: A → C is also bijective.
5If f: A → B and g: B → C are two functions, then (gof)⁻¹ = f⁻¹og⁻¹.
6The composition of functions is associative, i.e., ho(gof) = (hog)of, whenever the compositions are defined.

Worked examples

Example 1

Let R be a relation in the set A = {1, 2, 3, ..., 14} defined as R = {(x, y) : 3x - y = 0}. Determine if R is reflexive, symmetric, and transitive.

IFirst, let's list the elements of the relation R:\nGiven 3x - y = 0, so y = 3x.\nFor x = 1, y = 3 ⇒ (1, 3) ∈ R\nFor x = 2, y = 6 ⇒ (2, 6) ∈ R\nFor x = 3, y = 9 ⇒ (3, 9) ∈ R\nFor x = 4, y = 12 ⇒ (4, 12) ∈ R\n(For x = 5, y = 15, which is not in A, so we stop here).\nSo, R = {(1, 3), (2, 6), (3, 9), (4, 12)}.

IICheck for Reflexivity:\nA relation R on set A is reflexive if (a, a) ∈ R for every a ∈ A.\nHere, for a = 1, (1, 1) ∉ R (since 3(1) - 1 = 2 ≠ 0).\nThus, R is not reflexive.

IIICheck for Symmetry:\nA relation R on set A is symmetric if (a, b) ∈ R implies (b, a) ∈ R.\nWe have (1, 3) ∈ R.\nFor R to be symmetric, (3, 1) must be in R. But 3(3) - 1 = 9 - 1 = 8 ≠ 0, so (3, 1) ∉ R.\nThus, R is not symmetric.

IVCheck for Transitivity:\nA relation R on set A is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R.\nWe have (1, 3) ∈ R and (3, 9) ∈ R.\nFor R to be transitive, (1, 9) must be in R. But 3(1) - 9 = 3 - 9 = -6 ≠ 0, so (1, 9) ∉ R.\nThus, R is not transitive.

Answer

The relation R is neither reflexive, nor symmetric, nor transitive.

To prove a relation is NOT reflexive, symmetric, or transitive, it is sufficient to provide a single counterexample.

Example 2

Show that the function f: N → N, given by f(x) = 2x, is one-one but not onto.

ICheck for One-one (Injectivity):\nLet x₁, x₂ ∈ N such that f(x₁) = f(x₂).\nThen 2x₁ = 2x₂.\nDividing by 2, we get x₁ = x₂.\nSince f(x₁) = f(x₂) implies x₁ = x₂, the function f is one-one.

IICheck for Onto (Surjectivity):\nA function f: A → B is onto if for every y ∈ B, there exists an x ∈ A such that f(x) = y.\nHere, the codomain is N (natural numbers).\nConsider an element y = 1 in the codomain N.\nWe need to find an x ∈ N such that f(x) = 1, i.e., 2x = 1.\nThis gives x = 1/2.\nHowever, 1/2 ∉ N (the domain).\nSince there is an element (e.g., 1) in the codomain that has no pre-image in the domain, the function f is not onto.

Answer

The function f: N → N, given by f(x) = 2x, is one-one but not onto. Hence proved.

When proving a function is not onto, it is crucial to show an element in the codomain that has no pre-image in the domain.

Example 3

Let f: R → R be defined by f(x) = 4x + 3. Show that f is invertible and find f⁻¹.

ITo show f is invertible, we must prove it is bijective (one-one and onto).

IICheck for One-one:\nLet x₁, x₂ ∈ R such that f(x₁) = f(x₂).\nThen 4x₁ + 3 = 4x₂ + 3.\nSubtracting 3 from both sides: 4x₁ = 4x₂.\nDividing by 4: x₁ = x₂.\nThus, f is one-one.

IIICheck for Onto:\nLet y be an arbitrary element in the codomain R.\nWe need to find an x ∈ R (domain) such that f(x) = y.\nSo, 4x + 3 = y.\n4x = y - 3.\nx = (y - 3) / 4.\nSince y ∈ R, (y - 3) / 4 will also be a real number. So, x ∈ R.\nThus, for every y in the codomain, there exists an x in the domain such that f(x) = y. Hence, f is onto.

IVSince f is both one-one and onto, it is a bijective function, and therefore, it is invertible.

VTo find f⁻¹:\nLet y = f(x).\nSo, y = 4x + 3.\nWe need to express x in terms of y.\n4x = y - 3\nx = (y - 3) / 4.\nNow, replace y with x to get the expression for f⁻¹(x).\nf⁻¹(x) = (x - 3) / 4.

Answer

The function f is invertible, and its inverse is f⁻¹(x) = (x - 3) / 4.

Always verify that the calculated 'x' belongs to the domain of the original function when proving 'onto'.

Common mistakes

✗Confusing the conditions for reflexive, symmetric, and transitive relations, especially when providing counterexamples.
✗Assuming a function is onto without explicitly showing that every element in the codomain has a pre-image in the domain.
✗Incorrectly finding the inverse of a function without first verifying that it is bijective.
✗Mistaking the order of composition, i.e., assuming fog is the same as gof (they are generally not).
✗Not specifying the domain and codomain when defining a function or its inverse, which can lead to incorrect conclusions about its type or invertibility.

Exam tips

★Always state the domain and codomain of relations and functions clearly at the beginning of your solution.
★For proving a function is one-one, start with f(x₁) = f(x₂) and algebraically show that x₁ = x₂.
★For proving a function is onto, let 'y' be an arbitrary element in the codomain and find 'x' in the domain such that f(x) = y. Ensure this 'x' is valid in the domain.
★When checking for types of relations, if a property fails, provide a specific counterexample with elements from the given set.
★Practice finding the inverse of various types of functions, especially those involving algebraic manipulation, and always verify your inverse by checking f(f⁻¹(x)) = x and f⁻¹(f(x)) = x.

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