# Vidyarthi Academy

Home NCERT Solutions Chapter Notes Test Papers Contact Us

CBSE NOTES CLASS 12 MATHEMATICS CHAPTER 1

RELATIONS AND FUNCTIONS

Cartesian Product

Given two non-empty sets P and Q, the cartesian product P × Q is the set of all ordered pairs of elements from P and Q, i.e.,

P × Q = { (p,q) : p ∈ P, q ∈ Q }

If either P or Q is the null set, then P × Q will also be empty set, or, P × Q = Φ

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. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B. The second element is called the image of the first element.

Types of Relations

1. Empty or Void Relation (Φ)

A relation R in a set A is called empty relation, if no element of A is related to any element of A, i.e., R = Φ ⊂ A × A.

2. Universal Relation

A relation R in a set A is called universal relation, if each element of A is related to every element of A, i.e., R = A × A.

Both the empty relation and the universal relation are called trivial relations.

3. Identity Relation

The relation IA = {(a, a) : a ∈ A} is called the identity relation on A.

4. Reflexive Relation

A relation R is said to be reflexive, if every element of A is related to itself.

Thus, (a, a) ∈ R, ∀ a ∈ A = R is reflexive.

5. Symmetric Relation

A relation R is said to be symmetric, iff

(a, b) ∈ R ⇒ (b, a) ∈ R, ∀ a, b ∈ A

i.e., a R b ⇒ b R a, ∀ a, b ∈ A

6. Transitive Relation

A relation R is said to be transitive, iff (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A

7. Anti-symmetric Relation:

A binary relation R on a set X is anti-symmetric iff for all a and b in X

if R(a,b) and R(b,a), then a = b,

or, equivalently,

if R(a,b) with a ≠ b, then R(b,a) must not hold.

8. Equivalence Relation

A relation R is said to be an equivalence relation if it is reflexive, symmetric and transitive on A.

9. Inverse Relation

If A and B are two non-empty sets and R be a relation from A to B, such that R = {(a, b) : a ∈ A, b ∈ B}, then the inverse of R, denoted by R-1, is a relation from B to A and is defined by

R-1 = {(b, a) : (a, b) ∈ R}

Equivalence Classes of an Equivalence Relation

Let R be equivalence relation in A (≠ Φ). Let a ∈ A, then, the equivalence class of a denoted by [a] or {a} is defined as the set of all those points of A which are related to ‘a’ under the relation R.

Given an arbitrary equivalence relation R in an arbitrary set X, R divides X into mutually disjoint subsets Ai called partitions or subdivisions of X satisfying:

(i) all elements of Ai are related to each other, for all i.

(ii) no element of Ai is related to any element of Aj , i ≠ j.

(iii) ∪ Aj = X and Ai ∩ Aj = Φ, i ≠ j.

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.

In other words, a function f is a relation from a non-empty set A to a non-empty set B such that the domain of f is A and no two distinct ordered pairs in f have the same first element.

Domain

The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of the relation R (it is not always A)

Codomain

The set of all second elements in a relation R from a set A to a set B is called the range of the relation R. The whole set B is called the codomain of the relation R. Note that range ⊆ codomain.

Thus, domain of R = {a : (a , b) ∈ R} and range of R = {b : (a, b) ∈ R}

• A relation may be represented algebraically either by the Roster method or by the Set-builder method.

• An arrow diagram is a visual representation of a relation.

To find domain of a function

Case 1: If the rational function is of the form $\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}$. The denominator g(x) should not be zero.

Case 2: If the function involves square roots, then the expression within the square root should not be negative.

Solve the equations and write the answer in interval notation.

To find range of a function

(i) Write y = f (x)

(ii) Solve x in terms of y, let x = g(y)

(iii) Find the values of y for which the values of x are real and in the domain of f.

(iv) This set is the range of f (x).

In case of complicated functions, it is better to draw graph of the function

Different Types of Functions

One-One and Many-One Function

A function f : X → Y is defined to be one-one (or injective), if the images of distinct elements of X under f are distinct, i.e., for every x1, x2 ∈ X, f (x1) = f (x2) implies x1 = x2. Otherwise, f is called many-one.

Onto Functions

A function f : X → Y is said to be onto (or surjective), if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an element x in X such that f (x) = y.

f : X → Y is onto if and only if Range of f = Y.

One-one and onto (bijective) functions

A function f : X → Y is said to be one-one and onto (or bijective), if f is both one-one and onto.

Methods to Test One-One

(i) Analytically, if x1, x2 ∈ A, then,

f(x1) = f(x2) => x1 = x2

or equivalently

x1 ≠ x2 => f(x1) ≠ f(x2)

(ii) Graphically, if any line parallel to x-axis cuts the graph of the function atmost at one point, then the function is one-one.

(iii) Monotonically, any function f(x), which is strictly increasing or decreasing in the entire domain, is one-one.

To find if a function is Onto

• Range of f(x) is equal to the domain of y

• When the range of the function is equal to codomain of the function then function is said to be onto or surjective function and if range is completely a subset of codomain then it is said to be into function.

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)), ∀ x ∈ A.

Example - Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as f (2) = 3, f (3) = 4, f(4) = f (5) = 5 and g (3) = g (4) = 7 and g (5) = g (9) = 11.

Then

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 (5) = 11.

Important Points to be Remembered

(i) If f and g are injective, then fog and gof are injective.

(ii) If f and g are surjective, then fog is surjective.

(iii) If f and g are bijective, then fog is bijective.

(iv) If f : X → Y is a bijective function then that there exists a function g : Y → X such that gof = Ix and fog = Iy.

(v) If f : X → Y is a function such that there exists a function g : Y → X such that gof = Ix and fog = Iy, then f must be one-one and onto.

Inverse of a Function

(i) A function f : X → Y is defined to be invertible, if there exists a function g : Y → X such that,

gof = Ix

and fog = Iy.

The function g is called the inverse of f and is denoted by f –1.

(ii) Thus, if f is invertible, then f must be one-one and onto and conversely, if f is one-one and onto, then f must be invertible, i.e., a function f : X → Y is invertible if and only if f is a bijective function.

(iii) If f : X → Y, g : Y → Z and h : Z → S are functions, then

h o (g o f) = (h o g) o f.

(iv) Let f : X → Y and g : Y → Z be two invertible functions. Then g o f is also invertible with (g o f)–1 = f –1 o g–1.

Example:

Let f : N → Y be a function defined as f (x) = 4x + 3, where, Y = {y N: y = 4x + 3 for some x N}. Show that f is invertible. Find the inverse.

Solution

Consider an arbitrary element y of Y. By the definition of Y, y = 4x + 3, for some x in the domain N.

$\mathrm{Define g : Y \to N by g\left(y\right) =}\frac{\mathrm{y}-3}{4}$

Now,

gof (x) = g (f (x)) = g (4x + 3)

$=\frac{\left[\left(4\mathrm{x}+3\right)-3\right]}{4}=\mathrm{x}={\mathrm{I}}_{\mathrm{x}}$

fog (y) = f (g (y)) = f ((y-3)/4)

$=\left[\frac{4\left(\mathrm{y}-3\right)}{4}+3\right]=\mathrm{y}={\mathrm{I}}_{\mathrm{y}}$

That is gof = Ix

and fog = Iy,

which implies that f is invertible and g is the inverse of f.

Theorem 1

If f : X → Y, g : Y → Z and h : Z → S are functions, then

h o (g o f ) = (h o g) o f.

Proof

We have,

h o (g o f ) (x) = h (g o f (x)) = h( g(f (x))), ∀ x in X

and (h o g) o f (x) = hog(f (x)) = h(g(f (x))), ∀ x in X.

Hence, h o (g o f) = (h o g) o f.

Theorem 2

Let f : X → Y and g : Y → Z be two invertible functions. Then gof is also invertible with (g o f)–1 = f –1 o g–1.

Proof

To show that gof is invertible with (g o f)–1 = f –1 og–1, it is enough to show that

(f –1 o g–1) o (g o f) = Ix and

(g o f) o (f –1 o g–1) = Iz.

Now,

(f–1 o g–1) o (g o f) = ((f –1 o g–1) o g) o f, [by Theorem 1]

= (f–1 o (g–1 o g)) o f, [by Theorem 1]

= (f –1 o Iy) o f, by definition of g–1 = Ix.

Similarly, it can be shown that (g o f) o (f–1 o g–1) = Iz.

Binary Operations

A binary operation * on a set A is a function

∗ : A × A → A

We denote *(a, b) by a * b.

• A binary operation * on the set X is called commutative, if a b = b a, for every a, b ∈ X.

• A binary operation ∗ : A × A → A is said to be associative if

(a b) ∗ c = a ∗ (b c), ∀ a, b, c, ∈ A.

• Given a binary operation ∗: A × A → A, an element e ∈ A, if it exists, is called identity for the operation ∗, if

a e = a = e a, ∀ a ∈ A.

• Given a binary operation ∗ : A × A → A with the identity element e in A, an element a ∈ A is said to be invertible with respect to the operation ∗, if there exists an element b in A such that a b = e = b a. In this case b is called the inverse of a and is denoted by a–1.