By going through these CBSE Class 12 Maths Notes Chapter 1 Relations and Functions, students can recall all the concepts quickly.

## Relations and Functions Notes Class 12 Maths Chapter 1

**RELATION**

**1. Types of Relations**

→ Empty Relation: A relation in a set A is known as empty relation, if no element of A is related to any element of A, i.e., R = Φ ⊆ A × A. e.g.

Let the set A = {1, 2,3,4,5) and R is given by

R= {(a,b): a – b = 20}

There is no pair (a, b) that satisfies the condition

a – b = 20.

⇒ The relation R is the empty relation.

→ Universal Relation: A relation R in a set A is called a universal relation, if each element of A is related to every element of A, i.e.,

R = A × A. e.g.

Let the set A = {1, 2,3, 4,5} and R is given by R = {(a, b): ab > 0}

Here, R = {(a, b): ab > 0} is the whole set A × A as all pairs (a, b) in A × A satisfy ab > 0.

Thus, this is the universal relation.

→ A relation R in a set A is called

(a) reflexive: if (a, a) ∈ R for every a ∈ A.

(b) symmetric: ii (a, b) ∈ R implies that (b, a) ∈ R for all a, b ∈ A.

(c) transitive: if (a, b) ∈ R and (b, c) e R implies that (a, c) ∈ R for all a,b,c ∈ A.

→ Equivalence Relation: A relation R in A is an equivalence relation if R is reflexive, symmetric, and transitive. For example:

(1) Let T be the set of all triangles in a plane with R a relation in T given by

R = ((T_{1}, T_{2}): T_{1} is similar to T_{2})

(a) R is reflexive since every triangle is similar to itself.

(b) (T_{1}, T_{2}) ∈ R ⇒ T_{1} is similar to T_{2}.

(T_{2}, T_{3}) ∈ R ⇒ T_{2} is similar to T_{1}

Therefore, R is symmetric.

(c) (T_{1}, T_{2}) and (T_{2}, T_{3}) lies in R

⇒ T_{1 }is similar to T_{2} and T_{2} is similar to T_{3}, which means T_{1} is similar to T_{3},

i.e., (T_{1}, T_{3}) lies in R.

∴ R is transitive.

Now R is reflexive, symmetric, and transitive, therefore R is an equivalence relation.

(2) Consider the set A = {1,2,3,4} and the relation R = {(1,1), (2, 2), (3,3), (4, 4), (1, 2), (2, 3), (3, 4)}.

(a) Now (1,1), (2, 2), (3, 3), (4, 4) lie in R. Relation R is reflexive.

(b) (1, 2) lies in R but (2,1) does not lie in it.

∴ It is not symmetric.

(c) (1,2), (2, 3) lie in R but (1, 3) does not lie in it. Therefore, R is not transitive.

Here, R is reflexive but neither symmetric nor transitive. Therefore, R is not an equivalence relation.

**2. Equivalence Class [a] containing a**

For an arbitrary equivalence relation R in an arbitrary set X, R divides X into mutually disjoint subsets A_{i}, which are known as partitions or sub-divisions of X satisfying:

(a) All elements of A_{i} are related to each other for all i.

(b) No element of A_{i} is related to any element of A_{j}, i ≠ j.

(c) ∪ A_{j} = X and A_{i} ∩ A. = Φ, i ≠ j.

The subsets Af are said to be equivalence classes.

Example: Let R be the relation defined in the set A = {p, q, s, t, e, o, u} by

R = {(a, b): both a and b are either consonants or vowels,

Here, R is an equivalence relation.

(a) Any element ∈ A is either consonant or vowel,

i.e., (a, a) ∈ R ⇒ R is reflexive.

(b) If (a, b) ∈ R ⇒ a and b both are either consonants or vowels ⇒ (b, a) e R.

∴ R is symmetric.

(c) If (a, b) ∈ R and (b, c) ∈ R, then a, b; b, c both pairs are either consonants or vowels.

i.e., a, b, c all are either consonants or vowels.

⇒ (a, c) ∈ R.

∴ R is transitive.

Thus, R is an equivalence relation.

Further, all the elements of (p, q, s, t) are related to each other as all the elements of this subset are consonants.

Similarly, all the elements of {e, i, o, u } are related to each other as all of them are vowels. But no element of {p, q, s, t} can be related to any element of {e, i, o, u}, since the elements of {p, q, s, t} are all consonants and the elements of {e, i, o, u} are all vowels. {p, q, s, t} is an equivalence class.denoted by an element as {p}. Similarly, {e, i, o, u} is an equivalence class denoted by an element [e).

**FUNCTIONS**

**1. Types of Functions**

→ One-one (or Injective): A function f: X → Y is said to be one-one (or injective), if the images of the distinct elements of X under/are distinct, i.e., for every x_{1}, x_{2} ∈ X, if f(x_{1}) = f(x_{2}) implies that x_{1} = x_{2}.

Each element of X has a distinct image in Y. Such a function or a mapping is one-one.

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

Corresponding to each element of Y, there is a pre-image in X. Such a mapping is onto.

→ One-one and Onto (Bijective): A function f: X to Y is known as one-one and onto (or bijective), if f is both one-one and onto.

Here,f is both one-one and onto. Therefore,f is said to be one-one onto function or bijective function.

**2. Composition of Functions**

Let f: A → B and g: B → C be the two functions. The composition of f and g is defined as. gof: A → C, such that

gof(x) = g{f(x)}, for all x ∈ A.

A function f: X → Y is said to be invertible if there exists a function g: Y → X such that gof = I_{x} and fog = I_{y}. The function g is called the inverse of f. It is denoted by f^{-1}.

Inverse or composite function: If f: X →Y and g: Y → Z be the two invertible functions, then gof is also invertible such that (gof)^{-1} = f^{-1}og^{-1}

**BINARY OPERATION**

→ Binary Operation: A binary operation on a set A is a function X: A × A → A, defined by × (a,b) = a × b, e.g., ×: R × R → R is given by (a, b) → a + b. Here +, — and x are the functions but + : R × R →, R, written as (a, b) → \(\frac{a}{b}\) is not a function. It is not a binary operation, since it is not defined for b = O.

→ Commutative Binary Operation: A binary operation × on the set A is commutative,if for every a,b ∈ A, a × b = b × a.

→ Associative Binary Operation: A binary operation × on the set A is associative, if (a × b) × c = a × (b × c).

It may be noted that associative property, a × b × c × d, … is not defined unless brackets are used.

→ An Identity Element e for Binary Operation: Let ×: A × A → A be a binary operation. There exists an element e ∈ A such that a × e = a = e × a, for all a ∈ A.

The element e is known as the identity element. It should be noted that 0 is the identity element for addition but not for natural numbers N, since 0 ∉ N.

→ The inverse of an element a: Let ×: A × A → A be a binary operation with identity element e in A. An element a ∈ A is invertible w.r.t. binary operation ×, if there exists an element b in A such that a × b = e = b × a. The element b is said to be the inverse of a. It is denoted by a^{-1}, e.g.,

– a is the inverse of a for the operation of addition +.

\(\frac{1}{a}\) (a ≠ 0) is the inverse of a for multiplication.

1. RELATIONS

(i) Relation. A relation R from a set A to a set B is a subset of A x B.

(ii) Classification of Relations : a

(a) Reflexive Relation. A relation R in a set E is said to be reflexive if xRx ∀ x ∈ E.

(b) Symmetric Relation. A relation R in a set E is said to be symmetric if:

xRy = yRx ∀ x, y ∈ E.

(c) Transitive Relation. A relation R in a set E is said to be transitive if:

vRy and yRz ⇒ xRz ∀ x, y, z ∈ E.

(d) Equivalence Relation. A relation R in a set E is said to be an equivalence relation if it is :

- reflexive
- symmetric and
- transitive.

2. FUNCTIONS

(i) Let X and Y be two non-empty sets. Then ‘f’ is a rule, which associates to each element x in X . a unique element y in Y.

(a) The unique element y of Y is called the value of f at x.

(b) The element x of X is called pre-image of y.

(c) The set X is called the domain of f

(d) The set of images of elements of X under f is called the range of f.

(ii) (a) D_{f} = {x : x ∈ R, f(x) ∈ R}

(b) R_{f} = {f(x):x ∈ D_{f}}

(c) f is one-one iff x_{1} = x_{2}

⇒ f(x_{1}) = f(x_{2}) for x_{1}, x_{2} ∈ D_{f}

or iff x_{1} ≠ x_{2}

⇒ f(x_{1}) ≠ f(x_{2}) for x_{1}, x_{2} ∈ D_{f}

(d) f is invertible iff f is one-one onto and D_{f-1} = R_{f}, R_{f-1}= DR_{f}.

3. ALGEBRA OF FUNCTIONS

Let f and g be two functions. Then

(i) (f+g) (x) =f(x) + g(x); D_{f+g} = D_{f} ∩ D_{g}

(ii) (f- g) (x) = f(x) – g(x); D_{f-g} = D_{f} ∩ D_{g}

(iii) (fg) (x) =f(x) g(x); D_{fg} = D_{f} ∩ D_{g}

(iv) \(\left(\frac{f}{g}\right) x=\frac{f(x)}{g(x)}\); D_{f/g} = D_{f} ∩ D_{g} – {x:x∈D_{g}, g(x) = 0}