**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 I_{A} = {(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 A_{i} are related to each other, for all i.

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

(iii) ∪ A_{j} = X and A_{i} ∩ A_{j }= 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

*x*

_{1},

*x*

_{2}∈ X,

*f*(

*x*

_{1}) =

*f*(

*x*

_{2}) implies

*x*

_{1}=

*x*

_{2}. 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 x_{1}, x_{2} ∈ A, then,

f(x_{1}) = f(x_{2}) => x_{1} = x_{2}

or equivalently

x_{1} ≠ x_{2} => f(x_{1}) ≠ f(x_{2})

**(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 *= I_{x} and *fog *= I_{y}.

(v) If *f *: X → Y is a function such that there exists a function *g *: Y → X such that *gof *= I_{x} and *fog *= I_{y}, 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 *= I_{x}

and *fog *= I_{y}.

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*) = 4*x *+ 3, where, Y = {*y *∈ **N**: *y *= 4*x *+ 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 *= 4*x *+ 3, for some *x *in the domain **N**.

$$\mathrm{}\mathrm{x}\mathrm{}=\frac{\mathrm{y}-3}{4}$$

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

Now,

*gof *(*x*) = *g *(*f *(*x*)) = *g *(4*x *+ 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 *= I_{x}

and *fog *= I_{y},

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 *g*o*f *is also invertible with (*g o f*)^{–1} = *f *^{–1 }o *g*^{–1}.

**Proof **

To show that *g*o*f *is invertible with (*g o f*)^{–1} = *f*^{ –1} o*g*^{–1}, it is enough to show that

(*f *^{–1 }*o g*^{–1}) *o* (*g o f*) = I_{x} and

(*g o f*) *o* (*f *^{–1 }*o g*^{–1}) = I_{z}.

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 *I_{y}) *o f*, by definition of *g*^{–1 }= I_{x}.

Similarly, it can be shown that (*g o f*) o (*f*^{–1 }*o* *g*^{–1}) = I_{z}.

**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
, if*commutative*, for every*a*∗*b*=*b*∗*a**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**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*i**nvertible**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}.