### CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

### Relation and Functions Class XII Chapter 1

Class 12  chapter 1

Relations and functions

Basic concepts of topic relations and functions class XII chapter 1 of mathematics. Equivalence relations, different types of functions, composition and inverse of functions.
Introduction
In class 11 we have studied about Cartesian product of two sets, relations, functions, domain, range and co-domains. Now in this chapter we have studied about the different types of relations, different types of functions, composition of functions and invertible functions.

Relations
A relation R from set A to set B is the subset of Cartesian product A x B.
In Cartesian product A x B number of relations equal to the number of subsets.

Domain
Set of all first elements of the ordered pairs in the relation R is called its domain.

Range
Set of all second elements of the ordered pairs in the relation R is called its range.
In any ordered pair second element is also called the image of first element.

Types of relations
i) Empty Relation:
A relation R in set A is called empty relation, if no element of A is related to any element of A, i.e., R = φ  ⊂ A x A
Example: Let A = {1, 2, 3} and R be a relation on the set A defined as R = {(a, b) : a + b = 10; a, b ∈ A}
Here R is empty relation. R = Φ

ii) Identity Relation
A relation R in set A is called identity relation, if every element of A is related to itself, i.e., R = {a, a)  ⊂ A x A

A relation R is an identity relation in set A if for all a ∈ A, (a, a) ∈ R.

iii) 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 x A

iv) Reflexive relation
A relation R in a set A is called reflexive, if (a, a) ∈ R,  for every a ∈ A

v) Symmetric Relation
A relation R in a set A is called symmetric, if (a, b) ∈R implies that (b, a) ∈ R, for all a, b ∈ A.

vi) Transitive Relation
A relation R in a set A is called transitive, if (a, b) ∈R and (b, c) ∈R, implies that (a, c) ∈ R, for all a, b, c ∈ A

vii) Equivalence relation:
A relation R in set A is said to be equivalence relation if it is reflexive, symmetric  and transitive relation.

Viii) Equivalence Relations:
Let R be an equivalence relation on a set A and let a ∈ A. Then we define equivalence class of a as
[a] ={b ∈ A : b is related to a} = {b ∈ A : (b, a) ∈ R}

Note : Both empty relation and universal relations are sometimes called equivalence relation of trivial relation.

Functions
One-One functions (or injective functions):
A function f : A →B is said to  be one –one or injective function if different elements of set A has different images in set B
Mathematically:
Case 1: For all a, b ∈ A, we have: If  f(a) = f(b) a =  b
Case 2 :  For a, b ∈ A, such that : If ≠ b ⇒ f(a) ≠ f(b)

In the above figure different elements of set A have different images in set B. So this function is called one-one function.

Set A has 3 elements and the set B has 4 elements. Then the number of injective functions that can be defined from set A to set B is  = 4P3 = 4! = 24.
Generalization
Set A has m elements and the set B has n elements (n > m). Then the number of injective functions that can be defined from set A to set B is  =  nPm  = n!

If n < m then Then the number of injective functions that can be defined from set A to set B is = 0

If a function has m elements and is defined from A to A then number of injective relations from A to A = n!

Many One functions:
If two or more elements of set A have same image in set B then it is called many one function.
If a function is not one-one then it is called many one.

Here two different elements a and b of set A has same image in set B. So it is not an one-one function. It is only a many one function.

Onto function ( or surjective function)
A function f : A →B is said to be onto (or surjective) function, if every element of set B is the image of some element in set A.

Let y ∈ B, then for every  element  y ∈ B,  ∃ (there exist) an element   x ∈ A such that  f(x) = y

In the above figure every element of set B is the image of some element in set A so it is a onto function. It is neither  many one nor   one-one.

If n(A) = p, then number of bijective functions from set A to A are p!
Because If n(A) = p and n(A) = n(B) then the number of bijective functions from A to B is p!

Other definitions of onto functions are as follows
A function f : A →B is said to be onto (or surjective) function, if every element of set B has pre - image in set A.

If range of the function is equal to the co-domain(or set B), then it is called onto function.
Exercise 1.2
Q No. 9
$f:N\rightarrow N,\; \; f(n)=\left\{\begin{matrix} \frac{n+1}{2}, if\; \; n\; \; is\; \; odd\\ \frac{n}{2}, if\; \; n \; \; is\; \; even \end{matrix}\right.$
One-One
$Let\; (1,2)\epsilon N\; then$$f(1)=\frac{n+1}{2}=\frac{1+1}{2}=\frac{2}{2}=1$$f(2)=\frac{n}{2}=\frac{2}{2}=1$$f(1)=f(2)\: but\: \: 1\neq 2$⇒ f is not one one function.
Onto
Case 1: If n = odd = 2q+1=4q+1$f(n)=\frac{n+1}{2}$$f(4q+1)=\frac{4q+1+1}{2}=\frac{4q+2}{2}=2q+1$⇒ 2q + 1 is also an odd function ∈ N
Case 2: If n = even = 2q = 4q, $f(n)=\frac{n}{2}=\frac{4q}{2}=2q$⇒ 2q  is also an even function ∈ N
From case 1 and case 2 we conclude that :
*For all odd numbers in the domain we have odd numbers in the co-domain .
*For all even numbers in the domain we have even numbers in the co-domain .
⇒ Rf = Odd natural number and Even natural numbers.
⇒ Rf = All Natural Numbers = Co-domain
⇒ f(x) is an onto function.
Now f(x) is an onto function but not one-one.
So f(x) is not a bijective function.

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Composition of two 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

Composition of two functions exists only if co-domain of first function is equal to the domain of second function

In the above figure function f is defined on the element of set A (i.e.  x), function g is defined on the elements of set B (i.e f(x) ) so composition from set A to set C becomes gof or g[f(x)]

Bijective functions:
The functions which are one-one and onto both are called bijective functions.

Invertible functions and existence of inverse
If a function f is one-one and onto then it is an invertible function.

If a function is an invertible function then inverse of f exist and is denoted by f-1.

If f is invertible the f must be one-one and onto, conversely if f is one-one and onto  then f must be invertible. If f is invertible then f-1 exists.

Let f(x) = y is an invertible function then f-1 exists and we can write f-1(y) = x

Example : f(x) = { (a, 1), (b, 2), (c, 3), (d, 4)}, this can be written as
f(a) = 1, f(b) = 2,  f(c) = 3,  f(d) = 4

If f is invertible then we can write
f-1(1) = a,  f-1(2) = b,  f-1(3) = c,  f-1(4) = d
If f and g are two invertible functions then their composition gof is also invertible.

Give an example of a relation which is
i) Symmetric but neither reflexive nor transitive
Ans  R = { (1,2), (2,1)}
ii) Transitive but neither reflexive nor symmetric
Ans:  R = { (1,1), (2,2), (1,2), (2,1), (3,4)}
iii) Reflexive and symmetric but not transitive.
Ans: R = {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)}
iv) Reflexive and transitive but not symmetric.
Ans: R = {(1,1), (2,2), (1,2), (2,1), (2,3), (3,3)}
v) Symmetric and transitive but not reflexive

{(5,6), (6,5),(5,5) }