ALLEN MATHS MODULE

CHAPTER
1 FUNCTION

Recap of Early Classes

In earlier classes, we have studied about sets, a collection of elements. We also had a brief idea about functions
and relations. In this chapter, we will systematically discuss “Functions”. Their properties and
application will lead us into further understanding differential and integral calculus and their wider practical
applications.




1.0 CARTESIAN PRODUCT OF TWO SETS
2.0 RELATION
3.0 REPRESENTATION OF A RELATION :
4.0 INVERSE RELATION
5.0 IDENTITY RELATIONS
6.0 CLASSIFICATION OF RELATIONS
7.0 EQUIVALENCE RELATION
8.0 FUNCTION
9.0 DOMAIN, CO-DOMAIN & RANGE OF A FUNCTION
10.0 ALGEBRAIC OPERATIONS ON FUNCTIONS
11.0 IMPORTANT TYPES OF FUNCTION
12.0 EQUAL OR IDENTICAL FUNCTION
13.0 ODD & EVEN FUNCTIONS
14.0 CLASSIFICATION OF FUNCTIONS
15.0 BASIC TRANSFORMATIONS ON GRAPHS
16.0 COMPOSITE OF UNIFORMLY & NON-UNIFORMLY DEFINED FUNCTION
17.0 HOMOGENEOUS FUNCTIONS
18.0 BOUNDED FUNCTION
19.0 IMPLICIT & EXPLICIT FUNCTION
20.0 INVERSE OF A FUNCTION
21.0 PERIODIC FUNCTION
22.0 GENERAL
23.0 BASIC FUNCTIONAL EQUATIONS
EXERCISE-1
EXERCISE-2
EXERCISE-3
EXERCISE-4(A)
EXERCISE-4(B)
EXERCISE-5

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FUNCTION
This chapter deals with establishing binary relation between elements of one set and elements of another set
according to some particular rule of relationship.

1.0 CARTESIAN PRODUCT OF TWO SETS
SL AL

The Cartesian product of two sets A, B is a non-void set of all ordered pair (a, b) where a Î A and b Î B. This
is denoted by A × B
\ A × B = {(a, b) " a Î A and b Î B}
e.g. A = {1, 2} B = {a, b}
A × B = {(1, a), (1, b), (2, a), (2, b)}
Note :
(i) A × B ¹ B × A (Non-commutative)
(ii) n (A × B) = n(A) n(B) and n(P(A × B))= 2n(A)n(B)
(iii) A = f and B = f Û A × B = f
(iv) If A and B are two non-empty sets having n elements in common, then (A × B) and (B × A) have n 2
elements in common.
(v) A × (B È C) = (A × B) È (A × C)
(vi) A × (B Ç C) = (A × B) Ç (A × C)
(vii) A × (B – C) = (A × B) – (A × C)



Illustration 1. If n(A) = 7, n(B) = 8 and n(A Ç B) = 4 then match the following column-1 with column-II

Column-1 Column-II

(i) n(A È B) (a) 56
(ii) n(A × B) (b) 16
(iii) n(B × A) × A) (c) 392
(iv) n((A × B) Ç (B × A)) (d) 96
(v) n((A × B) È (B × A)) (e) 11

Solution (i) n(A È B) = n(A) + n(B) – n(A Ç B) = 7 + 8 – 4 = 11
(ii) n(A × B) = n(A) n(B) = 7 × 8 = 56 = n(B× A)
(iii)n((B × A) × A) = n(B×A). n(A) = 56 × 7 = 392
(iv) n((A × B) Ç (B× A)) = (n(AÇB)2) = 42=16
(v) n((A × B) È (B× A)) = n (A× B) + n(B×A) – n(A×B) Ç (B× A)
= 56 + 56 – 16 = 96
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Illustration 2. If A = {2, 4} and B = {3, 4, 5}, then (A Ç B) × (A È B) is :
(1) {(2, 2), (3, 4), (4, 2), (5, 4)}
(2) {(2, 3), (4, 3), (4, 5)}
(3) {(2, 4), (3, 4), (4, 4), (4, 5)}
(4) {(4, 2), (4, 3), (4, 4), (4, 5)}
Solution (A Ç B) = {4} and A È B = {2, 3, 4, 5}
(A Ç B) × (A È B) = {(4, 2), (4, 3), (4, 4), (4, 5)}


1

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2.0 RELATION
SL AL


Every subset of A × B defined a realtion from set A to set B. If R is relation from A ® B
R : {(a, b) |(a, b) Î A × B and a R b}
Highlights :
Let A and B be two non empty sets and R : A ® B be a relation such that R : {(a, b) | (a, b) Î R, a Î A and
b Î B}
(i) ‘b’ is called image of ‘a’ under R
(ii) ‘a’ is called pre-image of ‘b’ under R.
(iii) Domain of R : Collection of all elements of A which has a image in B.
(iv) Range of Collection of all elements of B which has a pre-image in A
Note :
(1) It is not necessary that each and every element of set A has a image in Set B and each and every element
of set B has preimage in set A
(2) Elements of set A having image in B is not necessarily unique.
(3) Basically relation is the number of subsets of A × B
Number of non empty relations = No. of ways of selecting a non zero subset of A × B
= mnC1 +mnC1 +..................mnCmn= 2mn –1
Total number of relation = 2mn (including void relation)



Illustration 3. A = {1, 2, 3, 4, 5} and B = {2, 4, 5}
aRb Þ a and b are relatively prime or co-prime (i.e. HCF is 1)
Solution R = {(1, 2), (1, 4), (1, 5), (2, 5), (3, 2), (3, 4), (3, 5), (4, 5), (5, 2), (5, 4)}
Domain of R{1, 2, 3, 4, 5}
Range of R{2, 4, 5}

Illustration 4. A = {Jaipur, Patna, Kanpur, Lucknow} and B = {Rajasthan, Uttar Pradesh, Bihar}
aRb Þ a is capital of b, a Î A and b Î B
Solution R = {(Jaipur, Rajasthan), (Patna, Bihar), (Lucknow, Uttar Pradesh)}

Illustration 5. If A = {1, 3, 5, 7}, B = {2, 4, 6, 8}
Relation is aRb Þ a > b, aÎA, bÎB
Solution R = {(3, 2), (5, 2), (5, 4), (7, 2), (7, 4), (7, 6)}
Domain = {3, 5, 7}
Range = {2, 4, 6}


3.0 REPRESENTATION OF A RELATION :
SL AL

1. Roster form : In this form we represents set of all ordered pairs (a, b) suc that (a, b) Î R
where a Î A, b Î B
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2. Set builder notation : Here we denote the relation by the rule which co relates the two set.
3. Arrow-diagram (Mapping) : This is the pictorial notation of any relation.
Ex. Let A = {–2, –1, 4}, B = {1, 4, 9}
A relation from A to B i.e. a R b is defined as a is less than b.
This can be represented in the following ways.
1. Roster form :
R = {(–2, 1) (–2, 4), (–2, 9), (–1, 1), (–1, 4), (–1, 9), (4, 9)}
2. Set builder notation :
R = {(a, b) : a Î A and b Î B, a is less than b}

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