vMathematics is the indispensable instrument of
all physical research. – BERTHELOT v
2.1 Introduction
Much of mathematics is about finding a pattern – a
recognisable link between quantities that change. In our
daily life, we come across many patterns that characterise
relations such as brother and sister, father and son, teacher
and student. In mathematics also, we come across many
relations such as number m is less than number n, line l is
parallel to line m, set A is a subset of set B. In all these, we
notice that a relation involves pairs of objects in certain
order. In this Chapter, we will learn how to link pairs of
objects from two sets and then introduce relations between
the two objects in the pair. Finally, we will learn about
special relations which will qualify to be functions. The
concept of function is very important in mathematics since it captures the idea of a
mathematically precise correspondence between one quantity with the other.
2.2 Cartesian Products of Sets
Suppose A is a set of 2 colours and B is a set of 3 objects, i.e.,
A = {red, blue}and B = {b, c, s},
where b, c and s represent a particular bag, coat and shirt, respectively.
How many pairs of coloured objects can be made from these two sets?
Proceeding in a very orderly manner, we can see that there will be 6
distinct pairs as given below:
(red, b), (red, c), (red, s), (blue, b), (blue, c), (blue,
s).
Thus, we get 6 distinct objects (Fig 2.1).
Let us recall from our earlier classes that an ordered pair of elements
taken from any two sets P and Q is a pair of elements written in small
Fig 2.1
Chapter
2
RELATIONS AND FUNCTIONS
G . W. Leibnitz
(1646–1716)
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RELATIONS AND FUNCTIONS 31
brackets and grouped together in a particular order, i.e., (
p,q), p ∈ P and q ∈ Q .
This
leads to the following definition:
Definition 1
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, i.e., P × Q = φ
From the illustration given above we note that
A × B = {(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)}.
Again, consider the two sets:
A = {DL, MP, KA}, where DL, MP, KA represent Delhi,
Madhya Pradesh and Karnataka, respectively and B = {01,02,
03}representing codes for the licence plates of vehicles issued
by DL, MP and KA .
If the three states, Delhi, Madhya Pradesh and Karnataka
were making codes for the licence plates of vehicles, with the
restriction that the code begins with an element from set A,
which are the pairs available from these sets and how many such
pairs will there be (Fig 2.2)?
The available pairs are:(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03),
(KA,01), (KA,02), (KA,03) and the product of set A and set B is given by
A
× B = {(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02),
(KA,03)}.
It can easily be seen that there will be 9 such pairs in the Cartesian product, since
there are 3 elements in each of the sets A and B. This gives us 9 possible codes. Also
note that the order in which these elements are paired is crucial. For example, the code
(DL, 01) will not be the same as the code (01, DL).
As a final illustration, consider the two sets A= {a
1
, a
2
} and
B = {b
1
, b
2
, b
3
, b
4
} (Fig 2.3).
A × B = {( a
1
, b
1
), (a
1
, b
2
), (a
1
, b
3
), (a
1
, b
4
), (a
2
, b
1
), (a
2
, b
2
),
(a
2
, b
3
), (a
2
, b
4
)}.
The 8 ordered pairs thus formed can represent the position of points in
the plane if A and B are subsets of the set of real numbers and it is
obvious that the point in the position (a
1
, b
2
) will be distinct from the point
in the position (b
2
, a
1
).
Remarks
(i) Two ordered pairs are equal, if and only if the corresponding first elements
are equal and the second elements are also equal.
DL MP
KA
03
02
01
Fig 2.2
Fig 2.3
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32 MATHEMATICS
(ii) If there are p elements in A and q elements in B, then there will be pq
elements in A
× B, i.e., if
n(A) = p and n(B) = q, then n(A × B) = pq.
(iii) If A and B are non-empty sets and either A or B is an infinite set, then so is
A × B.
(iv) A × A × A = {(a, b, c) : a, b, c ∈ A}. Here (a, b, c) is called an ordered
triplet.
Example
1 If (x + 1, y – 2) = (3,1), find the values of x and y.
Solution
Since the ordered pairs are equal, the corresponding elements are equal.
Therefore x + 1 = 3 and
y – 2 = 1.
Solving we get x = 2 and y = 3.
Example 2 If P = {a, b, c} and Q = {r}, form the sets P × Q and Q × P.
Are these two products equal?
Solution By the definition of the cartesian product,
P × Q = {(a, r), (b, r), (c, r)} and Q × P = {(r, a), (r, b), (r, c)}
Since, by the definition of equality of ordered pairs, the pair (a, r) is not equal to the pair
(r, a), we conclude that P × Q ≠ Q × P.
However, the number of elements in each set will be the same.
Example 3 Let A = {1,2,3}, B = {3,4} and C = {4,5,6}. Find
(i) A × (B ∩ C) (ii) (A × B) ∩ (A × C)
(iii) A × (B ∪ C) (iv) (A × B) ∪ (A × C)
Solution
(i) By the definition of the intersection of two sets, (B ∩ C) = {4}.
Therefore, A × (B ∩ C) = {(1,4), (2,4), (3,4)}.
(ii) Now (A × B) = {(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)}
and (A × C) = {(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)}
Therefore, (A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}.
(iii) Since, (B ∪ C) = {3, 4, 5, 6}, we have
A × (B ∪ C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,3),
(3,4), (3,5), (3,6)}.
(iv) Using the sets A × B and A × C from part (ii) above, we obtain
(A × B) ∪ (A × C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6),
(3,3), (3,4), (3,5), (3,6)}.
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RELATIONS AND FUNCTIONS 33
Example 4 If P = {1, 2}, form the set P × P × P.
Solution We have, P × P × P = {(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1),
(2,2,2)}.
Example 5 If
R is the set of all real numbers, what do the cartesian products R ×
R
and R × R × R represent?
Solution
The Cartesian product R × R represents the set R × R={(x, y) : x, y ∈
R}
which represents the coordinates of all the points in two dimensional space and the
cartesian product R × R × R represents the set R × R × R ={(x, y, z) : x, y, z ∈ R}
which represents the coordinates of all the points in three-dimensional space.
Example 6
If A × B ={(p, q),(p, r), (m, q), (m, r)}, find A and B.
Solution A = set of first elements = {p, m}
B = set of second elements = {q, r}.
EXERCISE 2.1
1. If
2 5 1
1
3 3 3 3
x
, y – ,
+ =
, find the values of x and y.
2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of
elements in (A×B).
3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
4. State whether each of the following statements are true or false. If the statement
is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered
pairs (x, y) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A
× (B ∩ φ) = φ.
5. If A = {–1, 1}, find A × A
× A.
6. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.
7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A × (B ∩ C) = (A × B) ∩ (A × C). (ii) A
× C is a subset of B × D.
8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have?
List them.
9. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1)
are in A × B, find A and B, where x, y and z are distinct elements.
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34 MATHEMATICS
10. The Cartesian product A ×
A has 9 elements among which are found (–1, 0) and
(0,1). Find the set A and the remaining elements of A
× A.
2.3 Relations
Consider the two sets P = {a, b, c} and Q = {Ali, Bhanu, Binoy, Chandra, Divya}.
The cartesian product of
P and Q has 15 ordered pairs which
can be listed as P × Q = {(a, Ali),
(a,Bhanu), (a, Binoy), ..., (c, Divya)}.
We can now obtain a subset of
P × Q by introducing a relation R
between the first element x and the
second element y of each ordered pair
(x, y) as
R= { (x,y): x is the first letter of the name y, x ∈ P, y ∈ Q}.
Then R = {(a, Ali), (
b, Bhanu), (b, Binoy), (
c, Chandra)}
A visual representation of this relation R (called an arrow diagram) is shown
in Fig 2.4.
Definition 2
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.
Definition 3 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.
Definition 4 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.
Remarks (i) A relation may be represented algebraically either by the Roster
method or by the Set-builder method.
(ii) An arrow diagram is a visual representation of a relation.
Example 7 Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by
R = {(x, y) : y = x + 1 }
(i) Depict this relation using an arrow diagram.
(ii) Write down the domain, codomain and range of R.
Solution (i) By the definition of the
relation,
R = {(1,2), (2,3), (3,4), (4,5), (5,6)}.
Fig 2.4
2020-21
RELATIONS AND FUNCTIONS 35
The corresponding arrow diagram is
shown in Fig 2.5.
(ii)
We can see that the
domain ={1, 2, 3, 4, 5,}
Similarly, the range = {2, 3, 4, 5, 6}
and the codomain = {1, 2, 3, 4, 5, 6}.
Example 8
The Fig 2.6 shows a relation
between the sets P and Q. Write this relation (i) in set-builder form, (ii) in roster form.
What is its domain and range?
Solution
It is obvious that the relation R is
“x is the square of y”.
(i) In set-builder form, R = {(x, y): x
is the square of y, x
∈ P, y ∈
Q}
(ii) In roster form, R = {(9, 3),
(9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}
The domain of this relation is {4, 9, 25}.
The range of this relation is {– 2, 2, –3, 3, –5, 5}.
Note that the element 1 is not related to any element in set P.
The set Q is the codomain of this relation.
A
Note The total number of relations that can be defined from a set A to a set B
is the number of possible subsets of A × B. If n(A ) = p and n(B) = q, then
n (A × B) =
pq and the total number of relations is 2
pq
.
Example 9 Let A = {1, 2} and B = {3, 4}. Find the number of relations from A to B.
Solution We have,
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}.
Since n (A×B ) = 4, the number of subsets of A×B is 2
4
. Therefore, the number of
relations from A into B will be 2
4
.
Remark A relation R from A to A is also stated as a relation on A.
EXERCISE 2.2
1. Let A = {1, 2, 3,...,14}. Define a relation R from A to A by
R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and
range.
Fig 2.5
Fig 2.6
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36 MATHEMATICS
2. Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5,
x is a natural number less than 4; x, y ∈N}. Depict this relationship using roster
form. Write down the domain and the range.
3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by
R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in
roster form.
4. The Fig2.7 shows a relationship
between the sets P and Q. Write this
relation
(i) in set-builder form (ii) roster form.
What is its domain and range?
5. Let A = {1, 2, 3, 4, 6}. Let R be the
relation on A defined by
{(a, b): a , b ∈A,
b is exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.
6. Determine the domain and range of the relation R defined by
R = {(x, x + 5) :
x ∈ {0, 1, 2, 3, 4, 5}}.
7. Write the relation R = {(x, x
3
) :
x is a prime number less than 10} in roster form.
8. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
9. Let R be the relation on Z defined by R = {(a,b): a, b ∈ Z, a – b is an integer}.
Find the domain and range of R.
2.4 Functions
In this Section, we study a special type of relation called function.
It is one of the most
important concepts in mathematics. We can, visualise a function as a rule, which produces
new elements out of some given elements. There are many terms such as ‘map’ or
‘mapping’ used to denote a function.
Definition 5 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.
If f is a function from A to B and (a, b) ∈ f, then f (a) = b, where b is called the
image of a under f and a is called the preimage of b under f.
Fig 2.7
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RELATIONS AND FUNCTIONS 37
The function f from A to B is denoted by f: A
à B.
Looking at the previous examples, we can easily see that the relation in Example 7 is
not a function because the element 6 has no image.
Again, the relation in Example 8 is not a function because the elements in the
domain are connected to more than one images. Similarly, the relation in Example 9 is
also not a function. (Why?) In the examples given below, we will see many more
relations some of which are functions and others are not.
Example 10
Let N be the set of natural numbers and the relation R be defined on
N such that R = {(x, y) : y = 2
x, x, y ∈ N}.
What is the domain, codomain and range of R? Is this relation a function?
Solution
The domain of R is the set of natural numbers N. The codomain is also N.
The range is the set of even natural numbers.
Since every natural number n has one and only one image, this relation is a
function.
Example 11 Examine each of the following relations given below and state in each
case, giving reasons whether it is a function or not?
(i) R = {(2,1),(3,1), (4,2)}, (ii) R = {(2,2),(2,4),(3,3), (4,4)}
(iii) R = {(1,2),(2,3),(3,4), (4,5), (5,6), (6,7)}
Solution (i) Since 2, 3, 4 are the elements of domain of R having their unique images,
this relation R is a function.
(ii) Since the same first element 2 corresponds to two different images 2
and 4, this relation is not a function.
(iii) Since every element has one and only one image, this relation is a
function.
Definition 6
A function which has either R or one of its subsets as its range is called
a real valued function. Further, if its domain is also either R or a subset of R, it is
called a real function.
Example 12 Let N be the set of natural numbers. Define a real valued function
f : Nà N by f (x) = 2x + 1. Using this definition, complete the table given below.
x 1 2 3 4 5 6
7
y f (1) = ... f (2) = ... f (3) = ... f (4) = ... f (5) = ... f (6) = ... f (7) = ...
Solution The completed table is given by
x 1 2 3 4 5 6
7
y f (1) = 3 f (2) = 5 f (3) = 7 f (4) = 9 f (5) = 11 f (6) = 13 f (7) =15
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38 MATHEMATICS
2.4.1 Some functions and their graphs
(i) Identity function Let R be the set of real numbers. Define the real valued
function
f : R → R by y = f(x) = x for each
x ∈ R. Such a function is called the
identity function
. Here the domain and range of f are R. The graph is a straight line as
shown in Fig 2.8. It passes through the origin.
Fig 2.9
Fig 2.8
(ii) Constant function Define the function f: R → R by y = f (x) = c, x ∈ R where
c is a constant and each x ∈ R. Here domain of f is R and its range is {c}.
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RELATIONS AND FUNCTIONS 39
The graph is a line parallel to x-axis. For example, if f(x)=3 for each x∈R, then its
graph will be a line as shown in the Fig 2.9.
(iii) Polynomial function
A function f : R → R is said to be polynomial function if
for each x in
R, y = f (x) = a
0
+ a
1
x + a
2
x
2
+ ...+ a
n
x
n
, where n is a non-negative
integer and
a
0
, a
1
, a
2
,...,a
n
∈R.
The functions defined by f(x) =
x
3
– x
2
+ 2, and g(x) = x
4
+
2
x are some examples
of polynomial functions, whereas the function h defined by h(x) =
2
3
x
+ 2x is not a
polynomial function.(Why?)
Example 13 Define the function f: R → R by y = f(x) = x
2
, x ∈ R. Complete the
Table given below by using this definition. What is the domain and range of this function?
Draw the graph of f.
x – 4 –3 –2 –1 0 1
2 3 4
y = f(x) = x
2
Solution The completed Table is given below:
x – 4 –3 –2 –1 0 1 2 3
4
y = f (x) = x
2
16 9 4 1 0 1 4 9 16
Domain of f = {x : x∈R}. Range of f = {x
2
: x ∈ R}. The graph of f is given
by Fig 2.10
Fig 2.10
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40 MATHEMATICS
Example 14 Draw the graph of the function f :R → R defined by
f (x) = x
3
, x∈R.
Solution
We have
f(0) = 0,
f(1) = 1,
f(–1) = –1, f(2) = 8, f(–2) = –8, f(3) = 27; f(–3) = –27, etc.
Therefore, f = {(x,x
3
): x∈R}.
The graph of f is given in Fig 2.11.
Fig 2.11
(iv) Rational functions are functions of the type
( )
( )
f x
g x
, where f(x) and g(x) are
polynomial functions of x defined in a domain, where g(x) ≠ 0.
Example 15 Define the real valued function f : R – {0} → R defined by
1
( ) =
f x
x
,
x ∈ R –{0}. Complete the Table given below using this definition. What is the domain
and range of this function?
x –2 –1.5 –1 –0.5 0.25 0.5 1 1.5 2
y =
1
x
... ... ... ... ... ... ... ... ...
Solution The completed Table is given by
x –2 –1.5 –1 –0.5 0.25 0.5 1 1.5 2
y =
1
x
– 0.5 – 0.67 –1 – 2 4 2 1 0.67 0.5
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RELATIONS AND FUNCTIONS 41
The domain is all real numbers except 0 and its range is also all real numbers
except 0. The graph of f is given in Fig 2.12.
Fig 2.13
(v) The Modulus function The function
f: R→R defined by f(x) = |x| for each
x ∈R is called modulus function. For each
non-negative value of x, f(x) is equal to x.
But for negative values of x, the value of
f(x) is the negative of the value of x, i.e.,
0
( )
0
x,x
f x
x,x
≥
=
− <
The graph of the modulus function is given
in Fig 2.13.
(vi) Signum function The function
f:R→R defined by
1 if 0
( ) 0 if 0
1 if 0
, x
f x , x
, x
>
= =
− <
Fig 2.12
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42 MATHEMATICS
is called the signum function. The domain of the signum function is R and the range is
the set {–1, 0, 1}. The graph of the signum function is given by the Fig 2.14.
Fig 2.14
(vii) Greatest integer function
The function f: R →
R defined
by f(x) = [x], x ∈R assumes the
value of the greatest integer, less
than or equal to
x. Such a function
is called the greatest integer
function.
From the definition of [x], we
can see that
[x] = –1 for –1 ≤ x < 0
[x] = 0 for 0
≤ x < 1
[x] = 1 for 1
≤ x < 2
[x] = 2 for 2
≤ x < 3 and
so on.
The graph of the function is
shown in Fig 2.15.
2.4.2 Algebra of real functions In this Section, we shall learn how to add two real
functions, subtract a real function from another, multiply a real function by a scalar
(here by a scalar we mean a real number), multiply two real functions and divide one
real function by another.
(i) Addition of two real functions Let f : X → R and g : X → R be any two real
functions, where X ⊂ R. Then, we define (f + g): X → R by
(f + g) (x) = f (x) + g (x), for all x ∈ X.
Fig 2.15
f(x) =
x
x
, x x‘ 0 and 0 for = 0
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RELATIONS AND FUNCTIONS 43
(ii) Subtraction of a real function from another Let
f : X → R and
g: X → R be
any two real functions, where X
⊂
R. Then, we define (f – g) : X→R by
(f–g) (x) = f(x) –g(x), for all
x ∈ X.
(iii) Multiplication by a scalar Let f : X→R be a real valued function and α be a
scalar. Here by scalar, we mean a real number. Then the product
α f is a function from
X to R defined by (
α f ) (x) = α f (x), x ∈X.
(iv) Multiplication of two real functions
The product (or multiplication) of two real
functions f:X→R
and g:X→R is a function fg:X→R defined by
(fg) (x) = f(x) g(x), for all x ∈ X.
This is also called pointwise multiplication.
(v) Quotient of two real functions Let f and g be two real functions defined from
X→R, where X
⊂
R. The quotient of f by g denoted by
f
g
is a function defined by ,
( )
( )
( )
f f x
x
g g x
=
, provided g(x) ≠ 0, x ∈ X
Example 16 Let f(x) = x
2
and g(x) = 2x + 1 be two real functions.Find
(f + g) (x), (f –g) (x), (fg) (x),
( )
f
x
g
.
Solution We have,
(f + g) (x) = x
2
+ 2x + 1, (f –g) (x) = x
2
– 2x – 1,
(fg) (x) = x
2
(2x + 1) = 2x
3
+ x
2
,
( )
f
x
g
=
2
2 1
x
x
+
, x ≠
1
2
−
Example 17 Let f(x) =
x
and g(x) = x be two functions defined over the set of non-
negative real numbers. Find (f + g) (x), (f – g) (x), (fg) (x) and
f
g
(x).
Solution We have
(f + g) (x) =
x
+ x, (f – g) (x) =
x
– x ,
(fg) x =
3
2
x( x ) x
=
and
( )
f
x
g
1
2
0
–
x
x , x
x
= = ≠
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44 MATHEMATICS
EXERCISE 2.3
1. Which of the following relations are functions? Give reasons. If it is a function,
determine its domain and range.
(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
(iii) {(1,3), (1,5), (2,5)}.
2. Find the domain and range of the following real functions:
(i) f(x) = –
x
(ii) f(x) =
2
9
x
−
.
3. A function f is defined by f(x) = 2x –5. Write down the values of
(i) f (0), (ii) f (7), (iii) f (–3).
4. The function ‘t’ which maps temperature in degree Celsius into temperature in
degree Fahrenheit is defined by t(C) =
9C
5
+ 32.
Find (i) t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212.
5. Find the range of each of the following functions.
(i) f (x) = 2 – 3x, x ∈ R, x > 0.
(ii) f (x) = x
2
+ 2, x is a real number.
(iii) f (x) = x, x is a real number.
Miscellaneous Examples
Example 18 Let R be the set of real numbers.
Define the real function
f: R→R by f(x) = x + 10
and sketch the graph of this function.
Solution Here f(0) = 10, f(1) = 11, f(2) = 12, ...,
f(10) = 20, etc., and
f(–1) = 9, f(–2) = 8, ...,
f(–10) = 0 and so on.
Therefore, shape of the graph of the given
function assumes the form as shown in Fig 2.16.
Remark The function f defined by f(x) = mx + c ,
x ∈ R, is called linear function, where m and c are
constants. Above function is an example of a linear
function.
Fig 2.16
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RELATIONS AND FUNCTIONS 45
Example 19 Let R be a relation from Q to Q defined by R = {(a,b): a,b ∈ Q and
a – b ∈
Z}. Show that
(i) (a,a) ∈ R for all a ∈ Q
(ii) (a,b) ∈ R implies that (
b, a) ∈ R
(iii) (a,b) ∈ R and (b,c) ∈ R implies that (a,c) ∈R
Solution (i) Since,
a – a = 0 ∈ Z, if follows that (a, a) ∈ R.
(ii) (a,b) ∈ R implies that a –
b ∈ Z. So, b – a ∈ Z. Therefore,
(b, a) ∈ R
(iii) (a, b) and (
b, c) ∈ R implies that a – b ∈ Z. b – c ∈ Z. So,
a – c = (a – b) + (b – c) ∈ Z. Therefore, (a,c) ∈ R
Example 20 Let f = {(1,1), (2,3), (0, –1), (–1, –3)} be a linear function from Z into Z.
Find f(x).
Solution Since f is a linear function, f (x) = mx + c. Also, since (1, 1), (0, – 1) ∈ R,
f (1) = m + c = 1 and f (0) = c = –1. This gives m = 2 and f(x) = 2x – 1.
Example 21 Find the domain of the function
2
2
3 5
( )
5 4
x x
f x
x x
+ +
=
− +
Solution Since x
2
–5x + 4 = (x – 4) (x –1), the function f is defined for all real numbers
except at x = 4 and x = 1. Hence the domain of f is R – {1, 4}.
Example 22 The function f is defined by
f (x) =
1 0
1 0
1 0
x, x
, x
x , x
− <
=
+ >
Draw the graph of f (x).
Solution Here, f(x) = 1 – x, x < 0, this gives
f(– 4) = 1 – (– 4) = 5;
f(– 3) =1 – (– 3) = 4,
f(– 2) = 1 – (– 2) = 3
f(–1) = 1 – (–1) = 2; etc,
and f(1) =
2, f (2) = 3, f (3) = 4
f(4) = 5 and so on for f(x) = x + 1, x > 0.
Thus, the graph of f is as shown in Fig 2.17
Fig 2.17
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46 MATHEMATICS
Miscellaneous Exercise on Chapter 2
1. The relation f is defined by
2
0 3
( ) =
3 3 10
x , x
f x
x, x
≤ ≤
≤ ≤
The relation g is defined by
2
, 0 2
( )
3 , 2 10
x x
g x
x x
≤ ≤
=
≤ ≤
Show that f is a function and g is not a function.
2. If f (x) = x
2
, find
(1 1) (1)
(1 1 1)
f . – f
. –
.
3. Find the domain of the function f (x)
2
2
2 1
8 12
x x
x – x
+ +
=
+
.
4. Find the domain and the range of the real function f defined by f (x) =
( 1)
x
−
.
5. Find the domain and the range of the real function f defined by f (x) =
–1
x
.
6. Let
2
2
, :
1
x
f x x
x
= ∈
+
R
be a function from R into R. Determine the range
of f.
7. Let f, g : R→R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find
f + g, f – g and
f
g
.
8. Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by
f(x) = ax + b, for some integers a, b. Determine a, b.
9. Let R be a relation from N to N defined by R = {(a, b) : a, b ∈N and a = b
2
}. Are
the following true?
(i) (a,a) ∈ R, for all a ∈ N (ii) (a,b) ∈ R, implies (b,a) ∈ R
(iii) (a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R.
Justify your answer in each case.
10. Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)}
Are the following true?
(i) f is a relation from A to B (ii) f is a function from A to B.
Justify your answer in each case.
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RELATIONS AND FUNCTIONS 47
11. Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a
function from
Z to Z? Justify your answer.
12. Let A = {9,10,11,12,13} and let f : A→N be defined by f (n) = the highest prime
factor of
n. Find the range of f.
Summary
In this Chapter, we studied about relations and functions.The main features of
this Chapter are as follows:
®
Ordered pair A pair of elements grouped together in a particular order.
® Cartesian product
A × B of two sets A and B is given by
A × B = {(
a, b): a ∈ A, b ∈ B}
In particular R × R = {(x, y): x, y ∈ R}
and R × R ×
R = (x, y, z): x, y, z ∈ R}
® If (a, b) = (x, y), then a = x and b = y.
® If n(A) = p and n(B) = q, then n(A × B) = pq.
® A × φ = φ
® In general, A × B ≠ B × A.
® Relation A relation R from a set A to a set B is a subset of the cartesian
product A × B obtained by describing a relationship between the first element
x and the second element y of the ordered pairs in A × B.
® The image of an element x under a relation R is given by y, where (x, y) ∈ R,
® The domain of R is the set of all first elements of the ordered pairs in a
relation R.
® The range of the relation R is the set of all second elements of the ordered
pairs in a relation R.
® Function A function f from a set A to a set B is a specific type of relation for
which every element x of set A has one and only one image y in set B.
We write f: A→B, where f(x) = y.
® A is the domain and B is the codomain of f.
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48 MATHEMATICS
® The range of the function is the set of images.
® A real function has the set of real numbers or one of its subsets both as its
domain and as its range.
® Algebra of functions For functions f : X → R and
g : X → R, we have
(f + g) (x) = f (x) + g(x), x ∈ X
(f – g) (x) = f (x) – g(x), x ∈ X
(f.g) (x) =
f (x) .g (x), x ∈ X
(kf) (x) = k (f (x) ), x ∈
X, where k is a real number.
( )
f
x
g
=
( )
( )
f x
g x
, x ∈ X, g(x) ≠ 0
Historical Note
The word FUNCTION first appears in a Latin manuscript “Methodus
tangentium inversa, seu de fuctionibus” written by Gottfried Wilhelm Leibnitz
(1646-1716) in 1673; Leibnitz used the word in the non-analytical sense. He
considered a function in terms of “mathematical job” – the “employee” being
just a curve.
On July 5, 1698, Johan Bernoulli, in a letter to Leibnitz, for the first time
deliberately assigned a specialised use of the term function in the analytical
sense. At the end of that month, Leibnitz replied showing his approval.
Function is found in English in 1779 in Chambers’ Cyclopaedia: “The
term function is used in algebra, for an analytical expression any way compounded
of a variable quantity, and of numbers, or constant quantities”.
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