Page 24 - Mathematics_Form_3

Relations

Example 1.17 To find the range of the relation R,
Find the domain and range of the write x in terms of y for the given
relation equation.
1
, which implies that
R = {(x, y ) : y is divisible by x, where Given y = _
x + 3
FOR ONLINE READING ONLY
x and y are integers} . x = − 3.
1 _
y
Solution For the equation x = − 3 to be
1 _
y
If and y are integers and y is divisible defined, y must not be equal to zero.
y
by x, then = k , where k is an integer. All other values of y satisfy the equation
_
x
If the fraction is not defined when the x = − 3. Thus, the
1 _
y
denominator is zero, the domain of the range = {y : y ∈ ℝ, y ≠ 0} .
relation R is a set of all integers except
zero, that is, Therefore, the
domain = { x : x ∈ ℤ , x ≠ 0 }. domain = {x : x ∈ ℝ, x ≠ − 3} and
range = { y : y ∈ ℝ, y ≠ 0} .
To find the range of the relation R, note
that y as a numerator takes all integer
values. Thus, the range of R is the set Exercise 1.6
of all integers, that is,
1. Find the domain and range of each
range = {y : y ∈ ℤ} . of the following relations:

Therefore, the domain = { x : x ∈ ℤ, (a) R = {(x, y ) : x and y are real
_
x ≠ 0} and range = {y : y ∈ ℤ} . numbers and y = √ x } .

(b) R = {(x, y ) : x and y are natural
numbers and y is a multiple of x} .
Example 1.18
2. Let R = {(x, y ) : y = 3x + 1} be a
Let R be a relation on the set of real relation on the set of real numbers.
1
. Find the
numbers given by y = _ Find:
x + 3
domain and range of R. (a) The set of ordered pairs which
belong to R from the set Mathematics for Secondary Schools
Solution  15 
{(1, 2), (0, 1),−  ,  ,
1
, the
Given the relation y = _  2 2 
x + 3
−−
denominator needs to be a non-zero ( 2, 5), (3,10)}
number. So, x ≠ − 3. All other values (b) The domain of the set obtained
of x satisfy the equation. Therefore, in (a).
(c) The range of the set obtained in
domain = {x : x ∈ ℝ, x ≠ − 3} .
(a).

Student\s Book Form Three 17

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MATHEMATIC F3 SB.indd 17 18/09/2025 09:58:42
MATHEMATIC F3 SB.indd 17