Page 28 - Mathematics_Form_3

Relations

(b) The domain of R = {x : x ≥ 0}
−1
1
10
equation y = , ( x −≠ ), thus
_
and the range of R = {y : y ∈ ℝ} . x − 1
−1
x ¹ 1 . Therefore, the domain of
R= {x : x ∈ ℝ, x ≠1} .
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Example 1.24
(b) To find the range of R, first express x
Given the relation R = {(x, y ) : y < x + 4} , in terms of y, that is, from y = ,
1
_
find the: 1 x − 1
it follows that =x + 1.
(a) Inverse of R. y
1 _

(b) Domain and range of R . Observe that y ≠ 0 for x = + 1
−1
y
Solution to be defined. Hence, the range of
(a) From the inequality y < x + 4, R = {y : y ∈ ℝ, y ≠ 0} .
interchange the variables x and

y to obtain x < y + 4. This is (c) For the inverse of R, given that
1
, upon interchanging
equivalent to x − 4 < y which can y = _
x − 1
1
.
be written as y > x − 4. Hence, x and y, it becomes x = _
y − 1
R = {(x, y ): y > x − 4} . Making y the subject of the equation
−1
(b) The domain of R = {x : x ∈ ℝ} gives y = + 1. Therefore,
1 _
−1
x
−1
and the range of R = {y : y ∈ ℝ}
1 _
R = {(x, y ) : y = + 1} .
−1
x
Example 1.25 (d) From the equation for R , that is,
−1
1 _
Given the relation y = + 1 , observe that x should
x
1
, find the:
R = (x, y ) : y = _ } not be zero. Thus, the domain of
{
x − 1
R = {x : x ∈ ℝ, x ≠ 0} .
−1
(a) Domain of R.
(b) Range of R. (e) Using the fact that the domain of Mathematics for Secondary Schools
(c) Inverse of R. R is the range of R , the range of
−1
(d) Domain of R −1. R = {y : y ∈ ℝ, y ≠ 1} or { : is
−1
(e) Range of R . all real numbers except 1}.
−1
Solution
(a) Since the denominator of a fraction
cannot be equal to zero, then in the

Student\s Book Form Three 21

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MATHEMATIC F3 SB.indd 21 18/09/2025 09:58:45
MATHEMATIC F3 SB.indd 21