LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

CONDITION FOR RELATION TO BE A FRACTION

**Condition for a
relation to be a function –**

**1) When the relation is expressed in the roster from – every element of A must appear as
first element of an ordered pair. Also, the first element of the ordered pairs
must be different.**

**In other
words, two or more ordered pairs of the relation must not have the same first
element. However, the second element of the ordered pairs may be repeated.**

**2) when the relation is represented by an arrow diagram – every element of A must have one
and only one image in B. However, two or more elements of A may have the same
image in B**

**There are
some example are given below -**

**Example.1) Determine whether the
given relation can be a function.**

**a) R = {(5,
8), (6, 8), (7, 8)},
**

**the first
elements of the ordered pairs of R are different. So, this relation can be a
function **

**b) R = {(7, 2), (0, 0), (7, -2)}**

**this relation
cannot be a function because the number ‘7’ is the first elements in more than
one ordered pair.**

**Example.2) Does the arrow diagram
represent a function? If, so express the function in the roster form and in the
equation form. Also, find the domain and range of the function**

**Ans.) The relation is a function from set A to set B,**

**because each
element of A has a unique image in B the
function f in the roster form is **

** So, f = {(-2, 4), (-1, 1), (1, 1), (2, 4),
(3, 9)}**

**Since, f (-2)
= 4 = (-2)²,**

** f(-1) = 1 = (-1)²,**

** f(1) = 1 = 1²,**

** f(2) = 4 = 2²**

**and, f(3) = 9 = 3², we have f(x) = x², x Є A,**

**so, the
equation form of function f is –**

**f(x) =
x² or
f : x → x², x Є A**

**the domain of
f = {-2, -1, 1, 2, 3} and the range of f = {1, 4, 9}**

**Example.3) Let A = {2, 4, 7}, B =
{4, 16, 49, 81} and f : A → B where f(x) = x²,
x Є A.**

**a) Draw an arrow diagram for this mapping**

**b) Represent f in the roster form**

**c)
Find the domain and range of ‘f’**

** Ans.)
here (x) = x², so f(2) = 2² = 4 **

** f(4) =
4² = 16**

** f(7) =
7² = 49**

**a) the arrow diagram is shown upside**

**b) the ordered pairs of
f = (2, 4), (5, 25), (7, 49)**

** so, in the roster
form f = {(2, 4), (5, 25), (7, 49)}**

**c) The domain of f = {2, 5, 7}**

** the range of f = {4,
25, 49}**

**Example.4) Let A = {1, 3, 5, 7}, B =
{2, 3, 5, 8} and R be the relation “is less than or equal to” from A to B.**

**a) Write R in the roster form**

**b) Represent by an arrow diagram**

**c) find the domain and the range of
R**

**d) is R a mapping ? give reasons.**

**Answer) 1 < 2, 1
< 3, 1 < 5, 1 < 8, 3 = 3, 3 < 5, 3 < 8, 5 = 5, 5 < 8, 7 <
8**

**So, the (1, 2), (1, 3), (1, 5), (1, 8), (3, 3), (3, 5), (3,
8), (5, 5), (5, 8), (7, 8)**

**a)
the relation in the roster form is R = {(1, 2), (1, 3), (1, 5), (1, 8), (3, 3),
(3, 5), (3, 8), (5, 5), (5, 8), (7, 8)}**

**b) the adjoining arrow diagram represents R**

**c) the domain of R = {1, 3, 5, 7}, the range of R = {2, 3,
5, 8}**

**d) this relation is not a mapping because the members 3 and
5 of the set A are matched with more than one member of the set B.**

**Example.5) Draw an arrow diagram for
the relation R = {(-1, -3), (0, 0), (1, 3), (2, 6)}, is this relation a
function? If so, express it in the form of an equation, also find the domain
and the range of function.**

**Ans.) The arrow
diagram has shown alongside –**

**The relation is a function from set A to set B, because each
element of A has a unique image in B. Since the image of each element of A is
thrice the element, the function can be expressed as **

**So, f(x) = 3x, x Є A or f : x → 3x, x Є A**

**The domain of f = {-1, 0, 1, 2} and the range of f = {-3, 0,
3, 6}**

**Example.6) Let, f : x → 3x + 2,
where A = {1, 2, 3, 4} and B is the set of images of the elements of A.**

**a) Find B & f**

**b) Draw an arrow diagram**

**c) Fin the domain and range of f**

**Answer) Here, f(x) =
3x + 2 …………………….(1)**

**So now, A = {1, 2, 3, 4}, substituting x = 1, 2, 3, 4
respectively in (1), we get**

**f(1) = 3x + 2 = 3.1 + 2 = 5**

**f(2) = 3x + 2 = 3.2 + 2 = 8**

**f(3) = 3x + 2 = 3.3 + 2 = 11**

**f(4) = 3x + 2 = 3.4 + 2 = 14**

**so, the images of the elements 1, 2, 3, & 4 of A are 5,
8, 11, & 14 respectively.**

**a) B = set of images
of elements of A = {5, 8, 11, 14}**

**so, f = {(1, 5), (2, 8), (3, 11), (4, 14)}**

**b) the arrow diagram is shown above**

**c) the domain of f = {1, 2, 3, 4} and the range is f = {5,
8, 11, 14}**