Question

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)}

Answer 1

To solve this question, we will first of all define a function and its range and domain. Then we will consider all the options separately and see which one lies and matches with the definition of the function. The one that matches its domain and range can be calculated. If $f:X\to Y$ is a function, then the domain is \left\\{ x\in X:f\left( x \right)=y \right\\} and range is \left\\{ y\in Y:f\left( x \right)=y \right\\}.

Complete step-by-step answer:
Let us first define what a function, its domain and range are.
Function: A function is a binary relation between two sets that associates every element of the first set to exactly one element of the second set. A mapping having two images for a single preimage is not a set. If $f:A\to B$ is a function where A and B are sets, then the elements of set A are called preimages and the elements of set B are called images. Example: The mapping is a function as one preimage has one image

But the mapping is not a function as one pre-image ‘b’ has two images 2 and 3. That f(b) = 2 and f(b) = 3 are not acceptable.

The domain of a function: The domains are the set of preimages of the function if $f:X\to Y$ is a function then the domain is \left\\{ x\in X:f\left( x \right)=y \right\\}. So, the domain is a set of all x values such that f(x) = y for $x\in X$ and $y\in Y.$
Range of a function: A range of a function is the set of all ranges of the function if $f:X\to Y$ us a function then the range is given by \left\\{ y\in Y:f\left( x \right)=y \right\\}. So, the range is a set of all y values such that f(x) = y for all $x\in X$ and $y\in Y.$
Now, let us consider our question.
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
This function as by the above definition one pre-image cannot have more than one image but one image can have more than one pre-image. Therefore it is a function.
Domain = {2, 5, 8, 11, 14, 17} as defined from the above definition.
Range = {1} as all are mapping {1} only.
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Clearly, it is a function because not a single preimage has more than one image. Hence, this is a function.
\text{Domain}=\left\\{ 2,4,6,8,10,12,14 \right\\}
\text{Range}=\left\\{ 1,2,3,4,5,6,7 \right\\}
(iii) {(1, 3), (1, 5), (2, 5)}
Here, we observe that one pre-image ‘1’ has two images 3 and 5 and hence it is not a function. So, we cannot define the domain and range of it.

Note: The function of the below type is possible.

\Rightarrow f:\left\\{ \left( a,1 \right),\left( b,1 \right),\left( c,1 \right),\left( d,1 \right) \right\\} is possible in the definition of the function. But the case of the below type is not possible.

i.e. {(a, 1),(a,2),(a,3),(a,4)} is not a function.