Question

Which of the following does not function?
(a) $(x,y)\left| y=x+3,x\in N,4\le x\le 8 \right.$
(b) \[(x,y)\left| {{y}^{2}}={{x}^{2}}+3,x\in Z,-1\le x\le 1 \right.\]
(c) $(x,y)\left| y=2x-3,x\in N,2\le x\le 6 \right.$
(d) $(x,y)\left| y=3-x,x\in N,5\le x\le 10 \right.$

Answer 1

Hint: Check all the options one by one and confirm that, for no value of x there are multiple values of y. If in any option, we get that there are multiple values of y for a single value of x then that relation will not be a function and the options in which there is only one value of y for each value of x then that relation will be a function.

Complete step-by-step answer:
First we should understand what is a function and its difference with relation.
A function is a relation which describes that there should be only one output for each input. Now, a relation is a set of inputs and outputs that are related in some way. The basic difference between a relation and a function is that, when each input in a relation has exactly one output, the relation is said to be a function. However one output may be related to several inputs. If we want to check if a relation is a function or not, we check that no input has more than one output.
Let us check the options one by one.
(a) $(x,y)\left| y=x+3,x\in N,4\le x\le 8 \right.$
Here, we can see that the relation between x and y is linear. Therefore, for each value of x there will be only one value of y. Hence, it is a function.
(b) \[(x,y)\left| {{y}^{2}}={{x}^{2}}+3,x\in Z,-1\le x\le 1 \right.\]
Here, we can see that the relation is a polynomial of degree 2. Now, if we will substitute x = 1 in the above relation then:
$\begin{align}
  & {{y}^{2}}={{1}^{2}}+3 \\
 & \Rightarrow {{y}^{2}}=4 \\
 & \Rightarrow y=\pm 2 \\
\end{align}$
Clearly, for x = 1, we have two different values of y. Hence, it is not a function.
(c) $(x,y)\left| y=2x-3,x\in N,2\le x\le 6 \right.$
Here, we can see that the relation between x and y is linear. Therefore, for each value of x there will be only one value of y. Hence, it is a function.
(d) $(x,y)\left| y=3-x,x\in N,5\le x\le 10 \right.$
Here, we can see that the relation between x and y is linear. Therefore, for each value of x there will be only one value of y. Hence, it is a function.

Therefore, option (b) is the correct answer.

Note: You may note that, if there is a relation in which, for several values of x there is only a single value of y, then the relation will be a function. It will be a type of function called many-one function. So, do not get confused in the theory. Always remember that a function is defined when, for no value of x there are multiple values of y.