Question

Which of the following is not function
A) $(x,y)\left| {y = x + 3,x \in N,4 \leqslant x \leqslant 8} \right.$
B) $(x,y)\left| {{y^2} = {x^2} + 3,x \in Z, - 1 \leqslant x \leqslant 1} \right.$
C) \[(x,y)\left| {y = 2x - 3,x \in N,2 \leqslant x \leqslant 6} \right.\]
D) $(x,y)\left| {y = 3 - x,x \in N,5 \leqslant x \leqslant 10} \right.$

Answer 1

Hint: According to the question we have to determine which of the following options is correct by solving each of the given options. But first of all we have to understand the condition of an expression to be a function as explained below.
Functions are relations that derive one output for each input, or one value of y for any value of x inserted into the given expression/equation. In graphical terms, a function is a relation where the first number in the ordered pair has one and only one value as it’s second number, the other part of the ordered pair.
So, first of all we have to solve the given option (A) which is $(x,y)\left| {y = x + 3,x \in N,4 \leqslant x \leqslant 8} \right.$ So to determine the given expression \[\left| {y = x + 3,x \in N,4 \leqslant x \leqslant 8} \right.\] by substituting the value of x which is a natural number between 4 to 8.

Complete step-by-step answer:
Step 1: First of all we will check the option (A) which is $(x,y)\left| {y = x + 3,x \in N,4 \leqslant x \leqslant 8} \right.$
By substituting the value of x in between $4 \leqslant x \leqslant 8$as given in the expression so, first of all we will substitute the value of x = 4 in the expression,
$
   \Rightarrow y = 4 + 3 \\
   \Rightarrow y = 7
 $
Now, on substituting y = 8 in the expression,
$
   \Rightarrow y = 8 + 3 \\
   \Rightarrow y = 11
 $
Hence, by substituting the value of y = 4 and y = 8 we can say that option (A) is true.
Step 2: Same as the step 1 we will determine the option (B) which is $(x,y)\left| {{y^2} = {x^2} + 3,x \in Z, - 1 \leqslant x \leqslant 1} \right.$ Hence,
By substituting the value of x in between $ - 1 \leqslant x \leqslant 1$as given in the expression so, first of all we will substitute the value of x = -1 in the expression,
$
   \Rightarrow {y^2} = {( - 1)^2} + 3 \\
   \Rightarrow {y^2} = 1 + 3 \\
   \Rightarrow {y^2} = 4 \\
   \Rightarrow y = \pm 2
 $
Now, on substituting y = 1 in the expression,
$
   \Rightarrow {y^2} = 1 + 3 \\
   \Rightarrow y = \sqrt 4 \\
\Rightarrow y = \pm 2
 $
Hence, by substituting the value of y = -1 and y = 1 there are 2 roots or solutions obtained hence we can say that option (B) is not true.
Means (B) is not a function.
Step 3: Now, we will check the option (A) which is \[(x,y)\left| {y = 2x - 3,x \in N,2 \leqslant x \leqslant 6} \right.\]
By substituting the value of x in between \[2 \leqslant x \leqslant 6\]as given in the expression so, first of all we will substitute the value of x = 2 in the expression,
\[
   \Rightarrow y = 2x - 3 \\
   \Rightarrow y = 4 - 3 \\
   \Rightarrow y = 1
 \]
Now, on substituting y = 8 in the expression,
\[
   \Rightarrow y = 2 \times 6 - 3 \\
   \Rightarrow y = 12 - 3 \\
   \Rightarrow y = 9
 \]
Hence, by substituting the value of y = 2 and y = 6 only one value of y is obtained so we can say that option (C) is true and it is a function.
Step 4: Now, we will check the option (A) which is $(x,y)\left| {y = 3 - x,x \in N,5 \leqslant x \leqslant 10} \right.$
By substituting the value of x in between $5 \leqslant x \leqslant 10$as given in the expression so, first of all we will substitute the value of x = 5 in the expression,
\[
   \Rightarrow y = 3 - 5 \\
   \Rightarrow y = - 2
 \]
Now, on substituting y = 10 in the expression,
\[
   \Rightarrow y = 3 - 10 \\
   \Rightarrow y = - 7
 \]
Hence, by substituting the value of y = 5 and y = 10 only one value of y is obtained so we can say that option (D) is true and it is a function.

Hence, by solving each of the options we have determined that the correct option is (B) $(x,y)\left| {{y^2} = {x^2} + 3,x \in Z, - 1 \leqslant x \leqslant 1} \right.$ which is not a function.

Note: An order of a pair is a point on x-y coordinate graph with an x and y value for example (3, -3) is an order pair with 3 as the value of x and -3 as the value of y.
It is relatively easy to determine whether an equation is a function by solving for y. When you are going an equation and specific value for x, there should only be one corresponding value of y for that value of x.