Question

$A=[-1,1],B=[0,1],C=[-1,0]$

$S_1=\{($$\frac{(x,y)}{x^2+y^2}$$=1,x\in A,y\in A\}$

$S_2$$=\left\{\right($$\frac{(x,y)}{x^2+y^2}$$=1,x\in A,y\in B\}$

$S_3$$=\left\{\right($$\frac{(x,y)}{x^2+y^2}$$=1,x\in A,y\in C\}$

$S_4$ $=\left\{\right($$\frac{(x,y)}{x^2+y^2}$$=1,x\in B,y\in C\}$

• A. $S_1$ is not a graph of a function
• B. $S_2$ is not a graph of a function
• C. $S_3$ is not a graph of a function
• D. $S_4$is not a graph of a function

Answer 1

The correct option is A

$S_1$ is not a graph of a function

Explanation for correct answer:

The concept of graph is that "A set of points in the plane is the graph of a function if and only if no vertical line intersects the graph in more than one point."

Here $S_1$​ is not a graph of a function as there is a vertical line which cuts the circle in more than one points these points are $(0,1)$and $(0,-1)$and the vertical line is the $\mathrm{y}-\mathrm{axis}$.

(This method is called vertical line test)But $S_2,S_3,S_4$ are graphs of a function by vertical line test.

$0^2+1^2=1$

$0^2+{(-1)}^2=1$

$$\begin{gathered} (0,1)\in S_1 \\ (0,-1)\in S_1 \end{gathered}$$

Hence, $S_1$is not a graph of a function.

Hence Option(A) is correct.