Analyzing the Set S: Function Determination

The question asks us to determine if the set S represents a function. The set S is defined as: $$S = \{(x, y): x^2 + y^2 = 1, -1 \le x \in \mathbb{R} \le 1 \text{ and } -1 \le y \in \mathbb{R} \le 1\}$$

The condition $x^2 + y^2 = 1$ describes all points $(x, y)$ that lie on a circle centered at the origin (0,0) with a radius of 1. The additional conditions $-1 \le x \le 1$ and $-1 \le y \le 1$ define the range of possible x and y values. For the equation $x^2 + y^2 = 1$, the x-values are automatically within $[-1, 1]$ if y is real, and y-values are automatically within $[-1, 1]$ if x is real. So, the set S is simply the set of all points on the unit circle $x^2 + y^2 = 1$.

Understanding the Definition of a Function

A set of ordered pairs $(x, y)$ is considered a function if for every value of $x$ in the domain, there is exactly one corresponding value of $y$. In other words, for any given input $x$, there must be a unique output $y$. If a single input $x$ can lead to two or more different outputs $y$, then the set does not represent a function.

Testing the Set S for Function Properties

Let's look at the equation $x^2 + y^2 = 1$ that defines the points $(x, y)$ in S. We can try to express $y$ in terms of $x$:

For the set S to be a function, for each valid $x$ value (where $-1 \le x \le 1$), there must be only one $y$ value. Let's pick an $x$ value within the domain $(-1, 1)$, for example, $x = 0$.

If $x = 0$, the equation becomes:

So, for $x = 0$, we have two possible $y$ values: $y = 1$ and $y = -1$. This means both points $(0, 1)$ and $(0, -1)$ are in the set S. Since a single input value ($x=0$) corresponds to two different output values ($y=1$ and $y=-1$), the set S does not satisfy the definition of a function.

Evaluating the Options

Based on our analysis, let's examine the given options:

• Option 1: S is a one - one function. A one-one function requires each unique output to correspond to a unique input. Since S is not even a function, it cannot be a one-one function.
• Option 2: S is a many - one function. A many-one function allows multiple inputs to map to the same output, but still requires each input to map to only one output. Since a single input in S maps to multiple outputs, S is not a many-one function.
• Option 3: S is a bijective mapping. A bijective mapping is both one-one and onto. Since S is not a function, it cannot be a bijective mapping.
• Option 4: S is not a function. As we demonstrated by finding an x-value ($x=0$) that corresponds to two distinct y-values ($y=1$ and $y=-1$), the set S does not satisfy the fundamental definition of a function.

Conclusion on Set S and Functions

The set S, defined by $x^2 + y^2 = 1$, does not represent a function because for many values of $x$ in the domain $(-1, 1)$, there are two corresponding values of $y$. Therefore, S is not a function.

Revision Table: Function Concepts

Additional Information: Graphs and Functions

Visually, a set of points represents a function if and only if any vertical line drawn through the graph intersects the graph at most once. This is known as the Vertical Line Test. The graph of $x^2 + y^2 = 1$ is a circle. If you draw a vertical line through the circle (for example, at $x=0$), it intersects the circle at two points: $(0, 1)$ and $(0, -1)$. Because a vertical line intersects the graph more than once, the set S, the unit circle, does not represent a function of $x$.

While $x^2 + y^2 = 1$ does not define $y$ as a function of $x$, it does define a relation. It's also possible to define $x$ as a function of $y$ for restricted domains/ranges. For example, $x = \sqrt{1 - y^2}$ for $y \in [-1, 1]$ defines the right semi-circle, which is a function of $y$. Similarly, $y = \sqrt{1 - x^2}$ for $x \in [-1, 1]$ defines the upper semi-circle, which is a function of $x$. However, the entire set S, the full circle, is not a function of $x$ or $y$ in the standard definition.