Question

If \[ R=\{(x,y)\mid x,y\in \mathbb{R},\ x^2+y^2=1\} \] is a relation in \(\mathbb{R}\), then \(R\) is: 

• A. \( \{0,1,2\} \)
• B. \( \{0,-1,-2\} \)
• C. \( \{-2,-1,0,1,2\} \)
• D. \( \{-1,0,2\} \)

Hint:

To find the domain of a relation defined by an equation or inequality involving \(x\) and \(y\):
• Focus on the first coordinate \(x\).
• Find all values of \(x\) for which at least one valid value of \(y\) exists.
• For inequalities involving squares, use the fact that: \[ x^2\geq 0,\qquad y^2\geq 0 \] to simplify the range quickly. In integer-based relations, always check whether the required values are integers.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A relation R on a set A is a subset of the Cartesian product \(A \times A\).
The domain of a relation is the set of all first coordinates (the \(x\)-values) of the ordered pairs that belong to the relation.
In this specific problem, the relation is defined over the set of integers \(\mathbb{Z}\) and is constrained by the algebraic inequality \(x^2 + y^2 \leq 4\).
This represents a disk centered at the origin with a radius of 2.
Because we are working in the set of integers, the domain consists of all integer values of \(x\) for which there exists at least one integer value of \(y\) satisfying the given inequality.
Step 2: Key Formula or Approach:
The inequality is \(x^2 + y^2 \leq 4\).
Since \(y^2\) must be non-negative for any real or integer \(y\) (\(y^2 \geq 0\)), the maximum possible value for \(x^2\) occurs when \(y^2\) is at its minimum.
The minimum value of \(y^2\) is 0 (when \(y = 0\)).
Substituting \(y = 0\) gives the upper bound for \(x^2\): \(x^2 \leq 4\).
This simplifies to finding all integers \(x\) such that \(|x| \leq 2\).
Step 3: Detailed Explanation:
Let's test potential integer values for \(x\) to see if a valid integer \(y\) exists:
If \(x^2 \leq 4\), the possible integer values for \(x\) are \(-2, -1, 0, 1, 2\).
We now verify each value:
1. For \(x = -2\): \((-2)^2 + y^2 \leq 4 \implies 4 + y^2 \leq 4 \implies y^2 \leq 0\). The only integer solution is \(y = 0\). Since a \(y\) exists, \(-2\) is in the domain.
2. For \(x = -1\): \((-1)^2 + y^2 \leq 4 \implies 1 + y^2 \leq 4 \implies y^2 \leq 3\). Integer solutions for \(y\) are \(\{-1, 0, 1\}\). Since \(y\) values exist, \(-1\) is in the domain.
3. For \(x = 0\): \(0^2 + y^2 \leq 4 \implies y^2 \leq 4\). Integer solutions for \(y\) are \(\{-2, -1, 0, 1, 2\}\). Thus, 0 is in the domain.
4. For \(x = 1\): \(1^2 + y^2 \leq 4 \implies y^2 \leq 3\). Integer solutions for \(y\) are \(\{-1, 0, 1\}\). Thus, 1 is in the domain.
5. For \(x = 2\): \(2^2 + y^2 \leq 4 \implies 4 + y^2 \leq 4 \implies y^2 \leq 0\). Only \(y = 0\) works. Thus, 2 is in the domain.
If we were to try \(x = 3\), we would get \(3^2 + y^2 \leq 4 \implies 9 + y^2 \leq 4 \implies y^2 \leq -5\). No real or integer \(y\) exists.
Therefore, the collection of all valid first components is \(\{-2, -1, 0, 1, 2\}\).
Step 4: Final Answer:
The domain of the relation R is the set \(\{-2, -1, 0, 1, 2\}\).
This corresponds precisely with Option (C).