Question

If

\[ R = \{(x, y) : x, y \in \mathbb{Z}, \, x^2 + 3y^2 \leq 7 \} \] is a relation on the set of integers \( \mathbb{Z} \), then the range of the relation \( R \) is:

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

Hint:

To find the range of a relation, determine all possible values of the second element in the ordered pairs that satisfy the given condition. In this case, solve for \( y \) for different integer values of \( x \).

Answer 1

Answer: (E)

We are given the relation: \[ R = \{(x, y) : x, y \in \mathbb{Z}, x^2 + 3y^2 \leq 7 \}. \] This is a set of pairs of integers \( (x, y) \) where \( x^2 + 3y^2 \leq 7 \). 
Step 1: We will find the possible values for \( y \) for different integer values of \( x \). 
- For \( x = 0 \): \[ 0^2 + 3y^2 \leq 7 \quad \Rightarrow \quad 3y^2 \leq 7 \quad \Rightarrow \quad y^2 \leq \frac{7}{3} \approx 2.33. \] Therefore, \( y^2 \leq 2 \), so \( y \in \{-1, 0, 1\} \). 
- For \( x = \pm 1 \): \[ 1^2 + 3y^2 \leq 7 \quad \Rightarrow \quad 1 + 3y^2 \leq 7 \quad \Rightarrow \quad 3y^2 \leq 6 \quad \Rightarrow \quad y^2 \leq 2. \] Therefore, \( y \in \{-1, 0, 1\} \). 
- For \( x = \pm 2 \): \[ 2^2 + 3y^2 \leq 7 \quad \Rightarrow \quad 4 + 3y^2 \leq 7 \quad \Rightarrow \quad 3y^2 \leq 3 \quad \Rightarrow \quad y^2 \leq 1. \] Therefore, \( y \in \{-1, 0, 1\} \). - For \( x = \pm 3 \): \[ 3^2 + 3y^2 \leq 7 \quad \Rightarrow \quad 9 + 3y^2 \leq 7 \quad \Rightarrow \quad 3y^2 \leq -2. \] This results in no valid solutions for \( y \).
Step 2: From the above analysis, we see that the possible values for \( y \) are \( \{-1, 0, 1\} \).
Therefore, the range of the relation \( R \), which consists of the set of all possible values of \( y \), is: \[ \{0, -1, 1\}. \] 
Thus, the correct answer is option (E).