Check Which Relation Defines a Function
🎥 Video Explanation
📝 Question
Let \(A = B = \{x \in \mathbb{R} : -1 \le x \le 1\}\) and \(C = \{x \in \mathbb{R} : x \ge 0\}\).
\[ S = \{(x, y) \in A \times B : x^2 + y^2 = 1\} \]
\[ S_0 = \{(x, y) \in A \times C : x^2 + y^2 = 1\} \]
- A. S defines a function from A to B
- B. S₀ defines a function from A to C
- C. S₀ defines a function from A to B
- D. S defines a function from A to C
✅ Solution
🔹 Step 1: Equation Meaning
\[ x^2 + y^2 = 1 \] represents a unit circle.
\[ y = \pm \sqrt{1 – x^2} \]
—🔹 Step 2: Relation S
Here \(y \in [-1,1]\).
For each \(x \in [-1,1]\), two values:
\[ +\sqrt{1-x^2}, \quad -\sqrt{1-x^2} \]
❌ Not a function (one input → two outputs)
—🔹 Step 3: Relation S₀
Here \(y \in [0,\infty)\).
Only positive root allowed:
\[ y = \sqrt{1-x^2} \]
✔️ Function (one output per input)
—🔹 Final Answer
\[ \boxed{\text{Option B is correct}} \]