Step 1: Understanding the Concept:
A set \(S\) in \(\mathbb{R}^2\) is convex if for any two points \(P_1, P_2 \in S\), the line segment connecting them, \(\lambda P_1 + (1-\lambda)P_2\) for \(0 \le \lambda \le 1\), is entirely contained within \(S\). We can analyze this geometrically or by testing with points.
Step 2: Detailed Explanation:
A. \( S = \{(x, y) | xy \le 1 \ \).}
This set is not convex. Consider two points in the set, for example, \(P_1 = (2, 0.1)\) and \(P_2 = (0.1, 2)\). Both are in \(S\) because their product is \(0.2 \le 1\). However, their midpoint \(M = (1.05, 1.05)\) is not in \(S\), because \(1.05 \times 1.05 = 1.1025>1\). The line segment connecting \(P_1\) and \(P_2\) bows out of the set.
B. \( S = \{(x, y) | x^2 + 4y^2 \le 12 \ \).}
This inequality describes the area on and inside an ellipse centered at the origin. The interior of an ellipse is a classic example of a convex set. Any line segment connecting two points inside an ellipse remains entirely inside.
C. \( S = \{(x, y) | y^2 - 4x \le 0 \ \).}
This inequality can be written as \(x \ge \frac{y^2}{4}\). This describes the region to the right of (and including) the parabola \(x = y^2/4\), which opens to the right. This region is also a convex set.
D. \( S = \{(x, y) | x^2 + 4y^2 \ge 12 \ \).}
This inequality describes the region on and outside the ellipse from part B. This set is not convex. Consider the points \(P_1 = (4, 0)\) and \(P_2 = (-4, 0)\). Both are in \(S\) as \(16 \ge 12\). The line segment connecting them is the interval on the x-axis from -4 to 4. This segment includes the origin \((0,0)\), which is not in \(S\) since \(0 \ge 12\) is false.
Step 3: Final Answer:
Only the sets described in statements B and C are convex. Therefore, the correct option is (A).