To determine which subsets of \(\R^2\) are bounded, we need to analyze each given subset with respect to the definition of boundedness. A subset \(S \subseteq \R^2\) is bounded if there exists some \(M > 0\) such that for all \((x, y) \in S\), the inequalities \(|x| \leq M\) and \(|y| \leq M\) hold. Let's evaluate each subset:
\({(x, y) \in \R^2 : e^{x^2} + y^2 \le 4}\)
In this case, the term \(\displaystyle e^{x^2}\) increases rapidly as \(|x|\) increases, surpassing any constant bound. For the inequality to hold, \(\displaystyle e^{x^2}\) must remain small, leading to small values of \(|x|\). As a result, this set is bounded because both \(x\) and \(y\) are constrained by the inequality.
\({(x, y) \in \R^2 : x^4 + y^2 \le 4}\)
Here, the growth of \(\displaystyle x^4\) is similar to or faster than quadratic growth. Additionally, for \(x^4\) to remain less than or equal to 4, \(|x|\) must be limited. Similarly, \(y^2\\) being bounded implies \(|y|\) is limited as well. Therefore, this set is also bounded.
\({(x, y) \in \R^2 : |x| + |y| \le 4}\)
This set is clearly bounded since both \(|x|\) and \(|y|\) are explicitly limited by 4. The constraint guarantees that neither coordinate exceeds 4 in absolute value, thus bounding the set.
\({(x, y) \in \R^2 : e^{x^3} + y^2 \le 4}\)
The function \(\displaystyle e^{x^3}\) grows extremely fast, and unlike \(\displaystyle e^{x^2}\), there's no quadratic or linear behavior to counterbalance it when \(|x|\) is large. Therefore, it's possible for \(|x|\) to become arbitrarily large while still satisfying the inequality, especially for negative large values, making the set unbounded.
Based on the evaluations, the bounded subsets are: