Let \(\mathrm{A}=\left\\{(\mathrm{x}, \mathrm{y}): \mathrm{y}=\mathrm{e}^{\mathrm{x}}, \mathrm{x} \in \mathrm{R}\right\\}, \mathrm{B}=\left\\{(\mathrm{x}, \mathrm{y}): \mathrm{y}=\mathrm{e}^{-\mathrm{x}}, \mathrm{x} \in \mathrm{R}\right\\}\) then (a) \(\mathrm{A} \cap \mathrm{B}=\Phi\) (b) \(\mathrm{A} \cap \mathrm{B} \neq \Phi\) (c) \(A \cup B=R\) (d) \(A \cup B=A\)

) First, we need to find the intersection of A and B (A ∩ B). The intersection of A and B consists of all the points (x, y) that belong to both A and B. For a point (x, y) to belong to A, it must satisfy the condition y = e^x. For a point (x, y) to belong to B, it must satisfy the condition y = e^(-x). We will now set e^x = e^(-x) to find if there's any x for which this equation holds true. \(e^x = e^{-x}\) \(e^{2x} = 1\) As we know, \(e^0 = 1\). So, if x = 0, then both exponents are 0, and the equation holds \(e^{2(0)} = 1\) \(e^0 = 1\) Now, let's check for the corresponding y value when x = 0. For A: y = e^x = e^0 = 1 For B: y = e^(-x) = e^(-0) = e^0 = 1 So, (0, 1) is the point that belongs to both A and B, and therefore, A ∩ B ≠ Φ. Hence, option (b) is correct.

) Now, let's find the union of sets A and B (A ∪ B). The union of A and B consists of all the points (x, y) that belong to either A or B or both sets. Since both A and B include exponential functions with base e, their union represents all points (x, y) in ℝ where x can be any real number. However, notice that while A represents the set of all points with positive y values (y = e^x, where x ∈ R), B represents the set of all points with positive y values as well (y = e^(-x), where x ∈ R). Therefore, the union of A and B doesn't cover negative y values. Since A ∪ B doesn't represent all real numbers (it doesn't include negative y values), option (c) is incorrect. Now let's check the union of the sets. Since set A contains all positive y values for x ∈ R, and set B also contains all positive y values for x ∈ R, A ∪ B includes all the points that belong to A and all the points that belong to B. However, A and B are not the same sets, so A ∪ B ≠ A. Option (d) is incorrect. #Conclusion#: Based on the results of our analysis: - The correct answer for the intersection is A ∩ B ≠ Φ (option b). - The correct answer for the union is A ∪ B ≠ R and A ∪ B ≠ A (options c and d are incorrect).