The problem requires us to determine the nature of the function \( f: A \to \mathbb{R} \) defined as \( f(x) = \dfrac{2x}{x-1} \), where the domain \( A \) excludes positive integers. We need to classify \( f \) as injective, surjective, both (bijective), or neither.
- Understanding Injectivity:
- A function is injective (one-to-one) if different inputs map to different outputs.
- Assume \( f(x_1) = f(x_2) \) for \( x_1, x_2 \in A \), then \( \dfrac{2x_1}{x_1-1} = \dfrac{2x_2}{x_2-1} \).
- Cross-multiplying gives: \( 2x_1(x_2 - 1) = 2x_2(x_1 - 1) \).
- Simplify to \( 2x_1x_2 - 2x_1 = 2x_2x_1 - 2x_2 \), leading to \( -2x_1 = -2x_2 \).
- This simplifies to \( x_1 = x_2 \), confirming injectivity.
- Understanding Surjectivity:
- A function is surjective (onto) if every element of the codomain \( \mathbb{R} \) has a preimage in the domain \( A \).
- We need to determine if for every \( y \in \mathbb{R} \), there exists an \( x \in A \) such that \( f(x) = y \).
- Setting \( \dfrac{2x}{x-1} = y \) leads to \( 2x = y(x - 1) \) and simplifying gives \( x = \dfrac{y}{y - 2} \).
- For \( x \) to be in the domain \( A \), it cannot be a positive integer.
- If \( y = 2 \), the denominator in \( \dfrac{y}{y - 2} \) becomes zero, making \( x \) undefined, thus \( f \) is not surjective since 2 has no preimage.
- Conclusion:
- Since \( f \) is injective but not surjective, the correct classification is that \( f \) is "Injective but not surjective".
The correct option is: Injective but not surjective.