Question

Let A = { x : x ∈ R; x is not a positive integer } Define f : A → R as f(x) = \(\frac{2x}{x-1}\), then f is

• A. injective but not surjective
• B. surjective but not injective
• C. bijective
• D. neither injective nor surjective

Answer 1

The Correct Option is A

The function \( f(x) = \frac{2x}{x-1} \) is defined for all \( x \in \mathbb{R} \) except for \( x = 1 \) (since the denominator cannot be zero). Now, let's analyze whether the function is injective (one-to-one) and surjective (onto).
Injective (One-to-one): A function is injective if, for every pair of distinct elements \( x_1 \) and \( x_2 \), \( f(x_1) \neq f(x_2) \). To check if \( f \) is injective, we solve for \( x_1 \) and \( x_2 \) such that: \[ f(x_1) = f(x_2) \] \[ \frac{2x_1}{x_1 - 1} = \frac{2x_2}{x_2 - 1} \] Cross-multiply and solve: \[ 2x_1(x_2 - 1) = 2x_2(x_1 - 1) \] Simplifying: \[ x_1x_2 - x_1 = x_2x_1 - x_2 \] \[ -x_1 = -x_2 \] \[ x_1 = x_2 \] Therefore, the function is injective. **Surjective (Onto)**: A function is surjective if for every element \( y \in \mathbb{R} \), there exists an \( x \in A \) such that \( f(x) = y \). Let's examine if the function covers all real numbers: \[ f(x) = \frac{2x}{x-1} \] We can see that there are values of \( x \) for which the function will not reach certain values in \( \mathbb{R} \). For example, the function cannot take the value \( 2 \) since: \[ \frac{2x}{x-1} = 2 \quad \Rightarrow \quad 2x = 2(x-1) \quad \Rightarrow \quad 2x = 2x - 2 \] This results in a contradiction, meaning the function is not surjective.

Therefore, \( f \) is injective but not surjective.