Question

Let $A = \{x : x \in \mathbb{R}, \, x \text{ is not a positive integer}\}$. Define $f: A \to \mathbb{R}$ as $f(x) = \dfrac{2x}{x-1}$, then $f$ is:

• A. Injective but not surjective
• B. Surjective but not injective
• C. Bijective
• D. Neither injective nor surjective

Hint:

For rational functions of the form \(\frac{ax+b}{cx+d}\), the horizontal asymptote \(y = a/c\) is always excluded from the range. Since the codomain here is \(R\), the exclusion of \(y=2\) immediately proves the function is not surjective.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To determine the nature of the function, we check for injectivity (one-to-one) by seeing if \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\). We check for surjectivity (onto) by determining if the range of the function is equal to the codomain (\(R\)).

Step 2: Detailed Explanation:
1. Injectivity: Assume \(f(x_1) = f(x_2)\): \[ \frac{2x_1}{x_1-1} = \frac{2x_2}{x_2-1} \] \[ 2x_1(x_2-1) = 2x_2(x_1-1) \] \[ x_1x_2 - x_1 = x_2x_1 - x_2 \implies -x_1 = -x_2 \implies x_1 = x_2 \] Thus, the function is Injective. 2. Surjectivity: Let \(y = \frac{2x}{x-1}\). We solve for \(x\) in terms of \(y\): \[ y(x-1) = 2x \implies yx - y = 2x \implies x(y-2) = y \] \[ x = \frac{y}{y-2} \] For \(x\) to be defined, \(y \neq 2\). Additionally, the domain \(A\) excludes positive integers. If we pick a value of \(y\) such that \(x\) results in a positive integer (e.g., if \(y=4\), then \(x = 4/(4-2) = 2\)), that \(y\) has no pre-image in set \(A\). Furthermore, \(y=2\) is never reached because the horizontal asymptote is \(y=2\). Since the range does not cover all of \(R\), it is not surjective.

Step 3: Final Answer
The function is injective but not surjective.