Question

For the mapping $f: \mathbb{R} \setminus \{1\} \to \mathbb{R} \setminus \{2\}$ given by $f(x) = \frac{2x}{x - 1}$, which of the following is correct?

• A. f is one-one but not onto
• B. f is onto but not one-one
• C. f is neither one-one nor onto
• D. f is both one-one and onto

Answer 1

The Correct Option is D

Given: The mapping \( f: \mathbb{R} \setminus \{1\} \to \mathbb{R} \setminus \{2\} \) is defined by \( f(x) = \frac{2x}{x - 1} \). We are tasked with determining whether the function is one-one (injective), onto (surjective), or both. Step 1: Check if the function is one-one (injective). A function \( f \) is one-one (injective) if for all \( x_1, x_2 \in \mathbb{R} \setminus \{1\} \), whenever \( f(x_1) = f(x_2) \), we must have \( x_1 = x_2 \). Let's suppose \( f(x_1) = f(x_2) \), i.e., $$ \frac{2x_1}{x_1 - 1} = \frac{2x_2}{x_2 - 1}. $$ Cross-multiplying: $$ 2x_1(x_2 - 1) = 2x_2(x_1 - 1) $$ Simplifying: $$ x_1 x_2 - x_1 = x_2 x_1 - x_2 $$ This simplifies further to: $$ -x_1 = -x_2 $$ Thus, \( x_1 = x_2 \). Therefore, the function \( f \) is injective (one-one). Step 2: Check if the function is onto (surjective). A function \( f \) is onto (surjective) if for every \( y \in \mathbb{R} \setminus \{2\} \), there exists an \( x \in \mathbb{R} \setminus \{1\} \) such that \( f(x) = y \). We need to solve for \( x \) in the equation \( f(x) = y \): $$ \frac{2x}{x - 1} = y $$ Multiplying both sides by \( x - 1 \): $$ 2x = y(x - 1) $$ Expanding the right-hand side: $$ 2x = yx - y $$ Rearranging: $$ 2x - yx = -y $$ Factoring out \( x \) on the left-hand side: $$ x(2 - y) = -y $$ Solving for \( x \): $$ x = \frac{-y}{2 - y} $$ For \( x \) to be defined, the denominator \( 2 - y \) must not be 0, which means \( y \neq 2 \). Thus, for every \( y \in \mathbb{R} \setminus \{2\} \), we can find an \( x \in \mathbb{R} \setminus \{1\} \) such that \( f(x) = y \). Therefore, the function \( f \) is surjective (onto). Step 3: Conclusion. Since the function \( f \) is both injective (one-one) and surjective (onto), the correct answer is: f is both one-one and onto.