Relations & Functions
One-One and Onto Functions
Grade 12
Question:
<p>Consider the function \(f : \mathbb{R} - \{1\} \to \mathbb{R} - \{2\}\) given by \(f(x) = \frac{2x}{x-1}\). Then:</p>
<p>(a) \(f\) is one-one but not onto</p>
<p>(b) \(f\) is onto but not one-one</p>
<p>(c) \(f\) is neither one-one nor onto</p>
<p>(d) \(f\) is both one-one and onto</p>
Step-by-Step Solution
Key Concept: Check one-one by solving \(f(x_1) = f(x_2)\) and verifying \(x_1 = x_2\). Check onto by solving \(f(x) = y\) for \(x\) in terms of \(y\).
<p>To verify one-one: Suppose \(f(x_1) = f(x_2)\). Then \(\frac{2x_1}{x_1-1} = \frac{2x_2}{x_2-1}\). Cross-multiplying and simplifying gives \(x_1 = x_2\), so \(f\) is one-one.</p><p>To verify onto: For any \(y \in \mathbb{R} - \{2\}\), we need to find \(x \in \mathbb{R} - \{1\}\) such that \(\frac{2x}{x-1} = y\). Solving for \(x\): \(2x = y(x-1)\) gives \(x = \frac{y}{y-2}\). Since \(y \neq 2\), this \(x\) is well-defined and \(x \neq 1\) (since \(y \neq 2\)). Thus \(f\) is onto.</p><p>∴ Answer is D.</p>
Correct Answer: d