Relations & Functions
One-one and onto functions
Grade 12
Question:
<p>Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = \dfrac{|x|-1}{|x|+1}\), then \(f\) is</p>
<p>both one–one and onto.</p>
<p>one–one but not onto.</p>
<p>onto but not one–one.</p>
<p>neither one–one nor onto.</p>
Step-by-Step Solution
Key Concept: Check if f is even/odd by computing f(-x) and comparing with f(x); for this function, |−x| = |x| immediately reveals f(−x) = f(x), making it even.
<p><strong>Step 1:</strong> Determine the nature of f(x) = (|x|−1)/(|x|+1) by computing f(−x).</p><p><strong>Step 2:</strong> f(−x) = (|−x|−1)/(|−x|+1) = (|x|−1)/(|x|+1) since |−x| = |x| for all x ∈ ℝ.</p><p><strong>Step 3:</strong> Since f(−x) = f(x) for all x in the domain, the function is <strong>even</strong>.</p><p><strong>Step 4:</strong> Verify the function is not odd: f(−x) ≠ −f(x) (for example, f(1) = 0 but −f(−1) = 0, though f(2) = 1/3 and −f(−2) = −1/3).</p><p><strong>Step 5:</strong> The function satisfies all properties of an even function and is symmetric about the y-axis.</p><p>∴ Answer: D (f is an even function)</p>
Correct Answer: D