Sets, Relations & Functions
Types of functions
Grade 11
Question:
<p>The function \(f : \mathbb{R} \to \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]\) defined as \(f(x) = \dfrac{x}{1+x^2}\) is</p>
<p>injective but not surjective.</p>
<p>surjective but not injective.</p>
<p>neither injective nor surjective.</p>
<p>invertible.</p>
Step-by-Step Solution
Key Concept: Analyze both injectivity (one-to-one) and surjectivity (onto) by finding the range of f(x) = x/(1+x²) and checking if different x-values map to the same y-value. Use calculus to find extrema and verify if the function attains all values in [-1/2, 1/2].
<p><strong>Step 1: Check Surjectivity (onto)</strong></p><p>Find f'(x) = (1+x²-x·2x)/(1+x²)² = (1-x²)/(1+x²)²</p><p>f'(x) = 0 when x = ±1</p><p>f(1) = 1/2, f(-1) = -1/2, f(0) = 0</p><p>As x→±∞, f(x)→0. The range is exactly [-1/2, 1/2]. ✓ Surjective</p><p><strong>Step 2: Check Injectivity (one-to-one)</strong></p><p>For x ≠ y, we need f(x) ≠ f(y):</p><p>f(x) = f(y) ⟹ x/(1+x²) = y/(1+y²)</p><p>⟹ x(1+y²) = y(1+x²)</p><p>⟹ x + xy² = y + x²y</p><p>⟹ x - y = xy(x-y)</p><p>⟹ (x-y)(1-xy) = 0</p><p>If x ≠ y, then xy = 1, so y = 1/x</p><p>But f(x) = x/(1+x²) and f(1/x) = (1/x)/(1+1/x²) = (1/x)/((x²+1)/x²) = x/(1+x²) = f(x)</p><p>So f(2) = f(1/2), meaning f is NOT injective. ✗ Not injective</p><p><strong>Step 3: Conclusion</strong></p><p>The function is surjective but not injective (many-to-one onto function).</p><p>∴ Answer: B (Surjective but not Injective)</p>
Correct Answer: B