<p><strong>135.</strong> The function \(f:[0,\infty)\to[0,\infty)\) defined by \(f(x)=\dfrac{2x}{1+2x}\) is:</p>
<p>(a) one-one onto</p>
<p>(b) one-one but not onto</p>
<p>(c) onto but not one-one</p>
<p>(d) neither one-one nor onto</p>
Step-by-Step Solution
Key Concept: Check if f is bijective by verifying both injectivity (one-to-one via monotonicity or algebraic manipulation) and surjectivity (range equals codomain [0,∞)). For surjectivity, solve y = f(x) for x and verify y ∈ [0,∞) implies x ∈ [0,∞).
<p><strong>Step 1: Check Injectivity</strong></p><p>f'(x) = d/dx[2x/(1+2x)] = [2(1+2x) - 2x(2)]/(1+2x)² = 2/(1+2x)² > 0 for all x ≥ 0</p><p>Since f'(x) > 0, f is strictly increasing on [0,∞), hence <strong>injective</strong>.</p><p><strong>Step 2: Check Surjectivity</strong></p><p>From y = 2x/(1+2x), solve for x:</p><p>y(1+2x) = 2x</p><p>y + 2xy = 2x</p><p>y = 2x - 2xy = 2x(1-y)</p><p>x = y/[2(1-y)]</p><p><strong>Step 3: Verify Range</strong></p><p>For y ∈ [0,∞) to have x ∈ [0,∞):</p><p>• When y = 0: x = 0 ✓</p><p>• When y ∈ (0,1): x = y/[2(1-y)] > 0 ✓</p><p>• When y → 1⁻: x → ∞ ✓</p><p>• When y ≥ 1: denominator 2(1-y) ≤ 0, but numerator y > 0, so x would be negative or undefined ✗</p><p>Range of f = [0,1), NOT [0,∞). Therefore f is <strong>injective but NOT surjective</strong>.</p><p>∴ Answer: B (Injective but not Surjective)</p>
Correct Answer: B