<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 both injectivity (one-to-one) and surjectivity (onto) by analyzing the function's behavior: injectivity via monotonicity/derivative, and surjectivity by examining the range relative to codomain [0,∞).
<p><strong>Step 1: Check Injectivity</strong></p><p>Find f'(x): f(x) = 2x/(1+2x)</p><p>f'(x) = [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, hence <strong>injective ✓</strong></p><p><strong>Step 2: Check Surjectivity</strong></p><p>Find range: As x → ∞, f(x) = 2x/(1+2x) → 2x/2x = 1</p><p>At x = 0, f(0) = 0</p><p>Since f is strictly increasing and continuous on [0,∞): Range = [0,1)</p><p>The codomain is [0,∞), but range is [0,1) ⊂ [0,∞)</p><p>Not all elements of [0,∞) have preimages, so <strong>NOT surjective ✗</strong></p><p><strong>Step 3: Conclusion</strong></p><p>The function is <strong>injective but not surjective</strong></p><p>∴ Answer: B</p>
Correct Answer: B