<p>Let \(f(x) = \frac{\ln g(x)}{g(x)}\). Find the value of \(x\) (in km/h) at which \(f'(x) = 0\), given \(g(x) = \left(\frac{e-1}{50}\right)x + 1\).</p>
Step-by-Step Solution
Key Concept: To find where f'(x) = 0, use the quotient rule on f(x) = ln(g(x))/g(x), then recognize that the numerator equals zero when ln(g(x)) = 1, meaning g(x) = e.
<p><strong>Step 1:</strong> Apply quotient rule to f(x) = ln(g(x))/g(x):</p><p>f'(x) = [g(x) · (1/g(x)) · g'(x) − ln(g(x)) · g'(x)] / [g(x)]²</p><p><strong>Step 2:</strong> Simplify the numerator:</p><p>f'(x) = [g'(x) − ln(g(x)) · g'(x)] / [g(x)]²</p><p>f'(x) = g'(x)[1 − ln(g(x))] / [g(x)]²</p><p><strong>Step 3:</strong> For f'(x) = 0, either g'(x) = 0 or 1 − ln(g(x)) = 0</p><p>Since g'(x) = (e−1)/50 ≠ 0, we need: ln(g(x)) = 1</p><p>Therefore: g(x) = e</p><p><strong>Step 4:</strong> Solve for x:</p><p>((e−1)/50)x + 1 = e</p><p>((e−1)/50)x = e − 1</p><p>x = 50(e − 1)/(e − 1) = 50</p><p>∴ Answer: x = 50 km/h (Option A)</p>
Correct Answer: A