<p>Let \(f: R \to R\) be a differentiable function and \(f(1) = 4\). Then the value of \(\lim_{x \to 1} \int_4^{f(x)} \frac{2t}{x-1} dt\) is</p>
Step-by-Step Solution
Key Concept: Use L'Hôpital's rule on the limit of an integral by recognizing the indeterminate form 0/0, then apply the Fundamental Theorem of Calculus to differentiate the integral with respect to x.
<p><strong>Step 1:</strong> Recognize the form. As x → 1, the numerator ∫₄^f(x) 2t dt → ∫₄^f(1) 2t dt = ∫₄⁴ 2t dt = 0 (since f(1) = 4), and denominator (x-1) → 0. This is a 0/0 indeterminate form.</p><p><strong>Step 2:</strong> Apply L'Hôpital's rule. Differentiate numerator and denominator with respect to x:</p><p>lim_{x→1} [d/dx ∫₄^f(x) 2t dt] / [d/dx(x-1)]</p><p><strong>Step 3:</strong> By Fundamental Theorem of Calculus with chain rule, the derivative of ∫₄^f(x) 2t dt with respect to x is:</p><p>2f(x) · f'(x)</p><p>The derivative of (x-1) is 1.</p><p><strong>Step 4:</strong> Substitute the limit:</p><p>lim_{x→1} [2f(x) · f'(x)] / 1 = 2f(1) · f'(1) = 2(4) · f'(1) = 8f'(1)</p><p>∴ Answer: <strong>8f'(1)</strong> (Option A)</p>
Correct Answer: A