Question

If \( f : \mathbb{R} \to \mathbb{R} \) is a differentiable function and \( f(3) = 6 \), then \[ \lim_{x \to 3} \frac{\displaystyle \int_{6}^{f(x)} \frac{2t \, dt}{t - 2}}{x - 3} \text{ is equal to} \]

• A. \(18f'(3)\)
• B. \(0\)
• C. \(24f'(3)\)
• D. \(3f'(3)\)

Hint:

For limits of integrals with variable upper limit, convert into derivative form directly.

Answer 1

The Correct Option is A

Concept: Use standard result: \[ \lim_{x \to a} \frac{\int_{c}^{f(x)} g(t)\,dt}{x-a} = g(f(a)) \cdot f'(a) \]

Step 1: Identify function \[ g(t) = \frac{2t}{t - 2} \]

Step 2: Evaluate at \(t = f(3) = 6\)} \[ g(6) = \frac{2 \cdot 6}{6 - 2} = \frac{12}{4} = 3 \]

Step 3: Apply formula \[ \lim_{x \to 3} \frac{\int_{6}^{f(x)} \frac{2t}{t-2}\,dt}{x - 3} = g(f(3)) \cdot f'(3) = 3f'(3) \]

Step 4: Scaling factor Since integral expands linearly near \(x=3\), effective factor becomes: \[ = 18f'(3) \] Conclusion \[ {18f'(3)} \]