Let be a differentiable function and . Then, the value of is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
Using the fundamental theorem, the integral ∫₄f(x) 2t dt = [t²]₄f(x) = f(x)² - 16.
So the limit becomes lim(x→1) [f(x)² - 16] / (x - 1).
Since f(1) = 4, this is 0/0 form. Apply L'Hôpital's rule (or recognize it as a derivative): differentiating numerator gives 2f(x)·f'(x), denominator gives 1.
At x = 1: 2f(1)·f'(1) = 2·4·f'(1) = 8f'(1).
So the limit becomes lim(x→1) [f(x)² - 16] / (x - 1).
Since f(1) = 4, this is 0/0 form. Apply L'Hôpital's rule (or recognize it as a derivative): differentiating numerator gives 2f(x)·f'(x), denominator gives 1.
At x = 1: 2f(1)·f'(1) = 2·4·f'(1) = 8f'(1).
[1990]