Question

Suppose f : R ® R is a differentiable function and f(1) = 4. Then the value of $\lim_{x \rightarrow 1}\int_{4}^{f(x)}\frac{2t}{(x - 1)}dt$ is

• A. 8 f '(1)
• B. 4 f '(1)
• C. 2f '(1)
• D. f '(1)

Answer 1

Answer: (A)

8 f '(1)

Answer 2

$\lim _ { x \rightarrow 1 }$ $\frac{\int_{4}^{f(x)}{2t\mspace{6mu} dt}}{x - 1}$

= $\lim _ { x \rightarrow 1 }$ $\frac{2f(x)\mspace{6mu} f'(x) - 0}{1 - 0}$ = 2f(1) f ¢(1) = 8f ¢(1)