Question

Let $ f:R\to R $ be a differentiable function and $ f(1)=4. $ Then the value of $ \underset{x\to 1}{\mathop{\lim }}\,\int_{4}^{f(x)}{\frac{2t}{x-1}}dt, $ if $ f'(1)=2 $ is:

• A. 16
• B. 8
• C. 4
• D. 2
• E. 1

Answer 1

Answer: (A)

$ \underset{x\to 1}{\mathop{\lim }}\,\frac{\int_{4}^{f(x)}{2t\,\,dt}}{x-1} $
$ =\underset{x\to 1}{\mathop{\lim }}\,\frac{2f(x).f(x)}{1} $
$ =2f(1).f(1)=2.4.2=16 $