Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Let $f:R\rightarrow R$ be a differentiable function and $f(1)=4$, then the value of $\large \int_4^{f(x)}\large\frac{2t\;dt}{x-1}$ is


Can you answer this question?

1 Answer

0 votes
We have $f:R\rightarrow R$ a differentiable function and $f(1)=4$
$\lim\limits_{x\to 1}\int\limits_4^{f(x)}\large\frac{2t}{x-1}$$dt=\lim\limits_{x\to 1}\big[\large\frac{t^2}{x-1}\big]_4^{f(x)}$
$\Rightarrow \lim\limits_{x\to 1}\large\frac{(f(x))^2-16}{x-1}$
$\Rightarrow \lim\limits_{x\to 1}\large\frac{f(x)-4}{x-1}.$$\lim\limits_{x\to 1}f(x)+4$
Using $f(1)=4$
$\Rightarrow f'(1).f((1)+4)=8f'(1)$
Hence (a) is the correct answer.
answered Dec 31, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App