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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


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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

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