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Suppose $f:R\rightarrow R$ is a differentiable function and $ f(1)=4$.Then value of $\lim\limits_{x\to 1}\int\limits_4^{f(x)}\large\frac{2t}{x-1}$$dt$ is


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$\lim\limits_{x\to 1}\int_4^{f(x)}\large\frac{2t}{x-1}$$dt$
$\Rightarrow \lim\limits_{x\to 1}\large\frac{1}{x-1}$$\big[t^2\big]_4^{f(x)}$
$\Rightarrow \lim\limits_{x\to 1}\large\frac{1}{x-1}$$[(f(x))^2-16]$
$\Rightarrow \large\frac{2f(x)f'(x)}{1}$
$\Rightarrow 2f(1).f'(1)$
$\Rightarrow 2.4.f'(1)$
$\Rightarrow 8f'(1)$
Hence (a) is the correct option.
answered Jan 2, 2014 by sreemathi.v

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