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If $f$ is twice differentiable and $f''(0)=2$ then $\lim\limits_{x\to 0}\large\frac{2f(x)-3f(2x)+f(4x)}{x^2}$=

$(a)\;6\qquad(b)\;3\qquad(c)\;12\qquad(d)\;None\;of\;these$

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limit=$\lim\limits_{x\to 0}\large\frac{2f'(x)-6f'(2x)+4f'(4x)}{2x}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{2f''(x)-12f''(2x)+16f''(4x)}{2}$
$\Rightarrow \large\frac{6f''(0)}{2}$
$\Rightarrow 6$
Hence (a) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 

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