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If $x$ is so small that $x^{2}$ and higher power of $x$ are neglected then $\frac{{(1 + 2 x)}^{\frac{1}{2}} (3 + 4 x)}{\sqrt[3]{1 - 2 x}} =$

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$3 - 5 x$

$3 + 7 x$

$3 - 9 x$

$3 + 9 x$

answer is D.

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Detailed Solution

$\frac{{(1 + 2 x)}^{\frac{1}{2}} (3 + 4 x)}{\sqrt[3]{1 - 2 x}} = {(1 + 2 x)}^{\frac{1}{2}} (3 + 4 x) {(1 - 2 x)}^{- \frac{1}{3}}$

$= (\frac{1 + 2 x}{2}) (3 + 4 x) (1 + \frac{2 x}{3})$

$= (3 + 4 x + 3 x) (1 + \frac{2 x}{3})$

$\Rightarrow 3 + 7 x + 2 x = 3 + 9 x$

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