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

1 + \frac{11 x}{12}

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2 + \frac{35 x}{6}

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1 - \frac{5 x}{12}

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1 + \frac{5 x}{12}

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Solution

The correct option is B $2 + \frac{35 x}{6}$
According to fractional index pension neglecting $x^{2}$ & Ligher terms
$\frac{\sqrt{1 + x} + \sqrt[3]{1 + 4 x}}{(1 + x^{2}) \sqrt[3]{{(1 + 3 x)}^{2}}} = \frac{1 + \frac{1}{2} x + 1 + \frac{4}{3} x}{(1 + x^{2}) (1 - 2 x)} = \frac{2 + \frac{11}{6} x}{(1 - 2 x)}$
$= \frac{(1 + \frac{11}{6} x) (1 + 2 x)}{1 + 4 x^{2}} = 2 + \frac{35}{6} x$
$B$ is correct

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