If x is small so that $x^{2}$ and higher powers of x can be neglected then the approximate value for $\frac{(1 - 2 x)^{- 1} (1 - 3 x)^{- 2}}{(1 - 4 x)^{- 3}}$

1 - 2 x

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1 - 3 x

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1 - 5 x

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1 - 4 x

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Solution

The correct option is D $1 - 4 x$
$\frac{(1 - 2 x)^{- 1} (1 - 3 x)^{- 2}}{(1 - 4 x)^{- 3}}$

$\Rightarrow$ $(1 - 2 x)^{- 1} (1 - 3 x)^{- 2} (1 - 4 x)^{3}$
Since, $x$ is very small, then $(1 + x)^{n} = 1 + n x$

$\Rightarrow$ $(1 + 2 x) (1 + 6 x) (1 - 12 x)$
$\Rightarrow$ $(1 + 6 x + 2 x + 12 x^{2}) (1 - 12 x)$
$\Rightarrow$ $(1 + 8 x + 12 x^{2}) (1 - 12 x)$
$\Rightarrow$ $1 + 8 x + 12 x^{2} - 12 x - 96 x^{2} - 144 x^{3}$
$\Rightarrow$ $1 - 4 x - 84 x^{2} - 144 x^{3}$
Since, we have to neglect higher power then, $- 84 x^{2} - 144 x^{3} = 0$
$\Rightarrow$ $1 - 4 x$

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