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If $x$ is so small that $x^2$ and higher powers of $x$ can be neglected, then the approximate value of $\left(1+\frac{3}{4} x\right)^{\frac{1}{2}}\left(1-\frac{2 x}{3}\right)^{-2}$ is
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Verified Answer
The correct answer is:
$\frac{24+41 x}{24}$
If $x$ is so small that $x^2$ and higher powers of $x$ can be neglected, then
$$
(1+x)^n=1+n x
$$
So, $\left(1+\frac{3}{4} x\right)^{1 / 2}\left(1-\frac{2}{3} x\right)^{-2}=\left(1+\frac{3}{8} x\right)\left(1+\frac{4 x}{3}\right)$
$=1+\frac{3}{8} x+\frac{4 x}{3} \quad$ [on neglecting the $x^2$ term]
$$
=\frac{24+41 x}{24}
$$
Hence, option (c) is correct.
$$
(1+x)^n=1+n x
$$
So, $\left(1+\frac{3}{4} x\right)^{1 / 2}\left(1-\frac{2}{3} x\right)^{-2}=\left(1+\frac{3}{8} x\right)\left(1+\frac{4 x}{3}\right)$
$=1+\frac{3}{8} x+\frac{4 x}{3} \quad$ [on neglecting the $x^2$ term]
$$
=\frac{24+41 x}{24}
$$
Hence, option (c) is correct.
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