If $x$ is so small that $x^2$ and higher powers of $x$ can be neglected, then the approximate value of…
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
$\frac{41+24 x}{41}$
$\frac{41-24 x}{41}$
$\frac{24+41 x}{24}$
$\frac{24-41 x}{24}$
Solution
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