Question

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

• A. $\frac{41+24 x}{41}$
• B. $\frac{41-24 x}{41}$
• C. $\frac{24+41 x}{24}$
• D. $\frac{24-41 x}{24}$

Answer 1

Answer: (C)

If $x$ is so small that $x^{2}$ and higher powers of $x$ can be neglected, then
$( l +x)^{n}= 1 +n x$
So, $\left(1+\frac{3}{4} x\right)^{1 / 2}\left(1-\frac{2}{3} x\right)^{-2}=\left(1+\frac{3}{8} x\right)\left(1+\frac{4 x}{3}\right)$
$=1+\frac{3}{8} x+\frac{4 x}{3}\,\,\,$[on neglecting the $x^{2}$ term $]$
$=\frac{24+41 x}{24}$