Question

If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected, then $\frac{{\left(1+x\right)}^{\frac{3}{2}}-{\left(1+\frac{1}{2}x\right)}^3}{{\left(1-x\right)}^{\frac{1}{2}}}$ may be approximated as

• A. $1-\frac{3}{8}{x}^2$
• B. $3x+\frac{3}{8}{x}^2$
• C. $-\frac{3}{8}{x}^2$
• D. $\frac{x}{2}-\frac{3}{8}{x}^2$

Answer 1

Answer: (C)

$\because x^3$ and higher powers of x may be neglected
$\therefore \frac{\left(1+x\right)\frac{3}{2}-{\left(1+\frac{x}{2}\right)}^3}{\left(1-x^{\frac{1}{2}}\right)}$
$={\left(1-x\right)}^{\frac{-1}{2}}\left[\left(1+\frac{3}{2}x+\frac{\frac{3}{2}.\frac{1}{2}}{2!}{x}^2\right)-\left(1+\frac{3x}{2}+\frac{3.2}{2!}\frac{x^2}{4}\right)\right]$
$=\left[1+\frac{x}{2}+\frac{\frac{1}{2}.\frac{3}{2}}{2!}{x}^2\right]\left[\frac{-3}{8}{x}^2\right]=\frac{-3}{8}{x}^2$
(as $x^3$ and higher powers of x can be neglected)