Question

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

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

Answer 1

The correct option is B $-\frac{3}{8}{x}^2$

$\frac{(1+x{)}^{3/2}-{(1+\frac{1}{2}x)}^3}{(1-x{)}^{1/2}}$

$\Rightarrow$ $\frac{(1+x{)}^{\frac{3}{2}}-{(1+\frac{x}{2})}^3}{(1-x{)}^{\frac{1}{2}}}$

$\Rightarrow$ $(1-x{)}^{\frac{-1}{2}}[(1+\frac{3}{2}x+\frac{\frac{3}{2}.\frac{1}{2}}{2!}{x}^2)-(1+\frac{3x}{2}+\frac{3.2}{2!}\frac{x^2}{4})]$

$\Rightarrow$ $[1+\frac{x}{2}+\frac{\frac{1}{2}.\frac{3}{2}}{2!}{x}^2]\left[\frac{-3}{8}{x}^2\right]$

It is given that $x$ is small so $x^3$ and higher power can be neglected.

$\Rightarrow$ $\frac{-3}{8}{x}^2$