Question

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

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

Answer 1

Answer: (D)

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

$\frac{{(1+x)}^{\frac{3}{2}}-{(1+\frac{1}{2x})}^3}{{(1-x)}^{\frac{1}{2}}}$ $=\frac{1+\frac{3}{2}x+\frac{\frac{3}{2}-\frac{1}{2}}{2}{x}^2-(1-\frac{3x}{2}+\frac{3.2}{2}.\frac{x^2}{4})}{{(1-x)}^{\frac{1}{2}}}$

$=\frac{-\frac{3}{8}{x}^2}{{(1-x)}^{\frac{1}{2}}}=\frac{-3}{8}{x}^2{(1-x)}^{\frac{-1}{2}}$$=\frac{3}{8}{x}^2(1+\frac{x}{2}+...)=\frac{3}{8}{x}^2$