Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$(1+x{)}^{\frac{3}{2}}=1+\frac{3}{2}x+\frac{\frac{3}{2}\times \frac{1}{2}}{2!}{x}^2$

$(1-x{)}^{\frac{-1}{2}}$ = $1+\frac{1}{2}x+\frac{1}{2}\times \frac{3}{2\times 2!}\times x^2$

${(1+\frac{1}{2}x)}^3$ = $1+3\frac{1}{2}x+\frac{3\times 2}{2!}\times \frac{x^2}{4}$

$\Rightarrow (1+x{)}^{\frac{3}{2}}-{(1+\frac{1}{2}x)}^3=x^2(\frac{3}{8}-\frac{3}{4})$

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

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

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

= $\frac{-3}{8}{x}^2-\frac{3}{16}{x}^3-\frac{9}{64}{x}^4$

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