Question

If x is so small that $x^3$ and higher powers of x may be neglected, then $\frac{(1+x{)}^{3/2}-{\left(1+\frac{1}{2}x\right)}^3}{(1-x{)}^{1/2}}$ may be approximately 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 $-\frac{3}{8}{x}^2$

This question uses binomial expansion

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

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

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

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

$=-\frac{3x^2}{8}$ (neglecting $x^3$ and higher powers)

This is the required answer.