Q. $If\ x\ is\ so\ small\ that\ x^{3}\ and\ higher\ powers\ of\ x\ may\ be\ neglected,\ then$ $\frac{{(1 + x)}^{3/2} - \left( 1 + \frac{1}{2}x \right)^{3}}{{(1 - x)}^{1/2}}\ may\ be\ approximated\ as$
A) $\frac{x}{2} - \frac{3}{8}x^{2}$
B) $- \frac{3}{8}x^{2}$
C) $3x + \frac{3}{8}x^{2}$
D) $1 - \frac{3}{8}x^{2}$
Explanation
(b)
$\frac{(1 + x)^{3/2} - \left( 1 + \frac{1}{2}x \right)^{3}}{(1 - x)^{\frac{1}{2}}}$
$= \frac{\left( 1 + \frac{3}{2}x + \frac{\frac{3}{2} \bullet \frac{1}{2}}{2}x^{2} \right) - \left( 1 + \frac{3x}{2} + \frac{3 \bullet 2}{2} \bullet \frac{x^{2}}{4} \right)}{{(1 - x)}^{1/2}}$
[neglecting higher powers of $x$]
$= - \frac{{3x}^{2}}{8}{(1 - x)}^{- 1/2}$
$= - \frac{{3x}^{2}}{8}\left( 1 + \frac{1}{2}x + \frac{\frac{1}{2} \bullet \frac{3}{2}}{2} \bullet x^{2} \right) = - \frac{{3x}^{2}}{8}$
[neglecting higher powers of $x$]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.