Question
If $$x$$ is so small that $${x^3}$$ and higher powers of $$x$$ may be neglected, then $$\frac{{{{\left( {1 + x} \right)}^{\frac{3}{2}}} - {{\left( {1 + \frac{1}{2}x} \right)}^3}}}{{{{\left( {1 - x} \right)}^{\frac{1}{2}}}}}$$ may be approximated as
A.
$$1 - \frac{3}{8}{x^2}$$
B.
$$3x + \frac{3}{8}{x^2}$$
C.
$$ - \frac{3}{8}{x^2}$$
D.
$$\frac{x}{2} - \frac{3}{8}{x^2}$$
Answer :
$$ - \frac{3}{8}{x^2}$$
Solution :
$$\because \,\,{x^3}$$ and higher powers of $$x$$ may be neglected
$$\eqalign{
& \therefore \,\,\frac{{{{\left( {1 + x} \right)}^{\frac{3}{2}}} - {{\left( {1 + \frac{x}{2}} \right)}^3}}}{{{{\left( {1 - x} \right)}^{\frac{1}{2}}}}} \cr
& = {\left( {1 - x} \right)^{\frac{{ - 1}}{2}}}\left[ {\left( {1 + \frac{3}{2}x + \frac{{\frac{3}{2}.\frac{1}{2}}}{{2!}}{x^2}} \right) - \left( {1 + \frac{{3x}}{2} + \frac{{3.2}}{{2!}}\frac{{{x^2}}}{4}} \right)} \right] \cr
& = \left[ {1 + \frac{x}{2} + \frac{{\frac{1}{2}.\frac{3}{2}}}{{2!}}{x^2}} \right]\left[ {\frac{{ - 3}}{8}{x^2}} \right] = \frac{{ - 3}}{8}{x^2} \cr} $$
(as $${x^3}$$ and higher powers of $$x$$ can be neglected)