Question

If $c$ is very small when compared to $k$, then ${\left(\frac{k}{k+c}\right)}^{\frac{1}{2}}+{\left(\frac{k}{k-c}\right)}^{\frac{1}{2}}=$

• A. $2+\frac{c}{k}$
• B. $2-\frac{c}{k}$
• C. $2+\frac{3c^2}{4k^2}$
• D. $2-\frac{3c^2}{4k^2}$

Answer 1

The correct option is C $2+\frac{3c^2}{4k^2}$

Given, $(\frac{k}{k+c}{)}^{1/2}+(\frac{k}{k-c}{)}^{1/2}$
$=(\frac{k+c}{k}{)}^{-1/2}+(\frac{k+c}{k}{)}^{-1/2}$
$=(1+\frac{c}{k}{)}^{-1/2}+(1-\frac{c}{k}{)}^{-1/2}$
$=(1-\frac{c}{2k}+\frac{3c^2}{2!4k^2}...)+(1+\frac{c}{2k}+\frac{3c^2}{2!4k^2}...)$ ...(neglecting higher terms as $c$ is very small)
$=2+\frac{\left(2\right)3c^2}{8k^2}$
$=2+\frac{3c^2}{4k^2}$