Question

If $x$ is very small in magnitude compared with $a$, then ${\left(\frac{a}{a+x}\right)}^{\frac{1}{2}}+{\left(\frac{a}{a-x}\right)}^{\frac{1}{2}}$ can be approximately equal to

• A. $1+\frac{1}{2}\frac{x}{a}$
• B. $\frac{x}{a}$
• C. $1+\frac{3}{4}\frac{x^2}{a^2}$
• D. $2+\frac{3}{4}\frac{x^2}{a^2}$

Answer 1

Answer: (D)

${\left(\frac{a}{a+x}\right)}^{\frac{1}{2}}+{\left(\frac{a}{a-x}\right)}^{\frac{1}{2}}={\left(\frac{a+x}{a}\right)}^{-\frac{1}{2}}+{\left(\frac{a-x}{a}\right)}^{-\frac{1}{2}}$
$={\left(1+\frac{x}{a}\right)}^{-\frac{1}{2}}+{\left(1-\frac{x}{a}\right)}^{-\frac{1}{2}}$
$=\left[1-\frac{1}{2}\frac{x}{a}+\frac{3}{8}\frac{x^2}{a^2}\right]+\left[1+\frac{1}{2}\frac{x}{a}+\frac{3}{8}\frac{x^2}{a^2}\right]$
$=2+\frac{3}{4}.\frac{x^2}{a^2}\left[\because x<<a,\therefore \frac{x}{a}<<1\right]$