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)

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