Question

If $x$ is very small in magnitude compared to $a$ such that,
${\left(\frac{a}{a+x}\right)}^{\frac{1}{2}}+{\left(\frac{a}{a-x}\right)}^{\frac{1}{2}}=2+k\frac{x^2}{a^2}$, then the value of $k$ :

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{3}{2}$

${\left(\frac{a}{a+x}\right)}^{\frac{1}{2}}+{\left(\frac{a}{a-x}\right)}^{\frac{1}{2}}={\left(\frac{1}{1+\frac{x}{a}}\right)}^{\frac{1}{2}}+{\left(\frac{1}{1-\frac{x}{a}}\right)}^{\frac{1}{2}}={(1+\frac{x}{a})}^{-\frac{1}{2}}+{(1-\frac{x}{a})}^{-\frac{1}{2}}$

Now, 3rd term of ${(1+\frac{x}{a})}^{-\frac{1}{2}}=\frac{\left(\frac{-1}{2}\right)(\frac{-1}{2}-1)}{1}{\left(\frac{x}{a}\right)}^2=\frac{3}{4}\frac{x^2}{a^2}$

Similarly, 3rd term of ${(1-\frac{x}{a})}^{-\frac{1}{2}}=\frac{\left(\frac{-1}{2}\right)(\frac{-1}{2}-1)}{1}{\left(\frac{x}{a}\right)}^2=\frac{3}{4}\frac{x^2}{a^2}$

Hence, expanding the question, ${(1+\frac{x}{a})}^{-\frac{1}{2}}+{(1-\frac{x}{a})}^{-\frac{1}{2}}$

we get $1+\frac{-1}{2}\frac{x}{a}+\frac{3}{4}\frac{x^2}{a^2}+1+\frac{1}{2}+\frac{3x^2}{4a^2}=2+\frac{6x^2}{4a^2}$

On, equating it with $2+\frac{k{x}^2}{a^2}$, we get, $k=\frac{3}{2}$