If 'c' is small in comparison with l then `((l)/(l+c))^(1//2) + ((l)/(l-c))^(1//2)` =

Options

A. `2 + (3c)/(4l)`
B. `2 + (3c^2)/(4l^2)`
C. `l + (3c^2)/(4l^2)`
D. `l + (3c)/(4l)`

Answer

To solve the problem \(\left(\frac{l}{l+c}\right)^{1/2} + \left(\frac{l}{l-c}\right)^{1/2}\), we will use the binomial expansion for small values of \(c\) in comparison with \(l\).

### Step-by-Step Solution:

1. **Rewrite the Terms**:
We start by rewriting the terms in a more manageable form:
\[ \left(\frac{l}{l+c}\right)^{1/2} = \left(\frac{1}{1+\frac{c}{l}}\right)^{1/2} \]
and
\[ \left(\frac{l}{l-c}\right)^{1/2} = \left(\frac{1}{1-\frac{c}{l}}\right)^{1/2}. \]

2. **Apply Binomial Expansion**:
For small \(x\), we can use the binomial expansion:
\[ (1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2 + \ldots \]
In our case, we have \(n = -\frac{1}{2}\) for both expansions:
- For \(\left(1+\frac{c}{l}\right)^{-1/2}\):
\[ \left(1+\frac{c}{l}\right)^{-1/2} \approx 1 - \frac{1}{2}\frac{c}{l} + \frac{3}{8}\left(\frac{c}{l}\right)^2 + \ldots \]
- For \(\left(1-\frac{c}{l}\right)^{-1/2}\):
\[ \left(1-\frac{c}{l}\right)^{-1/2} \approx 1 + \frac{1}{2}\frac{c}{l} + \frac{3}{8}\left(\frac{c}{l}\right)^2 + \ldots \]

3. **Combine the Expansions**:
Now, we can add the two expansions:
\[ \left(\frac{l}{l+c}\right)^{1/2} + \left(\frac{l}{l-c}\right)^{1/2} \approx \left(1 - \frac{1}{2}\frac{c}{l} + \frac{3}{8}\left(\frac{c}{l}\right)^2\right) + \left(1 + \frac{1}{2}\frac{c}{l} + \frac{3}{8}\left(\frac{c}{l}\right)^2\right). \]

4. **Simplify the Expression**:
When we combine the two expansions, the terms involving \(\frac{c}{l}\) will cancel out:
\[ 1 + 1 + \left(-\frac{1}{2}\frac{c}{l} + \frac{1}{2}\frac{c}{l}\right) + \left(\frac{3}{8}\left(\frac{c}{l}\right)^2 + \frac{3}{8}\left(\frac{c}{l}\right)^2\right) = 2 + \frac{3}{4}\left(\frac{c}{l}\right)^2. \]

5. **Final Result**:
Thus, the final result of the expression is:
\[ 2 + \frac{3}{4}\left(\frac{c}{l}\right)^2. \]