Let e be the eccentricity of a hyperbola and f(e ) be the eccentricity of its conjugate hyperbola then `int_(1)^(3)ubrace(fff...f(e))_("n times")` de is equal to

Options

A. 4 if n is even
B. 4 if n is odd
C. 2 if n is even
D. `2 sqrt2` if n is odd

Answer

To solve the problem, we need to evaluate the integral

\[ I = \int_{1}^{3} f(f(...f(e))) \, de \quad \text{(n times)} \]

where \( e \) is the eccentricity of a hyperbola, and \( f(e) \) is the eccentricity of its conjugate hyperbola.

### Step 1: Understand the relationship between \( e \) and \( f(e) \)

The eccentricity \( f(e) \) of the conjugate hyperbola is related to \( e \) by the formula:

\[ \frac{1}{e^2} + \frac{1}{f(e)^2} = 1 \]

From this, we can derive:

\[ \frac{1}{f(e)^2} = 1 - \frac{1}{e^2} \]

Taking the reciprocal gives:

\[ f(e)^2 = \frac{e^2}{e^2 - 1} \]

Taking the square root, we have:

\[ f(e) = \sqrt{\frac{e^2}{e^2 - 1}} \]

### Step 2: Evaluate \( f(f(e)) \)

Now, we need to find \( f(f(e)) \):

Using the same relationship, we substitute \( e \) with \( f(e) \):

\[ f(f(e)) = f\left(\sqrt{\frac{e^2}{e^2 - 1}}\right) \]

Applying the formula again, we find:

\[ f(f(e)) = \sqrt{\frac{f(e)^2}{f(e)^2 - 1}} = \sqrt{\frac{\frac{e^2}{e^2 - 1}}{\frac{e^2}{e^2 - 1} - 1}} = \sqrt{\frac{\frac{e^2}{e^2 - 1}}{\frac{e^2 - (e^2 - 1)}{e^2 - 1}}} = \sqrt{\frac{e^2}{1}} = e \]

### Step 3: Generalize for \( n \) times

Continuing this process, we can see that:

- If \( n \) is odd, \( f(f(...f(e))) \) will return to \( e \).
- If \( n \) is even, \( f(f(...f(e))) \) will return to \( f(e) \).

Thus, we can summarize:

\[ f(f(...f(e))) = \begin{cases} e & \text{if } n \text{ is odd} \\ f(e) & \text{if } n \text{ is even} \end{cases} \]

### Step 4: Evaluate the integral

Now we can evaluate the integral:

1. **For odd \( n \)**:

\[ I = \int_{1}^{3} e \, de = \left[ \frac{e^2}{2} \right]_{1}^{3} = \frac{3^2}{2} - \frac{1^2}{2} = \frac{9}{2} - \frac{1}{2} = \frac{8}{2} = 4 \]

2. **For even \( n \)**:

\[ I = \int_{1}^{3} f(e) \, de = \int_{1}^{3} \sqrt{\frac{e^2}{e^2 - 1}} \, de \]

This integral is more complex and would require further evaluation, but the key point is that we have established the relationship based on the parity of \( n \).

### Final Result

Thus, the final answer is:

- \( I = 4 \) if \( n \) is odd.
- The integral needs further evaluation if \( n \) is even.