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:

\[ \int_{1}^{3} f(f(...f(e)) \, (n \text{ times}) \, dE \]

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

### Step 1: Understand the relationship between eccentricities
The eccentricity \( e \) of a hyperbola and the eccentricity \( f(e) \) of its conjugate hyperbola are related by the equation:

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

From this, we can derive:

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

This simplifies to:

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

Taking the square root gives us:

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

### Step 2: Determine the form of \( f(f(...f(e))) \)
We need to evaluate \( f(f(...f(e))) \) for \( n \) times.

1. If \( n \) is even, we can express it as:

\[ f(f(...f(e))) = e \]

2. If \( n \) is odd, we have:

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

### Step 3: Set up the integral based on the parity of \( n \)
Now we can set up the integral based on whether \( n \) is even or odd.

- **If \( n \) is even**:
\[ \int_{1}^{3} e \, dE = \left[ \frac{e^2}{2} \right]_{1}^{3} = \frac{3^2}{2} - \frac{1^2}{2} = \frac{9}{2} - \frac{1}{2} = \frac{8}{2} = 4 \]

- **If \( n \) is odd**:
We need to evaluate:

\[ \int_{1}^{3} \sqrt{\frac{e^2}{e^2 - 1}} \, dE \]

Let \( u = e^2 \), then \( du = 2e \, dE \) or \( dE = \frac{du}{2\sqrt{u}} \). The limits change accordingly:
- When \( E = 1 \), \( u = 1^2 = 1 \)
- When \( E = 3 \), \( u = 3^2 = 9 \)

The integral becomes:

\[ \int_{1}^{9} \sqrt{\frac{u}{u - 1}} \cdot \frac{du}{2\sqrt{u}} = \frac{1}{2} \int_{1}^{9} \sqrt{\frac{1}{u - 1}} \, du \]

This integral can be evaluated using a substitution or integral table, and it results in:

\[ = 2\sqrt{2} \text{ (after evaluating the integral)} \]

### Final Result
Thus, the final result is:

- For \( n \) even: \( 4 \)
- For \( n \) odd: \( 2\sqrt{2} \)