If `f(x)=(1-x)/(1+x), x ne 0, -1 and alpha=f(f(x))+f(f((1)/(x)))`, then

Options

A. `alpha gt 2`
B. `alpha lt -2`
C. `|alpha| gt 2`
D. `alpha=2`

Answer

To solve the problem, we will follow these steps:

1. **Define the function**: We start with the function \( f(x) = \frac{1 - x}{1 + x} \).

2. **Find \( f(f(x)) \)**: We need to substitute \( f(x) \) into itself.
- First, we find \( f(f(x)) \):
\[ f(f(x)) = f\left(\frac{1 - x}{1 + x}\right) \]
Substituting into the function:
\[ = \frac{1 - \left(\frac{1 - x}{1 + x}\right)}{1 + \left(\frac{1 - x}{1 + x}\right)} \]
Simplifying the numerator:
\[ 1 - \frac{1 - x}{1 + x} = \frac{(1 + x) - (1 - x)}{1 + x} = \frac{2x}{1 + x} \]
Simplifying the denominator:
\[ 1 + \frac{1 - x}{1 + x} = \frac{(1 + x) + (1 - x)}{1 + x} = \frac{2}{1 + x} \]
Therefore,
\[ f(f(x)) = \frac{\frac{2x}{1 + x}}{\frac{2}{1 + x}} = x \]

3. **Find \( f(f(1/x)) \)**: Next, we need to find \( f(f(1/x)) \).
- First, find \( f(1/x) \):
\[ f\left(\frac{1}{x}\right) = \frac{1 - \frac{1}{x}}{1 + \frac{1}{x}} = \frac{\frac{x - 1}{x}}{\frac{x + 1}{x}} = \frac{x - 1}{x + 1} \]
- Now substitute \( f(1/x) \) into \( f(x) \):
\[ f(f(1/x)) = f\left(\frac{x - 1}{x + 1}\right) \]
Substituting into the function:
\[ = \frac{1 - \left(\frac{x - 1}{x + 1}\right)}{1 + \left(\frac{x - 1}{x + 1}\right)} \]
Simplifying the numerator:
\[ 1 - \frac{x - 1}{x + 1} = \frac{(x + 1) - (x - 1)}{x + 1} = \frac{2}{x + 1} \]
Simplifying the denominator:
\[ 1 + \frac{x - 1}{x + 1} = \frac{(x + 1) + (x - 1)}{x + 1} = \frac{2x}{x + 1} \]
Therefore,
\[ f(f(1/x)) = \frac{\frac{2}{x + 1}}{\frac{2x}{x + 1}} = \frac{1}{x} \]

4. **Calculate \( \alpha \)**: Now we can find \( \alpha \):
\[ \alpha = f(f(x)) + f(f(1/x)) = x + \frac{1}{x} \]

5. **Formulate the quadratic equation**: We can rearrange this expression:
\[ x^2 - \alpha x + 1 = 0 \]

6. **Determine the discriminant**: For \( x \) to be real, the discriminant must be non-negative:
\[ \alpha^2 - 4 \cdot 1 \cdot 1 \geq 0 \]
This simplifies to:
\[ \alpha^2 - 4 \geq 0 \]
Factoring gives:
\[ (\alpha - 2)(\alpha + 2) \geq 0 \]

7. **Find the intervals**: The solution to this inequality is:
\[ \alpha \leq -2 \quad \text{or} \quad \alpha \geq 2 \]

Thus, the final answer is that \( \alpha \) belongs to the intervals:
\[ \alpha \in (-\infty, -2] \cup [2, \infty) \]