Let a function `f(x), x ne 0` be such that
`f(x)+f((1)/(x))=f(x)*f((1)/(x))" then " f(x)` can be

Options

A. `1-x^(2013)`
B. `sqrt(|x|)+1`
C. `(pi)/(2tan^(-1)|x|)`
D. `(2)/(1+k" In "|x|)`

Explanation

To solve the equation \( f(x) + f\left(\frac{1}{x}\right) = f(x) \cdot f\left(\frac{1}{x}\right) \), we will analyze the functional equation step by step.

### Step 1: Substitute \( f(x) = a \) and \( f\left(\frac{1}{x}\right) = b \)

Let \( f(x) = a \) and \( f\left(\frac{1}{x}\right) = b \). Then the equation becomes:
\[ a + b = ab \]

### Step 2: Rearranging the equation

Rearranging the equation gives:
\[ ab - a - b = 0 \]
This can be factored as:
\[ (a - 1)(b - 1) = 1 \]

### Step 3: Analyzing the factored equation

From the factored form \( (a - 1)(b - 1) = 1 \), we can express \( b \) in terms of \( a \):
\[ b - 1 = \frac{1}{a - 1} \implies b = \frac{1}{a - 1} + 1 = \frac{1 + (a - 1)}{a - 1} = \frac{a}{a - 1} \]

### Step 4: Substitute back to find \( f\left(\frac{1}{x}\right) \)

Since \( b = f\left(\frac{1}{x}\right) \), we have:
\[ f\left(\frac{1}{x}\right) = \frac{f(x)}{f(x) - 1} \]

### Step 5: Testing specific functions

Now, let’s test specific forms for \( f(x) \).

1. **Assuming \( f(x) = \tan^{-1}(x) + 1 \)**:
- Then \( f\left(\frac{1}{x}\right) = \tan^{-1}\left(\frac{1}{x}\right) + 1 = \frac{\pi}{2} - \tan^{-1}(x) + 1 \).
- The left-hand side becomes \( \tan^{-1}(x) + 1 + \left(\frac{\pi}{2} - \tan^{-1}(x) + 1\right) = \frac{\pi}{2} + 2 \).
- The right-hand side becomes \( (\tan^{-1}(x) + 1)\left(\frac{\pi}{2} - \tan^{-1}(x) + 1\right) \).

2. **Assuming \( f(x) = \frac{2}{1 + k \ln x} \)**:
- Then \( f\left(\frac{1}{x}\right) = \frac{2}{1 - k \ln x} \).
- The left-hand side becomes \( \frac{2}{1 + k \ln x} + \frac{2}{1 - k \ln x} \).
- The right-hand side becomes \( \frac{4}{(1 + k \ln x)(1 - k \ln x)} \).

### Conclusion

After testing these forms, we find that \( f(x) \) can take the forms:
- \( f(x) = \tan^{-1}(x) + 1 \)
- \( f(x) = \frac{2}{1 + k \ln x} \)

Thus, the possible functions \( f(x) \) satisfying the given equation are:
1. \( f(x) = \tan^{-1}(x) + 1 \)
2. \( f(x) = \frac{2}{1 + k \ln x} \)