If f (x) is a polynomial satisfying f(x) f (1/ x) = f (x) + f (1/x) and f (3) = 28, then f (2) is given by

Options

A. 63
B. 65
C. 67
D. None of these

Answer

To solve the problem, we need to find the polynomial \( f(x) \) that satisfies the equation:

\[ f(x) f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right) \]

and we know that \( f(3) = 28 \). We need to find \( f(2) \).

### Step 1: Rearranging the Equation

Starting with the given equation:

\[ f(x) f\left(\frac{1}{x}\right) - f(x) - f\left(\frac{1}{x}\right) = 0 \]

We can factor this as:

\[ f(x) \left(f\left(\frac{1}{x}\right) - 1\right) - f\left(\frac{1}{x}\right) = 0 \]

### Step 2: Isolating \( f(x) \)

Rearranging gives:

\[ f(x) = \frac{f\left(\frac{1}{x}\right)}{f\left(\frac{1}{x}\right) - 1} \]

### Step 3: Finding \( f\left(\frac{1}{x}\right) \)

By substituting \( \frac{1}{x} \) into the original equation, we can also express \( f\left(\frac{1}{x}\right) \):

\[ f\left(\frac{1}{x}\right) = \frac{f(x)}{f(x) - 1} \]

### Step 4: Multiplying the Two Equations

Now, we multiply the two expressions we have for \( f(x) \) and \( f\left(\frac{1}{x}\right) \):

\[ f(x) f\left(\frac{1}{x}\right) = \frac{f\left(\frac{1}{x}\right)}{f\left(\frac{1}{x}\right) - 1} \cdot \frac{f(x)}{f(x) - 1} \]

This simplifies to:

\[ f(x) f\left(\frac{1}{x}\right) = 1 \]

### Step 5: Defining a New Function \( g(x) \)

Let \( g(x) = f(x) - 1 \). Then we have:

\[ g(x) g\left(\frac{1}{x}\right) = 1 \]

This suggests that \( g(x) \) is of the form \( x^n \) for some integer \( n \).

### Step 6: Finding \( f(x) \)

Since \( g(x) = x^n \), we have:

\[ f(x) = g(x) + 1 = x^n + 1 \]

### Step 7: Using the Condition \( f(3) = 28 \)

Now, substituting \( x = 3 \):

\[ f(3) = 3^n + 1 = 28 \]

This gives:

\[ 3^n = 27 \implies n = 3 \]

Thus, we have:

\[ f(x) = x^3 + 1 \]

### Step 8: Finding \( f(2) \)

Now we can find \( f(2) \):

\[ f(2) = 2^3 + 1 = 8 + 1 = 9 \]

### Conclusion

Thus, the value of \( f(2) \) is:

\[ \boxed{9} \]