A polynomial function f (x) satisfies `f(x) f((1)/(x))=f(x)=f((1)/(x))`. If f(10) =1001, then what is the value of f(20)?

Options

A. 2002
B. 8004
C. 8001
D. None of these

Answer

To solve the problem, we start with the given polynomial function \( f(x) \) that satisfies the equation:

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

We are also given that \( f(10) = 1001 \) and we need to find \( f(20) \).

### Step 1: Analyze the functional equation
We can rewrite the functional equation as:

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

This suggests that \( f(x) \) and \( f\left(\frac{1}{x}\right) \) might be related in a specific way.

### Step 2: Consider a general form for \( f(x) \)
Based on the analysis, we can assume a general form for \( f(x) \):

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

where \( k \) is a constant and \( n \) is a positive integer.

### Step 3: Substitute \( f(x) \) into the functional equation
Substituting \( f(x) \) and \( f\left(\frac{1}{x}\right) \) into the equation gives:

\[ (1 + k x^n)(1 + k \frac{1}{x^n}) = (1 + k x^n) + (1 + k \frac{1}{x^n}) \]

Expanding both sides:

\[ 1 + k x^n + k \frac{1}{x^n} + k^2 = 1 + k x^n + 1 + k \frac{1}{x^n} \]

This simplifies to:

\[ 1 + k x^n + k \frac{1}{x^n} + k^2 = 2 + k x^n + k \frac{1}{x^n} \]

### Step 4: Solve for \( k \)
From the equation, we can see that:

\[ k^2 = 1 \]

Thus, \( k = 1 \) or \( k = -1 \).

### Step 5: Determine the correct value of \( k \)
Since we know \( f(10) = 1001 \), we can substitute \( x = 10 \) into our general form:

1. If \( k = 1 \):
\[ f(10) = 1 + 10^n = 1001 \implies 10^n = 1000 \implies n = 3 \]
Therefore, \( f(x) = 1 + x^3 \).

2. If \( k = -1 \):
\[ f(10) = 1 - 10^n = 1001 \implies -10^n = 1000 \implies n \text{ is not valid.} \]

Thus, we conclude that \( k = 1 \) and \( n = 3 \).

### Step 6: Find \( f(20) \)
Now we can find \( f(20) \):

\[ f(20) = 1 + 20^3 = 1 + 8000 = 8001 \]

### Final Answer:
Thus, the value of \( f(20) \) is:

\[ \boxed{8001} \]