A polynomial function f(x) satisfies the condition
`f(x)f((1)/(x))=f(x)+f((1)/(x))` .
If f(10)=1001, then f(20)=

Options

A. 2002
B. 8008
C. 8001
D. none of these

Answer

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

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

Given that \( f(10) = 1001 \), we need to find \( f(20) \).

### Step 1: Identify the form of \( f(x) \)

We can assume that the polynomial function \( f(x) \) has the form:

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

This form is chosen because it satisfies the condition for polynomial functions.

### Step 2: Verify the condition

Let's verify if this form satisfies the given condition:

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

Now, calculate \( f(x)f\left(\frac{1}{x}\right) \):

\[ f(x)f\left(\frac{1}{x}\right) = \left(1 + x^n\right)\left(1 + \frac{1}{x^n}\right) \]
\[ = 1 + x^n + \frac{1}{x^n} + 1 = 2 + x^n + \frac{1}{x^n} \]

Now calculate \( f(x) + f\left(\frac{1}{x}\right) \):

\[ f(x) + f\left(\frac{1}{x}\right) = \left(1 + x^n\right) + \left(1 + \frac{1}{x^n}\right) = 2 + x^n + \frac{1}{x^n} \]

Since both expressions are equal, the form \( f(x) = 1 + x^n \) satisfies the condition.

### Step 3: Use the given value \( f(10) = 1001 \)

Now we can use the information given in the problem:

\[ f(10) = 1 + 10^n = 1001 \]
\[ 10^n = 1000 \implies n = 3 \]

Thus, the function is:

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

### Step 4: Find \( f(20) \)

Now we can find \( f(20) \):

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

### Final Answer

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

\[ \boxed{8001} \]