Let f(x) be a polynomial function satisfying `f(x)+f((1)/(x))=f(x)f((1)/(x))" for all "xne0.` If `f(5)=126" and a,b,c are in G.P., then"f'(a),f'(b),f'(c)` are in

Options

A. AP
B. GP
C. HP
D. none of these

Explanation

To solve the problem, we need to find the polynomial function \( f(x) \) that satisfies the equation \( f(x) + f\left(\frac{1}{x}\right) = f(x) f\left(\frac{1}{x}\right) \) for all \( x \neq 0 \), and then determine the derivatives \( f'(a) \), \( f'(b) \), and \( f'(c) \) when \( a, b, c \) are in geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
We start with the functional equation:
\[ f(x) + f\left(\frac{1}{x}\right) = f(x) f\left(\frac{1}{x}\right) \]
This suggests a relationship between \( f(x) \) and \( f\left(\frac{1}{x}\right) \).

2. **Assuming a Polynomial Form**:
We assume \( f(x) \) is a polynomial of the form:
\[ f(x) = x^n + 1 \]
where \( n \) is a non-negative integer. This form is chosen because it satisfies the symmetry required by the functional equation.

3. **Finding \( n \)**:
We know from the problem that:
\[ f(5) = 126 \]
Substituting \( x = 5 \) into our assumed polynomial:
\[ f(5) = 5^n + 1 = 126 \]
This simplifies to:
\[ 5^n = 125 \]
Recognizing that \( 125 = 5^3 \), we find:
\[ n = 3 \]

4. **Final Form of the Polynomial**:
Thus, we have:
\[ f(x) = x^3 + 1 \]

5. **Calculating the Derivative**:
Now we find the derivative \( f'(x) \):
\[ f'(x) = 3x^2 \]

6. **Finding \( f'(a), f'(b), f'(c) \)**:
For \( a, b, c \) in G.P., we calculate:
\[ f'(a) = 3a^2, \quad f'(b) = 3b^2, \quad f'(c) = 3c^2 \]

7. **Using the Property of G.P.**:
Since \( a, b, c \) are in G.P., we can express \( b \) as \( b = ar \) and \( c = ar^2 \) for some common ratio \( r \). Therefore:
\[ f'(b) = 3(ar)^2 = 3a^2r^2, \quad f'(c) = 3(ar^2)^2 = 3a^2r^4 \]
Thus, we have:
\[ f'(a), f'(b), f'(c) = 3a^2, 3a^2r^2, 3a^2r^4 \]

8. **Conclusion**:
The terms \( f'(a), f'(b), f'(c) \) can be factored as:
\[ 3a^2(1), 3a^2(r^2), 3a^2(r^4) \]
Since \( 1, r^2, r^4 \) are in G.P. (as \( r^2 \) is the geometric mean of \( 1 \) and \( r^4 \)), we conclude that:
\[ f'(a), f'(b), f'(c) \text{ are in G.P.} \]