Question

Let \( f(x) \) be a polynomial function satisfying \[ f(x) \cdot f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right). \] If \( f(4) = 65 \) and \( I_1, I_2, I_3 \) are in GP, then \( f'(I_1), f'(I_2), f'(I_3) \) are in:

• A. AP
• B. GP
• C. Both
• D. None of these

Hint:

For polynomial functions satisfying specific functional equations, determining the form of the polynomial is often the first step. Derivatives of polynomials can then be analyzed to determine the nature of sequences.

Answer 1

The Correct Option is B

Step 1: Determine the form of \( f(x) \). The given functional equation is: \[ f(x) \cdot f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right). \] Rearranging, we get: \[ (f(x) - 1)(f\left(\frac{1}{x}\right) - 1) = 1. \] This suggests that \( f(x) - 1 \) is of the form \( x^n \), where \( n \) is an integer. Let: \[ f(x) = x^n + 1. \]
Step 2: Use the given condition \( f(4) = 65 \). \[ f(4) = 4^n + 1 = 65 \quad \Rightarrow \quad 4^n = 64 \quad \Rightarrow \quad n = 3. \] Thus, \( f(x) = x^3 + 1 \).
Step 3: Compute the derivative \( f'(x) \). \[ f'(x) = 3x^2. \]
Step 4: Analyze \( f'(I_1), f'(I_2), f'(I_3) \). Given \( I_1, I_2, I_3 \) are in GP, let: \[ I_2 = I_1 r, \quad I_3 = I_1 r^2, \] where \( r \) is the common ratio. Then: \[ f'(I_1) = 3I_1^2, \quad f'(I_2) = 3I_2^2 = 3I_1^2 r^2, \quad f'(I_3) = 3I_3^2 = 3I_1^2 r^4. \] Thus, the sequence \( f'(I_1), f'(I_2), f'(I_3) \) is: \[ 3I_1^2, 3I_1^2 r^2, 3I_1^2 r^4, \] which is a geometric progression (GP).
Step 5: Determine the correct option. The sequence \( f'(I_1), f'(I_2), f'(I_3) \) is in GP.