<p>Let \(f(x)\) be a polynomial function of second degree. If \(f(1) = f(-1)\) and \(a, b, c\) are in AP, then \(f'(a)\), \(f'(b)\) and \(f'(c)\) are in</p>
Step-by-Step Solution
Key Concept: Since f(x) is quadratic with f(1) = f(-1), the polynomial must be even (of the form f(x) = px² + q). The derivative f'(x) = 2px is a linear function, which maps arithmetic progressions to arithmetic progressions.
<p><strong>Step 1:</strong> Since f(x) is a second-degree polynomial with f(1) = f(-1), we have:</p><p>f(x) = Ax² + Bx + C where A ≠ 0</p><p>f(1) = A + B + C and f(-1) = A - B + C</p><p>f(1) = f(-1) ⟹ A + B + C = A - B + C ⟹ B = 0</p><p><strong>Step 2:</strong> Therefore f(x) = Ax² + C, and f'(x) = 2Ax</p><p><strong>Step 3:</strong> Since a, b, c are in AP, we have: b - a = c - b (common difference d)</p><p><strong>Step 4:</strong> Now compute the differences:</p><p>f'(b) - f'(a) = 2Ab - 2Aa = 2A(b - a) = 2Ad</p><p>f'(c) - f'(b) = 2Ac - 2Ab = 2A(c - b) = 2Ad</p><p><strong>Step 5:</strong> Since f'(b) - f'(a) = f'(c) - f'(b) = 2Ad, the terms f'(a), f'(b), f'(c) have equal common differences.</p><p>∴ Answer: <strong>f'(a), f'(b), f'(c) are in AP</strong></p>
Correct Answer: A