<p>Let \(f(x) = px^2 + qx + r\). If \(f(1) = f(-1)\), then \(f'(a),\ f'(b),\ f'(c)\) are in AP for any \(a, b, c\) in AP. Which of the following is correct?</p>
<p>This is always true regardless of the condition</p>
<p>\(f'(a), f'(b), f'(c)\) are in AP as the nature of AP remains the same on multiplying by a real number</p>
<p>\(f'(a), f'(b), f'(c)\) are in GP</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: If f(1) = f(-1), then q = 0, making f(x) = px² + r. The derivative f'(x) = 2px is linear in x, so f'(a), f'(b), f'(c) are automatically in AP whenever a, b, c are in AP—this is always true regardless of the values of a, b, c.
<p><strong>Step 1:</strong> Apply the condition f(1) = f(-1).</p><p>f(1) = p + q + r and f(-1) = p - q + r</p><p>f(1) = f(-1) ⟹ p + q + r = p - q + r ⟹ 2q = 0 ⟹ q = 0</p><p><strong>Step 2:</strong> Simplify f(x) with q = 0.</p><p>f(x) = px² + r, so f'(x) = 2px</p><p><strong>Step 3:</strong> Check if f'(a), f'(b), f'(c) are in AP for any a, b, c in AP.</p><p>If a, b, c are in AP, then b - a = c - b (common difference d exists)</p><p>f'(a) = 2pa, f'(b) = 2pb, f'(c) = 2pc</p><p>For AP: 2f'(b) = f'(a) + f'(c) ⟹ 2(2pb) = 2pa + 2pc ⟹ 4pb = 2p(a + c)</p><p>Since a, b, c in AP: b = (a + c)/2 ⟹ 2b = a + c ✓</p><p>∴ The statement is always true for any a, b, c in AP. Answer: B</p>
Correct Answer: B