<p>If \(f(x)\) is twice differentiable and \(f''(0) = p\) then \(\lim_{x \to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}\) is</p>
Step-by-Step Solution
Key Concept: Use Taylor expansion of f around x=0 up to second order: f(x) = f(0) + f'(0)x + ½f''(0)x² + o(x²). Substitute for f(x), f(2x), and f(4x) to find the coefficient of x² in the numerator.
<p><strong>Step 1:</strong> Expand f(x), f(2x), and f(4x) using Taylor series around x = 0:</p><p>f(x) = f(0) + f'(0)x + ½f''(0)x² + o(x²)</p><p>f(2x) = f(0) + f'(0)(2x) + ½f''(0)(2x)² + o(x²) = f(0) + 2f'(0)x + 2f''(0)x² + o(x²)</p><p>f(4x) = f(0) + f'(0)(4x) + ½f''(0)(4x)² + o(x²) = f(0) + 4f'(0)x + 8f''(0)x² + o(x²)</p><p><strong>Step 2:</strong> Compute the numerator 2f(x) - 3f(2x) + f(4x):</p><p>= 2[f(0) + f'(0)x + ½f''(0)x²] - 3[f(0) + 2f'(0)x + 2f''(0)x²] + [f(0) + 4f'(0)x + 8f''(0)x²] + o(x²)</p><p>= 2f(0) + 2f'(0)x + f''(0)x² - 3f(0) - 6f'(0)x - 6f''(0)x² + f(0) + 4f'(0)x + 8f''(0)x² + o(x²)</p><p><strong>Step 3:</strong> Collect terms:</p><p>Constant: 2f(0) - 3f(0) + f(0) = 0</p><p>Coefficient of x: 2f'(0) - 6f'(0) + 4f'(0) = 0</p><p>Coefficient of x²: f''(0) - 6f''(0) + 8f''(0) = 3f''(0) = 3p</p><p><strong>Step 4:</strong> Evaluate the limit:</p><p>lim(x→0) [3f''(0)x² + o(x²)]/x² = 3f''(0) = 3p</p><p>∴ Answer: <strong>3p</strong> (Option B)</p>
Correct Answer: B