Question:

<div class="solution"><p><strong>Step 1:</strong> <span class="math-block">\[\frac{f(h)-f(-2h)}{h}=\frac{f(h)-f(0)}{h}+\frac{f(0)-f(-2h)}{h}\]</span></p><p><strong>Step 2:</strong> First term <span class="math-inline">\(\to f'(0)=1\)</span>. Second term: <span class="math-block">\[\frac{f(0)-f(-2h)}{h}=2\cdot\frac{f(-2h)-f(0)}{-2h}\to 2\cdot f'(0)=2\]</span></p><p><strong>Step 3:</strong> Sum <span class="math-inline">\(=1+2=3\)</span>.</p><p><strong>Answer: (A) 3</strong></p><div class="trap-box"><strong>Trap:</strong> Students forget to account for the chain rule factor of 2 in the second term.</div><div class="key-concept"><strong>Key Concept:</strong> Split limit into standard f'(0) form; track the factor from substitution</div></div>