Limits, Continuity & Differentiability
General
Grade 12
Question:
<p><span class="math-inline">\(f(x+y)+f(x-y)=2f(x)\ \forall x,y\in\mathbb{R}\)</span>, <span class="math-inline">\(f(0)=0\)</span> and <span class="math-inline">\(f\)</span> is differentiable. Which are correct?</p>
<strong>f is odd</strong>
f is even
<strong>f'(1)=f'(2)</strong>
<strong>f'(2)=f'(3)=f'(4)</strong>
Step-by-Step Solution
Key Concept: General
<div class="solution"><p><strong>Identify f:</strong> This is the Cauchy functional equation for cosine-type (or linear). With f(0)=0: put x=y=0: 2f(0)=2f(0) ✓. Put x=0: f(y)+f(-y)=0 ⟹ f is odd.</p><p>Differentiate w.r.t. y: f'(x+y)-f'(x-y)=0 ⟹ f'(x+y)=f'(x-y) for all y ⟹ f'=constant ⟹ f(x)=cx.</p><p>(A) f is odd: f(-x)=-f(x) ✓</p><p>(B) f is even: FALSE (f is odd)</p><p>(C) f'(1)=f'(2): since f'=constant ✓</p><p>(D) f'(2)=f'(3)=f'(4): ✓</p><p><strong>Answer: (A),(C),(D)</strong></p><div class="key-concept"><strong>Key Concept:</strong> f(x+y)+f(x-y)=2f(x) with f(0)=0 forces f to be linear (odd)</div></div>
Correct Answer: A,C,D