Question:

<div class="solution"><p><strong>Key Idea:</strong> Combine two shifted symmetries to extract period and parity.</p><p><strong>Step 1:</strong> From <span class="math-inline">\(f(10+x)=f(10-x)\)</span>: <span class="math-inline">\(f(t)=f(20-t)\)</span></p><p><strong>Step 2:</strong> From <span class="math-inline">\(f(20+x)=-f(20-x)\)</span>: <span class="math-inline">\(f(t)=-f(t-20)\)</span>, so <span class="math-inline">\(f(t+20)=-f(t)\)</span></p><p><strong>Step 3:</strong> <span class="math-inline">\(f(t+40)=f(t)\)</span> — period 40.</p><p><strong>Step 4:</strong> From Step 1 and 2: <span class="math-inline">\(f(-x)=-f(x)\)</span> — odd.</p><p><strong>Answer: Odd and periodic (period 40)</strong></p><div class="trap-box"><strong>Trap:</strong> Even-about-a is not the same as ordinary evenness. Shift the centre first.</div><div class="key-concept"><strong>Key Concept:</strong> Two reflection symmetries → periodicity; period = twice the distance between centres</div></div>