<p>Let \(f(2-x)=f(2+x)\) and \(f(20-x)=f(x)\) for all \(x\in\mathbb{R}\). If \(f(0)=5\), minimum solutions of \(f(x)=5\) on \([0,170]\):</p>
Step-by-Step Solution
<div class="solution"><p>Step 1: Two symmetries \to f(t)=f(t-16) \to period 16.</p><p>Step 2: f(0)=5 \to f(16k)=5 for k=0,1,...,10: that's 11 values (0,16,...,160).</p><p>Step 3: Symmetry about x=2 gives f(0)=f(4)=5 \to f(4+16k)=5 for k=0,...,10: another 11 values (4,20,...,164).</p><p>Total minimum = 22.</p><p><strong>Answer: (A) 22</strong></p><div class="trap-box"><strong>Trap:</strong> Symmetry about x=2 forces a second AP. Don't count only 16k.<div class="key-concept"><strong>Key Concept:</strong> Two reflections \to period; symmetry centre shifts by T/2
Correct Answer: 22