Question:

<div class="solution"><p><strong>Step 1:</strong> <span class="math-inline">\(|x^2-3x+2|=|(x-1)(x-2)|\)</span>: critical at x=1,2. <span class="math-inline">\(\cos|x|\)</span>: critical at x=0.</p><p><strong>Step 2:</strong> At x=0: <span class="math-inline">\((x^2-1)|x^2-3x+2|=(−1)(2)=−2\)</span>, smooth. <span class="math-inline">\(\cos|x|\)</span> is even and differentiable. So <span class="math-inline">\(f\)</span> is differentiable at x=0.</p><p><strong>Step 3:</strong> At x=1: <span class="math-inline">\((x^2-1)=(x-1)(x+1)\)</span> has factor (x-1) matching |x-1| in |(x-1)(x-2)|. Product is smooth at x=1.</p><p><strong>Step 4:</strong> At x=2: <span class="math-inline">\((x^2-1)|_{x=2}=3\neq 0\)</span>. The function has a corner from |(x-2)|. Non-differentiable at x=2.</p><p><strong>Answer: (D) x=2</strong></p><div class="trap-box"><strong>Trap:</strong> Missing that (x²-1) = (x-1)(x+1) cancels the corner at x=1.</div><div class="key-concept"><strong>Key Concept:</strong> Factor cancellation removes corner points</div></div>