Limits, Continuity & Differentiability
Rolle's and Mean Value Theorems
Grade 12
Question:
<p>Let <span class="math">f(x)</span> be a twice differentiable function defined on <span class="math">(-\infty, \infty)</span> such that <span class="math">f(x) = f(2-x)</span> and <span class="math">f'\left(\frac{1}{2}\right) = f'\left(\frac{1}{4}\right) = 0</span>.</p><p>The minimum number of values where <span class="math">f''(x)</span> vanishes on <span class="math">[0, 2]</span> is:</p>
<p>(a) 2</p>
<p>(b) 3</p>
<p>(c) 4</p>
<p>(d) 5</p>
Step-by-Step Solution
Key Concept: Use the symmetry condition <span class="math">f(x) = f(2-x)</span> to find all critical points, then apply Rolle's theorem between them to locate zeros of <span class="math">f''(x)</span>.
<p><strong>Solution:</strong> From the symmetry condition <span class="math">f(x) = f(2-x)</span>, the graph is symmetric about <span class="math">x = 1</span>. Differentiating the symmetry relation gives <span class="math">f'(x) = -f'(2-x)</span>. Given that <span class="math">f'\left(\frac{1}{2}\right) = 0</span> and <span class="math">f'\left(\frac{1}{4}\right) = 0</span>, by symmetry <span class="math">f'\left(\frac{3}{2}\right) = 0</span> and <span class="math">f'\left(\frac{7}{4}\right) = 0</span>. Also <span class="math">f'(1) = 0</span> (by symmetry). Between consecutive critical points, <span class="math">f''(x)</span> must vanish at least once by Rolle's theorem, giving a minimum of 3 zeros of <span class="math">f''(x)</span>.</p>
Correct Answer: b