Applications of Derivatives
Symmetry and Critical Points
Grade 12

Question:

<p>Let <span class="math">f(x)</span> be a non-constant twice derivable function defined on <span class="math">\mathbb{R}</span> such that <span class="math">f(2+x) = f(2-x)</span> and <span class="math">f'\left(\frac{1}{2}\right) = 0 = f'(1)</span>. Which of the following alternative(s) is/are correct?</p>
<p>(A) <span class="math">f(-4) = f(8)</span></p>
<p>(B) Minimum number of roots of the equation <span class="math">f''(x) = 0</span> in <span class="math">(0, 4)</span> are 4.</p>

Step-by-Step Solution

Key Concept: The function satisfies f(2+x) = f(2-x), making x=2 an axis of symmetry. Combined with f'(1/2) = 0 and f'(1) = 0, we can determine properties of f and f'' to verify equalities and count zeros of f''(x).
<p><strong>Step 1: Analyze the symmetry condition.</strong></p><p>Given: f(2+x) = f(2-x). This means f is symmetric about the line x = 2.</p><p>Differentiating both sides with respect to x: f'(2+x) = -f'(2-x).</p><p></p><p><strong>Step 2: Use the critical points.</strong></p><p>Given f'(1/2) = 0 and f'(1) = 0.</p><p>From f'(2+x) = -f'(2-x):</p><p>• At x = -3/2: f'(2-3/2) = -f'(2+3/2) ⟹ f'(1/2) = -f'(7/2)</p><p>Since f'(1/2) = 0, we have f'(7/2) = 0.</p><p>• At x = -1: f'(2-1) = -f'(2+1) ⟹ f'(1) = -f'(3)</p><p>Since f'(1) = 0, we have f'(3) = 0.</p><p></p><p><strong>Step 3: Verify option (A): f(-4) = f(8).</strong></p><p>Using symmetry f(2+x) = f(2-x):</p><p>Set 2+x = -4 ⟹ x = -6, so f(-4) = f(2-(-6)) = f(8). ✓</p><p>Option (A) is correct.</p><p></p><p><strong>Step 4: Analyze option (B): Minimum roots of f''(x) = 0 in (0,4).</strong></p><p>We have critical points at x = 1/2, 1, 3, 7/2 for f'(x).</p><p>In the interval (0, 4): critical points are 1/2, 1, 3, 7/2.</p><p>By Rolle's theorem, between consecutive critical points of f'(x), there exists at least one zero of f''(x).</p><p>Between: (1/2, 1), (1, 3), (3, 7/2) gives at least 3 zeros of f'' in (0, 4).</p><p>However, the minimum guaranteed is 3, not 4. Option (B) is not necessarily correct.</p><p></p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A

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