Limits, Continuity & Differentiability
Differentiability and symmetric functions
Grade 12
Question:
<p>Let \(y = f(x)\) be a differentiable function such that \(f(3 - x) = f(3 + x)\) \(\forall\, x \in \mathbb{R}\) and the equation \(f(x) = 0\) has exactly 5 distinct real roots \(x_1, x_2, x_3, x_4\) and \(x_5\). If \(x_1 < x_2 < x_3 < x_4 < x_5\), then which of the following is/are must be <b>correct</b>?</p>
<p>(a) \(x_1 + x_2 + x_3 + x_4 + x_5 = 15\)</p>
<p>(b) \(f'(x_3) = 0\)</p>
<p>(c) \(y = |f(x)|\) is not differentiable at \(x = x_1, x_2, x_4\) and \(x_5\).</p>
<p>(d) \(y = |f(x)|\) is a differentiable function.</p>
Step-by-Step Solution
Key Concept: A function satisfying f(3-x) = f(3+x) is symmetric about x=3, meaning roots must occur symmetrically about x=3 in pairs, with one root necessarily at x=3. Since f(x)=0 has exactly 5 roots, one must be at x=3 and the other 4 form 2 symmetric pairs.
<p><strong>Step 1:</strong> Analyze the symmetry condition. f(3-x) = f(3+x) means f is symmetric about the vertical line x = 3.</p><p><strong>Step 2:</strong> For a symmetric function about x=3, if r is a root, then 6-r must also be a root (since f(3+t) = f(3-t) implies if f(3+t)=0, then f(3-t)=0).</p><p><strong>Step 3:</strong> With exactly 5 roots and symmetry about x=3, one root must lie ON the axis of symmetry itself. Therefore x₃ = 3 (the middle root).</p><p><strong>Step 4:</strong> The remaining 4 roots form 2 symmetric pairs: if x₁ < x₃ and x₂ < x₃, then by symmetry x₄ = 6-x₂ and x₅ = 6-x₁, giving x₁ + x₅ = 6 and x₂ + x₄ = 6.</p><p><strong>Step 5:</strong> Verify x₁ < x₂ < 3 < x₄ < x₅ with x₁+x₅=6 and x₂+x₄=6:</p><p>• x₃ = 3 ✓</p><p>• x₁ + x₅ = 6 ✓</p><p>• x₂ + x₄ = 6 ✓</p><p>∴ Answer: <strong>ABC</strong> (all three statements are correct)</p>
Correct Answer: ABC