Limits, Continuity & Differentiability
Differentiability and Symmetry of 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 <strong>correct</strong>?</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: The symmetry condition f(3-x) = f(3+x) means f is symmetric about x=3, so roots must be symmetric about this axis: if r is a root, then 6-r is also a root. With exactly 5 roots and this symmetry, one root must lie on the axis of symmetry at x=3.
<p><strong>Step 1:</strong> Analyze the symmetry condition f(3-x) = f(3+x). This means f is symmetric about the vertical line x = 3.</p><p><strong>Step 2:</strong> If r is a root of f, then f(r) = 0. By symmetry, f(3-(r-3)) = f(3+(r-3)) implies f(6-r) = 0, so 6-r is also a root.</p><p><strong>Step 3:</strong> Roots come in symmetric pairs (r, 6-r) except when r = 6-r, which gives r = 3. With exactly 5 roots, we must have two symmetric pairs plus the middle root x₃ = 3.</p><p><strong>Step 4:</strong> Order the roots: x₁ < x₂ < x₃ < x₄ < x₅. By symmetry: x₁ + x₅ = 6 and x₂ + x₄ = 6, with x₃ = 3 at the center.</p><p><strong>Step 5:</strong> Differentiate f(3-x) = f(3+x): -f'(3-x) = f'(3+x). Setting x=0: f'(3) = -f'(3), so 2f'(3) = 0, thus f'(3) = 0.</p><p><strong>Step 6:</strong> Since x₃ = 3 is a root and f'(3) = 0, x₃ is a root where f has a critical point. This means x₃ = 3 is either a local extremum (touching the x-axis) or an inflection point with horizontal tangent.</p><p>∴ <strong>x₃ = 3</strong>, and the correct option contains statements that x₃ = 3 and f'(3) = 0</p>
Correct Answer: ABC