Limits, Continuity & Differentiability
Differentiability of piecewise functions
Grade 12
Question:
<p>Let \(f(x)\) be a function that is non-differentiable at \(x = 0\) and \([2x]\) is discontinuous at \(x = \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3\) (5 points). If \(f(x)\) should be differentiable and continuous at \(x = 2\), so \(k = 3\), and given \(f'(3^+) = f'(3^-)\) implies \(2(3 - a) = 0\), so \(a = 3\), and \(f(3^+) = f(3^-)\) implies \((3 - a)^2 + b = 5\), so \(b = 5\). Find the value of \(a \cdot b \cdot k\).</p>
Step-by-Step Solution
Key Concept: Piecewidth function continuity and differentiability require matching both function values and derivatives at boundary points; the greatest integer function [2x] is discontinuous at half-integer values where the jump occurs.
<p><strong>Step 1: Identify discontinuities of [2x]</strong></p><p>The greatest integer function [2x] is discontinuous when 2x is an integer, i.e., at x = 0, 1/2, 1, 3/2, 2, 5/2, 3, ... The problem specifies we consider points: 1/2, 1, 3/2, 2, 5/2 (5 points as stated).</p><p><strong>Step 2: Apply continuity condition at x = 2</strong></p><p>For f(x) to be continuous at x = 2: f(2⁻) = f(2⁺). Given k = 3 is determined from this condition.</p><p><strong>Step 3: Apply differentiability condition at x = 3</strong></p><p>For f(x) to be differentiable at x = 3, we need:<br/>• Left and right derivatives equal: f'(3⁻) = f'(3⁺)<br/>• Given: 2(3 - a) = 0 ⟹ <strong>a = 3</strong></p><p><strong>Step 4: Apply continuity condition at x = 3</strong></p><p>For f(x) to be continuous at x = 3: f(3⁻) = f(3⁺)<br/>Given: (3 - a)² + b = 5<br/>Substituting a = 3: (3 - 3)² + b = 5 ⟹ <strong>b = 5</strong></p><p><strong>Step 5: Calculate the product</strong></p><p>a · b · k = 3 × 5 × 3 = <strong>45</strong></p>
Correct Answer: 45