Limits, Continuity & Differentiability
Continuity and Differentiability
Grade 12
Question:
<p>Let \(f\) be a function defined by \(y = f(x)\) where \(x = 2t - |t|\) and \(y = t^2 + t|t|\) for \(t \in \mathbb{R}\), then:</p>
<p>\(f(x)\) is both continuous and differentiable at \(x = 0\).</p>
<p>\(f(x)\) is non differentiable at \(x = 0\).</p>
<p>\(f(x)\) is discontinuous at \(x = 0\).</p>
<p>\(f(x)\) is neither continuous nor differentiable at \(x = 0\).</p>
Step-by-Step Solution
Key Concept: The parametric function must be analyzed by considering t ≥ 0 and t < 0 separately due to the absolute value |t|, which creates a piecewise definition. This reveals the actual relationship between x and y, and whether f is continuous/differentiable at the junction point.
<p><strong>Step 1:</strong> For <strong>t ≥ 0</strong>: |t| = t</p><p>x = 2t - t = t, so t = x</p><p>y = t² + t·t = t² + t² = 2t² = 2x²</p><p><strong>Step 2:</strong> For <strong>t < 0</strong>: |t| = -t</p><p>x = 2t - (-t) = 3t, so t = x/3</p><p>y = t² + t·(-t) = t² - t² = 0</p><p><strong>Step 3:</strong> The piecewise function is:</p><p>f(x) = {2x² for x ≥ 0; 0 for x < 0}</p><p><strong>Step 4:</strong> Check continuity at x = 0:</p><p>lim(x→0⁺) f(x) = 0 and f(0) = 0, so continuous from right ✓</p><p>lim(x→0⁻) f(x) = 0 = f(0), so continuous from left ✓</p><p><strong>Step 5:</strong> Check differentiability at x = 0:</p><p>Right derivative: f'(0⁺) = 4x|ₓ₌₀ = 0</p><p>Left derivative: f'(0⁻) = 0</p><p>∴ f is continuous and differentiable everywhere</p>
Correct Answer: A