Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>If \(x + |y| = 2y\) then \(y\) as a function of \(x\) is</p>
<p>(a) defined for all \(x\)</p>
<p>(b) continuous at \(x = 0\)</p>
<p>(c) differentiable for all \(x\)</p>
<p>(d) such that \(\frac{dy}{dx} = \frac{1}{3}\) for \(x < 0\)</p>
Step-by-Step Solution
Key Concept: Rearrange the equation by considering the sign of y separately: if y ≥ 0, then |y| = y giving y = x/2; if y < 0, then |y| = -y giving y = -x. The domain restrictions determine which piece applies where.
<p><strong>Step 1:</strong> Start with x + |y| = 2y and rearrange: x = 2y - |y|</p><p><strong>Step 2:</strong> <strong>Case 1 (y ≥ 0):</strong> Then |y| = y, so x = 2y - y = y, giving <strong>y = x with x ≥ 0</strong></p><p><strong>Step 3:</strong> <strong>Case 2 (y < 0):</strong> Then |y| = -y, so x = 2y - (-y) = 2y + y = 3y, giving <strong>y = x/3 with x < 0</strong></p><p><strong>Step 4:</strong> Combine: y = {x (for x ≥ 0); x/3 (for x < 0)} or equivalently y = max(x/3, x). This is a piecewise function defined for all real x.</p><p>∴ Answer: D</p>
Correct Answer: D