Limits, Continuity & Differentiability
Differentiability of piecewise functions
Grade 12
Question:
<p>Let \(g(x) = \begin{cases} k\sqrt{x+1}; & 0 \leq x \leq 3 \\ mx + 2; & 3 < x \leq 5 \end{cases}\). If \(g\) is differentiable at \(x = 3\), then find \(k + m\).</p>
<p>\(\dfrac{10}{5}\)</p>
<p>\(2\)</p>
<p>\(4\)</p>
<p>\(\dfrac{8}{5}\)</p>
Step-by-Step Solution
Key Concept: For g(x) to be continuous at x=3, left and right limits must be equal: k√4 = 3m+2. Use differentiability condition at x=3 where left derivative of k√(x+1) equals right derivative of mx+2.
<p><strong>Step 1: Apply Continuity at x=3</strong></p><p>Left limit: k√(3+1) = k√4 = 2k</p><p>Right limit: m(3) + 2 = 3m + 2</p><p>For continuity: 2k = 3m + 2 ... (1)</p><p><strong>Step 2: Apply Differentiability at x=3</strong></p><p>For f(x) = k√(x+1), f'(x) = k/(2√(x+1))</p><p>Left derivative at x=3: f'(3⁻) = k/(2√4) = k/4</p><p>Right derivative at x=3: (mx+2)' = m</p><p>For differentiability: k/4 = m ... (2)</p><p><strong>Step 3: Solve the System</strong></p><p>From (2): m = k/4</p><p>Substitute in (1): 2k = 3(k/4) + 2</p><p>2k = 3k/4 + 2</p><p>8k = 3k + 8</p><p>5k = 8</p><p>k = 8/5, m = 2/5</p><p>∴ Answer: B</p>
Correct Answer: B