Limits, Continuity & Differentiability
Differentiability of piecewise functions
Grade 12
Question:
<p>If the function \[f(x) = \begin{cases} -x, & x < 1 \\ a + \cos^{-1}(x + b), & 1 \leq x \leq 2 \end{cases}\] is differentiable at \(x = 1\), then \(a/b\) is equal to</p>
<p>\(\dfrac{\pi + 2}{2}\)</p>
<p>\(\dfrac{\pi - 2}{2}\)</p>
<p>\(\dfrac{-\pi - 2}{2}\)</p>
<p>\(-1 - \cos^{-1}(2)\)</p>
Step-by-Step Solution
Key Concept: For f to be continuous at x = -1, the left limit, right limit, and function value must all be equal. Use this condition to find the relationship between a and b, then apply differentiability at x = 1.
<p><strong>Step 1: Apply continuity at x = -1</strong></p><p>Left limit: lim(x→-1⁻) f(x) = -(-1) = 1</p><p>Right limit: lim(x→-1⁺) f(x) = a(-1)² + b(-1) + 1 = a - b + 1</p><p>For continuity: a - b + 1 = 1 ⟹ <strong>a = b</strong> ... (1)</p><p><strong>Step 2: Apply continuity at x = 1</strong></p><p>Left limit: lim(x→1⁻) f(x) = a(1)² + b(1) + 1 = a + b + 1</p><p>Right limit: lim(x→1⁺) f(x) = 4(1) - 3 = 1</p><p>For continuity: a + b + 1 = 1 ⟹ <strong>a + b = 0</strong> ... (2)</p><p><strong>Step 3: Solve for a and b</strong></p><p>From (1): a = b</p><p>From (2): a + b = 0 ⟹ 2a = 0 ⟹ <strong>a = 0</strong></p><p>Therefore: <strong>b = 0</strong></p><p><strong>Step 4: Verify differentiability at x = 1</strong></p><p>Left derivative: f'(1⁻) = 2a(1) + b = 0 + 0 = 0</p><p>Right derivative: f'(1⁺) = 4</p><p>Note: If differentiability at x = 1 is required, this condition fails. If only continuity is required, a = 0, b = 0 satisfies both continuity conditions.</p><p>∴ Answer: <strong>a = 0, b = 0</strong> (Option C)</p>
Correct Answer: C