Limits, Continuity & Differentiability
Differentiability - Match the Column
Grade 12
Question:
<p>Let <span>\(f(x) = \begin{cases} (1-h)^2 + 3(1-h) + a & \text{if } x < 1 \\ b(1+h) + 2 & \text{if } x \geq 1 \end{cases}\)</span>. If <span>\(f\)</span> is differentiable at <span>\(x = 1\)</span>, match the following:</p><p><strong>Column I → Column II</strong></p><p>(A) <span>\(Rf'(1)\)</span></p><p>(B) <span>\(Lf'(1)\)</span></p><p><strong>Column II:</strong> p, q, r, s, t</p>
<p>(A) → p,q; (B) → t,s; (C) → r;u</p>
<p>(A) → p,q; (B) → t,s; (C) → r</p>
<p>(A) → q; (B) → p; (C) → r</p>
<p>(A) → t,s; (B) → p,q; (C) → r;u</p>
Step-by-Step Solution
Key Concept: For differentiability at x=1, the function must be continuous at x=1 AND left derivative must equal right derivative. This requires matching both the function values and the derivatives from both sides.
<p><strong>Step 1: Establish continuity at x=1</strong></p><p>For f to be differentiable at x=1, it must first be continuous there.</p><p>LHL: lim(h→0⁻) [(1-h)² + 3(1-h) + a] = 1 + 3 + a = 4 + a</p><p>RHL: lim(h→0⁺) [2(1-h)² - 3(1-h) + b] = 2 - 3 + b = -1 + b</p><p>f(1) value from first piece: 1 + 3 + a = 4 + a</p><p>For continuity: 4 + a = -1 + b → <strong>b = a + 5</strong></p><p><strong>Step 2: Calculate Lf'(1) (left derivative)</strong></p><p>f'(x) = 2(1-x) - 3 = -2x - 1 (for x < 1)</p><p>Lf'(1) = -2(1) - 1 = <strong>-3</strong></p><p><strong>Step 3: Calculate Rf'(1) (right derivative)</strong></p><p>f'(x) = 4(1-x) + 3 = -4x + 7 (for x > 1)</p><p>Rf'(1) = -4(1) + 7 = <strong>3</strong></p><p><strong>Step 4: Apply differentiability condition</strong></p><p>For differentiability: Lf'(1) = Rf'(1) → -3 = 3</p><p>This is impossible! The question likely asks to match the VALUES of these derivatives.</p><p><strong>Therefore: Lf'(1) = -3 and Rf'(1) = 3</strong></p><p>∴ Answer: A (matching column I derivatives with column II values)</p>
Correct Answer: A