Limits, Continuity & Differentiability
Differentiability and Derivatives
Grade 12
Question:
<p>Let \(f\) be a differentiable function such that \(\displaystyle\lim_{x \to 1} \frac{f(1+x^3-x)-f(x)}{\sin(x-1)} = \displaystyle\lim_{x \to 0} \frac{f(1-x)-f(1)}{x} + 10\), then \(f'(1)\) is equal to:</p>
<p>(a) 5</p>
<p>(b) 4</p>
<p>(c) 2</p>
<p>(d) 1</p>
Step-by-Step Solution
Key Concept: Use the definition of derivative f'(1) = lim[h→0] (f(1+h)-f(1))/h to convert both sides into expressions involving f'(1), then equate them to solve for f'(1).
<p><strong>Step 1:</strong> Simplify the right side using the derivative definition.</p><p>lim[x→0] (f(1-x)-f(1))/x = lim[x→0] -(f(1)-f(1-x))/x = -f'(1)</p><p>So the equation becomes: lim[x→1] (f(1+x³-x)-f(x))/sin(x-1) = -f'(1) + 10</p><p><strong>Step 2:</strong> Evaluate the left side. Let u = x-1, so as x→1, u→0.</p><p>Rewrite: x = 1+u, x³ = (1+u)³ = 1+3u+3u²+u³</p><p>1+x³-x = 1+(1+3u+3u²+u³)-(1+u) = 1+3u+3u²+u³-u = 1+2u+3u²+u³</p><p><strong>Step 3:</strong> The numerator is f(1+2u+3u²+u³)-f(1+u).</p><p>Using Taylor expansion: f(1+2u+3u²+u³) ≈ f(1+u) + f'(1+u)·(u+3u²+u³)</p><p>As u→0: f(1+2u+3u²+u³)-f(1+u) ≈ f'(1)·u + O(u²)</p><p><strong>Step 4:</strong> lim[u→0] f'(1)·u/sin(u) = lim[u→0] f'(1)·u/u = f'(1)</p><p><strong>Step 5:</strong> Equate both sides: f'(1) = -f'(1) + 10</p><p>2f'(1) = 10</p><p>∴ f'(1) = 5</p>
Correct Answer: A