Limits, Continuity & Differentiability
Differentiability of Functions
Grade 12
Question:
<p>The set of points where the function \(f(x) = \frac{1}{(1+|x|)}\left(x-1\right)\sin\frac{x}{x-1}\), if \(x \neq 1\) is differentiable, is</p>
<p>(a) Only \((-\infty, \infty)\)</p>
<p>(b) Only \([0, \infty)\)</p>
<p>(c) Only \((-\infty, 0) \cup (0, \infty)\)</p>
<p>(d) Only \((0, \infty)\)</p>
Step-by-Step Solution
Key Concept: We must analyze differentiability by examining continuity first, then checking if the derivative exists at potentially problematic points. The function involves absolute value and a term that appears undefined at x=1, but the limit as x→1 needs careful analysis.
<p><strong>Step 1: Analyze the domain and behavior near x=1</strong></p><p>For x ≠ 1, we have f(x) = 1/(1+|x|) · (x-1)sin(x/(x-1)).</p><p>Near x=1, let u = x-1 (so u→0). Then sin(x/(x-1)) = sin((u+1)/u) = sin(1/u + 1).</p><p>We need: lim(u→0) u·sin(1/u + 1).</p><p>Since |sin(1/u + 1)| ≤ 1, we have |u·sin(1/u + 1)| ≤ |u| → 0 as u→0.</p><p>Therefore: lim(x→1) (x-1)sin(x/(x-1)) = 0.</p><p><strong>Step 2: Check continuity at x=1</strong></p><p>At x=1: f(1) is undefined (given x ≠ 1), so x=1 is excluded from the domain. The function has a removable discontinuity here, but since x=1 is not in the domain, we don't consider differentiability there.</p><p><strong>Step 3: Check differentiability at x=0</strong></p><p>At x=0: |x| changes definition. We need to verify left and right derivatives exist.</p><p>For x > 0: f(x) = 1/(1+x) · (x-1)sin(x/(x-1))</p><p>For x < 0: f(x) = 1/(1-x) · (x-1)sin(x/(x-1))</p><p>Both expressions are differentiable in their respective domains. At x=0, both the left and right derivatives can be computed using standard differentiation rules (product rule, chain rule) since the denominator 1+|x| is differentiable at x=0 from both sides, and (x-1)sin(x/(x-1)) is differentiable everywhere except x=1.</p><p><strong>Step 4: Check differentiability for all other points</strong></p><p>For x ≠ 0, 1, the function is a composition and product of differentiable functions:</p><p>• 1/(1+|x|) is differentiable except at x=0</p><p>• (x-1)sin(x/(x-1)) is differentiable except at x=1</p><p>Since x=1 is excluded from the domain and x=0 is differentiable (checked above), f is differentiable on (-∞, ∞).</p><p><strong>Step 5: Verify x=0 differentiability rigorously</strong></p><p>At x=0, the function and its left/right derivatives exist and are equal because the absolute value doesn't create a sharp corner—the function (x-1)sin(x/(x-1)) is smooth enough to make the overall derivative continuous at x=0.</p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a