Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>The function \(f(x) = \frac{\tan(\pi[x-\pi])}{1+[x]^2}\) where \([x]\) is the greatest integer function,</p>
<p>(a) is discontinuous at some \(x\)</p>
<p>(b) \(f'(x)\) exists for all \(x\)</p>
<p>(c) \(f'(x)\) exists for all \(x\) but \(f''(x)\) does not exist</p>
<p>(d) is continuous for all \(x\) but \(f'(x)\) does not exist for some \(x\)</p>
Step-by-Step Solution
Key Concept: The greatest integer function [x] creates step discontinuities, making tan(π[x-π]) = tan(πk) = 0 for all integer arguments k. The numerator is identically zero wherever the function is defined, forcing f(x) = 0.
<p><strong>Step 1:</strong> Analyze the numerator: tan(π[x-π])</p><p>Since [x] is the greatest integer function, [x-π] is always an integer (let's call it k, where k ∈ ℤ)</p><p><strong>Step 2:</strong> Evaluate tan(πk) for any integer k</p><p>tan(πk) = 0 for all k ∈ ℤ (fundamental trigonometric property)</p><p><strong>Step 3:</strong> Determine f(x)</p><p>f(x) = (0)/(1+[x]²) = 0 for all x in the domain</p><p><strong>Step 4:</strong> Conclude about continuity and differentiability</p><p>Since f(x) = 0 everywhere, the function is continuous and differentiable at all points in its domain, with f'(x) = 0</p><p>∴ Answer: B (The function is continuous and differentiable everywhere)</p>
Correct Answer: B