Limits, Continuity & Differentiability
Continuity of functions
Grade 12
Question:
<p>Let <span>\(f(x) = [x]\cos\left(\dfrac{2x-1}{2}\right)\pi\)</span>, where <span>\([\cdot]\)</span> denotes the greatest integer function. Then <span>\(f\)</span> is:</p>
<p>Discontinuous at every integer</p>
<p>Continuous only at <span>\(x = 0\)</span></p>
<p>Continuous for every real <span>\(x\)</span></p>
<p>Discontinuous only at <span>\(x = 0\)</span></p>
Step-by-Step Solution
Key Concept: Analyze continuity and differentiability by examining the greatest integer function's jump discontinuities and how the oscillating cosine term behaves at integer points. The product of a step function with a continuous oscillating function creates specific failure points.
<p><strong>Step 1:</strong> Identify where [x] is discontinuous. The greatest integer function [x] is discontinuous at all integers n ∈ ℤ.</p><p><strong>Step 2:</strong> Check continuity at non-integer points. For x ∉ ℤ, [x] is constant in a neighborhood, so f(x) = k·cos((2x-1)π/2) is a product of constants and continuous functions, hence continuous.</p><p><strong>Step 3:</strong> Check continuity at integer points x = n. We have:</p><p>• lim(x→n⁻) f(x) = (n-1)·cos((2n-1)π/2)</p><p>• lim(x→n⁺) f(x) = n·cos((2n-1)π/2)</p><p>• f(n) = n·cos((2n-1)π/2)</p><p>Since (n-1) ≠ n, the left limit ≠ right limit (unless the cosine term equals 0, which it doesn't for all integers). Thus f is discontinuous at every integer.</p><p><strong>Step 4:</strong> Conclusion. f is continuous on ℝ \ ℤ (all non-integer points) and discontinuous at all integers. Therefore, f is neither continuous nor differentiable at integers.</p><p>∴ <strong>f is continuous on ℝ \ ℤ and discontinuous at every integer point</strong> (Answer: C)</p>
Correct Answer: C