Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>If <span>\( f : R \to R \)</span> is a function defined by <span>\[ f(x) = [x]\cos\left(\frac{2x-1}{2}\right)\pi, \]</span> where <span>\( [x] \)</span> denotes the greatest integer function, then <span>\( f \)</span> is</p>
<p>continuous for every real <span>\( x \)</span>.</p>
<p>discontinuous only at <span>\( x = 0 \)</span>.</p>
<p>discontinuous only at non-zero integral values of <span>\( x \)</span>.</p>
<p>continuous only at <span>\( x = 0 \)</span>.</p>
Step-by-Step Solution
Key Concept: Analyze continuity and differentiability separately at integer and non-integer points. At integers, the GIF has jump discontinuities, making the function discontinuous there. At non-integers, f is continuous but check differentiability by examining the derivative's behavior near integer boundaries.
<p><strong>Step 1: Analyze continuity at non-integer points</strong></p><p>For any x ∈ (n, n+1) where n ∈ ℤ, [x] = n (constant), so f(x) = n·cos((2x-1)π/2). Since cosine is continuous and [x] is constant on this interval, f is continuous on each open interval (n, n+1).</p><p><strong>Step 2: Check continuity at integer points</strong></p><p>At x = n (integer): lim(x→n⁻) f(x) = (n-1)·cos((2n-1)π/2) and lim(x→n⁺) f(x) = n·cos((2n-1)π/2). Since [x] jumps from n-1 to n, these one-sided limits are generally different (unless cosine term = 0). Thus f is <strong>discontinuous at all integers</strong>.</p><p><strong>Step 3: Analyze differentiability on (n, n+1)</strong></p><p>On each interval (n, n+1), [x] = n, so f(x) = n·cos((2x-1)π/2). Taking derivative: f'(x) = n·(-sin((2x-1)π/2))·2·(π/2) = -nπ·sin((2x-1)π/2). This exists for all x ∈ (n, n+1).</p><p><strong>Step 4: Check differentiability at integers</strong></p><p>Since f is discontinuous at integers, it cannot be differentiable there (continuity is necessary for differentiability).</p><p>∴ <strong>Answer: f is continuous on ℝ \ ℤ (continuous on all non-integer points) and differentiable on ℝ \ ℤ</strong></p>
Correct Answer: A