<p>If \(f(x) = \cos\left[\frac{\pi}{x}\right]\cos\left(\frac{\pi(x-1)}{2}\right)\), where \([\cdot]\) denotes the greatest integer function, then \(f(x)\) is continuous at</p>
Step-by-Step Solution
Key Concept: For continuity at a point, we need the function value and left/right limits to be equal. The greatest integer function [π/x] is discontinuous at most points, but we must check which specific values make the overall function continuous by examining both components.
<p><strong>Step 1: Analyze the structure</strong></p><p>We have f(x) = cos([π/x]) · cos(π(x-1)/2). For continuity at a point, we need lim_{x→a} f(x) = f(a).</p><p><strong>Step 2: Check x = 0</strong></p><p>As x → 0, π/x → ±∞, so [π/x] oscillates wildly. Also, f(0) is undefined. Not continuous at x = 0. ✗</p><p><strong>Step 3: Check x = 2</strong></p><p>At x = 2: π(x-1)/2 = π(2-1)/2 = π/2, so cos(π/2) = 0.</p><p>For x near 2: [π/x] jumps between different values as x crosses 2. Specifically, at x = 2⁻, [π/2 + ε] = 1, and at x = 2⁺, [π/2 - ε] = 1.</p><p>However, since cos(π/2) = 0, we have f(2) = cos([π/2]) · 0 = 0.</p><p>But the limit involves [π/x] which jumps, making the limit problematic. ✗</p><p><strong>Step 4: Check x = 3</strong></p><p>At x = 3: π/3 ≈ 1.047, so [π/3] = 1 (constant in a neighborhood of 3).</p><p>For x in (π/(1+ε), π/(1-ε)), [π/x] = 1. Since π/2 ≈ 1.57 and π ≈ 3.14, for x near 3, we have 1 < π/x < 1.5, so [π/x] = 1 consistently.</p><p>f(3) = cos(1) · cos(π(3-1)/2) = cos(1) · cos(π) = cos(1) · (-1) = -cos(1).</p><p>lim_{x→3} f(x) = lim_{x→3} cos(1) · cos(π(x-1)/2) = cos(1) · cos(π) = -cos(1).</p><p>Since [π/x] = 1 in a neighborhood of x = 3, both the function value and limit equal -cos(1). ✓</p><p><strong>∴ Answer:</strong> b</p>
Correct Answer: b