The function f(x) = <span class="math-inline">\( [x^2 - [x]^{2}] \)</span> (where [y] is the greatest integer less than or equal to y), is discontinuous at :
Step-by-Step Solution
Key Concept: For a function to be continuous at an integer n, we must check if lim(x→n⁻) f(x) = lim(x→n⁺) f(x) = f(n). For any integer n, [x]² jumps discontinuously, causing f(x) to be discontinuous at that point unless the discontinuity is canceled.
<p><strong>Step 1:</strong> Write f(x) = [x² - [x]²], where [y] denotes the greatest integer ≤ y.</p><p><strong>Step 2:</strong> Let n be any integer. We need to check continuity at x = n by finding left and right limits.</p><p><strong>Step 3:</strong> For x slightly less than n (say x = n - ε where 0 < ε < 1): [x] = n - 1, so [x]² = (n-1)². Thus f(x) = [x² - (n-1)²] = [(n-ε)² - (n-1)²] = [n² - 2nε + ε² - n² + 2n - 1] = [2n - 2nε + ε² - 1].</p><p><strong>Step 4:</strong> As ε → 0⁻: f(x) → [2n - 1] = 2n - 1.</p><p><strong>Step 5:</strong> For x slightly greater than n (say x = n + ε where 0 < ε < 1): [x] = n, so [x]² = n². Thus f(x) = [x² - n²] = [(n+ε)² - n²] = [2nε + ε²].</p><p><strong>Step 6:</strong> As ε → 0⁺: f(x) → [0] = 0.</p><p><strong>Step 7:</strong> At x = n: f(n) = [n² - [n]²] = [n² - n²] = [0] = 0.</p><p><strong>Step 8:</strong> Since lim(x→n⁻) f(x) = 2n - 1 and lim(x→n⁺) f(x) = 0 = f(n), we see that 2n - 1 ≠ 0 for all integers n. Therefore, f is discontinuous at every integer.</p><p><strong>Step 9:</strong> Verify for n = 0: lim(x→0⁻) f(x) = 2(0) - 1 = -1 ≠ 0. Verify for n = 1: lim(x→1⁻) f(x) = 2(1) - 1 = 1 ≠ 0. The pattern holds for all integers.</p><p><strong>∴ Answer:</strong> A</p>
Correct Answer: A