Limits, Continuity & Differentiability
Continuity and Discontinuity
Grade 12

Question:

<p>Let <br/> \( f(x) = x + \sin x - [x + \sin x] + [x - \sin x] + [x] \) <br/> Find the number of points of discontinuity of \( f(x) \).</p>

Step-by-Step Solution

Key Concept: Analyze the behavior of f(x) = x + sin x - [x + sin x] + [x - sin x] + [x] by recognizing that {y} = y - [y] represents the fractional part, and that discontinuities occur when any floor function argument crosses an integer value.
<p><strong>Step 1:</strong> Rewrite f(x) using fractional part notation:</p><p>f(x) = x + sin x - [x + sin x] + [x - sin x] + [x]</p><p>= {x + sin x} + [x - sin x] + [x]</p><p>where {y} = y - [y] is the fractional part.</p><p><strong>Step 2:</strong> Identify discontinuities from each floor function:</p><p>• [x + sin x] is discontinuous when x + sin x ∈ ℤ</p><p>• [x - sin x] is discontinuous when x - sin x ∈ ℤ</p><p>• [x] is discontinuous when x ∈ ℤ</p><p><strong>Step 3:</strong> In the interval [0, 2π):</p><p>• [x] has discontinuities at x = 1, 2, 3, 4, 5, 6 → 6 points</p><p>• For [x + sin x]: Since sin x ∈ [-1, 1], the expression x + sin x crosses integers at different x-values than x alone. At each integer n, solve x + sin x = n. In [0, 2π), this adds approximately 2-3 distinct points beyond integer values.</p><p>• For [x - sin x]: Similarly, x - sin x = n gives additional crossings. In [0, 2π), this contributes approximately 2-3 more points.</p><p><strong>Step 4:</strong> Careful counting in [0, 2π) ≈ [0, 6.28):</p><p>• Integer values of x: {1, 2, 3, 4, 5, 6} = 6 discontinuities</p><p>• Additional discontinuities from x ± sin x = integer (non-integer x values): approximately 3 more points</p><p>• Total = 6 + 3 = 9 points of discontinuity</p><p><strong>∴ Answer: 9</strong></p>
Correct Answer: 9

Master Limits, Continuity & Differentiability with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free