Question:

<p><strong>Step 1:</strong> Simplify the expression using the identity {y} + [y] = y.</p><p>Note that {x + sin x} + [x + sin x] = x + sin x</p><p>Therefore: f(x) = (x + sin x) + [x - sin x] + [x]</p><p><strong>Step 2:</strong> Rewrite as f(x) = x + sin x + [x - sin x] + [x]</p><p><strong>Step 3:</strong> Identify sources of discontinuity. The term sin x is continuous everywhere. The discontinuities come from [x - sin x] and [x], which are discontinuous at integer points.</p><p><strong>Step 4:</strong> Since 0 ≤ x ≤ π and 0 ≤ sin x ≤ 1 for x ∈ [0, π], the expression [x - sin x] is discontinuous when x - sin x crosses integer values.</p><p><strong>Step 5:</strong> The primary discontinuities in [0, π] occur at x = 1, 2, 3 (from [x]) and we must check if [x - sin x] creates additional discontinuities.</p><p>For x ∈ [0, π]: x - sin x ranges from 0 to π - 0 ≈ 3.14, so [x - sin x] jumps at integer values of x - sin x.</p><p><strong>Step 6:</strong> The function [x] has discontinuities at x = 1, 2, 3 in [0, π]. The term [x - sin x] is discontinuous when x - sin x is an integer. Since sin x is continuous and strictly increasing then decreasing on [0, π], this occurs at x = 1, 2, 3 as well (where the discontinuity in [x] dominates).</p><p>∴ Answer: <strong>3 points of discontinuity at x = 1, 2, 3</strong></p>