Question:

<p><strong>Step 1:</strong> Find dy/dx to analyze y = 2x tan⁻¹(x) - ln(1+x²).</p><p>dy/dx = 2tan⁻¹(x) + 2x·(1/(1+x²)) - 2x/(1+x²) = 2tan⁻¹(x) ✓</p><p><strong>Step 2:</strong> Since tan⁻¹(x) > 0 for x > 0 and tan⁻¹(x) < 0 for x < 0, we have dy/dx > 0 ∀x ∈ ℝ. Thus y is strictly increasing.</p><p><strong>Step 3:</strong> Find range of y:</p><p>• As x → -∞: tan⁻¹(x) → -π/2, so y → -∞</p><p>• As x → +∞: y → +∞</p><p>• At x = 0: y = 0</p><p>Therefore, Range of y = (-∞, +∞) = ℝ</p><p><strong>Step 4:</strong> For 5 - |[x]| to equal some value in the range of y (which is all real numbers), we need 5 - |[x]| ∈ ℝ.</p><p>This is always true. However, |[x]| ∈ {0, 1, 2, 3, 4, 5, ...}</p><p><strong>Step 5:</strong> The possible values of 5 - |[x]| are:</p><p>• |[x]| = 0 ⟹ 5 - |[x]| = 5</p><p>• |[x]| = 1 ⟹ 5 - |[x]| = 4</p><p>• |[x]| = 2 ⟹ 5 - |[x]| = 3</p><p>• |[x]| = 3 ⟹ 5 - |[x]| = 2</p><p>• |[x]| = 4 ⟹ 5 - |[x]| = 1</p><p>• |[x]| = 5 ⟹ 5 - |[x]| = 0</p><p>• |[x]| ≥ 6 ⟹ 5 - |[x]| ≤ -1</p><p><strong>Step 6:</strong> The number of non-negative integer values is: {0, 1, 2, 3, 4, 5} = 6 values.</p><p>∴ Answer: <strong>6</strong></p>