<p>If \(\alpha\) is an integer satisfying \(|\alpha| \leq 5 - |[x]|\), where \(x\) is a real number for which \(2x\tan^{-1}x\) is greater than or equal to \(\ln(1+x^2)\), then the number of maximum non-negative possible values of \(\alpha\) is (where \([.]\) denotes the greater integer function)</p>
Step-by-Step Solution
Key Concept: First determine the range of x from the inequality 2x·tan⁻¹(x) ≥ ln(1+x²), then find [x], and finally count integers α satisfying |α| ≤ 5 - |[x]|.
<p><strong>Step 1: Analyze the inequality 2x·tan⁻¹(x) ≥ ln(1+x²)</strong></p><p>Let f(x) = 2x·tan⁻¹(x) - ln(1+x²). We need f(x) ≥ 0.</p><p>Taking derivative: f'(x) = 2tan⁻¹(x) + 2x/(1+x²) - 2x/(1+x²) = 2tan⁻¹(x)</p><p>So f'(x) ≥ 0 when x ≥ 0, and f'(x) ≤ 0 when x ≤ 0.</p><p>At x = 0: f(0) = 0.</p><p>For x > 0: f(x) ≥ f(0) = 0 ✓</p><p>For x < 0: f(x) ≥ f(0) = 0 (by symmetry argument or direct verification) ✓</p><p>Therefore, the inequality holds for <strong>all real x</strong>.</p><p><strong>Step 2: Find maximum value of |α|</strong></p><p>Since the inequality holds for all x ∈ ℝ, we need to find which x gives the maximum count of valid α values.</p><p>For any real x: |[x]| ≥ 0</p><p>The constraint is: |α| ≤ 5 - |[x]|</p><p>To maximize the number of non-negative values of α, we minimize |[x]|.</p><p><strong>Step 3: Minimize |[x]|</strong></p><p>The minimum value of |[x]| is 0, which occurs when [x] = 0, i.e., when x ∈ [0, 1).</p><p><strong>Step 4: Count valid α values</strong></p><p>When |[x]| = 0 (minimum):</p><p>|α| ≤ 5 - 0 = 5</p><p>For non-negative values of α: α ∈ {0, 1, 2, 3, 4, 5}</p><p>This gives <strong>6 values</strong>.</p><p>∴ Answer: <strong>6</strong></p>
Correct Answer: 6