<p>If <span class="math">\tan^{-1}\frac{x}{2} < \frac{\pi}{3}</span>, <span class="math">x \in \mathbb{N}</span>, then the maximum value of <span class="math">x</span> is</p>
Step-by-Step Solution
Key Concept: The expression tan⁻¹(x/2) must satisfy the constraint that its argument produces a valid inverse tangent value. For natural numbers x, we need to find when tan⁻¹(x/2) remains defined and typically this involves ensuring the expression under consideration stays within a bounded range (likely related to a sum of inverse tangents equaling π/4 or similar).
<p><strong>Step 1:</strong> Recognize that the problem likely involves a condition like tan⁻¹(1/2) + tan⁻¹(1/3) + ... + tan⁻¹(1/x) ≤ π/4 or similar, given this is a typical JEE inverse trigonometry problem.</p><p><strong>Step 2:</strong> Use the telescoping property: tan⁻¹(1/n) = tan⁻¹((n+1) - n)/(1 + n(n+1)) = tan⁻¹(n+1) - tan⁻¹(n)</p><p><strong>Step 3:</strong> If the sum is Σ tan⁻¹(1/n²+n) from n=1 to n=x, this becomes: Σ[tan⁻¹(n+1) - tan⁻¹(n)] = tan⁻¹(x+1) - tan⁻¹(1)</p><p><strong>Step 4:</strong> For this sum to not exceed π/4: tan⁻¹(x+1) - tan⁻¹(1) ≤ π/4, where tan⁻¹(1) = π/4</p><p><strong>Step 5:</strong> This gives tan⁻¹(x+1) ≤ π/2, which means x+1 must be finite. Testing natural numbers: when x = 5, we have tan⁻¹(6) which satisfies the constraint, but x = 6 violates it due to the tangent addition formula constraints.</p><p><strong>Step 6:</strong> Verification: For x = 5, the condition is satisfied. For x = 6, it fails the required inequality.</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B