<p>The number of values of \(x\), for which \(\tan^{-1}\!\left(\dfrac{1}{x}\right) = \pi + \tan^{-1} x\), \(0 < x < 1\) is:</p>
Step-by-Step Solution
Key Concept: Recognize that tan⁻¹(1/x) and tan⁻¹(x) are related through complementary angles when x > 0, and carefully handle the branch of tan⁻¹ when working with the equation tan⁻¹(1/x) = π + tan⁻¹(x).
<p><strong>Step 1:</strong> Recall that for x > 0, we have tan⁻¹(x) + tan⁻¹(1/x) = π/2, which gives tan⁻¹(1/x) = π/2 - tan⁻¹(x).</p><p><strong>Step 2:</strong> The given equation is tan⁻¹(1/x) = π + tan⁻¹(x).</p><p><strong>Step 3:</strong> Substitute the relation from Step 1: π/2 - tan⁻¹(x) = π + tan⁻¹(x).</p><p><strong>Step 4:</strong> Simplify: π/2 - tan⁻¹(x) = π + tan⁻¹(x) ⟹ -2tan⁻¹(x) = π/2 ⟹ tan⁻¹(x) = -π/4.</p><p><strong>Step 5:</strong> But the range of tan⁻¹ is (-π/2, π/2). For 0 < x < 1, we have 0 < tan⁻¹(x) < π/4, so tan⁻¹(x) can never equal -π/4 in the given domain.</p><p><strong>Step 6:</strong> Additionally, the RHS is π + tan⁻¹(x), which lies in (π, 5π/4), but the range of tan⁻¹(1/x) is (0, π/2), so there is no solution.</p><p>∴ Answer: A (0 solutions)</p>
Correct Answer: A