<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: Use the identity tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1-ab)) + nπ, recognizing that tan⁻¹(1/x) and tan⁻¹(x) differ by π when their sum equals π, which occurs only when 1 + x² < 0 (impossible) or when x < 0 in the restricted domain.
<p><strong>Step 1:</strong> Identify the range of tan⁻¹. The principal range of tan⁻¹(y) is (-π/2, π/2) for all real y.</p><p><strong>Step 2:</strong> Analyze the left side: tan⁻¹(1/x) ∈ (-π/2, π/2) for all x ≠ 0.</p><p><strong>Step 3:</strong> Analyze the right side: For 0 < x < 1, we have tan⁻¹(x) ∈ (0, π/4), so π + tan⁻¹(x) ∈ (π, 5π/4).</p><p><strong>Step 4:</strong> Compare ranges: The left side is always in (-π/2, π/2), while the right side is in (π, 5π/4) for the given domain. These intervals are disjoint.</p><p><strong>Step 5:</strong> Conclusion: No value of x satisfies the equation.</p><p>∴ Answer: <strong>A (0 solutions)</strong></p>
Correct Answer: A