<p>The number of points on \(y = \tan^{-1} x\), \(\forall x \in (0, \pi)\), whose image in \(y = x\) is the centre of the circle with radius \(\dfrac{\pi}{2\sqrt{2}}\) units and which is at a minimum distance of \(\dfrac{\pi}{2\sqrt{2}}\) units from the circle.</p>
Step-by-Step Solution
Key Concept: A point on y = tan⁻¹(x) reflects across y = x to give a center of a circle. The minimum distance from this center to the circle equals the radius, meaning the center lies outside the circle at distance = 2r from the circle's center. Set up the distance equation between the reflected point and the original point.
<p><strong>Step 1:</strong> Let P(a, tan⁻¹(a)) be a point on y = tan⁻¹(x) where a ∈ (0, π).</p><p><strong>Step 2:</strong> The reflection of P across y = x is Q(tan⁻¹(a), a). This Q is the center of the circle with radius r = π/(2√2).</p><p><strong>Step 3:</strong> The minimum distance from Q to the circle is the distance from some point on the circle to Q. If a point is at minimum distance π/(2√2) from the circle, and the circle has radius π/(2√2), then the distance from Q to the circle's center is r + π/(2√2) = π/√2.</p><p><strong>Step 4:</strong> But Q itself IS the center. The problem states minimum distance FROM the circle is π/(2√2). This means we need another circle with center at some point and radius π/(2√2), and its minimum distance to our circle equals π/(2√2). This implies the distance between centers = 2r = π/√2.</p><p><strong>Step 5:</strong> For point P(a, tan⁻¹(a)) reflected to Q(tan⁻¹(a), a), the distance PQ = √2|a - tan⁻¹(a)| (using distance formula and simplifying).</p><p><strong>Step 6:</strong> Setting √2|a - tan⁻¹(a)| = π/√2, we get |a - tan⁻¹(a)| = π/2.</p><p><strong>Step 7:</strong> For a ∈ (0, π), since tan⁻¹(a) ∈ (0, π/2) and tan⁻¹(a) < a for all a > 0, we have a - tan⁻¹(a) = π/2.</p><p><strong>Step 8:</strong> Define f(a) = a - tan⁻¹(a). We need f(a) = π/2. Since f'(a) = 1 - 1/(1+a²) > 0, f is strictly increasing. Checking: f(π) = π - tan⁻¹(π) ≈ π - 1.262 ≈ 1.88, and we need π/2 ≈ 1.57. By intermediate value theorem, exactly one solution exists in (0, π).</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: 1