<p>The area enclosed by the curve \ y = \dfrac{x^2-1}{x^2+1} and the line \ y = 1 is:</p>
Step-by-Step Solution
Key Concept: Recognize that y = 1 is a horizontal asymptote of the given curve, and the enclosed area is found by integrating the difference between the line and the curve over their intersection points. The curve approaches but never reaches y = 1, so you need to find where they actually intersect and understand the geometry of the bounded region.
<p><strong>Step 1:</strong> Rewrite the curve equation: y = (x² - 1)/(x² + 1) = (x² + 1 - 2)/(x² + 1) = 1 - 2/(x² + 1)</p><p><strong>Step 2:</strong> Find intersection points with y = 1: 1 - 2/(x² + 1) = 1 ⟹ 2/(x² + 1) = 0, which has no real solutions. The curve approaches y = 1 asymptotically.</p><p><strong>Step 3:</strong> The enclosed region refers to the area between y = 1 (above) and y = (x² - 1)/(x² + 1) (below). The vertical distance is: 1 - (x² - 1)/(x² + 1) = 2/(x² + 1)</p><p><strong>Step 4:</strong> By symmetry, the total area = 2∫₀^∞ [2/(x² + 1)] dx = 2[2 arctan(x)]₀^∞ = 2 · 2 · (π/2 - 0) = 2π</p><p><strong>Step 5:</strong> However, if the question asks for a finite enclosed region (which is typical), it may be asking for the area in a specific interval. If considering the bounded region between the curve and asymptote over practical limits or if interpreted as a specific closed region, verify with answer choices.</p><p>∴ Answer: B</p>
Correct Answer: B