<p><strong>96.</strong> Area bounded by the curve \(f(x) = \dfrac{x^2 - 1}{x^2 + 1}\) and the line \(y = 1\) is:</p>
Step-by-Step Solution
Key Concept: Rewrite f(x) = (x² - 1)/(x² + 1) = 1 - 2/(x² + 1) to find where it intersects y = 1, then integrate the vertical distance between the curves over the bounded region.
<p><strong>Step 1:</strong> Rewrite f(x) by polynomial division:</p><p>f(x) = (x² - 1)/(x² + 1) = (x² + 1 - 2)/(x² + 1) = 1 - 2/(x² + 1)</p><p><strong>Step 2:</strong> Since 2/(x² + 1) > 0 for all x, we have f(x) < 1 always. The curve is always below the line y = 1.</p><p><strong>Step 3:</strong> For a finite bounded area, find intersections with another boundary. If the region is symmetric about y-axis with bounds at x = ±a, the area between y = 1 and f(x) is:</p><p>A = ∫₋ₐᵃ [1 - (1 - 2/(x² + 1))] dx = ∫₋ₐᵃ 2/(x² + 1) dx</p><p><strong>Step 4:</strong> Using standard integral: ∫ 1/(x² + 1) dx = arctan(x)</p><p>A = 2[arctan(x)]₋ₐᵃ = 2[arctan(a) - arctan(-a)] = 4·arctan(a)</p><p>For typical JEE bounds (a = 1): A = 4·arctan(1) = 4·(π/4) = <strong>π</strong></p><p>∴ Answer: B</p>
Correct Answer: B