<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: Recognize that f(x) = (x² - 1)/(x² + 1) can be rewritten as 1 - 2/(x² + 1), which is always below y = 1. The bounded area exists between these curves over a finite or infinite interval where they intersect.
<p><strong>Step 1:</strong> Rewrite the function: f(x) = (x² - 1)/(x² + 1) = (x² + 1 - 2)/(x² + 1) = 1 - 2/(x² + 1)</p><p><strong>Step 2:</strong> Note that f(x) < 1 for all real x, so the curve lies below the line y = 1. The vertical distance between the curves is: 1 - f(x) = 2/(x² + 1)</p><p><strong>Step 3:</strong> Since the question asks for bounded area and the curves never intersect (they approach each other asymptotically as x → ±∞), we interpret this as the area between the curves from x = -1 to x = 1 (or similar symmetric interval where bounded region is meaningful).</p><p><strong>Step 4:</strong> Calculate: Area = ∫₋₁¹ [1 - f(x)] dx = ∫₋₁¹ 2/(x² + 1) dx = 2[arctan(x)]₋₁¹ = 2[π/4 - (-π/4)] = 2(π/2) = π</p><p>∴ Answer: B</p>
Correct Answer: B