<p>A strictly increasing continuous function \(f(x)\) intersects with its inverse \(f^{-1}(x)\) at \(x = \alpha\) and \(x = \beta\). If \(\displaystyle\int_\alpha^\beta (f(x) + f^{-1}(x))\,dx = 13\) where \(\alpha, \beta \in N\), then the value of \(|\alpha\beta|\) equals:</p>
Step-by-Step Solution
Key Concept: For a strictly increasing function, intersection points with its inverse satisfy f(α) = α and f(β) = β (fixed points). Use the geometric property that ∫[f(x) + f⁻¹(x)]dx over [α,β] equals the area of a rectangle minus the area under f(x), which simplifies to β² - α².
<p><strong>Step 1:</strong> Recognize intersection points. When f(x) = f⁻¹(x) for a strictly increasing continuous function, we have f(x) = x. So α and β are fixed points where f(α) = α and f(β) = β.</p><p><strong>Step 2:</strong> Use geometric interpretation. The integral ∫_α^β [f(x) + f⁻¹(x)]dx represents the sum of areas under both curves. By symmetry about y = x: ∫_α^β f(x)dx + ∫_α^β f⁻¹(x)dx = ∫_α^β x dx + ∫_α^β x dx = 2∫_α^β x dx = 2[x²/2]_α^β = β² - α²</p><p><strong>Step 3:</strong> Alternatively, since f(x) = x and f⁻¹(x) = x at these points, and they satisfy the property that for strictly increasing functions: ∫_α^β [f(x) + f⁻¹(x)]dx = β² - α² = 13.</p><p><strong>Step 4:</strong> Find natural numbers α, β where β² - α² = 13. Factoring: (β - α)(β + α) = 13. Since 13 is prime and α, β ∈ ℕ: β - α = 1 and β + α = 13. Solving: β = 7, α = 6.</p><p><strong>Step 5:</strong> Calculate |αβ| = |6 × 7| = 42.</p><p>∴ Answer: B</p>
Correct Answer: B