Definite Integration Question - None

Question

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:

Accepted Answer

Step 1: At intersection points of f(x) and f⁻¹(x), we have f(α) = α and f(β) = β (both points lie on the line y = x where a function equals its inverse).

Step 2: For a strictly increasing function, the key property is: ∫_α^β f(x)dx + ∫_α^β f⁻¹(x)dx = βf(β) - αf(α) - [∫_α^β x·d(f(x))] + ∫_α^β x·d(f⁻¹(x)) simplifies using the reflection property.

Step 3: Using the geometric interpretation: ∫_α^β [f(x) + f⁻¹(x)]dx represents the sum of areas. Since f(α) = α and f(β) = β, the integral equals the area of rectangle with width (β - α) and height (α + β).

Step 4: Therefore: ∫_α^β [f(x) + f⁻¹(x)]dx = (β - α)(α + β) = β² - α² = 13

Step 5: Since α, β ∈ ℕ and β > α (strictly increasing), we need β² - α² = 13. Factoring: (β - α)(β + α) = 13 = 1 × 13.

Step 6: Setting β - α = 1 and β + α = 13 gives: β = 7, α = 6.

Step 7: Therefore |αβ| = |6 × 7| = 42.

∴ Answer: B (42)

Question:

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