Question:

Step 1: Identify the intersection points of f(x) and f⁻¹(x) Since $f(x)$ is strictly increasing and continuous, when $f(x)$ intersects with its inverse $f^{-1}(x)$, these intersection points must lie on the line $y = x$. This is because if $f(a) = f^{-1}(a)$ for some value $a$, then applying $f$ to both sides gives $f(f(a)) = a$, which combined with the strictly increasing property implies $f(a) = a$. Therefore, at the intersection points: $$f(\alpha) = \alpha \quad \text{and} \quad f(\beta) = \beta$$ Step 2: Apply the integration property for inverse functions We use the standard property for integrating inverse functions: $$\int_{\alpha}^{\beta} f^{-1}(x)\,dx = \beta f(\beta) - \alpha f(\alpha) - \int_{f(\alpha)}^{f(\beta)} f(x)\,dx$$ Substituting $f(\alpha) = \alpha$ and $f(\beta) = \beta$: $$\int_{\alpha}^{\beta} f^{-1}(x)\,dx = \beta \cdot \beta - \alpha \cdot \alpha - \int_{\alpha}^{\beta} f(x)\,dx$$ $$\int_{\alpha}^{\beta} f^{-1}(x)\,dx = \beta^2 - \alpha^2 - \int_{\alpha}^{\beta} f(x)\,dx$$ Step 3: Evaluate the given integral Now we compute: $$\int_{\alpha}^{\beta} (f(x) + f^{-1}(x))\,dx = \int_{\alpha}^{\beta} f(x)\,dx + \int_{\alpha}^{\beta} f^{-1}(x)\,dx$$ Substituting the result from Step 2: $$= \int_{\alpha}^{\beta} f(x)\,dx + \beta^2 - \alpha^2 - \int_{\alpha}^{\beta} f(x)\,dx$$ $$= \beta^2 - \alpha^2$$ Step 4: Use the given condition to find α and β We are given that: $$\int_{\alpha}^{\beta} (f(x) + f^{-1}(x))\,dx = 13$$ Therefore: $$\beta^2 - \alpha^2 = 13$$ Factoring: $$(\beta - \alpha)(\beta + \alpha) = 13$$ Since $13$ is prime and $\alpha, \beta \in \mathbb{N}$ with $\alpha < \beta$, the only factorization is: $$\beta - \alpha = 1 \quad \text{and} \quad \beta + \alpha = 13$$ Step 5: Solve for α and β Adding these equations: $$2\beta = 14 \implies \beta = 7$$ Subtracting the first from the second: $$2\alpha = 12 \implies \alpha = 6$$ Step 6: Calculate the final answer $$|\alpha\beta| = |6 \times 7| = 42$$ The answer is **Option 3: 42** <div class="key-concept"><strong>Key Concept:</strong> Use the property of inverse functions in integration: ∫f⁻¹(x)dx can be computed via integration by parts, and the intersections of f with f⁻¹ lie on y=x.</div> <div class="trap-box"><strong>Trap:</strong> The key insight is that ∫(f(x)+f⁻¹(x))dx from α to β simplifies to β²-α² when f(α)=α and f(β)=β.</div>