Question:

Step 1: Recognize the limit as a Riemann sum. The given limit can be interpreted as a Riemann sum. As $n \to \infty$, we have: $$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{e^{k/n}+e^{-k/n}}{n\sqrt{11-e^{2k/n}-e^{-2k/n}}} = \int_0^1 \frac{e^x+e^{-x}}{\sqrt{11-e^{2x}-e^{-2x}}}dx$$ Step 2: Perform a substitution to simplify the integrand. Let $u = e^x - e^{-x}$. Then: $$du = (e^x + e^{-x})dx$$ This means the numerator $e^x + e^{-x}$ becomes $du$, which will simplify our integral significantly. Step 3: Express the denominator in terms of $u$. We need to rewrite $e^{2x} + e^{-2x}$ in terms of $u$. Starting with: $$u^2 = (e^x - e^{-x})^2 = e^{2x} - 2 + e^{-2x}$$ Therefore: $$e^{2x} + e^{-2x} = u^2 + 2$$ Substituting into the denominator: $$\sqrt{11 - (e^{2x} + e^{-2x})} = \sqrt{11 - (u^2 + 2)} = \sqrt{9 - u^2}$$ Step 4: Determine the limits of integration in terms of $u$. When $x = 0$: $u = e^0 - e^0 = 0$ When $x = 1$: $u = e^1 - e^{-1} = e - e^{-1}$ Step 5: Evaluate the transformed integral. The integral becomes: $$\int_0^{e-1/e} \frac{du}{\sqrt{9-u^2}}$$ This is a standard arctangent integral of the form $\int \frac{du}{\sqrt{a^2-u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C$, where $a = 3$. Therefore: $$\int_0^{e-1/e} \frac{du}{\sqrt{9-u^2}} = \left[\sin^{-1}\frac{u}{3}\right]_0^{e-1/e} = \sin^{-1}\left(\frac{e-e^{-1}}{3}\right) - \sin^{-1}(0) = \sin^{-1}\left(\frac{e-e^{-1}}{3}\right)$$ Step 6: Identify the values of $a$ and $b$ and compute the final answer. Comparing with $\sin^{-1}\left(\frac{e^a - e^{-a}}{b}\right)$, we have: - $a = 1$ (since $e^1 - e^{-1} = e - e^{-1}$) - $b = 3$ Therefore: $$a + b = 1 + 3 = 4$$ The answer is **Option 3: 4**