Question:

Step 1: Identify the indeterminate form As $x \to \pi/4$, we have $\sec^2(\pi/4) = 2$, so the numerator approaches $\displaystyle\int_{2}^{2} f(t)\,dt = 0$. The denominator approaches $\left(\frac{\pi}{4}\right)^2 - \frac{\pi^2}{16} = 0$. This is a $\frac{0}{0}$ indeterminate form, so we apply L'Hôpital's Rule. Step 2: Differentiate the numerator using the Leibniz integral rule We need to find: $$\frac{d}{dx}\int_{2}^{\sec^2 x} f(t)\,dt$$ By the Leibniz integral rule: $$\frac{d}{dx}\int_{2}^{\sec^2 x} f(t)\,dt = f(\sec^2 x) \cdot \frac{d}{dx}(\sec^2 x)$$ Computing the derivative of $\sec^2 x$: $$\frac{d}{dx}(\sec^2 x) = 2\sec x \cdot \sec x \tan x = 2\sec^2 x \tan x$$ Therefore, the numerator derivative is: $$f(\sec^2 x) \cdot 2\sec^2 x \tan x$$ Step 3: Differentiate the denominator $$\frac{d}{dx}\left(x^2 - \frac{\pi^2}{16}\right) = 2x$$ Step 4: Apply L'Hôpital's Rule and evaluate at $x = \pi/4$ By L'Hôpital's Rule: $$\lim_{x \to \pi/4} \frac{\displaystyle\int_{2}^{\sec^2 x} f(t)\,dt}{x^2 - \frac{\pi^2}{16}} = \lim_{x \to \pi/4} \frac{f(\sec^2 x) \cdot 2\sec^2 x \tan x}{2x}$$ At $x = \pi/4$: - $\sec^2(\pi/4) = 2$ - $\tan(\pi/4) = 1$ - The limit becomes: $$\frac{f(2) \cdot 2 \cdot 2 \cdot 1}{2 \cdot \frac{\pi}{4}} = \frac{4f(2)}{\frac{\pi}{2}} = \frac{8f(2)}{\pi}$$ Step 5: Match with the given form and identify $k$ and $a$ We are given that: $$\lim_{x \to \pi/4} \frac{\displaystyle\int_{2}^{\sec^2 x} f(t)\,dt}{x^2 - \frac{\pi^2}{16}} = \frac{k}{\pi}f(a)$$ Comparing $\dfrac{8f(2)}{\pi} = \dfrac{k}{\pi}f(a)$, we get: $$k = 8 \quad \text{and} \quad a = 2$$ Step 6: Calculate $k^a$ $$k^a = 8^2 = 64$$ **Final Answer:** The value of $k^a$ is equal to **64**, which corresponds to **Option 3**.