Question:

Step 1: Find the intersection points of the line and parabola. We need to find where $y = kx + k$ intersects $y = x^2$. Setting them equal: $$x^2 = kx + k$$ $$x^2 - kx - k = 0$$ Using the quadratic formula: $$x = \frac{k \pm \sqrt{k^2 + 4k}}{2}$$ Let $x_1 = \frac{k - \sqrt{k^2+4k}}{2}$ and $x_2 = \frac{k + \sqrt{k^2+4k}}{2}$ be the two roots, where $x_1 < x_2$. Step 2: Set up the integral for the area between the curves. The area $A_k$ bounded by the line and parabola is: $$A_k = \int_{x_1}^{x_2} (kx + k - x^2)\, dx$$ Since the line is above the parabola in this region, we integrate the difference. Step 3: Apply the standard formula for area between a line and parabola. For a parabola $y = x^2$ and a line intersecting it at two points, the area between them is given by: $$A_k = \frac{(x_2 - x_1)^3}{6}$$ Step 4: Calculate the distance between the roots. From the quadratic formula: $$x_2 - x_1 = \frac{k + \sqrt{k^2+4k}}{2} - \frac{k - \sqrt{k^2+4k}}{2} = \sqrt{k^2 + 4k}$$ Step 5: Express the area in terms of $k$. Substituting into the area formula: $$A_k = \frac{(\sqrt{k^2+4k})^3}{6} = \frac{(k^2+4k)^{3/2}}{6}$$ Step 6: Form the ratio $\frac{A_k}{k^3}$ and simplify. $$\frac{A_k}{k^3} = \frac{(k^2+4k)^{3/2}}{6k^3}$$ Factor out $k^2$ from inside the parentheses: $$(k^2+4k)^{3/2} = \left[k^2\left(1 + \frac{4}{k}\right)\right]^{3/2} = k^3\left(1 + \frac{4}{k}\right)^{3/2}$$ Therefore: $$\frac{A_k}{k^3} = \frac{k^3\left(1+\frac{4}{k}\right)^{3/2}}{6k^3} = \frac{\left(1+\frac{4}{k}\right)^{3/2}}{6}$$ Step 7: Evaluate the limit as $k \to \infty$. $$\lim_{k\to\infty} \frac{A_k}{k^3} = \lim_{k\to\infty} \frac{\left(1+\frac{4}{k}\right)^{3/2}}{6}$$ As $k \to \infty$, we have $\frac{4}{k} \to 0$, so: $$\lim_{k\to\infty} \frac{A_k}{k^3} = \frac{(1+0)^{3/2}}{6} = \frac{1}{6}$$ **Final Answer:** The value of $\displaystyle\lim_{k \to \infty} \dfrac{A_k}{k^3} = \dfrac{1}{6}$, which corresponds to **Option 3**.