UPSC NDA II 2024 Mathematics Paper

Concept:The area bounded by the parabola $y^{2} = kx$ and the vertical line $x = k$ is given as $\frac{4}{3}$ square units. We use integration to set up the area and solve for $k$.Explanation:The parabola $y^{2} = kx$ opens to the right for $k > 0$. The line $x = k$ is a vertical line. The bounded region is symmetric about the x-axis. The area is twice the area in the first quadrant.We express $y$ from the parabola as $y = \sqrt{kx}$ (positive branch). The limits of $x$ are from $0$ to $k$.Area = $2 \int_{0}^{k} y \, dx = 2 \int_{0}^{k} \sqrt{kx} \, dx$.Simplify: $\sqrt{kx} = \sqrt{k} \sqrt{x}$. So area = $2\sqrt{k} \int_{0}^{k} x^{1/2} \, dx$.Integrate: $\int x^{1/2} \, dx = \frac{2}{3} x^{3/2}$. Evaluate from $0$ to $k$: $\frac{2}{3} k^{3/2}$.Thus area = $2\sqrt{k} \cdot \frac{2}{3} k^{3/2} = \frac{4}{3} k^{2}$.Set this equal to given area: $\frac{4}{3} k^{2} = \frac{4}{3}$.Cancel $\frac{4}{3}$: $k^{2} = 1$. Since $k > 0$, we get $k = 1$.Answer:$k = 1$ (Option B).