JEE Advanced 2024 Paper 2 · Q02 · Area Under Curves

Question

Let $S = \{(x,y) \in \mathbb{R} 	imes \mathbb{R} : x \ge 0,\ y \ge 0,\ y^{2} \le 4x,\ y^{2} \le 12 - 2x\ 	ext{and}\ 3y + \sqrt{8}\, x \le 5\sqrt{8}\}$. If the area of the region $S$ is $\alpha\sqrt{2}$, then $\alpha$ is equal to

SolveKar AI · Accepted Answer

Since $\sqrt{8} = 2\sqrt{2}$, the third constraint reads $3y + 2\sqrt{2}\,x \le 10\sqrt{2}$. The two parabolas $y^{2} = 4x$ and $y^{2} = 12 - 2x$ meet at $(2, 2\sqrt{2})$ in the first quadrant; the line passes through this point too. Integrate along $y$: for $y \in [0, 2\sqrt{2}]$ the left edge is the parabola $x = y^{2}/4$, while the line $x = 5 - \dfrac{3y}{2\sqrt{2}}$ is to the left of $x = (12 - y^{2})/2$ throughout, so the line is the right boundary.
$	ext{Area} = \int_{0}^{2\sqrt{2}}\!\left(5 - \dfrac{3y}{2\sqrt{2}} - \dfrac{y^{2}}{4}ight)dy = \left[5y - \dfrac{3y^{2}}{4\sqrt{2}} - \dfrac{y^{3}}{12}ight]_{0}^{2\sqrt{2}} = 10\sqrt{2} - 3\sqrt{2} - \dfrac{4\sqrt{2}}{3} = \dfrac{17\sqrt{2}}{3}.$
Hence $\alpha = 17/3$.

JEE Advanced 2024 Paper 2 · Q02 · Area Under Curves

Solution