Exam

JEE Advanced

Year

2024

Paper

Paper 2

Subject

Mathematics

Chapter

Area Under Curves

JEE Advanced 2024 · Paper 2 · Mathematics · Area Under Curves

Let S=\ (x, y) ∈ ℝ × ℝ: x ≥ 0, y ≥ 0, y^2 ≤ 4 x, y^2 ≤ 12-2 x. and...

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

Step-by-step solution

Point of intersection of all curves is $(2,2 \sqrt{2})$ Area $=\mathrm{A}_1+\mathrm{A}_2 \alpha \sqrt{2}=\int_0^2 2 \sqrt{\mathrm{x}} \mathrm{dx}+\frac{1}{2} \times 3 \times 2 \sqrt{2}$$\begin{aligned} & \alpha \sqrt{2}=2\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^2+3 \sqrt{2} \\ & \alpha \sqrt{2}=\frac{17 \sqrt{2}}{3} \\ & \alpha=\frac{17}{3}\end{aligned}$$$