Let S= \ (x, y) in mathbb R × mathbb R : x geq 0, y geq 0, y 2 leq 4 x, y 2 leq 12-2 x . and .3 y+√(8) x leq 5 √(8) \ . If the area of the region S is α √(2) , then α is equal to

Mathematics · Area under the Curves

JEE Advanced 2024 — Paper 2 — Question 2

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

Option A:
$\frac{17}{2}$
Option B:
$\frac{17}{3}$
Correct
Option C:
$\frac{17}{4}$
Option D:
$\frac{17}{5}$

Answer: B

Step-by-step solution

$\quad x \geq 0, y \geq 0, y^{2} \leq 4 x, y^{2} \leq 12-2 x$

3 y+\sqrt{8} x \leq 5 \sqrt{8}

Area $=\int_{0}^{2} \sqrt{4 x} \mathrm{dx}+\frac{1}{2} \times 3 \times 2 \sqrt{2}$

$=\left[2 \times \frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2}+3 \sqrt{2}=2 \cdot \frac{2 \sqrt{2} \times 2}{3}+3 \sqrt{2}$

$=2 \cdot \frac{2 \sqrt{2} \times 2}{3}+3 \sqrt{2}=\frac{17 \sqrt{2}}{3} \Rightarrow \alpha=\frac{17}{3}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 2
Subject: Mathematics
Chapter: Area under the Curves
Topic: Area under the Curves