JEE Main 2022 · 27 Jun · Shift 1 · Mathematics Q26

← JEE Main 2022 · 27 Jun · Shift 1

MathematicsInteger+4 / -1Question 26 of 90

Let $A_1=\left\{(x,y):|x|\le y^2,\ |x|+2y\le8\right\}$ and $A_2=\left\{(x,y):|x|+|y|\le k\right\}$ . If $27\,(\text{Area }A_1)=5\,(\text{Area }A_2)$ , then $k$ is equal to :

Your answer

Answer and worked solution

Official answer: 6

Region. Read literally, $A_1$ is unbounded below ( $y\to-\infty$ ). The key uses the bounded part with $y\ge0$ : between $y=\sqrt{|x|}$ and $y=\frac{8-|x|}2$ , which meet at $|x|=4$ .

Area. $2\displaystyle\int_0^4\left(\frac{8-x}2-\sqrt x\right)dx=2\left(12-\frac{16}3\right)=\frac{40}3$ .

Square. Area $A_2=2k^2$ : $27\cdot\frac{40}3=5\cdot2k^2$ gives $k^2=36$ , $k=6$ .

Answer: 6

Topics:Area between curves

← Previous question Next question →

JEE Main 2022 · 27 Jun · Shift 1, Mathematics question 26