Mathematics · 2022
JEE Main · 27 June 2022, Shift 1 · Q26
Let A_1={(x, y):|x| ≤ y^2,|x|+2 y ≤ 8} and A_2={(x, y):|x|+|y| ≤ k }. If 27 (Area. A_1)=5 (Area A_2 ), then k is equal to:
Let
$\displaystyle \mathrm{A}_1=\left\{(x, y):|x| \leqslant y^2,|x|+2 y \leqslant 8\right\}$ and
$\displaystyle \mathrm{A}_2=\{(x, y):|x|+|y| \leqslant \mathrm{k}\}$. If $\displaystyle 27$ (Area $\displaystyle \left.\mathrm{A}_1\right)=5$ (Area $\displaystyle \mathrm{A}_2$ ), then k is equal to :
Official answer
6
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.