Mathematics · 2026
JEE Main · 24 January 2026, Shift 1 · Q20
Let A_1 be the bounded area enclosed by the curves y=x^2+2, x+y=8 and y -axis that lies in the first quadrant. Let A_2 be the bounded area enclosed…
Let $\displaystyle \mathrm{A}_1$ be the bounded area enclosed by the curves $\displaystyle y=x^2+2, x+y=8$ and $\displaystyle y$-axis that lies in the first quadrant. Let $\displaystyle \mathrm{A}_2$ be the bounded area enclosed by the curves $\displaystyle y=x^2+2, y^2=x, x=2$, and $\displaystyle y$-axis that lies in the first quadrant. Then $\displaystyle \mathrm{A}_1-\mathrm{A}_2$ is equal to
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle \frac{2}{3}(2 \sqrt{2}+1)$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.