Mathematics · 2026
JEE Main · 4 April 2026, Shift 1 · Q20
Let y=y(x) be the solution of the differential equation (d y)/(d x)=(1+x+x^2)(1-y+y^2), y(0)=1/2. Then (2 y(1)-1) is equal to
Let $\displaystyle y=y(x)$ be the solution of the differential equation $\displaystyle \frac{d y}{d x}=\left(1+x+x^2\right)\left(1-y+y^2\right), y(0)=\frac{1}{2}$. Then $\displaystyle (2 y(1)-1)$ is equal to
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle \sqrt{3} \tan \left(\frac{11 \sqrt{3}}{12}\right)$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.