Mathematics · 2024
JEE Main · 1 February 2024, Shift 1 · Q11
Let y=y(x) be the solution of the differential equation (d y)/(d x)=2 x(x+y)^3-x(x+y)-1, y(0)=1. Then, (1/(√ 2)+y(1/(√ 2)))^2 equals:
Let $\displaystyle y=y(x)$ be the solution of the differential equation $\displaystyle \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x(x+y)^3-x(x+y)-1, y(0)=1$.
Then, $\displaystyle \left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2$ equals :
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle \frac{1}{2-\sqrt{\mathrm{e}}}$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.