Mathematics · 2025
JEE Main · 22 January 2025, Shift 1 · Q20
Let x=x(y) be the solution of the differential equation y^2 d x+(x-1/y) d y=0. If x(1)=1, then x(1/2) is:
Let $\displaystyle x=x(y)$ be the solution of the differential equation $\displaystyle y^2 \mathrm{~d} x+\left(x-\frac{1}{y}\right) \mathrm{d} y=0$. If $\displaystyle x(1)=1$, then $\displaystyle x\left(\frac{1}{2}\right)$ is :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 3-e$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.