Mathematics · 2024
JEE Main · 30 January 2024, Shift 1 · Q12
Let y=y(x) be the solution of the differential equation sec x d y+{2(1-x) tan x+x(2-x)} d x=0 such that y(0)=2. Then y(2) is equal to:
Let $\displaystyle y=y(x)$ be the solution of the differential equation $\displaystyle \sec x \mathrm{~d} y+\{2(1-x) \tan x+x(2-x)\} \mathrm{d} x=0$ such that $\displaystyle y(0)=2$. Then $\displaystyle y(2)$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 2$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.