Mathematics · 2024
JEE Main · 8 April 2024, Shift 2 · Q13
Let y=y(x) be the solution curve of the differential equation sec y (d y)/(d x)+2 x sin y=x^3 cos y, y(1)=0. Then y(√ 3 ) is equal to:
Let $\displaystyle y=y(x)$ be the solution curve of the differential equation $\displaystyle \sec y \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 x \sin y=x^3 \cos y, y(1)=0$.
Then $\displaystyle y(\sqrt{3})$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle \frac{\pi}{4}$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.