Mathematics · 2025
JEE Main · 29 January 2025, Shift 2 · Q20
If for the solution curve y=f(x) of the differential equation (d y)/(d x)+( tan x) y=(2+ sec x)/((1+2 sec x)^2), x ∈((-π)/2, (π)/2), f((π)/3)=(√…
If for the solution curve $\displaystyle y=f(x)$ of the differential equation $\displaystyle \frac{\mathrm{d} y}{\mathrm{~d} x}+(\tan x) y=\frac{2+\sec x}{(1+2 \sec x)^2}$, $\displaystyle x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}$, then $\displaystyle f\left(\frac{\pi}{4}\right)$ is equal to:
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \frac{4-\sqrt{2}}{14}$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.