JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumDifferential Equations163× since 2002If for the solution curve y=f(x)y=f(x)y=f(x) of the differential equation dydx+(tanx)y=2+secx(1+2secx)2\frac{dy}{dx}+(\tan x)y=\frac{2+\sec x}{(1+2\sec x)^2}dxdy+(tanx)y=(1+2secx)22+secx, x∈(−π2,π2)x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)x∈(−2π,2π), f(π3)=310f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}f(3π)=103, then f(π4)f\left(\frac{\pi}{4}\right)f(4π) is equal to:AOption A. 3+110(4+3)\frac{\sqrt{3}+1}{10(4+\sqrt{3})}10(4+3)3+1BOption B. 5−322\frac{5-\sqrt{3}}{2\sqrt{2}}225−3COption C. 93+310(4+3)\frac{9\sqrt{3}+3}{10(4+\sqrt{3})}10(4+3)93+3DOption D. 4−214\frac{4-\sqrt{2}}{14}144−2CheckSkip · show solution