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JEE MAIN 2025
29-01-2025 SHIFT-2
Question
If for the solution curve $y=f(x)$ of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+(\tan x) y=\frac{2+\sec x}{(1+2 \sec x)^{2}}$, $x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}$, then $f\left(\frac{\pi}{4}\right)$ is equal to :
Select the correct option:
A
$\frac{\sqrt{3}+1}{10(4+\sqrt{3})}$
B
$\frac{9 \sqrt{3}+3}{10(4+\sqrt{3})}$
C
$\frac{5-\sqrt{3}}{2 \sqrt{2}}$
D
$\frac{4-\sqrt{2}}{14}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Main
Mathematics
Medium
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