Mathematics · 2019
JEE Main · 10 January 2019, Shift 1 · Q77
If (d y)/(d x)+3/(cos^2 x) y=1/(cos^2 x), x ∈((-π)/3, (π)/3), and y((π)/4)=4/3, then y(-(π)/4) equals:
If $\displaystyle \frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{3}{\cos ^2 x} y=\frac{1}{\cos ^2 x}, x \in\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)$,
and $\displaystyle y\left(\frac{\pi}{4}\right)=\frac{4}{3}$, then $\displaystyle y\left(-\frac{\pi}{4}\right)$ equals :
Official answer
(2)
$\displaystyle \frac{1}{3}+e^6$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.