Mathematics · 2019
JEE Main · 10 April 2019, Shift 2 · Q77
Let y=y(x) be the solution of the differential equation, (d y)/(d x)+y tan x=2 x+x^2 tan x, x ∈(-(π)/2, (π)/2), such that y(0)=1. Then:
Let $\displaystyle y=y(x)$ be the solution of the differential equation, $\displaystyle \frac{\mathrm{d} y}{\mathrm{~d} x}+y \tan x=2 x+x^2 \tan x$, $\displaystyle x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that $\displaystyle y(0)=1$. Then :
Official answer
(3)
$\displaystyle y^{\prime}\left(\frac{\pi}{4}\right)-y^{\prime}\left(-\frac{\pi}{4}\right)=\pi-\sqrt{2}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.