Mathematics · 2022
JEE Main · 24 June 2022, Shift 2 · Q17
If y= tan^-1( sec x^3- tan x^3), (π)/2<x^3<(3 π)/2, then
If $\displaystyle y=\tan ^{-1}\left(\sec x^3-\tan x^3\right), \frac{\pi}{2}<x^3<\frac{3 \pi}{2}$, then
Official answer
(2)
$\displaystyle x^2 y^{\prime \prime}-6 y+\frac{3 \pi}{2}=0$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.