Mathematics · 2022
JEE Main · 26 June 2022, Shift 2 · Q9
If y=y(x) is the solution of the differential equation x (d y)/(d x)+2 y=x e^x, y(1)=0 then the local maximum value of the function z(x)=x^2…
If $\displaystyle y=y(x)$ is the solution of the differential equation $\displaystyle x \frac{d y}{d x}+2 y=x e^x, y(1)=0$ then the local maximum value of the function $\displaystyle z(x)=x^2 y(x)-e^x, x \in \mathbb{R}$ is :
Official answer
(4)
$\displaystyle \frac{4}{e}-e$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.