Mathematics · 2019
JEE Main · 12 January 2019, Shift 1 · Q71
For x>1, if (2 x)^2 y=4 e^2 x-2 y, then (1+ log_e 2 x)^2 (d y)/(d x) is equal to:
For $\displaystyle x>1$, if $\displaystyle (2 x)^{2 y}=4 e^{2 x-2 y}$, then $\displaystyle \left(1+\log _e 2 x\right)^2 \frac{d y}{d x}$ is equal to :
Official answer
(3)
$\displaystyle \frac{x \log _e 2 x-\log _e 2}{x}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.