Mathematics · 2019
JEE Main · 12 April 2019, Shift 1 · Q71
If e^y+x y= e, the ordered pair ((d y)/(d x), (d^2 y)/(d x^2)) at x=0 is equal to:
If $\displaystyle \mathrm{e}^y+x y=\mathrm{e}$, the ordered pair $\displaystyle \left(\frac{\mathrm{d} y}{\mathrm{~d} x}, \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)$
at $\displaystyle x=0$ is equal to:
Official answer
(1)
$\displaystyle \left(-\frac{1}{\mathrm{e}}, \frac{1}{\mathrm{e}^2}\right)$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.