Mathematics · 2015
JEE Main · 11 April 2015, Shift 1 · Q77
The solution of the differential equation y d x-(x+2 y^2) d y=0 is x=f(y). If f(-1)=1, then f(1) is equal to:
The solution of the differential equation $\displaystyle y \mathrm{~d} x-\left(x+2 y^2\right) \mathrm{d} y=0$ is $\displaystyle x=f(y)$.
If $\displaystyle f(-1)=1$, then $\displaystyle f(1)$ is equal to :
Official answer
(2)
$\displaystyle 3$
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JEE Main 2015 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.