Mathematics · 2020
JEE Main · 2 September 2020, Shift 2 · Q60
The equation of the normal to the curve y=(1+x)^2 y+ cos^2( sin^-1 x) at x=0 is:
The equation of the normal to the curve $\displaystyle y=(1+x)^{2 y}+\cos ^2\left(\sin ^{-1} x\right)$ at $\displaystyle x=0$ is :
Official answer
(1)
$\displaystyle x+4 y=8$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.