Mathematics · 2020
JEE Main · 7 January 2020, Shift 1 · Q58
Let x^k +y^k = a^k,( a, k >0) and (d y)/(d x)+(y/x)^(1/3)=0, then k is:
Let $\displaystyle x^{\mathrm{k}}+y^{\mathrm{k}}=\mathrm{a}^{\mathrm{k}},(\mathrm{a}, \mathrm{k}>0)$ and $\displaystyle \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{y}{x}\right)^{\frac{1}{3}}=0$, then k is :
Official answer
(3)
$\displaystyle \frac{2}{3}$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.