Mathematics · 2021
JEE Main · 27 August 2021, Shift 2 · Q65
If 0<x<1 and y=1/2 x^2+2/3 x^3+3/4 x^4+…, then the value of e^1+y at x=1/2 is:
If $\displaystyle 0<x<1$ and $\displaystyle y=\frac{1}{2} x^2+\frac{2}{3} x^3+\frac{3}{4} x^4+\ldots$, then the value of $\displaystyle \mathrm{e}^{1+y}$ at $\displaystyle x=\frac{1}{2}$ is :
Official answer
(4)
$\displaystyle \frac{1}{2} \mathrm{e}^2$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.