Mathematics · 2020
JEE Main · 5 September 2020, Shift 1 · Q61
If ∫( e^2 x+2 e^x- e^-x-1) e^( e^x+ e^-x) d x = g (x) e^( e^x+ e^-x) + c, where c is a constant of integration, then g (0) is equal to:
If $\displaystyle \int\left(\mathrm{e}^{2 x}+2 \mathrm{e}^x-\mathrm{e}^{-x}-1\right) \mathrm{e}^{\left(\mathrm{e}^x+\mathrm{e}^{-x}\right)} \mathrm{d} x$
$\displaystyle =\mathrm{g}(x) \mathrm{e}^{\left(\mathrm{e}^x+\mathrm{e}^{-x}\right)}+\mathrm{c}$, where c is a constant of integration, then $\displaystyle \mathrm{g}(0)$ is equal to :
Official answer
(2)
$\displaystyle 2$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.