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Mathematics · 2020

JEE Main · 7 January 2020, Shift 2 · Q61

The value of α for which 4 α ∫_-1^2 e^-α|x| d x=5, is:

The value of $\displaystyle \alpha$ for which $\displaystyle 4 \alpha \int_{-1}^2 \mathrm{e}^{-\alpha|x|} \mathrm{d} x=5$, is :
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.