Mathematics · 2020
JEE Main · 6 September 2020, Shift 1 · Q61
If I_1=∫_0^1(1-x^50)^100 d x and I_2=∫_0^1(1-x^50)^101 d x such that I_2=α I_1 then α equals to:
If $\displaystyle \mathrm{I}_1=\int_0^1\left(1-x^{50}\right)^{100} \mathrm{~d} x$ and $\displaystyle \mathrm{I}_2=\int_0^1\left(1-x^{50}\right)^{101} \mathrm{~d} x$ such that $\displaystyle \mathrm{I}_2=\alpha \mathrm{I}_1$ then $\displaystyle \alpha$ equals to :
Official answer
(1)
$\displaystyle \frac{5050}{5051}$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.