Mathematics · 2024
JEE Main · 5 April 2024, Shift 2 · Q11
Let β( m, n )=∫_0^1 x^m -1 (1-x)^n -1 d x, m, n >0. If ∫_0^1(1-x^10)^20 d x= a × β( b, c ), then 100( a + b + c ) equals ____.
Let $\displaystyle \beta(\mathrm{m}, \mathrm{n})=\int_0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x, \mathrm{~m}, \mathrm{n}>0$. If $\displaystyle \int_0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x=\mathrm{a} \times \beta(\mathrm{b}, \mathrm{c})$, then $\displaystyle 100(\mathrm{a}+\mathrm{b}+\mathrm{c})$ equals $\displaystyle \_\_\_\_$.
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 2120$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.