Mathematics · 2024
JEE Main · 5 April 2024, Shift 2 · Q9
If the constant term in the expansion of ((√[5] 3)/x+(2 x)/(√[3] 5))^12, x ≠ 0, is α × 2^8 × √[5] 3, then 25 α is equal to:
If the constant term in the expansion of $\displaystyle \left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$, is $\displaystyle \alpha \times 2^8 \times \sqrt[5]{3}$, then $\displaystyle 25 \alpha$ is equal to:
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 693$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.