Mathematics · 2024
JEE Main · 8 April 2024, Shift 2 · Q6
If the term independent of x in the expansion of (√ a x^2+1/(2 x^3))^10 is 105, then a^2 is equal to:
If the term independent of $\displaystyle x$ in the expansion of $\displaystyle \left(\sqrt{\mathrm{a}} x^2+\frac{1}{2 x^3}\right)^{10}$ is $\displaystyle 105$ , then $\displaystyle \mathrm{a}^2$ is equal to:
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 4$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.