Mathematics · 2026
JEE Main · 4 April 2026, Shift 2 · Q9
In the expansion of (9 x-1/(3 √ x))^18, x>0, if the term independent of x is (221)k, then k is equal to:
In the expansion of $\displaystyle \left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}, x>0$, if the term independent of $\displaystyle x$ is ($\displaystyle 221$)k, then k is equal to:
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 84$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.