Mathematics · 2019
JEE Main · 12 April 2019, Shift 2 · Q67
The term independent of x in the expansion of (1/60-x^8/81) ·(2 x^2-3/x^2)^6 is equal to:
The term independent of $\displaystyle x$ in the expansion of $\displaystyle \left(\frac{1}{60}-\frac{x^8}{81}\right) \cdot\left(2 x^2-\frac{3}{x^2}\right)^6$ is equal to :
Official answer
(3)
$\displaystyle -36$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.