Mathematics · 2019
JEE Main · 10 January 2019, Shift 1 · Q67
If the third term in the binomial expansion of (1+x^log_2 x)^5 equals 2560, then a possible value of x is:
If the third term in the binomial expansion of $\displaystyle \left(1+x^{\log _2 x}\right)^5$ equals $\displaystyle 2560$, then a possible value of $\displaystyle x$ is :
Official answer
(2)
$\displaystyle \frac{1}{4}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.