Mathematics · 2022
JEE Main · 25 July 2022, Shift 1 · Q23
If the maximum value of the term independent of t in the expansion of ( t^2 x^(1/5)+((1-x)^(1/10))/t)^15, x ≥ 0, is K, then 8 K is equal to \ \ \ \.
If the maximum value of the term independent of $\displaystyle t$ in the expansion of $\displaystyle \left(\mathrm{t}^2 x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{\mathrm{t}}\right)^{15}, x \geqslant 0$, is K , then $\displaystyle 8$ K is equal to \_\_\_\_.
Official answer
6006
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.