Mathematics · 2022
JEE Main · 30 June 2022, Shift 1 · Q6
For two positive real numbers a and b such that 1/a^2+1/b^3=4, the minimum value of the constant term in the expansion of ( a x^(1/8)+ b x^-1/12)^10…
For two positive real numbers $\displaystyle a$ and $\displaystyle b$ such that $\displaystyle \frac{1}{a^2}+\frac{1}{b^3}=4$, the minimum value of the constant term in the expansion of $\displaystyle \left(\mathrm{a} x^{\frac{1}{8}}+\mathrm{b} x^{-\frac{1}{12}}\right)^{10}$ is :
Official answer
(3)
$\displaystyle \frac{105}{8}$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.