Mathematics · 2022
JEE Main · 28 July 2022, Shift 1 · Q17
If the minimum value of f(x)=(5 x^2)/2+(α)/x^5, x>0, is 14, then the value of α is equal to:
If the minimum value of $\displaystyle f(x)=\frac{5 x^2}{2}+\frac{\alpha}{x^5}, x>0$, is $\displaystyle 14$ , then the value of $\displaystyle \alpha$ is equal to:
Official answer
(3)
$\displaystyle 128$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.