Mathematics · 2021
JEE Main · 25 February 2021, Shift 2 · Q67
The minimum value of f(x)= a^a^x + a^1- a^x, where a, x ∈ R and a >0, is equal to:
The minimum value of $\displaystyle f(x)=\mathrm{a}^{\mathrm{a}^x}+\mathrm{a}^{1-\mathrm{a}^x}$, where $\displaystyle \mathrm{a}, x \in \mathbf{R}$ and $\displaystyle \mathrm{a}>0$, is equal to:
Official answer
(3)
$\displaystyle 2 \sqrt{a}$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.