QJEEM21Ma762c: JEE Main 2021 Mathematics - Class 11 JEEM21Ma762c Solution: The minimum value of , where a, and a > 0, is equal to :

Question

Verified Solution • 2025–26 Session moderate

The minimum value of

f(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}

, where a,

x \in R

and a > 0, is equal to :

Accepted Answer

We know,

AM \ge GM

\therefore

{{{a^{a^x}} + {a \over {{a^{a^x}}}}} \over 2} \ge {\left( {{a^{a^x}}\,.\,{a \over {{a^{a^x}}}}} \right)^{1/2}}

\Rightarrow {a^{a^x}} + {a^{1 - a^x}} \ge 2\sqrt a

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