The minimum value of $f(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}$, where a, $x \in R$ and a > 0, is equal to :
Solution
We know, $AM \ge GM$<br><br>$\therefore$ $${{{a^{a^x}} + {a \over {{a^{a^x}}}}} \over 2} \ge {\left( {{a^{a^x}}\,.\,{a \over {{a^{a^x}}}}} \right)^{1/2}} $$
<br><br>$\Rightarrow {a^{a^x}} + {a^{1 - a^x}} \ge 2\sqrt a$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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