JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 11)
The minimum value of $$f(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}$$, where a, $$x \in R$$ and a > 0, is equal to :
$$a + {1 \over a}$$
2a
a + 1
$$2\sqrt a $$
Explanation
We know, $$AM \ge GM$$
$$ \therefore $$ $${{{a^{a^x}} + {a \over {{a^{a^x}}}}} \over 2} \ge {\left( {{a^{a^x}}\,.\,{a \over {{a^{a^x}}}}} \right)^{1/2}} $$
$$\Rightarrow {a^{a^x}} + {a^{1 - a^x}} \ge 2\sqrt a $$
$$ \therefore $$ $${{{a^{a^x}} + {a \over {{a^{a^x}}}}} \over 2} \ge {\left( {{a^{a^x}}\,.\,{a \over {{a^{a^x}}}}} \right)^{1/2}} $$
$$\Rightarrow {a^{a^x}} + {a^{1 - a^x}} \ge 2\sqrt a $$
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