Question

The minimum value of $f\left(x\right)=a^{a^x}+a^{1-a^x}$where $a,$ $x\in R$ and $a>0$, is equal to

• A. $a+\frac{1}{a}$
• B. $a+1$
• C. $2a$
• D. $2\sqrt{a}$

Answer 1

The correct option is D

$2\sqrt{a}$

Find the minimum value of the given function

Given : $f\left(x\right)=a^{a^x}+a^{1-a^x}$

Using A.M.$\ge$G.M. inequality,

$\begin{array}{rcl}\frac{a^{a^x}+{\displaystyle \frac{a}{a^{a^x}}}}{2}& \ge & {\left(a^{a^x}.\frac{a}{a^{a^x}}\right)}^{\frac{1}{2}}\\ & \Rightarrow & a^{a^x}+\frac{a}{a^{a^x}}\ge 2\sqrt{a}\\ & \Rightarrow & a^{a^x}+a^{1-a^x}\ge 2\sqrt{a}\end{array}$

Therefore the minimum value of $f\left(x\right)=a^{a^x}+a^{1-a^x}$is $2\sqrt{a}$.

Hence option $\left(D\right)$ is the correct option.