JEE Main 2021 (Online) 25th February Evening ShiftQuestion 1
The minimum value of $f(x)=a^{a^{x}}+a^{1-a^{x}}$, where $a,x\in\mathbf{R}$ and $a>0$, is equal to :
Differentiation and its Application
Solution
Concept. With $u=a^{a^x}$, the second term is $\frac au$; AM-GM gives the minimum. Formulas used. (1) $u+\frac au\ge2\sqrt a$ for $u>0$, with equality at $u=\sqrt a$Step 1 - Rewrite. $a^{1-a^x}=\frac{a}{a^{a^x}}=\frac au$, so $f=u+\frac au$. Step 2 - Minimum. By (1), $f\ge2\sqrt a$. Step 3 - Attained. $u=\sqrt a$ means $a^{a^x}=a^{1/2}$, i.e. $a^x=\frac12$, which has a solution (for $a\ne1$; for $a=1$, $f=2=2\sqrt a$ anyway). Answer: (C)