Question

Let $f:R\to R$ be such that for all $x\in R,(2^{1+x}+2^{1-x}),f\left(x\right)$ and $(3^x+3^{-x})$ are in A.P., then the minimum value of $f\left(x\right)$ is:

• A. $0$
• B. $4$
• C. $3$
• D. $2$

Answer 1

The correct option is C $3$

$2^{1-x}+2^{1+x},f\left(x\right),3^x+3^{-x}$ are in A.P.

$\therefore f\left(x\right)=\frac{3^x+3^{-x}+2^{1+x}+2^{1-x}}{2}=\frac{3^x+3^{-x}}{2}+\frac{2^{1+x}+2^{1-x}}{2}$

Applying $A.M.\ge G.M$. inequality, we get
$\frac{(3^x+3^{-x})}{2}\ge \sqrt{3^x\cdot 3^{-x}}$
$\Rightarrow \frac{(3^x+3^{-x})}{2}\ge 1...\left(1\right)$

$\frac{2^{1+x}+2^{1-x}}{2}\ge \sqrt{2^{1+x}\cdot 2^{1-x}}$
$\Rightarrow \frac{2^{1+x}+2^{1-x}}{2}\ge 2...\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$, we get
$f\left(x\right)\ge 1+2=3$
Also, $f\left(0\right)=3$
Thus, the minimum value of $f\left(x\right)$ is $3$.