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

$3$

Explanation for the correct option:

Determine the value of $f\left(x\right)$

Step 1: Apply the condition of A.M and find a relation.

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

If $a_1,a_2,a_3,...$ are in A.P, then there is a common difference between any two consecutive terms.

Also we know that,

$a_2=\frac{a_1+a_3}{2}$,

Then

$$\begin{gathered} f\left(x\right)=\frac{\left(3^x+3^{-x}\right)+\left(2^{1+x}+2^{1-x}\right)}{2} \\ 2f\left(x\right)=\left(3^x+3^{-x}\right)+\left(2^{1+x}+2^{1-x}\right)\to \left(i\right) \end{gathered}$$

Step 2: Apply the condition $A.M\ge G.M$

Let $3^x$ and $3^{-x}$ be two numbers , then

A.M=$\frac{\left(3^x+3^{-x}\right)}{2}$

G.M=$\sqrt{3^x{3}^{-x}}$

We know that $A.M\ge G.M$ applying this inequality we get:

$$\begin{gathered} \Rightarrow \frac{\left(3^x+3^{-x}\right)}{2}\ge \sqrt{3^x{3}^{-x}} \\ \Rightarrow 3^x+3^{-x}\ge 2\sqrt{1} \\ \Rightarrow 3^x+3^{-x}\ge 2\to \left(ii\right) \end{gathered}$$

Similarly,

Let $2^{1+x}$ and $2^{1-x}$ be two numbers , then

A.M=$\frac{2^{1+x}+2^{1-x}}{2}$

G.M=$\sqrt{2^{1+x}{2}^{1-x}}$

We know that $A.M\ge G.M$ applying this inequality we get:

$$\begin{gathered} \Rightarrow \frac{2^{1+x}+2^{1-x}}{2}\ge \sqrt{2^{1+x}{2}^{1-x}} \\ \Rightarrow 2^{1+x}+2^{1-x}\ge 2\sqrt{2^2} \\ \Rightarrow 2^{1+x}+2^{1-x}\ge 2\times 2\to \left(iii\right) \end{gathered}$$

Step 3: Find the required result

Adding the equations $\left(ii\right)\text{and}\left(iii\right)$ we get,

$$\begin{gathered} \left(3^x+3^{-x}\right)+\left(2^{1+x}+2^{1-x}\right)\ge 2+\left(2\times 2\right) \\ \Rightarrow \left(3^x+3^{-x}\right)+\left(2^{1+x}+2^{1-x}\right)\ge 6 \end{gathered}$$

Substitute this value in equation $\left(i\right)$

$$\begin{gathered} 2f\left(x\right)=\left(3^x+3^{-x}\right)+\left(2^{1+x}+2^{1-x}\right)\ge 6 \\ \Rightarrow 2f\left(x\right)\ge 6 \\ \Rightarrow f\left(x\right)\ge \frac{6}{2} \\ \Rightarrow f\left(x\right)\ge 3 \end{gathered}$$

Hence, the minimum value of $f\left(x\right)$ is $3$.

Hence, option (C) is the correct answer.